Proceedings

of the Tomsk Technological Institute

of Emperor Nicholas II

Vol. 13. 1909. No. 1.

I.

A. V. Ugarov.

On the Study of Crank Mechanisms.

The Coupler Pole and Some Applications of Its Properties.

With three plates of drawings.

I—VI, 1—106.

Noted Errata

Page	Line	Printed	Should read
25	8 from bottom	$\delta$	$\sigma$
30	3 from bottom	before the radical the factor $(n-a)$ has been omitted
58	11 from top	$(m^2+n^2)^2$	$(m^2+k^2)^2$
60	15 from top	$4n^2k^2\operatorname{Cos}\vartheta$	$4n^2k^2\operatorname{Cos}^2\vartheta$
72	13 from top	$\vartheta=90^\circ$	$\vartheta=0^\circ$

On this basis, instead of the following three lines of text there should be:

the consequently produced mechanism turns out to be the basic one.

Contents

	Pages
Preface	V—VI
Chapter One.
General concept of crank mechanisms. Their classification	1—5
Chapter Two.
The basic connecting-rod-and-crank mechanism. The property of its conjugate chords. The connecting-rod pole	5—21
Chapter Three.
Rectilinear-generating crank mechanism. Conditions for the possibility of its existence	21—26
Chapter Four.
Properties of the chords of a rectilinear-generating crank mechanism	26—40
Chapter Five.
Investigation of the formulas expressing conjugations of the chord of a rectilinear-generating crank mechanism	40—57
Chapter Six.
Determination of the geometric locus of the points of intersection of conjugate chords by an approximate method. A numerical example	58—80
Chapter Seven.
Curvilinear-generating crank mechanism; conditions for the possibility of its existence. Properties of conjugate chords	81—85
Chapter Eight.
General conclusions	85—86
Chapter Nine.
Properties of the connecting-rod pole of the basic crank mechanism	86—103
Chapter Ten.
Conclusion	103—104
Index of literature	105—106

PREFACE.

The article “The Coupler Pole,” offered below for the attention of the technical public, has as its subject the study of crank mechanisms, by which are understood, in addition to the connecting-rod-and-crank mechanism, also its modifications: the eccentric and the slotted-link drives.

The study of connecting-rod-and-crank mechanisms has the aim of obtaining an exact and clear conception of the character of the changes that may occur in the mutual position of the parts of the mechanism.

Most vividly, all these properties are clarified on models of the mechanism; they are sometimes also made use of in designing linkages.

Another method of study consists in drawing the successive positions of the mechanism and in analyzing the mutual displacements of its links that occur as a result of a change in the position of some one link.

This link is usually the radius of the crank or the eccentricity of the eccentric, which describes a simple circular path.

Having drawn, at the chosen scale, the circle representing this path, and having marked on it a number of arbitrary points, we are able, by a simple geometrical construction, to obtain the corresponding positions of the mechanism and, consequently, to obtain a clear conception of the character of the mutual displacements of its links.

By connecting, in a known manner, the successive positions of some part of the mechanism selected, we obtain on the drawing a series of lines. These lines change their position on the drawing solely as a function of the changes in the position of the individual links, and consequently the geometrical properties of these lines depend on the properties of the mechanism under study.

Thus, the study of the geometrical properties of auxiliary lines, which do not exist in the mechanism, can give us an indication of the particular rules to which the mutual changes in the positions of the links of the mechanism are subject.

The auxiliary lines laid at the basis of the proposed investigation of crank mechanisms are suitably chosen chords of the crank circle.

These chords possess geometrical properties that depend on the dimensions and mutual arrangement of the parts of the mechanism; consequently, the study of the properties of these chords makes it possible to take into account the influence of the finite...

the length of the connecting rod or eccentric rod on the character of the motion produced by these organs.

The point of intersection of certain chords, which we have called the coupler pole, by its position within the crank circle characterizes the corresponding crank mechanism and determines its properties, and consequently may be useful in designing a mechanism to be newly constructed, for example, a link-block drive.

Without claiming to introduce anything entirely new into the extensive field of the study of crank mechanisms, the author will gratefully welcome indications of shortcomings in the method set forth.

A. Ugarov.

Tomsk.
1907–1908.

CHAPTER ONE.

General concept of crank mechanisms, and their classification.

From the kinematic point of view, a steam engine is a mechanism for transforming the rectilinear reciprocating motion of the piston into the continuous rotary motion of the machine’s main shaft.

The members of the kinematic chain that accomplish this transformation are, in succession: the cylinder—fixed relative to the frame, the piston, the rod, the crosshead, the guide—attached to the frame, the connecting rod, the crank, the main shaft, the main bearing—forming one whole with the frame.

Some of the enumerated members, being immovable relative to one another, must evidently, in the kinematic sense, be regarded as a single link.

The frame, cylinder, guide, and main bearing make one link: the frame.

The piston, rod, and crosshead form another link: the slider.

The crank and the main shaft are the third link, which we shall simply call the crank.

Thus we see that the steam engine is a four-link kinematic chain: frame, slider, connecting rod, and crank—the links of this chain.

A connecting-rod-and-crank mechanism of this kind we shall call the principal one.

To make the reciprocating motion of the slider possible, it is necessary alternately to admit and exhaust steam on both one side and the other of the piston; this is accomplished by steam-distribution organs.

For all their constructive variety, steam-distribution systems are divided into three groups:

I. Slide-valve steam distribution:
    a) simple slide valves,
    b) double slide valves,
    c) link-motion distribution.

II. Corliss distribution—which is the same slide-valve distribution, with oscillatory-circular motion of the slide valve.

III. Valve steam distribution.

A. Ugarov.

The distributions of the first and third groups are characterized, in distinction from the distributions of the second group, by the rectilinear-oscillatory motion of the organs that directly control the admission and exhaust of steam.

In the vast majority of cases, the oscillatory motion of steam-distribution organs of one kind or another is obtained from the transformed continuous-rotary motion of the main shaft of the engine. The mechanism producing this transformation is, most often, a circular eccentric, which, from the kinematic point of view, may be regarded as a crank-and-connecting-rod mechanism.

In one and the same engine, depending on the type of its steam distribution, there may be several eccentric mechanisms operating under various conditions. In what follows we shall call these mechanisms secondary.

Calling the end of the connecting rod adjoining the crank and describing, together with the latter, a circle the rotating end, we may call its other end, which performs oscillatory motion, the slider.

Relying on this terminology and on what has been said above, we arrive at the following conclusion: the difference between the main and the secondary crank-and-connecting-rod mechanisms consists in the fact that, in the former, the crank receives rotation from an oscillating slider, whereas in the latter the slider is set into oscillatory motion by a rotating crank. Obviously, this difference is determined only by the role which a given mechanism performs in the engine; in essence, both the main and the secondary crank-and-connecting-rod mechanisms are entirely identical. Therefore, in what follows, we shall make no distinction between them, and only in special cases of the investigation of crank-and-connecting-rod mechanisms shall we indicate the role performed by them in the engine.

Examining the motion of the individual links of the crank-and-connecting-rod mechanism, we shall at once see that this mechanism admits certain variations. Since the axis of rotation of the crank is the axis of the main shaft, fixed with respect to the frame, variations of the mechanism are possible by changing the path of the slider in the steam engine. Let us consider these variations in somewhat greater detail.

Let (Plate I, Fig. 1) the center of rotation of the crank be at \(O\), and let the midpoint of the slider’s oscillations be the point \(M\); then the distance \(OM=n\) must necessarily be equal to the length of the connecting rod, \(CD=L\). If the radius of the crank circle is \(R\), then the path traversed by the slider from one of its extreme positions to the other, \(AB=S\), will be equal to the diameter of the crank circle, or \(S=2R\). For the possibility of motion it is necess—

The Coupler Pole.

It is evident that \(R\) must be less than \(L\), or \(\dfrac{R}{L}<1\), i.e., the midpoint of oscillation of the slider must lie outside the crank circle.

The points \(A\) and \(B\), at which the velocity of the slider is zero, since at them it changes the direction of its motion, are called dead points.

Let us see what modifications the mechanism just described may undergo.

If we make the crank rotate about the former axis and move the path of the slider parallel to itself (Plate I, Fig. 2) to one side or the other of its former position, we obtain a connecting-rod-and-crank mechanism with properties different from that shown in Fig. 1. As we shall see below, when the separate variants are considered in detail, a change in the path of the slider entails a change in the lengths of the crank and of the connecting rod, and also a change in the conditions of motion of the separate links of the kinematic chain.

Let now (Plate I, Fig. 3) the center \(O\) of rotation of the crank and the midpoint \(M\) of oscillation of the slider remain unchanged, while the path of the slider is inclined, to one side or the other, at some angle to its original position. We have a new variant.

Under the condition that the path of the slider be rectilinear, all the other possible modifications of the mechanism are only combinations of the two just considered.

Let us now make the slider move not along a rectilinear path. There are innumerably many curved paths of the slider, and it is impossible to bring these curves under one general law; therefore here, and in what follows, we shall consider only the motion of the slider along a circle, as the case most often encountered in secondary mechanisms.

Leaving the center \(O\) of rotation of the crank and the dead points \(A\) and \(B\) of the path of the slider in their former places, and making the slider move along the curve \(AMB\), concave toward its original path (Fig. 4, Plate I), we obtain a fourth variant of the connecting-rod-and-crank mechanism.

A fifth variant is obtained if we leave unchanged the position of the point \(M\)—the midpoint of oscillation of the slider (Fig. 5, Plate I)—and make the slider move along the curve \(AMB\), convex toward its former path.

Variants 6 and 8 are obtained from variants 2 and 3 if the slider is made to move along curves with the former dead points; and variants 7 and 9 are obtained from 2 and 3 if the slider is made to oscillate along curves without changing the position of the midpoint \(M\).

The radius of the circle along which the slider moves can obviously be taken infinitely large—then we have variants 1, 2, and 3;

but it cannot be infinitely small, since then the length of the slider’s path becomes zero; in other words, the slider is immobile and the four-link kinematic chain becomes a rigid triangular system.

Let us assume, for the 4th case, that the slider makes definite swings between points $A$ and $B$ along a circle of finite radius. By decreasing the radius of this circle and leaving unchanged the dead points of the slider’s path, we shall find that, with the crank radius unchanged, the slider will swing along arcs subtending ever larger central angles.

When, finally, the arc path of the slider becomes equal to $180^\circ$, and this will occur, evidently, when the radius of the circle of the slider’s path becomes equal to the radius of the crank circle, we obtain the limiting case for all variants with curvilinear paths (Fig. 10, Plate I).

This variant is characterized by the fact that the oscillating end of the connecting rod—the slider—becomes rotating, as though under the action of a new crank with its axis of rotation at point $M$; the connecting rod moves parallel to itself; in a word, there is obtained a special case of a hinged quadrilateral, the so-called mechanism of parallel cranks.

From kinematics we know that such a mechanism, at the dead points, can pass into a mechanism of antiparallel cranks and, consequently, lose its essence—the property of strict definiteness of the relative motion of the individual links.

Since the subject of our investigation is connecting-rod-and-crank mechanisms with precisely defined conditions of motion of their links, the 10th variant cannot serve as an object of consideration for us; all the more so because, in essence, it does not constitute a mechanism for the transformation of any one kind of motion into another.

Let us proceed to the consideration of the remaining nine variants.

In order to have a definite factor characterizing one or another variant, we shall introduce the concept of the principal line of a connecting-rod-and-crank mechanism.

By the principal line we shall understand the imaginary line connecting the center $O$ of rotation of the crank (see Plate I) with point $M$—the midpoint of the slider’s oscillations.

With respect to this principal line we shall determine the form and direction of the slider’s path.

The first variant of the connecting-rod-and-crank mechanism, from which all the others are derived, is characterized by the fact that the slider’s path completely coincides with the principal line.

We shall call such a mechanism the principal connecting-rod-and-crank mechanism, and all the others we shall call derived.

Turning to the 2nd and 3rd variants, we see that, with respect to the principal line, both of these variants are characterized by one and the same definition: slider-crank mechanisms with the path of the slider inclined at some angle to the principal line; in other words, these two variants differ in no way from one another and represent one and the same derived mechanism.

Variants 4, 6, 7, 8, and 9 are likewise characterized, with respect to the principal line, by one common property: the path of the slider is along an arc of a circle, while the imagined tangent to the mean point of oscillation of the slider is inclined at some angle to the principal line; if the angle of inclination of this tangent is equal to zero, then we obtain, as a special case of the general one, the 5th variant.

From what has been said we conclude that the last six variants essentially represent, again, one and the same derived slider-crank mechanism.

Thus, the introduction of the concept of the principal line of a slider-crank mechanism shows us that, despite the apparent difference among the nine variants, we may assert that they represent only three kinds of mechanisms:

A) the principal slider-crank mechanism,
B) a derived one, with the path of the slider along a straight line,
C) a derived one, with the path of the slider along an arc of a circle.

In accordance with this division we shall also conduct the investigation of the properties of the slider-crank mechanism that interest us.

CHAPTER TWO.

The principal slider-crank mechanism. The property of its conjugate chords. The coupler pole.

As was already mentioned in the first chapter, a steam engine produces work by means of a main (one or several) and a secondary (one or several) slider-crank mechanism. For simplicity let us assume that the machine consists only of one main and one secondary mechanism.

Since the secondary mechanism receives its motion from the main one, it is evident that there must exist between the two a wholly exact kinematic connection, expressed in the fact that to a definite possible displacement of the links of the main slider-crank mechanism with respect to one another there must correspond

a definite (and only one) relative displacement of the links of the secondary mechanism.

On the other hand, a steam engine must often perform work that varies in accordance with the variation of the useful resistances—or, in other words, the secondary mechanism—the steam-distribution apparatus—must, for one and the same displacement of the principal crank-and-connecting-rod mechanism, undergo displacements that vary according to the required conditions*); for this purpose the kinematic connection between the principal and the secondary mechanisms must be constructed so that, while the machine is running, it can be varied either by hand (Meyer valve gear, link motions) or automatically (centrifugal or other governors). Such a kinematic connection we shall call an intermediate one.

In order to determine the variable working conditions of any machine, whether being designed or investigated, it is necessary to represent accurately and clearly the mutual connection of three kinematic chains—the principal, the intermediate, and the secondary.

For every particular case, given known data, this connection can be determined and expressed analytically; but far more quickly, more graphically, and more clearly the sought relation among the three chains is revealed by the graphical method of investigation—by drawing the so-called steam-distribution diagrams.

The principal constituent part of most diagrams is the circle described by the crank pin—we shall call it the crank circle.

To each point of the crank circle—when the length of the connecting rod is constant—there corresponds a definite position of the piston in the cylinder, or of the slide block in the guides, as of a part inseparably connected with the piston.

From the consideration of the conditions of crank motion, set forth in all manuals of kinematics, it is known that, when the slide block traverses equal paths in a given time, the rotating end of the connecting rod describes arcs of unequal magnitude; or, what is the same thing, with uniform rotation of the crank the slide block moves non-uniformly. Indeed, we know that at the dead points of its path the slide block changes the direction of its motion and its velocity becomes equal to zero—the slide block moves with variable velocity and, consequently, traverses unequal paths in equal intervals of time.

The middle of the oscillations of the slide block we shall call the mean point.

*) Regulation of the operation of the machine by means of a throttle valve, as not affecting the kinematic connection between the principal and secondary mechanisms, is not considered here.
A. U.

The Coupler Pole.

Positions of the slider in the drawing, equally distant on either side from the mean point, we shall call symmetric points.

Two symmetric points passed through by the slider during one revolution of the crank, as it moves from each of the dead points to the mean, we shall hereafter call equidistant.

The positions of the rotating end of the connecting rod on the crank circle, corresponding to equidistant points of the path of the slider, we shall call conjugate.

In the study of steam engines, for simplicity, the length of the connecting rod is often taken to be infinitely large. It is clear that then the conjugate points will be diametrically opposite.

With a finite length of the connecting rod, however, the lines connecting conjugate points of the crank circle will not be its diameters, but chords.

Let the point \(O\) be the center of the crank circle (see Drawing II); then its diameter \(M_1M_2\) will represent the magnitude of the stroke of the piston; \(M_1\) and \(M_2\) are the dead points. The line \(AB\) represents the path of the slider corresponding to a rotation of the crank through \(180^\circ\) from one of its dead positions to the other. The point \(M\) on the line \(AB\) is the mean point.

Let us mark on the line \(AB\) several symmetric points and find the corresponding positions of the rotating end of the connecting rod, or—which is the same—the crank pin on the crank circle.

As is known, the positions of the crank pin for given positions of the slider are determined graphically as follows: from each given point of the slider path the crank circle is intersected by a radius equal to the length of the connecting rod.

If the point taken is not one of the dead points of the slider path, then, obviously, under the indicated construction the crank circle is intersected at two points. For the dead points, instead of intersection there will be tangency; the point of tangency may be counted as two infinitely near points of intersection.

Thus we arrive at the conclusion that to each pair of equidistant points of the slider path there correspond four conjugate points on the crank circle.

In Drawing II, to the equidistant points \(a\) and \(a_1\) there correspond the points \(A_{\mathrm{I}}—A_{\mathrm{II}}\) and \(A'_{\mathrm{I}}—A'_{\mathrm{II}}\).

The four conjugate points obtained on the crank circle for two equidistant points we shall call a system of points.

Since a system of conjugate points lies on a circle, it is clear that these points, when connected by straight lines, will give us

the inscribed quadrilateral with its diagonals. As we noted above, these diagonals, in the general case, are chords; we shall therefore call them paired conjugate chords.

For the midpoint \(M\), both conjugate chords merge into one—this chord we shall call the middle chord; the inscribed quadrilateral degenerates, for the point \(M\), into a chord.

For the extreme points of the path of the slider \(A\) and \(B\), the system of conjugate points degenerates into two points \(M_1\) and \(M_2\); we shall call this line the chord of the dead points. For the case under consideration—the slider-crank mechanism of the basic type—the chord of the dead points is a diameter.

If on the line \(AB\) we take a series of equidistant points \(a, b\), etc., and find their corresponding conjugate chords, then, if all the geometrical constructions are carried out carefully, we shall see that all the conjugate chords intersect at one point.

Let us denote this point by the letter \(\lambda\), and its distance from the center \(O\) by \(\xi\).

Since the system of conjugate points consists of two pairs symmetrically situated with respect to the line \(M_1M_2\) (see the drawing), it is clear that all conjugate chords, whatever they may be, will intersect in pairs on the line \(M_1M_2\); in other words, the geometric locus of the points of intersection of paired conjugate chords is a straight line—the line of the dead points.

If, however, as has occurred in the drawing, all the conjugate chords intersect at one point—this point being the locus—then it is clear that the distance of this point from the center is a constant quantity, independent of the position of the chosen equidistant points \(a, b\), etc.

Let us see whether this is so.

Let \(AB\) represent the path of the slider (see Drawing III). Let us take on the line \(AB\) two equidistant points and find for them the corresponding one pair of conjugate points. To do this, let us notch the crank circle, with a radius equal to the length of the connecting rod, from the points \(L\) and \(N\)). As a result of the notches, we find the points \(L_1\) and \(N_1\). Let us denote by \(\alpha_1\) the angle of rotation of the crank corresponding to the path \(AN\) traversed by the slider from position \(A\), and by \(\alpha_2\) the angle through which the crank must rotate so that, during the reverse* motion of the slider, the point \(L\) coincides with the point \(A\). Let us denote by the letter \(c_1\) the distance of the point \(N\) from \(O\)—the center of the crank circle—and by the letter \(c_2\) the distance of the point \(L\)

*) In order to obtain a larger value of \(\xi\), the drawing is to be made without observing scale.
A. U.

The numerator is the sine of the difference of the two angles

\[ \left(\frac{\alpha_1-\alpha_2}{2}\right) \text{ and } \left(\frac{\alpha_1+\alpha_2}{2}\right), \]

whence

\[ \frac{\alpha_1-\alpha_2}{2}-\frac{\alpha_1+\alpha_2}{2} = -\frac{2\alpha_2}{2} = -\alpha_2 . \]

Consequently:

\[ \frac{\partial \xi}{\partial \alpha_1} = \frac{\sin(-\alpha_2)} {2\cos^2\frac{\alpha_1-\alpha_2}{2}} = -\frac{\sin\alpha_2} {2\cos^2\frac{\alpha_1-\alpha_2}{2}} . \]

In the same way, we find that

\[ \frac{\partial \xi}{\partial \alpha_2} = \frac{ -\sin\frac{\alpha_1+\alpha_2}{2}\cos\frac{\alpha_1-\alpha_2}{2} -\sin\frac{\alpha_1-\alpha_2}{2}\cos\frac{\alpha_1+\alpha_2}{2} } {2\cos^2\frac{\alpha_1-\alpha_2}{2}} = -\frac{\sin\alpha_1} {2\cos^2\frac{\alpha_1-\alpha_2}{2}} . \]

Let us now determine \(d\alpha_1\) and \(d\alpha_2\).

From the triangle \(NN_1O\) we have:

\[ \cos\alpha_1 = \frac{\overline{ON_1}^{\,2}-\overline{NN_1}^{\,2}+\overline{ON}^{\,2}} {2\cdot\overline{ON_1}\cdot\overline{ON}} = \frac{b^2-a^2+c_1^2}{2bc_1}. \]

From the triangle \(LL_1O\) we obtain:

\[ \cos\alpha_1 = \frac{\overline{OL_1}^{\,2}-\overline{LL_1}^{\,2}+\overline{OL}^{\,2}} {2\cdot\overline{OL_1}\cdot\overline{OL}} = \frac{b^2-a^2+c_2^2}{2bc_2}. \]

Denoting, for brevity,

\[ \frac{b^2-a^2}{2b}=m \quad\text{and}\quad \frac{1}{2b}=l, \]

we obtain:

\[ \cos\alpha_1=\frac{m+lc_1^2}{c_1} \quad\text{and}\quad \cos\alpha_2=\frac{m+lc_2^2}{c_2}. \]

Differentiating these expressions, we have:

\[ [2]\quad -\sin\alpha_1\,d\alpha_1 = \frac{2lc_1^2-m-lc_1^2}{c_1^2}\,dc_1 = \frac{lc_1^2-m}{c_1^2}\,dc_1 \quad\text{and}\quad -\sin\alpha_2\,d\alpha_2 = \frac{lc_2^2-m}{c_2^2}\,dc_2 . \]

The points \(N\) and \(L\) are symmetric with respect to the point \(M\)—the midpoint of the path of the slider. Therefore, if we denote the distance \(MO\) by \(n\), where \(n\) is some constant segment,* then we may write:

\[ ON=n+k \quad \text{and} \quad OL=n-k, \]

where \(k\) is a variable quantity.

Thus, we have:

\[ ON=c_1=n+k \]

\[ OL=c_2=n-k. \]

Adding these two expressions, we obtain:

\[ c_1+c_2=2n,\quad \text{i.e.,} \]

the sum of the distances of two equidistant points from the center of the crank circle is a constant quantity, equal to twice the distance of the midpoint to the center of rotation.

Differentiating this last expression, we have:

\[ dc_1+dc_2=0,\quad \text{whence,} \]

\[ dc_2=-dc_1. \]

Substituting these values into formulas \([2]\), we obtain

\[ -\sin\alpha_1\,d\alpha_1=\frac{lc_1^2-m}{c_1^2}\,dc_1, \]

\[ -\sin\alpha_2\,d\alpha_2=\frac{lc_2^2-m}{c_2^2}\,dc_2 = -\frac{lc_2^2-m}{c_2^2}\,dc_1. \]

Whence we have:

\[ d\alpha_1=-\frac{lc_1^2-m}{c_1^2\sin\alpha_1}\,dc_1, \]

\[ d\alpha_2=+\frac{lc_2^2-m}{c_2^2\sin\alpha_2}\,dc_1. \]

Introducing all the quantities found into formula \([1]\), we obtain the following expression for the total differential:

\[ [3]\qquad d\xi= \left\{ \frac{\sin\alpha_2}{2\cos^2\frac{\alpha_1-\alpha_2}{2}} \cdot \frac{lc_1^2-m}{c_1^2\sin\alpha_1} - \frac{\sin\alpha_1}{2\cos^2\frac{\alpha_1-\alpha_2}{2}} \cdot \frac{lc_2^2-m}{c_2^2\sin\alpha_2} \right\}dc_1=0. \]

Since \(dc_1 \ne 0\), it is necessary that the expression in braces be equal to \(0\), i.e.:

* With an unchanged length of the connecting rod, the distance of the midpoint of the oscillations of the slider from the center of the shaft is a constant quantity—equal to the length of the connecting rod.
A. U.

\[ \frac{\sin \alpha_2}{2\cos^2 \frac{\alpha_1-\alpha_2}{2}}\cdot \frac{lc_1^2-m}{c_1^2\sin \alpha_1} - \frac{\sin \alpha_1}{2\cos^2 \frac{\alpha_1-\alpha_2}{2}}\cdot \frac{lc_2^2-m}{c_2^2\sin \alpha_2} =0. \]

Whence

\[ \frac{\sin \alpha_2}{2\cos^2 \frac{\alpha_1-\alpha_2}{2}}\cdot \frac{lc_1^2-m}{c_1^2\sin \alpha_1} = \frac{\sin \alpha_1}{2\cos^2 \frac{\alpha_1-\alpha_2}{2}}\cdot \frac{lc_2^2-m}{c_2^2\sin \alpha_2}, \]

which, after cancellation and after some transformations, gives:

\[ [4]\qquad \frac{c_2^2\sin^2\alpha_2}{lc_2^2-m} = \frac{c_1^2\sin^2\alpha_1}{lc_1^2-m}. \]

Since \(\sin^2\alpha=1-\cos^2\alpha\), we may write, using the values of the cosines already known to us:

\[ \sin^2\alpha_1 = 1-\cos^2\alpha_1 = 1-\left(\frac{m+lc_1^2}{c_1}\right)^2 = \frac{c_1^2-(m+lc_1^2)^2}{c_1^2} \quad\text{and} \]

\[ \sin^2\alpha_2 = 1-\cos^2\alpha_2 = 1-\left(\frac{m+lc_2^2}{c_2}\right)^2 = \frac{c_2^2-(m+lc_2^2)^2}{c_2^2}. \]

Substituting the found values of the sines into formula [3]:

\[ \frac{c_2^2}{lc_2^2-m}\cdot \frac{c_2^2-(m+lc_2^2)^2}{c_2^2} = \frac{c_1^2}{lc_1^2-m}\cdot \frac{c_1^2-(m+lc_1^2)^2}{c_1^2}, \]

which gives:

\[ \frac{c_2^2-(m+lc_2^2)^2}{lc_2^2-m} = \frac{c_1^2-(m+lc_1^2)^2}{lc_1^2-m}. \]

Multiplying the means and extremes, we then obtain:

\[ \{c_2^2-(m+lc_2^2)^2\}(lc_1^2-m) = \{c_1^2-(m+lc_1^2)^2\}(lc_2^2-m). \]

Expanding the brackets, we have:

\[ lc_1^2c_2^2-mc_2^2-(m+lc_2^2)^2(lc_1^2-m) = lc_1^2c_2^2-mc_1^2-(m+lc_1^2)^2(lc^2-m). \]

Canceling like terms in both parts and carrying out the further operations, we obtain:

\[ -mc_2^2-(m^2+l^2c_2^4+2mlc_2^2)(lc_1^2-m) = -mc_1^2-(m^2+l^2c_1^4+2mlc_1^2)(lc_2^2-m). \]

Expanding the brackets and collecting like terms, we obtain the expression:

\[ m(c_1^2-c_2^2)-3m^2l(c_1^2-c_2^2)+l^3c_1^2c_2^2(c_1^2-c_2^2)-ml^2(c_1^4-c_2^4)=0, \]

which can be represented as the product of two factors

[5]

\[ (c_1^2-c_2^2)\{m-3m^2l+l^3c_1^2c_2^2-ml^2(c_1^2+c_2^2)\}=0. \]

Expression [5] resulted from replacing the variable quantities \(\alpha_1\) and \(\alpha_2\) in equation [3] by other variable quantities \(c_1\) and \(c_2\).

If the total differential of the quantity \(\xi\) is to be equal to zero for all possible values of \(\alpha_1\) and \(\alpha_2\), then it is obvious that equation [5] must hold for all values of \(c_1\) and \(c_2\); whence it follows that

[6]

\[ m-3m^2l+l^3c_1^2c_2^2-ml^2(c_1^2+c_2^2)=0, \]

since the factor \((c_1^2-c_2^2)\) becomes zero only in the particular case when \(c_1=\pm c_2\).

We shall therefore deal with equation [6].

We had earlier: \(c_1+c_2=2n\).

After squaring both sides of this equality, we obtain:

\[ c_1^2+c_2^2=4n^2-2c_1c_2. \]

Substitute this into equation [6]:

\[ m-3m^2l+l^3c_1^2c_2^2-ml^2(4n^2-2c_1c_2)=0, \]

\[ m-3m^2l+l^3c_1^2c_2^2-4mn^2l^2+2ml^2c_1c_2=0. \]

Take \(m\) outside the parentheses in the first, second, and fourth terms. Then we obtain:

[7]

\[ m(1-3ml-4l^2n^2)+2ml^2c_1c_2+l^3c_1^2c_2^2=0. \]

Transform the expression in parentheses, introducing the values of the quantities \(m\) and \(l\).

We know that \(m=\dfrac{b^2-a^2}{2b}\) and \(l=\dfrac{1}{2b}\), where \(b\) is the radius of the crank, taken by us, as already mentioned, as unity; \(a\) is the length of the connecting rod, equal, obviously, to \(n\)—the distance from the middle of the oscillations of the slide to the center of the main shaft.

Thus, we finally have:

\[ m=\frac{b^2-a^2}{2b}=\frac{1-n^2}{2}, \]

\[ l=\frac{1}{2b}=\frac{1}{2}, \]

\[ l^2=\frac{1}{4}. \]

Consequently:

\[ 1-3ml-4l^2n=1-\frac{3\cdot(1-n^2)}{2\cdot2}-\frac{4n^2}{4},\quad \text{or} \]

\[ 1 - 3ml - 4l^2 n = \frac{4 - 3(1 - n^2) - 4n^2}{4}, \]

\[ 1 - 3ml - 4l^2 n = \frac{1 - n^2}{4}, \]

\[ \frac{1 - n^2}{4} = \frac{1 - n^2}{2}\cdot \frac{1}{2} = m \cdot l, \]

i.e.

\[ 1 - 3ml - 4l^2 n = m \cdot l. \]

Substituting the obtained result into expression [7], we have:

\[ m^2 l + 2ml^2 c_1 c_2 + l^3 c_1^2 c_2^2 = 0, \]

which, upon division by \(l\), gives:

\[ \tag{8} m^2 + 2mlc_1 c_2 + l^2 c_1^2 c_2^2 = 0. \]

The left-hand side of the equation obtained is the square of the sum of two quantities; consequently, we may write:

\[ \tag{9} (m + lc_1 c_2)^2 = 0. \]

From expression [9] we conclude that the total differential of the quantity \(\xi\) is not equal to zero for all possible values of \(c_1\) and \(c_2\), since expression [9] vanishes only for particular values of \(c_1\) and \(c_2\), namely, when

\[ c_1 \cdot c_2 = -\frac{m}{l} = -(1 - n^2) = n^2 - 1. \]

Thus, the quantity \(\xi\) is not a constant quantity and varies together with the quantities \(\alpha_1\) and \(\alpha_2\).

Let us determine the nature of the variation of this quantity.

We had earlier:

\[ \xi = \frac{\cos \dfrac{\alpha_1 + \alpha_2}{2}} {\cos \dfrac{\alpha_1 - \alpha_2}{2}}. \]

Let us now find another expression for \(\xi\), showing the dependence of \(\xi\) on the radius of the crank and the length of the connecting rod.

Let (Fig. IV) the origin of coordinates coincide with the center of the crank circle.

Then the equation of this circle will be:

\[ x^2 + y^2 = 1. \]

Let us find any two conjugate points, whose coordinates, say, will be \((x_1, y_1)\) and \((x_2, y_2)\). These points, as we know, are beyond the crank circle by a radius equal to the length of ...

Obviously, \(d_1\), as the hypotenuse, will be determined by its legs, or

\[ d_1^2=(x_1-\xi)^2+y_1^2. \]

Expanding the brackets, we have:

\[ d_1^2=x_1^2+\xi^2-2x_1\xi+y_1^2. \]

Since the point \((x_1,y_1)\) belongs to the crank circle, for it the equality

\[ x_1^2+y_1^2=1, \]

will hold; and therefore,

\[ \tag{15} d_1^2=1+\xi^2-2x_1\xi. \]

Let us now determine \(d_2\).

As before, we shall have:

\[ d_2^2=(\xi-x_2)^2+y_2^2, \]

or, expanding the brackets,

\[ d_2^2=x_2^2+\xi^2-2x_2\xi+y_2^2. \]

Since

\[ x_2^2+y_2^2=1, \]

\[ \tag{16} d_2^2=1+\xi^2-2x_2\xi. \]

Multiplying [15] and [16], we then obtain:

\[ \tag{17} d_1^2\cdot d_2^2=(1+\xi^2-2x_1\xi)\cdot(1+\xi^2-2x_2\xi). \]

Squaring both sides of equation [14], we obtain:

\[ \tag{18} d_1^2\cdot d_2^2=(1-\xi^2)^2. \]

Combining equations [17] and [18], we obtain

\[ (1-\xi^2)^2=(1+\xi^2-2x_1\xi)\cdot(1+\xi^2-2x_2\xi), \]

or

\[ (1-\xi^2)^2=\{(1+\xi^2)-2x_1\xi\}\cdot\{(1+\xi^2)-2x_2\xi\}. \]

Expanding the large brackets, we obtain:

\[ (1-\xi^2)^2=(1+\xi^2)^2-2\xi(1+\xi^2)(x_1+x_2)+4\xi^2x_1x_2. \]

Transposing all terms to one side:

\[ (1+\xi^2)^2-(1-\xi^2)^2-2\xi(1+\xi^2)(x_1+x_2)+4\xi^2x_1x_2=0, \]

or, expanding the brackets in the first two terms and reducing, we have:

\[ 4\xi^2+4\xi^2x_1x_2-2\xi(1+\xi^2)(x_1+x_2)=0. \]

The Coupler Pole.

Taking \(2\xi\) outside the brackets, we then obtain:

\[ 2\xi \{2\xi+2\xi x_1x_2-(1+\xi^2)(x_1+x_2)\}=0. \]

Since \(\xi\) is not equal to zero, it is clear that it must be:

\[ 2\xi+2\xi x_1x_2-(1+\xi^2)(x_1+x_2)=0, \]

or

\[ 2\xi(1+x_1x_2)-(1+\xi^2)(x_1+x_2)=0. \]

This equation may be written as follows:

\[ \xi^2-2\cdot \frac{1+x_1x_2}{x_1+x_2}\,\xi+1=0. \]

Denoting the fraction \(\dfrac{1+x_1x_2}{x_1+x_2}\) by \(A\), we have:

\[ \tag{19} \xi^2-2A\xi+1=0. \]

Whence

\[ \tag{20} \xi=A\pm\sqrt{A^2-1}, \]

Let us find the value of the quantity \(A\). To do this, we shall first determine what the product \(x_1x_2\) will be equal to.

Previously we had:

\[ x_1=\frac{1+l^2+2nl}{2(n+l)}, \qquad x_2=\frac{1+l^2-2nl}{2(n-l)}. \]

Whence

\[ x_1x_2=\frac{1}{4(n^2-l^2)}\{(1+l^2)^2-4n^2l^2\}. \]

Consequently:

\[ 1+x_1x_2= \frac{4(n^2-l^2)+(1+l^2)^2-4n^2l^2}{4(n^2-l^2)}, \]

or

\[ 1+x_1x_2= \frac{4n^2-4l^2+1+l^4+2l^2-4n^2l^2}{4(n^2-l^2)}, \]

\[ 1+x_1x_2= \frac{4n^2(1-l^2)+(1-l^2)^2}{4(n^2-l^2)} = \frac{(1-l^2)(4n^2+1-l^2)}{4(n^2-l^2)}. \]

Now let us determine the value of the sum \(x_1+x_2\).

A. Ugarov.

We shall have:

\[ x_1+x_2=\frac{(1+l^2+2nl)(n-l)+(1+l^2-2nl)(n+l)}{2(n^2-l^2)}. \]

Whence

\[ x_1+x_2=\frac{(1+l^2)(n-l+n+l)+2nl(n-l-n-l)}{2(n^2-l^2)}, \]

or

\[ x_1+x_2=\frac{2n(1+l^2)-4nl^2}{2(n^2-l^2)}=\frac{n(1-l^2)}{n^2-l^2}. \]

Now we can write the value of the quantity \(A\):

\[ A=\frac{1+x_1x_2}{x_1+x_2} =\frac{(1-l^2)(4n^2+1-l^2)(n^2-l^2)}{4(n^2-l^2)\cdot n\cdot(1-l^2)} =\frac{4n^2+1-l^2}{4n}. \]

Thus, the coefficient \(A\) in equation [19] is a variable quantity depending on the quantity \(l\), and consequently the quantity \(\xi\) from [20] is likewise a variable quantity.

Let us see what values \(A\) takes for the limiting values of \(l\).

For \(l=0\) we obtain:

\[ A=\frac{4n^2+1}{4n}=n+\frac{1}{4n}. \]

For \(l=1\) we have \(A=n\).

Thus, the minimum value of the coefficient \(A\) is \(n\), while the maximum is \(n+\frac{1}{4n}\).

Corresponding to these two values, \(\xi\) too will have minimum and maximum values, between which all the other values possible for \(\xi\) will lie; in other words, the difference between the two extreme values of \(\xi\) will represent the path traversed by the point \(\lambda\) during a complete revolution of the crank.

From expression [20] it is seen that \(\xi\) is the algebraic sum of two quantities, \(A\) and \(\sqrt{A^2-1}\).

Since, in our consideration, when forming equations [10], [11], and [12], we took the radius of the crank to be equal to unity, and the length of the connecting rod to be equal to \(n\), it is evident that, for crank motion to be possible, the condition

\[ n\geq 1. \]

must be satisfied.

On the other hand, each pair of conjugate chords intersects inside the crank circle, and therefore the distance of the point \(\lambda\) from the center of the circle, i.e. \(\xi\), is always less than the radius; this gives us the right to assert that \(\xi\) is some proper fraction.

Hence we arrive at the conclusion that, in order to determine the value of \(\xi\) from expression [20], the radical must be taken with the minus sign. To determine at what values of the coefficient \(A\) the maximum and minimum values of the quantity \(\xi\) are obtained, we transform expression [20], taking \(A\) outside the brackets:

\[ \xi = A-\sqrt{A^2-1}=A\left(1-\sqrt{1-\frac{1}{A^2}}\right). \]

Considering this equality, we see that the expression in brackets decreases together with the increase of \(A\) and becomes zero when \(A\) is equal to infinity. Hence we conclude that the maximum value of the coefficient \(A\) corresponds to the minimum value of \(\xi\), and conversely.

Thus, we shall have:

\[ \xi \operatorname{minim}= \frac{4n^2+1-\sqrt{16n^4+1+8n^2-16n^2}}{4n} = \frac{4n^2+1-\sqrt{(4n^2-1)^2}}{4n}, \]

or

\[ \xi \min = \frac{4n^2+1-4n^2+1}{4n} = \frac{1}{2n}. \]

In exactly the same way we obtain:

\[ \xi \operatorname{mxm}=n-\sqrt{n^2-1}. \]

The length of the connecting rod \(n\) in machines of the normal type is usually taken equal to five times the radius of the crank.

We have taken the radius of the crank equal to unity; consequently, \(n=5\).

Let us compute, for this value of \(n\), the quantities \(\xi \operatorname{mxm}\) and \(\xi \min\):

\[ \xi \operatorname{mxm}=5-\sqrt{5^2-1}=5-\sqrt{24}=5-4.8990=0.101. \]

\[ \xi \min=\frac{1}{2.5}=0.10. \]

The difference between the maximum and minimum values of \(\xi\) will therefore be equal to one thousandth of the radius.

When drawing steam-distribution diagrams, the diameter of the crank circle is often taken equal to \(200\ \mathrm{m/m}\); for such a circle we shall have the displacement of point \(\lambda\) from one of its extreme positions \((\xi-\text{maximum})\) to the other \((\xi-\text{minimum})\) equal to one tenth of a millimeter, which, obviously, lies far beyond the limits of attainable accuracy of drawing. For a circle half a meter in diameter (this would already be a large circle for a Zeuner or Müller diagram), the displacement of point \(\lambda\) will constitute a segment of a straight line of length ...

a quarter of a millimeter, which is the thickness of an ordinary drafting line drawn with a not especially sharp pencil.

For a visual appreciation of the fact that, as \(n\)—the ratio of the length of the connecting rod \(L\) to the crank radius \(R\)—increases, the difference between the maximum and minimum values of \(\xi\) rapidly decreases, a table is given below. Since, from the kinematic point of view, the eccentric rod together with the eccentric disk must be regarded as a connecting-rod–crank mechanism, the last two columns of the table give data for the most frequently occurring ratios of the length of the eccentric rod to the radius of the eccentric.

Table.

\(n=\dfrac{L}{R}=\)	5	6	10	20
\(\xi\) maximum . .	0,101	0,0839	0,0501	0,02502
\(\xi\) minimum . .	0,100	0,0833	0,0500	0,02500
\(\xi\) max. — \(\xi\) min.	0,001	0,0006	0,0001	0,00002

Let us summarize the foregoing.

The geometrical locus of the points of intersection of conjugate chords for the basic connecting-rod–crank mechanism is a segment of a straight line coinciding with the line of dead centers. This segment has an extraordinarily small length, decreasing rapidly as the ratio of the length of the connecting rod to the crank increases; when this ratio changes from 5 to 20, the segment successively assumes dimensions from one thousandth to two hundred-thousandths parts of the radius.

Thus, even for a diagram with a diameter of the crank circle equal to \(500\ \mathrm{m}/\mathrm{m}\), this segment will have a length from one quarter to five thousandths of a millimeter.

On this basis one may assert: all conjugate chords intersect at one point, situated from the center of the crank circle, in the direction toward the slider, at a distance

\[ \xi=\frac{1}{2n}, \]

where \(n\) is the number showing the ratio of the length of the connecting rod to the radius of the crank.

This point—which in what follows we shall denote by the letter \(\lambda\)—we shall call the connecting-rod pole.

Since all conjugate chords intersect at one point, it is clear that, in order to find the connecting-rod pole of the basic connecting-rod–crank ...

of the crank mechanism it is sufficient to find the point of intersection of the middle chord with the line (chord) of the dead points.

Thus, the position of the coupler pole can be determined graphically by an extremely simple construction.

As will be seen further on, the coupler pole possesses properties that characterize the given crank mechanism and make it possible to determine the conditions of motion of its individual links.

CHAPTER THREE.

The rectilinear-generating crank mechanism. Conditions for the possibility of its existence.

Let us now pass to the consideration of a derived slider-crank mechanism with motion of the slider along a straight line—such a mechanism we shall, for short, call rectilinear-generating.

Let us first seek the conditions for the possibility of the existence of mechanisms of this kind.

Suppose that we are given the distance between the center of rotation of the crank and the midpoint of the oscillations of the slider, i.e. the length of the segment of the base line between these points; the length of the slider’s path and the angle of inclination \(\vartheta\) are given.

It is required to construct a slider-crank mechanism capable of satisfying the given conditions of motion of the slider.

In order to find the radius of the crank and the length of the connecting rod, we shall reason as follows: at the dead points of the slider’s path (see drawing V) the crank and the connecting rod lie on one straight line, and since the crank always passes through the center of its rotation, it is evident that the position of the said straight line is obtained by simply joining the corresponding dead point with the center of the crank circle.

Further: for the dead point nearest to the center of the crank circle, the connecting rod and the crank are superposed one upon the other; for the dead point farther from the center, the crank is the continuation of the connecting rod.

Denoting by \(d_1\) and \(d_2\) the distances of the dead points from the center of the crank circle, and by the letters \(r\) and \(l\) the radius of the crank and the length of the connecting rod, on the basis of what has been said we may write:

\[ d_1 = l - r, \]

\[ d_2 = l + r. \]

Adding and subtracting these two equalities term by term, we obtain successively:

\[ r=\frac{d_2-d_1}{2}\quad \text{and}\quad l=\frac{d_1+d_2}{2}, \]

i.e., for a rectilinear-generating crank-and-connecting-rod mechanism, the radius of the crank is equal to the half-difference, and the length of the connecting rod to the half-sum, of the distances of the dead points from the center of rotation of the crank.

Suppose that the path of the slider is perpendicular to the base line; then both dead points lie at the same distance from the center of rotation of the crank, forming an isosceles triangle. In this case the condition is \(d_1=d_2\), and, consequently, the radius of the crank is

\[ r=\frac{d_1-d_2}{2}=0. \]

In other words: a rectilinear-generating crank-and-connecting-rod mechanism with the path of the slider perpendicular to the base line is impossible.

Thus, the angle \(\vartheta\) cannot be equal to \(90^\circ\). If, however, the angle \(\vartheta\) is equal to zero degrees, then the path of the slider coincides with the base line and the derived crank-and-connecting-rod mechanism becomes the basic one.

Thus, we arrive at the condition for \(\vartheta\): \(0<\vartheta<90\). However, this condition is far from complete, since it does not give the final maximum limit for the angle \(\vartheta\), which, as we shall see below, exists for every given derived mechanism.

Let us find the functional dependence between the elements of the derived mechanism.

Let the length of the path of the slider \(AB\) be equal to \(2\sigma\), where \(\sigma\) is half the stroke, and let \(OM\)—the segment of the base line—be equal to \(n\).

The center \(O\) of rotation of the crank, the midpoint \(M\) of the path of the slider, and each of the dead points \(A\) and \(B\) form two triangles, \(OAM\) and \(OMB\), and the angles between the base line and the path of the slider in these triangles will be, respectively, \(OMA=\vartheta\) and \(OMB=180^\circ-\vartheta\). From these triangles we can determine \(d_1\) and \(d_2\), i.e. the sides \(OA\) and \(OB\). Thus, we have:

\[ d_1^2=(l-r)^2=n^2+\sigma^2-2n\sigma\cos\vartheta, \]

\[ d_2^2=(l+r)^2=n^2+\sigma^2+2n\sigma\cos\vartheta. \]

Expanding the brackets:

\[ l^2+r^2-2lr=n^2+\sigma^2-2n\sigma\cos\vartheta, \]

\[ l^2+r^2+2lr=n^2+\sigma^2+2n\sigma\cos\vartheta. \]

Subtracting the first equation from the second, we have:

\[ 4lr = 4n\sigma\cos\vartheta \quad \text{or} \]

\[ [21]\qquad lr = n\sigma\cos\vartheta . \]

Calling the quantity \(\sigma\cos\vartheta\)—the projection of half the path of the slider onto the base line—\(\rho\), we have from [21]:

\[ [22]\qquad l.r = n.\rho, \]

i.e. the product of the lengths of the connecting rod and the crank in a rectilinear-progressive mechanism is equal to the product of the lengths of the base line and the projection of the slider’s half-path onto this same line.

Expression [21] represents the sought functional dependence between the elements of the mechanism.

This expression contains five quantities. If three quantities \(n\), \(\sigma\), and \(\cos\vartheta\) are given, then they completely determine the triangles \(OMA\) and \(OMB\), and consequently the sides \(OA\) and \(OB\), equal to \(d_1\) and \(d_2\), are also determined.

But by the quantities \(d_1\) and \(d_2\), as we know, the dimensions of the connecting rod and crank of the given mechanism are determined; thus, by three given quantities we completely determine the mechanism, and by expression [21] we have the mutual relation of all the elements of the mechanism.

To solve the question we have posed concerning the limit for the angle \(\vartheta\), let us turn to an example.

Suppose that we are given the magnitude \(n\) and an angle \(\vartheta\) less than \(90^\circ\); it is required to find the mechanism for which the path of the slider would be the greatest of all possible paths for the given angle.

It is required to find the length of the slider’s path, in other words, to find the positions of the dead points.

Since the point \(M\) (see Drawing VI) is the midpoint, and it is given to us (given \(OM=n\)), it is enough for us to determine the position of one dead point—the other will be symmetrical to it with respect to the point \(M\). Thus, on the line \(AB\) it is required to find the point farthest from \(M\) which would be a dead point of the desired mechanism. We know that a dead point is such a point at which, first, the slider changes its direction and, second, the connecting rod coincides in direction with the crank, forming with it a straight line passing through the center of rotation of the crank. These two conditions make it possible to determine the dead point nearest to the center of rotation. It is evident that the slider will arrive at it when the rotating end of the connecting rod is farthest from the direction of the slider’s path. We know that the rotating end of the connecting rod moves along the crank circle; consequently, we must find on this circle the point farthest from

line \(AB\). Such a point is the intersection of the perpendicular to the line \(AB\) from the center \(O\) with the crank circle. We know only the position \(O\) of the center of this circle, but the radius is not known. Therefore, in order to find the elements of the mechanism from the given \(n\) and \(\vartheta\), we proceed as follows:

From \(O\) we drop a perpendicular to the line \(ML\); for this, on \(OM\), as on a diameter, we describe a semicircle; its intersection with \(ML\) gives the required nearest dead point, and thereby the whole mechanism is also determined.

We obtain:

\[ \sigma = n\cos\vartheta, \]

\[ d_1 = n\sin\vartheta, \]

\[ d_2 = \sqrt{n^2\sin^2\vartheta + 4n^2\cos^2\vartheta}, \]

and consequently we have \(r\) and \(l\) as the half-difference and half-sum of \(d_1\) and \(d_2\).

The length of the connecting rod is expressed thus:

\[ [23]\qquad l = r + n\sin\vartheta,\quad \text{or}\quad l-r = n\sin\vartheta. \]

Expression [23] shows the dependence between the lengths of the connecting rod and the crank for the greatest possible travel of the slider at a given angle \(\vartheta\).

It is clear from this that, for all other—smaller—lengths of the slider path, in order for crank motion to be possible, it is necessary that the length of the connecting rod, diminished by the magnitude of the crank radius, should be greater than the distance of the center of the crank circle from the line of the slider path.

The greatest travel of the slider is \(2\sigma = 2n\cos\vartheta\); consequently, the remaining possible travels are less than this quantity.

Let us consider another example.

A segment of the base line \(OM=n\) is given, and the length of the slider path is given as

\[ AB = A^1B^1 = 2r_g; \]

it is required to find the greatest angle of inclination \(\vartheta\). (See drawing VIII.)

Here we have the case of the formation of a derived mechanism, in which, for all changes of \(\vartheta\), the length of the slider path must remain the same as in the base mechanism, i.e. equal to \(2r_g\).

We found earlier that, for the limiting case, the nearest dead point is the foot of the perpendicular from the center of the crank circle to the direction of the slider path. Therefore the greatest angle \(\vartheta\) is the acute angle of a right triangle.

The whole mechanism is determined in the following way: on the line \(OM\), as on a diameter, construct a semicircle and intersect it with the radius \(r_g\) from the point \(M\) as from a center; the resulting point \(A\) will be the required nearest dead point; from it the remaining elements of the mechanism are also determined. The angle \(\vartheta\) is determined from the condition:

\[ \sigma = r_g = n\cos\vartheta. \]

The radius of the crank circle is still determined from \(d_1\) and \(d_2\)—the distances of the dead points from the center of the circle.

In the drawing, around the center \(O\), a circle is drawn in dotted line corresponding to the crank \(r_g\) of the basic mechanism from which the derived mechanism under consideration is formed, in order to illustrate the change, mentioned in the first chapter, of the radius of the crank and the length of the connecting rod with a change in the direction of the path of the slider.

The third variant of the rectilinear-derived mechanism is the mechanism in which the projection of the path of the slider onto the base line, for all changes in the magnitude of the angle \(\vartheta\), remains equal to the length of the slider path \(A'B' = 2r_g\) of the basic mechanism. (See drawing VIII).

As before, the greatest angle \(\vartheta\) is found from a right triangle whose hypotenuse is the segment \(OM\). The nearest dead point is determined by the intersection of the perpendicular at the point \(A'\) with the circle described on \(OM\) as diameter. The crank circle of the basic mechanism is shown on the drawing in dotted line.

We found earlier for rectilinear-derived mechanisms the expression [22]:

\[ l \cdot r = n \cdot \rho . \]

In the variant just considered, \(\rho\) is a constant quantity, and therefore \(l \cdot r = n \cdot \rho = \mathrm{const.}\)

In other words: for all derived mechanisms, with the same \(n\) and with equality of the projections of the half-path of the slider onto the base line, the product of the lengths of the connecting rod and the crank is a constant quantity.

On the basis of the examples analyzed, we can thus express the conditions for the possibility of existence of a rectilinear-derived slider-crank mechanism:

\[ [24]\qquad \begin{cases} l \cdot r = n \sigma \cdot \cos \vartheta,\\ l - r \ge n \cdot \sin \vartheta,\\ \delta \le n \cdot \cos \vartheta. \end{cases} \]

In conclusion, let us consider the limiting case of a rectilinear-derived mechanism in which the length of the connecting rod is equal to the diameter of the crank circle, i.e. \(l = 2r\). (See drawing IX).

In this case

\[ d_1 = l - r = 2r - r = r, \]

\[ d_2 = l + r = 2r + r = 3r. \]

The length of the path of the slider

\[ 2\sigma = \sqrt{2r^2 - r^2} = r\sqrt{8}, \]

whence

\[ \sigma = r\sqrt{2}. \]

The segment of the base line \(OM=n\) is determined as follows:

\[ n=\sqrt{r^2+(r\sqrt{2})^2}=r\sqrt{3}. \]

Substituting the values found into the first formula from [24], we have:

\[ 2r^2=r^2\sqrt{3}\cdot\sqrt{2}\cdot\cos\vartheta, \]
whence

\[ \cos\vartheta=\sqrt{\frac{2}{3}}=\mathrm{const}. \]

In other words: all rectilinear-producing mechanisms of the indicated limiting type are geometrically similar to one another.

The dead point \(A\) has, in these mechanisms, the property that the slider, having arrived at it, upon further rotation of the crank from the point \(D\), can move both in the direction toward \(B\) and in the opposite direction toward \(L\) and farther. Thus, the point \(A\) is the dead point of two mechanisms, symmetric with respect to the line \(AD\) and identical in dimensions.

CHAPTER FOUR.

Properties of the chords of a rectilinear-producing crank mechanism.

Having determined the conditions for the possibility of existence of a rectilinear-producing connecting-rod crank mechanism, let us pass to the study of some of its properties.

Let all the elements of the mechanism be given to us (see drawing V): the radius of the crank—\(r\), the length of the connecting rod—\(l\), the segment of the base line \(OM=n\), the path of the slider \(AB=2\sigma\), and the angle of inclination of the slider path \(\vartheta\).

Take on the line \(AB\) two symmetric points \(P\) and \(Q\) at a distance \(k\) from the midpoint \(M\), where \(k\) is a variable quantity that can take values from \(0\) to \(\sigma\). On the crank circle let us find the conjugate points corresponding to the symmetric points taken. For this, as we know, it is necessary to notch the crank circle from the points \(P\) and \(Q\) with a radius equal to the length of the connecting rod \(l\). Let us seek analytic expressions for the required points.

From the right triangle \(PP'M\) we have:

\[ P'M=k\cdot\cos\vartheta \quad\text{and}\quad PP'=k\cdot\sin\vartheta. \]

Let

\[ k\cdot\cos\vartheta=\alpha, \]
\[ k\cdot\sin\vartheta=\beta. \]

The Coupler Pole.

Then the coordinates of the center of the circle described from point \(P\)—or simply, the coordinates of point \(P\) in a rectangular system with origin of coordinates at \(O\), and with the \(x\)-axis coinciding in direction with the base line of the mechanism—will be: abscissa \(=(n-\alpha)\), ordinate \(=(\beta)\); for point \(Q\) we shall have, respectively: \((n+\alpha)\) and \((-\beta)\).

The equation of the crank circle, as coinciding in its center with the origin of coordinates, will have the following form:

\[ [25]\qquad x^2+y^2=r^2. \]

The equation of the first circle (drawn from point \(P\)) takes the form:

\[ [26]\qquad \{x-(n-\alpha)\}^2+(y-\beta)^2=l^2. \]

The equation of the second circle (from point \(Q\)) will accordingly be:

\[ [27]\qquad \{x-(n+\alpha)\}^2+(y+\beta)^2=l. \]

To find the points of intersection of the coupler circles with the crank circle, we shall find the equations of the radical axes.

The first radical axis is found by subtracting equation \([26]\) from \([25]\). Let us do this.

\[ x^2+y^2=r^2, \]

\[ x^2+(n-\alpha)^2-2(n-\alpha)x+y^2+\beta^2-2\beta y=l^2. \]

Subtracting, we have:

\[ [28]\qquad 2(n-\alpha)x+2\beta y-(n-\alpha)^2-\beta^2=r^2-l^2. \]

To find the first pair of conjugate points, we must solve equations \([25]\) and \([28]\) simultaneously. This we may do in the following manner: determine from \([28]\) the quantity \(y\), square it, and, substituting into \([25]\), obtain a quadratic equation containing only the unknown \(x\).

Thus, from \([28]\),

\[ y=\frac{r^2-l^2+(n-\alpha)^2+\beta^2-2(n-\alpha)x}{2\beta}. \]

On being squared, this expression gives an algebraic fraction with a fairly considerable number of terms in the numerator, and therefore we shall transform equation \([28]\) by introducing into it a certain conventional quantity. As such a conventional quantity we shall take the coordinates of the point of intersection of the first radical axis (see drawing V—the line \(HG\)) with the axis of abscissas.

To find this point, we assign to \(y\) the value equal to \(0\); then the value \(x\), determined from \([28]\), will give us the abscissa of the required point—which we shall denote by \(\xi_1\). Thus, we shall have:

\[ 2(n-\alpha)x+2\beta y-(n-\alpha)^2-\beta^2=r^2-l^2, \]

with

\[ y=0;\qquad x=\xi_1, \]

Consequently,

\[ 2(n-\alpha)\xi_1-(n-\alpha)^2-\beta^2=r^2-l^2. \]

Whence

\[ \tag{29} \xi_1=\frac{r^2-l^2+(n-\alpha)^2+\beta^2}{2(n-\alpha)}. \]

Transferring, in equation [28], the terms containing the unknowns to one side and those containing known quantities to the other, we obtain:

\[ 2(n-\alpha)x+2\beta y=r^2-l^2+(n-\alpha)^2+\beta^2. \]

Divide the equation by 2; then we have:

\[ (n-\alpha)x+\beta y=\frac{r^2-l^2+(n-\alpha)^2+\beta^2}{2}. \]

The right-hand side of the equation obtained is, obviously, equal to \((n-\alpha)\xi_1\), and therefore the whole equation [28] may be written as follows:

\[ \tag{30} (n-\alpha)x+\beta y=(n-\alpha)\xi_1. \]

We solve equation [30] simultaneously with [25].

From [30] we have:

\[ \beta y=(n-\alpha)\xi_1-(n-\alpha)x. \]

Squaring, we have:

\[ \tag{31} \beta^2 y^2=(n-\alpha)^2\xi_1^2+(n-\alpha)^2x^2-2(n-\alpha)^2x\xi_1. \]

We multiply equation [25] by \(\beta^2\).

We obtain:

\[ \beta^2x^2+\beta^2y^2=\beta^2r^2. \]

Whence

\[ \tag{32} \beta^2y^2=\beta^2r^2-\beta^2x^2. \]

Subtracting [32] from [31], we have:

\[ x^2\bigl[(n-\alpha)^2+\beta^2\bigr]-2(n-\alpha)^2x\xi_1+\xi_1^2(n-\alpha)^2-\beta^2r^2=0. \]

We give the expression obtained the usual form of a quadratic equation:

\[ \tag{33} x^2-2\frac{\xi_1(n-\alpha)^2}{(n-\alpha)^2+\beta^2}\,x+ \frac{\xi_1^2(n-\alpha)^2-\beta^2r^2}{(n-\alpha)^2+\beta^2}=0. \]

Solving [33], we have:

\[ x=\frac{\xi_1(n-\alpha)^2}{(n-\alpha)^2+\beta^2} \pm \sqrt{ \frac{\xi_1^2(n-\alpha)^4}{\bigl[(n-\alpha)^2+\beta^2\bigr]^2} - \frac{\xi_1^2(n-\alpha)^2-\beta^2r^2}{(n-\alpha)^2+\beta^2} }. \]

Transforming the expression under the radical, we have:

\[ x= \frac{ \xi_1(n-\alpha)^2 \pm \sqrt{ \xi_1^2(n-\alpha)^4- \bigl[\xi_1^2(n-\alpha)^2-\beta^2r^2\bigr]\bigl[(n-\alpha)^2+\beta^2\bigr] } }{ (n-\alpha)^2+\beta^2 }. \]

Expanding the brackets under the radical:

\[ x= \frac{\xi_1(n-\alpha)^2 \pm \sqrt{\xi_1^2(n-\alpha)^4-\xi_1^2(n-\alpha)^4-\xi_1^2(n-\alpha)^2\beta^2+\beta^2r^2(n-\alpha)^2+\beta^4r^2}} {(n-\alpha)^2+\beta^2}. \]

Combining like terms, we obtain finally:

\[ \tag{34} x=\frac{\xi_1(n-\alpha)^2\pm\beta\sqrt{(n-\alpha)^2(r^2-\xi_1^2)+\beta^2r^2}} {(n-\alpha)^2+\beta^2}. \]

Now determine \(y\).

From equation [30] we have:

\[ (n-\alpha)x=\xi_1(n-\alpha)-\beta y. \]

Squaring both sides of this expression:

\[ \tag{35} (n-\alpha)^2x^2=\xi_1^2(n-\alpha)^2+\beta^2y^2-2(n-\alpha)\beta\xi_1y. \]

Multiplying equation [25] by \((n-\alpha)^2\) and subtracting it from [35], we obtain:

\[ y^2\bigl[(n-\alpha)^2+\beta^2\bigr]-2\beta\xi_1(n-\alpha)y+\xi_1^2(n-\alpha)^2-r^2(n-\alpha)^2=0, \]

or

\[ \tag{36} y^2-2\cdot\frac{\beta\xi_1(n-\alpha)}{(n-\alpha)^2+\beta^2}\,y+ \frac{(n-\alpha)^2(\xi_1^2-r^2)}{(n-\alpha)^2+\beta^2}=0. \]

Solving [36], we have:

\[ y=\frac{\beta\xi_1(n-\alpha)}{(n-\alpha)^2+\beta^2} \pm \sqrt{ \frac{ \beta^2\xi_1^2(n-\alpha)^2-(n-\alpha)^2(\xi_1^2-r^2)\bigl[(n-\alpha)^2+\beta^2\bigr] } {\bigl[(n-\alpha)^2+\beta^2\bigr]^2} }, \]

which, after some transformations, gives:

\[ \tag{37} y=\frac{\beta\xi_1(n-\alpha)\pm(n-\alpha)\sqrt{(n-\alpha)^2(r^2-\xi_1^2)+\beta^2r^2}} {(n-\alpha)^2+\beta^2}. \]

Thus, we have obtained two values each for \(x\) and for \(y\), determining the two points of intersection of the curvilinear circle with the circle of radius \(l\) drawn from the point \(P\).

We shall now determine the second pair from the system of conjugate points—for the circle of radius \(l\) with center at the point \(Q\).

Let us find the equation of the second radical axis (see Drawing V—the line \(FE\)).

For this, subtract equation [27] from [25]—we obtain:

\[ \tag{38} 2(n+\alpha)x-2\beta y=r^2-l^2+(n+\alpha)^2+\beta^2. \]

Putting \(y\) equal to zero, we obtain the point of intersection of the second radical axis with the axis of abscissas.

Thus, we obtain:

\[ y=0,\qquad x=\xi_2. \]

\[ 2(n+\alpha)\xi_2=r^2-l^2+(n+\alpha)^2+\beta^2. \]

Whence

\[ \tag{39} \xi_2=\frac{r^2-l^2+(n+\alpha)^2+\beta^2}{2(n+\alpha)}. \]

Introducing into [38] the expression for \(\xi_2\), we obtain the equation of the second radical axis:

\[ \tag{40} (n+\alpha)x-\beta y=(n+\alpha)\xi_2. \]

Solving equation [40] together with [25], and carrying out operations analogous to the preceding ones, we obtain the values of the coordinates of the two sought points. These values will have the following form:

\[ \tag{41} x=\frac{\xi_2(n+\alpha)^2\pm \beta\sqrt{(n+\alpha)^2(r^2-\xi_2^2)+\beta^2r^2}}{(n+\alpha)^2+\beta^2}. \]

\[ \tag{42} y=\frac{-\beta\xi_2(n+\alpha)\pm (n+\alpha)\sqrt{(n+\alpha)^2(r^2-\xi_2^2)+\beta^2r^2}}{(n+\alpha)^2+\beta^2}. \]

Thus, we have obtained four conjugate points. Let us write down together the values of their coordinates and determine the position of each point in the drawing.

The first pair of points:

\[ \tag{43} \left\{ \begin{aligned} x_1&=\frac{\xi_1(n-\alpha)^2+\beta\sqrt{(n-\alpha)^2(r^2-\xi_1^2)+\beta^2r^2}}{(n-\alpha)^2+\beta^2},\\[6pt] y_1&=\frac{\xi_1\beta(n-\alpha)-(n-\alpha)\sqrt{(n-\alpha)^2(r^2-\xi_1^2)+\beta^2r^2}}{(n-\alpha)^2+\beta^2},\\[6pt] x'_1&=\frac{\xi_1(n-\alpha)^2-\beta\sqrt{(n-\alpha)^2(r^2-\xi_1^2)+\beta^2r^2}}{(n-\alpha)^2+\beta^2},\\[6pt] y'_1&=\frac{\xi_1\beta(n-\alpha)+(n-\alpha)\sqrt{(n-\alpha)^2(r^2-\xi_1^2)+\beta^2r^2}}{(n-\alpha)^2+\beta^2}. \end{aligned} \right. \]

The second pair of points:

\[ \left\{ \begin{aligned} x_2&=\frac{\xi_2(n+\alpha)^2+\beta\sqrt{(n+\alpha)^2(r^2-\xi_2^2)+\beta^2r^2}}{(n+\alpha)^2+\beta^2},\\[6pt] y_2&=\frac{-\xi_2\beta(n+\alpha)+\sqrt{(n+\alpha)^2(r^2-\xi_2^2)+\beta^2r^2}}{(n+\alpha)^2+\beta^2},\\[6pt] x'_2&=\frac{\xi_2(n+\alpha)^2-\beta\sqrt{(n+\alpha)^2(r^2-\xi_2^2)+\beta^2r^2}}{(n+\alpha)^2+\beta^2},\\[6pt] y'_2&=\frac{-\xi_2\beta(n+\alpha)-(n+\alpha)\sqrt{(n+\alpha)^2(r^2-\xi_2^2)+\beta^2r^2}}{(n+\alpha)^2+\beta^2}. \end{aligned} \right. \]

In these 8 expressions the signs before the radicals are taken with such a calculation that the condition for the point to lie on the crank circle is always satisfied, i.e.:

\[ x_i^2+y_i^2=r^2. \]

Let us now determine the place of each point on the drawing.

For this purpose let us first determine the positions of the crank corresponding to the dead points of the slider. Analytically, these positions are found as the points of tangency of the circles drawn from \(A\) and \(B\) with the crank circle.

To obtain the coordinates of the points of tangency it is sufficient in formula [43] to substitute, instead of \(k\), the magnitude of half the path of the slider, i.e. \(\sigma\).

With such a substitution the expressions under the radicals become zero, and, instead of eight, we obtain from [43] four expressions determining two points.

Indeed, for \(k=\sigma\), the expression \((n-\alpha)^2+\beta^2\) represents the square of the distance of the nearest dead point \(A\) from the center \(O\) of the crank circle. In the preceding chapter we showed that this distance is equal to the length of the connecting rod, diminished by the length of the crank radius,

\[ d_1=l-r. \]

Whence:

\[ (n-\alpha)^2+\beta^2=(l-r)^2. \]

Substituting in expression [29], we obtain, for the particular case under consideration:

\[ [44]\qquad \xi_1=\frac{r^2-l^2+(l-r)^2}{2(n-\alpha)} =\frac{r(r-l)}{n-\alpha}. \]

Let us take the quantity under the radical from [43] for \(x_i,\ x'_1\) and \(y_1,\ y'_1\), and substitute into it the value of \(\xi_1\) found from [44].

We shall have:

\[ (n-\alpha)^2(r^2-\xi_1^2)+\beta^2r^2 =(n-\alpha)^2r^2-(n-\alpha)^2\xi_1^2+\beta^2r^2= \]

\[ =\bigl[(n-\alpha)^2+\beta^2\bigr]r^2-(n-\alpha)^2\xi_1^2. \]

But

\[ \bigl[(n-\alpha)^2+\beta^2\bigr]=(l-r)^2, \]

and

\[ (n-\alpha)^2\xi_1^2=r^2(r-l)^2. \]

Whence

\[ \bigl[(n-\alpha)^2+\beta^2\bigr]r^2-(n-\alpha)^2\xi_1^2 =r^2(l-r)^2-r^2(r-l)^2=0, \]

as was required to prove.

Let us find the magnitude \(\xi_2\) for the given particular case.

As before, the expression \((n+\alpha)^2+\beta^2\) will represent the square of the distance of the far dead point \(B\) from the center \(O\) of the crank circle. We had earlier for this distance:

\[ d_2=r+l, \]

whence, by [39],

\[ [45]\qquad \xi_2=\frac{r^2-l^2+(r+l)^2}{2(n+\alpha)}=\frac{r(r+l)}{n+\alpha}. \]

Substituting the value found for \(\xi_2\) into the subradical expressions for \(x_2, x'_2\) and \(y_2, y'_2\) from [43], we obtain, as before, that these expressions vanish.

Thus, the coordinates of the points of tangency will have the form:

\[ x_a=\frac{\xi_1(n-\alpha)^2}{(n-\alpha)^2+\beta^2},\qquad y_a=\frac{\xi_1\beta(n-\alpha)}{(n-\alpha)^2+\beta^2}. \]

\[ x_b=\frac{\xi_2(n+\alpha)^2}{(n+\alpha)^2+\beta^2},\qquad y_b=\frac{-\xi_2\beta(n+\alpha)}{(n+\alpha)^2+\beta^2}. \]

Let us transform these expressions.

For \(k=\sigma,\)

\[ \alpha=\sigma\cos\vartheta \qquad\text{and}\qquad \beta=\sigma\sin\vartheta. \]

\[ (n-\alpha)^2+\beta^2=(l-r)^2,\qquad \xi_1=\frac{r(r-l)}{n-\alpha}. \]

\[ (n+\alpha)^2+\beta^2=(l+r)^2,\qquad \xi_2=\frac{r(l+r)}{n+\alpha}. \]

Therefore we shall finally have:

\[ [46]\qquad \left\{ \begin{aligned} x_a&=\frac{r(n-\sigma\cos\vartheta)}{r-l},& y_a&=-\frac{r\cdot\sigma\sin\vartheta}{\,l-r\,},\\[6pt] x_b&=\frac{r(n+\sigma\cos\vartheta)}{r+l},& y_b&=-\frac{r\cdot\sigma\sin\vartheta}{\,r+l\,}. \end{aligned} \right. \]

Since the length of the connecting rod is always greater than the radius of the crank—otherwise the mechanism is impossible (see the conditions of crankedness [24])—it is clear that the difference \(r-l\) is a negative quantity. In other words, the point \(D\) has both the abscissa \(x_a\) and the ordinate \(y_a\) negative, i.e., it lies in the third quadrant, reckoned in the direction of rotation of the crank; the point \(C\) has the abscissa \(x_b\) positive and the ordinate \(y_b\) negative, i.e., it is in the second quadrant.

Thus, we have obtained that the ordinates of the points of tangency of the connecting-rod circles with the crank circle are negative quantities, i.e., the points \(D\) and \(C\) always lie below the axis of abscissas.

Since, on the other hand, the points \(D\) and \(C\) are limiting points, we may say that for all positions of the crank between the points \(D\) and \(C\), lying below the line \(CD\)—the chord of the dead points—the ordinates are negative quantities; while for positions of the crank above the line \(CD\), the ordinates in the second and third quadrants are negative, and in the first and fourth are positive. In other words, the ordinates in certain definite positions of the crank change their sign, i.e., trans-

pass through zero. The positions of the crank pin with zero ordinates are determined, obviously, as the points of intersection of the crank circle with the axis of abscissas. It is clear that for these points the value \(\xi=\pm r\).

Therefore, in order to decide which of the expressions [43] gives for \(y\) a value changing sign, we must find \(y=0\) when \(\xi=r\).

We shall do this.

Substituting \(\xi=+r\) in the expressions for \(y_2\) and \(y'_2\), we obtain that the radical expression

\[ \sqrt{(n+\alpha)^2(r^2-\xi_2^2)+\beta^2 r^2} \]

is transformed into the expression \(\beta r\),

and therefore, for

\[ \xi_2=+r \]

\[ y_2= \frac{-\beta\cdot r\cdot(n+\alpha)+\beta\cdot r\cdot(n+\alpha)} {(n+\alpha)^2+\beta^2}=0, \]

\[ y'_2= \frac{-\beta\cdot r\cdot(n+\alpha)-\beta\cdot r\cdot(n+\alpha)} {(n+\alpha)^2+\beta^2} = -\frac{2\beta\cdot r(n+\alpha)} {(n+\alpha)^2+\beta^2}, \]

i.e. \(y_2\) passes through 0; consequently, the point determined by the coordinates \((x_2,y_2)\) lies above the line \(CD\).

In exactly the same way we find that, for \(\xi_1=-r\),

\[ y_1= \frac{-\beta\cdot r\cdot(n-\alpha)-\beta\cdot r\cdot(n-\alpha)} {(n-\alpha)^2+\beta^2} = -\frac{2\beta\cdot r\cdot(n-\alpha)} {(n+\alpha)^2+\beta^2}, \]

\[ y'_1= \frac{-\beta\cdot r\cdot(n-\alpha)+\beta\cdot r\cdot(n-\alpha)} {(n-\alpha)^2+\beta^2}=0, \]

i.e. \(y'_1\) passes through 0; consequently, the point determined by the coordinates \((x'_1,y'_1)\) lies above the line \(CD\).

Thus we have determined quite precisely the location of each of the four conjugate points whose analytic expressions for the coordinates are given by formulas [43].

We determine the location of the points with respect to the line \(CD\)—the chord of the dead points. We found above that this chord is always situated below the axis of abscissas—the base line. Let us see whether there is any further condition determining the position of the chord of the dead points.

We earlier determined the coordinates of the points \(C\) and \(D\) and obtained for them the expressions [46].

Let us form the equation of the straight line passing through these two points.

A. Ugarov.

The equation of this straight line in general form will be as follows:

\[ (y_a-y_b)x-(x_a-x_b)y+(x_ay_b-y_ax_b)=0. \]

Let us determine the values of the coefficients in this equation, substituting for \((x_a,y_a)\) and \((x_b,y_b)\) their values from [46]:

\[ y_a-y_b=-\frac{r.\sigma.\sin\vartheta}{l-r}+\frac{r.\sigma.\sin\vartheta}{l+r}, \]

\[ =\frac{-(l+r)r.\sigma.\sin\vartheta+(l-r)r.\sigma.\sin\vartheta}{l^2-r^2}, \]

\[ =\frac{-2r^2\sigma.\sin\vartheta}{l^2-r^2}. \]

\[ x_a-x_b=-\frac{r(n-\sigma.\cos\vartheta)}{l-r}-\frac{r(n+\sigma.\cos\vartheta)}{l+r}, \]

\[ =\frac{-r(l+r)(n-\sigma.\cos\vartheta)-r(l-r)(n+\sigma.\cos\vartheta)}{l^2-r^2}, \]

\[ =\frac{-2r(l.n-r.\sigma.\cos\vartheta)}{l^2-r^2}. \]

In exactly the same way we find:

\[ x_ay_b-y_ax_b= \frac{r^2\sigma.\sin\vartheta.(n-\sigma.\cos\vartheta)+r^2\sigma.\sin\vartheta.(n+\sigma.\cos\vartheta)} {l^2-r^2} = \frac{2r^2n\sigma.\sin\vartheta}{l^2-r^2}. \]

Dropping the common factor \(\dfrac{2r}{l^2-r^2}\), we finally obtain the equation of the chord of the dead points:

\[ \tag{47} -r.\sigma.\sin\vartheta.x+(l.n-r.\sigma.\cos\vartheta)y+r.n.\sigma.\sin\vartheta=0. \]

Since none of the coefficients of \(x\) and \(y\) in this equation is equal to zero, it is evident that the straight line expressed by this equation cannot be parallel to any of the coordinate axes, and therefore intersects them. Let us find the point of intersection of this straight line with the axis of abscissas. For this the ordinate \(y\) must be assigned the value zero. Then we shall have:

\[ -r.\sigma.\sin\vartheta.x_1+r.n.\sigma.\sin\vartheta=0. \]

Whence \(x_1=n\).

Thus: the chord of the dead points, when prolonged, passes through the midpoint of the path of the slider.

This property of the chord of the dead points may be used to find the second dead position of the crank for a given first one*). For this it is sufficient to connect the given dead position of the crank—

*) By the dead position of the crank here is meant the position of the crank radius corresponding to the dead points of the path of the slider.

A. U.

find with a straight line through the midpoint of the slider’s path—the point of intersection of this line with the crank circle will be the required second dead position of the crank.

Let us now turn to the middle chord. The points of its intersection with the crank circle are determined from formulas [43] for the case when \(k=0\). It is obvious that under this condition \(\alpha=\beta=0\), and we obtain, respectively:

\[ \xi_1=\xi_2=\frac{r^2-l^2+n^2}{2n}. \]

Whence:

\[ x_1=x'_1=x'_2=x_2=\frac{r^2-l^2+n^2}{2n}, \]

\[ y'_1=y_2=+\sqrt{\,r^2-\left(\frac{r^2-l^2+n^2}{2n}\right)^2\,}, \]

\[ y_1=y'_2=-\sqrt{\,r^2-\left(\frac{r^2-l^2+n^2}{2n}\right)^2\,}. \]

To form the equation of the middle chord, we shall find the expression for the radical axis for this case, since for the midpoint of the slider’s path both conjugate chords merge into one \((x_1=x'_1=x_2=x'_2)\).

As before, we shall have:

Equation of the crank circle:

\[ x^2+y^2=r^2. \]

Equation of the circle of radius \(l\), drawn from the point \(M\) as center:

\[ (x-n)^2+y^2=l^2. \]

Subtracting one equation from the other, we obtain the equation of the radical axis:

\[ \begin{aligned} 2nx-n^2-r^2+l^2&=0,\\ 2nx-(r^2-l^2+n^2)&=0. \end{aligned} \tag{48} \]

This equation contains no \(y\), and therefore represents a straight line drawn parallel to the \(y\)-axis at a distance from the origin of coordinates

\[ x=\frac{r^2-l^2+n^2}{2n}. \]

Thus: the middle chord is always perpendicular to the base line.

Both the middle chord and the chord of the dead points are particular cases of conjugate chords.

Let us now take up the construction of the equation of the conjugate chord for the general case.

Let us join the points \((x_1,y_1)\) and \((x_2,y_2)\) by a chord and write the equation of the line passing through the two points.

For brevity, in formulas [43] let us denote the radicals of the expressions \((x_1,x'_1)\) and \((y_1,y'_1)\) by \(R_1\), and their denominators by \(N_1\); for the expressions \((x_2,x'_2)\) and \((y_2,y'_2)\) we shall have the corresponding quantities \(R_2\) and \(N_2\). Thus, the formulas of expressions [43] will take the form:

\[ \tag{49} \left\{ \begin{aligned} x_1&=\frac{\xi_1(n-\alpha)^2+\beta R_1}{N_1},& y_1&=\frac{(n-\alpha)(\beta\xi_1-R_1)}{N_1},\\[6pt] x'_1&=\frac{\xi_1(n-\alpha)^2-\beta R_1}{N_1},& y'_1&=\frac{(n-\alpha)(\beta\xi_1+R_1)}{N_1},\\[6pt] x_2&=\frac{\xi_2(n+\alpha)^2+\beta R_2}{N_2},& y_2&=-\frac{(n+\alpha)(\beta\xi_2-R_2)}{N_2},\\[6pt] x'_2&=\frac{\xi_2(n+\alpha)^2-\beta R_2}{N_2},& y'_2&=-\frac{(n+\alpha)(\beta\xi_2+R_2)}{N_2}. \end{aligned} \right. \]

The general equation of the line passing through the two points \((x_1,y_1)\) and \((x_2,y_2)\) will have the form:

\[ (y_1-y_2)x-(x_1-x_2)y+(x_1y_2-y_1x_2)=0. \]

Let us determine the coefficients of this equation.

Form the expression \(y_1-y_2\).

\[ y_1-y_2=\frac{(n-\alpha)(\beta\xi_1-R_1)}{N_1} +\frac{(n+\alpha)(\beta\xi_2-R_2)}{N_2}, \]

\[ \tag{50} =\frac{(n-\alpha)(\beta\xi_1-R_1)N_2+(n+\alpha)(\beta\xi_2-R_2)N_1}{N_1N_2}. \]

In the same way we obtain:

\[ x_1-x_2=\frac{\xi_1(n-\alpha)^2+\beta R_1}{N_1} -\frac{\xi_2(n+\alpha)^2+\beta R_2}{N_2}, \]

\[ \tag{51} =\frac{\left[\xi_1(n-\alpha)^2+\beta R_1\right]N_2-\left[\xi_2(n+\alpha)^2+\beta R_2\right]N_1}{N_1N_2}. \]

And, finally,

\[ x_1y_2-y_1x_2= \]

\[ \tag{52} =-\frac{\left[\xi_1(n-\alpha)^2+\beta R_1\right](\beta\xi_2-R_2)(n+\alpha) +\left[\xi_2(n+\alpha)^2+\beta R_2\right](\beta\xi_1-R_1)(n-\alpha)} {N_1N_2}. \]

Discarding the common denominator \(N_1N_2\), we obtain the general equation of the conjugate chord in the following form:

The Coupler Pole

\[ \tag{53} \left\{ \begin{aligned} &\{(n-\alpha)(\beta \xi_1-R_1)N_2+(n+\alpha)(\beta \xi_2-R_2)N_2\}\,x \\ &\quad-\{[\xi_1(n-\alpha)^2+\beta R_1]N_2-[\xi_2(n+\alpha)^2+\beta R_2]N_1\}\,y \\ &\quad-\{[\xi_1(n-\alpha)^2+\beta R_1](\beta \xi_2-R_2)(n+\alpha) \\ &\qquad\quad+[\xi_2(n+\alpha)^2+\beta R_2](\beta \xi_1-R_1)(n-\alpha)\}=0 . \end{aligned} \right. \]

As a general equation, formula [53] must also include both of the previously considered particular cases of conjugate chords.

The middle chord is characterized by the fact that, for it, \(k\) is equal to zero. Let us introduce this condition into equation [53], to which, for simplicity, we give the form:

\[ A_1x+B_1y+C_1=0 . \]

For \(k=0\), evidently, \(\alpha=\beta=0\), and consequently

\[ N_1=(n-\alpha)^2+\beta^2 \]

will be equal to \(n^2\), just as also

\[ N_2=(n+\alpha)^2+\beta^2=n^2, \]

i.e.

\[ N_1=N_2=n^2 . \]

Under the same condition:

\[ \xi_1=\frac{r^2-l^2+(n-\alpha)^2+\beta^2}{2(n-\alpha)} \]

becomes equal to

\[ \xi_2=\frac{r^2-l^2+(n+\alpha)^2+\beta^2}{2(n+\alpha)}, \]

i.e.

\[ \xi_1=\xi_2=\frac{r^2-l^2+n^2}{2n}; \]

and further,

\[ R_1=R_2=\sqrt{\,n^2\left[r^2-\left(\frac{r^2-l^2+n^2}{2n}\right)^2\right]\,}. \]

Let us now determine the coefficients of equation [53] for the particular case under consideration.

\[ A_1=(n-\alpha)(\beta \xi_1-R_1)N_2+(n+\alpha)(\beta \xi_2-R_2)N_1, \]

for \(k=0\),

\[ A_1=-nR_1N_2-nR_2N_1=-2n^3R_1; \]

\[ B_1=[\xi_1(n-\alpha)^2+\beta R_1]N_2-[\xi_2(n+\alpha)^2+\beta R_2]N_1, \]

\(k=0\),

\[ B_1=\xi_1 n^2N_2-\xi_2 n^2N_1=0; \]

\[ \begin{aligned} C_1 ={}& -\{[\xi_1(n-\alpha)^2+\beta R_1](\beta\xi_2-R_2)(n+\alpha)+{}\\ &\quad +[\xi_2(n+\alpha)^2+\beta R_2](\beta\xi_1-R_1)(n-\alpha)\}, \end{aligned} \]

\[ \left. C_1 \right|_{k=0} = -\{-\xi_1 n^3 R_2-\xi_2 n^3 R_1\}, \]

\[ = +2\xi_1 n^3 R_1, \]

\[ = 2n^3 R_1 \frac{r^2-l^2+n^2}{2n}, \]

\[ = n^2 R_1(r^2-l^2+n^2). \]

Substituting the values found for the coefficients into equation [53], we obtain:

\[ -2n^3R_1x+n^2R_1(r^2-l^2+n^2)=0. \]

Dividing by \(n^2R_1\) and changing the sign, we arrive at equation [48]:

\[ 2nx-(r^2-l^2+n^2)=0. \]

Let us now turn to the second particular case—to the chord of the dead points. This chord must be obtained from the condition: \(k=\sigma\).

Under this condition \(\alpha=\sigma\cdot\cos\vartheta,\ \beta=\sigma\cdot\sin\vartheta,\)

\[ \xi_1=\frac{r(r-l)}{n-\alpha},\qquad \xi_2=\frac{r(r+l)}{n+\alpha}; \]

(see [44] and [45]),

\[ R_1=R_2=0, \]

and finally,

\[ N_1=(l-r)^2,\qquad N_2=(l+r)^2. \]

Let us determine the coefficients of equation [53] for the given particular case:

\[ \left. A_1=(n-\alpha)(\beta\xi_1-R_1)N_2+(n+\alpha)(\beta\xi_2-R_2)N_1 \right|_{k=\sigma} \]

\[ =(n-\alpha)\beta\xi_1N_2+(n+\alpha)\beta\xi_2N_1= \]

\[ =r\cdot(r-l)\cdot(l+r)^2\sigma\cdot\sin\vartheta +r(r+l)(r-l)^2\sigma\cdot\sin\vartheta= \]

\[ =-2r^2(l^2-r^2)\sigma\cdot\sin\vartheta. \]

\[ \left. B_1=-[\xi_1(n-\alpha)^2+\beta R_1]N_2+[\xi_2(n+\alpha)^2+\beta R_2]N_1 \right|_{k=\sigma} \]

\[ =-\xi_1(n-\alpha)^2N_2+\xi_2(n+\alpha)^2N_1= \]

\[ =-r\cdot(r-l)(n-\alpha)(l+r)^2 +r(r+l)(n+\alpha)(l-r)^2= \]

\[ =r(l^2-r^2)[(n-\alpha)(l+r)+(n+\alpha)(l-r)], \]

which, after expanding the brackets and substituting for \(\alpha\) its value, gives:

The Coupler Pole

\[ B_1=2r(l^2-r^2)(n\cdot l-r\cdot\sigma\cdot\cos\vartheta). \]

Let us now turn to the coefficient \(C_1\).

\[ C_1=-\{[\xi_1(n-\alpha)^2+\beta R_1](\beta\xi_2-R_2)(n+\alpha)+ [\xi_2(n+\alpha)^2+\beta R_2](\beta\xi_1-R_1)(n-\alpha)\}, \]

\[ C_1\big|_{k=\sigma} =-\{\xi_1(n-\alpha)^2\beta\xi_2(n+\alpha) +\xi_2(n+\alpha)^2\beta\xi_1(n-\alpha)\} \]

\[ =-2n\beta\xi_1\xi_2(n^2-\alpha^2), \]

or

\[ C_1\big|_{k=\sigma}=2(l^2-r^2)r^2n\sigma\cdot\sin\vartheta. \]

Substituting the values of the coefficients found into equation [53], we obtain:

\[ -2r^2(l^2-r^2)\sigma\cdot\sin\vartheta\cdot x +2r(l^2-r^2)(n\cdot l-r\cdot\sigma\cdot\cos\vartheta)y+ \]

\[ +2(l^2-r^2)r^2n\sigma\cdot\sin\vartheta=0, \]

which, upon division by \(2r(l^2-r^2)\), is transformed into equation [47]:

\[ -r\sigma\cdot\sin\vartheta\cdot x+(n\cdot l-r\cdot\sigma\cdot\cos\vartheta)y +rn\sigma\cdot\sin\vartheta=0. \]

Thus, we see that equation [53] indeed represents the general case of a conjugate chord. The verification of the formula carried out has shown us that, only for particular values of the quantity \(k\), the coefficients of equation [53] become equal either to the coefficients of equation [47], or to the coefficients of equation [48]; for all the remaining values of \(k\) between \(0\) and \(\sigma\), the coefficients of equation [53] are not equal to the coefficients of equations [47] and [48].

In the further consideration of the properties of the chords of the crank circle of a rectilinear-directing slider-crank mechanism, the following question naturally arises: does there not exist in this mechanism a true coupler pole—in other words: do not all conjugate chords intersect at a single point?

To answer this question, we shall reason as follows: if all the chords intersect at a single point, then it is obvious that any three arbitrarily chosen chords also intersect at this point, and therefore it is sufficient for us to consider only three chords.

It is clear that for this consideration we may take the chord of the dead points, the mean chord, and an arbitrary chord expressed by equation [53]. We shall do so.

From analytic geometry we know that three lines pass through one point when the determinant composed of the coefficients of the equations of these lines is equal to zero.

Let us therefore form the determinant from the coefficients of equations [47], [48], and [53].

We obtain:

\[ \begin{vmatrix} -r\cdot\sigma\cdot\sin\vartheta, & (l\cdot n-r\cdot\sigma\cdot\cos\vartheta), & r\cdot n\cdot\sigma\cdot\cos\vartheta\\ 2n, & 0, & -(r^{2}-l^{2}+n^{2})\\ A_{1}, & B_{1}, & C_{1} \end{vmatrix} =\Delta . \]

If we consider the columns of this determinant, we shall see that none of them contains elements successively equal to the elements of any other column; the same may be said also with respect to the rows, and consequently the determinant composed by us is in no case equal to zero; or, in other words, the conjugate chords do not intersect at one point, and an actual coupler pole does not exist in the mechanism under consideration.

Therefore another question arises before us: is it possible, without great error, to admit the existence of a coupler pole in straight-line generating mechanisms?

To answer this question we must determine the properties of the geometrical locus of the points of intersection with one another of pairs of conjugate chords.

CHAPTER FIVE.

Investigation of the formulae expressing the conjugate chords of a straight-line generating crank mechanism.

We obtained earlier, for the conjugate chord connecting the points \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\), equation [53]; in order to obtain the chord paired with it, we must join by a straight line the remaining two points from the system of points considered by us, i.e. the points \((x_{1}', y_{1}')\) and \((x_{2}', y_{2}')\). Proceeding analogously to the preceding case, we obtain for the second conjugate chord, relying on formulae [49], the following equation:

\[ \tag{54} \left\{ \begin{aligned} &\{(n-\alpha)(\beta\xi_{1}+R_{1})N_{2}+(n+\alpha)(\beta\xi_{2}+R_{2})N_{1}\}\,x\\ &\quad-\{[\xi_{1}(n-\alpha)^{2}-\beta R_{1}]N_{2}-[\xi_{2}(n+\alpha)^{2}-\beta R_{2}]N_{1}\}\,y\\ &\quad-\{[\xi_{1}(n-\alpha)^{2}-\beta R_{1}](\beta\xi_{2}+R_{2})(n+\alpha)\\ &\qquad+[\xi_{2}(n+\alpha)^{2}-\beta R_{2}](\beta\xi_{1}+R_{1})(n-\alpha)\}=0, \end{aligned} \right. \]

or, conventionally,

\[ A_2x+B_2y+C_2=0. \]

Let us write here as well, for comparison, equation [53]:

\[ \tag{53} \left\{ \begin{aligned} &\{(n-\alpha)(\beta \xi_1-R_1)N_2+(n+\alpha)(\beta \xi_2-R_2)N_1\}\,x\\ &-\{[\xi_1(n-\alpha)^2+\beta R_1]N_2-[\xi_2(n+\alpha)^2+\beta R_2]N_1\}\,y\\ &-\{[\xi_1(n-\alpha)^2+\beta R_1](\beta \xi_2-R_2)(n+\alpha)\\ &\quad+[\xi_2(n+\alpha)^2+\beta R_2](\beta \xi_1-R_1)(n-\alpha)\}=0, \end{aligned} \right. \]

or, conventionally,

\[ A_1x+B_1y+C_1=0. \]

We see that equations [53] and [54] differ from one another only in the signs of the quantities \(R_1\) and \(R_2\), which indeed must be the case, if we recall that formulas [49], which determine the coordinates of the entire system of points, have, in pairs, opposite signs before the quantities \(R_1\) and \(R_2\).

Now the question before us is: to find the geometric locus of the points of intersection of paired conjugate chords.

To solve this question we must eliminate from both equations [53] and [54] the variable quantity \(k\), which enters into both equations in hidden form as \(\alpha, \beta, N_1, N_2, R_1\), and \(R_2\).

Eliminating \(k\) from one equation and substituting the value found for \(k\) into the other equation, in view of their complexity, would require a very large number of operations; therefore, to determine the desired geometric locus, we shall proceed by another route.

As we have just noted, the two equations under consideration differ from one another only in the signs before the quantities \(R_1\) and \(R_2\) entering into them.

Therefore, if we expand the parentheses in the expressions for the coefficients of these equations, then each of these coefficients may be regarded by us as a binomial, the equations differing from one another only in the signs before the second terms of these binomials.

Conventionally, we may represent what has been said as follows:

In the equation \(Ax_1+B_1y+C_1=0\)

\[ A_1=A+a, \]

\[ B_1=B+b, \]

\[ C_1=C+c; \]

whereas in the equation \(A_2x+B_2y+C_2=0\),

\[ A_2=A-a, \]

\[ B_2=B-b, \]

\[ C_2=C-c. \]

Thus, the conditional forms of equations [53] and [54] will take the form:

\[ (A+a)x+(B+b)y+(C+c)=0, \]

\[ (A-a)x+(B-b)y+(C-c)=0. \]

It is clear that these two equations, after termwise addition and subtraction, will give us two new ones:

\[ \tag{55} Ax+By+C=0 \quad \text{and} \quad ax+by+c=0. \]

These new equations represent 2 new lines, which, however, pass through the point of intersection of the conjugate chords under consideration; and, consequently, for the solution of the question of the geometric locus we may operate with the equations of auxiliary lines from the same pencil, which are simpler than formulas [53] and [54].

This is what we shall do.

Let us first turn to the coefficients of equation [53].

\[ A_1=(n-\alpha)(\beta\xi_1-R_1)N_2+(n+\alpha)(\beta\xi_2-R_2)N_1. \]

Expanding the parentheses, we obtain:

\[ A_1=(n-\alpha)\beta\xi_1N_2+(n+\alpha)\beta\xi_2N_1-(n-\alpha)R_1N_2-(n+\alpha)R_2N_1 \]

or \(A_1=A+a\), where

\[ A=(n-\alpha)\beta\xi_1N_2+(n+\alpha)\beta\xi_2N_1, \]

\[ a=-\bigl[(n-\alpha)R_1N_2+(n+\alpha)R_2N_1\bigr]. \]

We pass to the coefficient \(B_1\).

\[ B_1=-\{[\xi_1(n-\alpha)^2+\beta R_1]N_2-[\xi_2(n+\alpha)^2+\beta R_2]N_1\}. \]

Expanding the small parentheses:

\[ B_1=-\{\xi_1(n-\alpha)^2N_2-\xi_2(n+\alpha)^2N_1+\beta R_1N_2-\beta R_2N_1\}. \]

Or \(B_1=B+b\), where

\[ B=\xi_2(n+\alpha)^2N_1-\xi_1(n-\alpha)^2N_2, \]

\[ b=\beta(R_2N_1-R_1N_2). \]

Proceeding in the same manner with the coefficient \(C_1\), we then obtain instead

\[ C_1=-\{\xi_1(n-\alpha)^2+\beta R_1\}(\beta\xi_2-R_2)(n+\alpha)+ \]

\[ +[\xi_2(n+\alpha)^2+\beta R_2](\beta\xi_1-R_1)(n-\alpha)\}, \]

$$ C_1=-\{\beta \xi_1\xi_2(n-\alpha)^2(n+\alpha)-\beta R_1R_2(n+\alpha)+\beta^2\xi_2R_1(n+\alpha)- $$

$$ -\xi_1(n-\alpha)^2(n+\alpha)R_2+\beta\xi_1\xi_2(n+\alpha)^2(n-\alpha)-\beta R_1R_2(n-\alpha)+ $$

$$ +\beta^2\xi_1R_2(n-\alpha)-\xi_2(n+\alpha)^2(n-\alpha)R_1\}. $$

Let us collect like terms:

$$ C_1=-\{2n\beta[\xi_1\xi_2(n^2-\alpha^2)-R_1R_2]-(n^2-\alpha^2-\beta^2)[\xi_1(n-\alpha)R_2+ $$

$$ +\xi_2(n+\alpha)R_1]\}. $$

Or

$$ C_1=C+c,\quad \text{where} $$

$$ C=2n\beta[R_1R_2-\xi_1\xi_2(n^2-\alpha^2)], $$

$$ c=(n^2-\alpha^2-\beta^2)[\xi_1(n-\alpha)R_2+\xi_2(n+\alpha)R_1]. $$

Let us proceed in the same way with equation [54].

$$ A_2=(n-\alpha)(\beta\xi_1+R_1)N_2+(n+\alpha)(\beta\xi_2+R_2)N_1,\quad \text{or} $$

$$ A_2=(n-\alpha)\beta\xi_1N_2+(n+\alpha)\beta\xi_2N_1+(n-\alpha)R_1N_2+(n+\alpha)R_2N_1, $$

which, conditionally, will be expressed as follows:

$$ A_2=A-a. $$

In exactly the same way we obtain:

$$ B_2=-\{\xi_1(n-\alpha)^2N_2-\xi_2(n+\alpha)^2N_1-\beta R_1N_2+\beta R_2N_1\} $$

or

$$ B_2=B-b. $$

Finally, with respect to \(C_2\), we have:

$$ C_2=-\{[\xi_1(n-\alpha)^2-\beta R_1](\beta\xi_2+R_2)(n+\alpha)+ $$

$$ +[\xi_2(n+\alpha)^2-\beta R_2](\beta\xi_1+R_1)(n-\alpha)\}. $$

Expanding the small brackets and collecting like terms, we obtain:

$$ C_2=-\{2n\beta[\xi_1\xi_2(n^2-\alpha^2)-R_1R_2]+(n^2-\alpha^2-\beta^2)[\xi_1(n-\alpha)R_2+\xi_2(n+\alpha)R_1]\} $$

or

$$ C_2=C-c. $$

Having determined the values of the quantities \(A\), \(B\), \(C\) and \(a\), \(b\), \(c\), we must substitute them into the conditional equations [55] and solve them with respect to the variable quantity \(k\).

Having performed this substitution, we obtain two equations of the following form:

$$ [(n-\alpha)\beta\xi_1N_2+(n+\alpha)\beta\xi_2N_1]x+[\xi_2(n+\alpha)^2N_1-\xi_1(n-\alpha)^2N_2]y+ $$

$$ [56]\qquad +2n\beta[R_1R_2-\xi_1\xi_2(n^2-\alpha^2)]=0, $$

and the second:

$$ [57]\qquad -[(n-\alpha)R_1N_2+(n+\alpha)R_2N_1]x+\beta(R_2N_1-R_1N_2)y+ $$

$$ +(n^2-\alpha^2-\beta^2)[\xi_1(n-\alpha)R_2+\xi_2(n+\alpha)R_1]=0. $$

Both equations contain within themselves the irrational quantities \(R_1\) and \(R_2\); we shall have to rid ourselves of this irrationality.

Noting that in equation [56] the quantities \(R_1\) and \(R_2\) enter in one term as a product, whereas in equation [57] these quantities enter in the form of an algebraic sum, we conclude that, in order to eliminate the irrationality, equation [56] must be raised to the square once, while equation [57] must be squared twice.

It is clear that such a double squaring will extremely complicate the form of equation [57] and thereby make it difficult to find the geometrical locus that interests us.

Therefore we shall replace the line expressed by equation [57] with another line, which, of course, passes through the point of intersection of the conjugate chords.

To decide which line, from the pencil passing through the point of intersection of the conjugate chords, is most convenient for us to introduce into the analysis, let us recall that the system of conjugate points forms a quadrilateral inscribed in a crooked circle.

If we extend the pairwise opposite sides of this quadrilateral to their intersections, then we obtain two new points, which together with the point of intersection of the diagonals of the quadrilateral form the so-called polar triangle.

The property of this triangle, as is known, is the following: each of the vertices of the triangle is the pole of the opposite side, and each side, consequently, serves as the polar of the opposite vertex.

Thus we see that through the point of intersection of the conjugate chords there must also pass the polars of the points of intersection of the extended sides of the quadrilateral.

We know that two (of the four) opposite sides of an inscribed quadrilateral are nothing other than radical axes, expressed by the simple equations [28] and [38]; and therefore we shall use these lines to find one of the missing vertices of the polar triangle.

Let us therefore find the point of intersection of the radical axes. We write here equations [28] and [38].

\[ [28]\qquad 2(n-\alpha)x+2\beta y-(n-\alpha)^2-\beta^2-r^2+l^2=0, \]

\[ [38]\qquad 2(n+\alpha)x-2\beta y-(n+\alpha)^2-\beta^2-r^2+l^2=0. \]

Solving these equations simultaneously, we obtain the following values of the coordinates of the point of intersection of the radical axes:

\[ x= \frac{ 2\beta\left[l^2-r^2-(n-\alpha)^2-\beta^2\right] + 2\beta\left[l^2-r^2-(n+\alpha)^2-\beta^2\right] }{ -4\beta(n-\alpha)-4\beta(n+\alpha) }, \]

II.

\[ y=\frac{2(n+\alpha)[l^2-r^2-(n-\alpha)^2-\beta^2]-2(n-\alpha)[l^2-r^2-(n+\alpha)^2-\beta^2]}{-4\beta(n-\alpha)-4\beta(n+\alpha)}, \]

which, upon expanding the brackets and reducing, gives:

\[ [58]\qquad \begin{cases} x=\dfrac{l^2-r^2-n^2-k^2}{2n},\\[6pt] y=-\operatorname{ctg}\vartheta\,\dfrac{l^2-r^2+n^2-k^2}{2n}. \end{cases} \]

In these expressions \(k^2=\alpha^2+\beta^2\) and \(\operatorname{ctg}\vartheta=\dfrac{\alpha}{\beta}\).

Now we must write the equation of the polar of the point determined by expressions [58].

This equation will have the following form:

\[ -\frac{l^2-r^2-n^2-k^2}{2n}\,x-\operatorname{ctg}\vartheta\cdot \frac{l^2-r^2+n^2-k^2}{2n}\,y-r^2=0, \]

or, finally:

\[ [59]\qquad (l^2-r^2-n^2-k^2)x+\operatorname{ctg}\vartheta(l^2-r^2+n^2-k^2)y+2nr^2=0. \]

It is precisely by this equation, which has a comparatively simple form, that we shall replace the complicated equation [57].

Thus, in order to find the locus of points of intersection of conjugate chords, expressed by the exceedingly complicated equations [53] and [54], we shall operate with the simpler equations [56] and [59].

Before proceeding to the simultaneous solution of equations [56] and [59], we shall investigate this latter equation.

For an indeterminate value of the quantity \(k\), equation [59] represents the general formula for the polars of all points of intersection of radical axes obtained by cutting the crank circle with circles whose radius is equal to the length of the coupler.

We shall call these circles coupler circles, and the radical axes obtained coupler-crank radical axes.

Since the crank circle is cut by two coupler circles from two different centers, it is evident that we are dealing here with a system of three circles and, consequently, have three radical axes.

The third radical axis is obtained from the intersection of the coupler circles with one another; we shall call it the coupler radical axis.

We know that in a system of three circles all three radical axes intersect at one point—the radical center of these circles.

Thus, we may say that equation [59] represents the polar of the radical center, whose coordinates are expressed by formulas [58].

Formula [58] contains the variable quantity \(k\), and therefore, for different \(k\) we shall obtain different values of the quantities \(x\) and \(y\). In other words, with a change in the magnitude \(k\), the radical center of the system of three circles does not remain in one place, but describes a certain path.

Let us determine this path.

For this purpose we shall find the equation of the connecting-rod radical axis. It is obtained by subtracting from one another the equations of the connecting-rod circles [26] and [27].

Let us do this.

\[ \tag{26} [x-(n-\alpha)]^2+(y-\beta)^2=l^2; \]

\[ \tag{27} [x-(n+\alpha)]^2+(y+\beta)^2=l^2. \]

Expanding the brackets, we have:

\[ x^2+(n-\alpha)^2-2(n-\alpha)x+y^2+\beta^2-2\beta y=l^2, \]

\[ x^2+(n+\alpha)^2-2(n+\alpha)x+y^2+\beta^2+2\beta y=l^2. \]

Subtracting one from the other gives:

\[ -4n\alpha+4\alpha x-4\beta y=0. \]

Since \(\alpha=k\cdot\cos\vartheta\) and \(\beta=k\cdot\sin\vartheta\), substituting, we obtain:

\[ 4k\cdot\cos\vartheta\cdot x-4k\cdot\sin\vartheta\cdot y-4nk\cdot\cos\vartheta=0. \]

Dividing by \(4k\), we have the final equation of the connecting-rod radical axis:

\[ \tag{60} x\cdot\cos\vartheta-y\cdot\sin\vartheta-n\cdot\cos\vartheta=0. \]

It is easy to see that this equation represents a straight line perpendicular to the path of the slide and passing through the midpoint of this path.

Furthermore, we see that equation [60] does not contain the variable quantity \(k\), whence we draw the conclusion that for all values of the quantity \(k\) the direction of the connecting-rod radical axis remains unchanged; in other words, the connecting-rod radical axis is fixed.

We saw earlier that the radical center of our system of three circles describes a certain path.

Since the radical center is the point of intersection of all three radical axes, then, on the basis of equation [60], we conclude that the radical center moves along the straight line determined by this equation.

On the other hand, we had earlier that the polar of the radical center is expressed by equation [59], which contains the variable quantity \(k\).

From the property of polars we know that, if a point moves along a certain straight line, then the polars of this point pass through the pole of that straight line.

Thus, we arrive at the conclusion that, independently of the value of the quantity \(k\), equation [59] must be satisfied by the coordinates of the pole of the line expressed by equation [60], i.e., must become an identity.

To find the pole of the coupler radical axis, expressed by equation [60], we need to know the length of the perpendicular from the center of the curvature circle (the origin of coordinates) to this line.

It is easy to see that the length of this perpendicular will be:

\[ p = n \cdot \cos \vartheta . \]

Knowing that the radius of a circle is the geometric mean between the distances from the center of any point and its polar, we find that the pole of the coupler radical axis will lie on the perpendicular to this axis at a distance from the center equal to

\[ \nu = \frac{r^{2}}{n \cdot \cos \vartheta}. \]

Since the perpendicular \(p\) is inclined at the angle \(\vartheta\) to the line \(n\)—the axis of abscissae—we may write that the coordinates of the pole of the coupler radical axis will be:

\[ \tag{61} \left\{ \begin{aligned} x_p &= \frac{r^{2}}{n \cdot \cos \vartheta}\cos \vartheta = \frac{r^{2}}{n},\\ y_p &= -\frac{r^{2}}{n \cdot \cos \vartheta}\sin \vartheta = -\operatorname{tg}\vartheta\,\frac{r^{2}}{n}. \end{aligned} \right. \]

These very values of the coordinates of the pole must transform equation [59] into an identity for all values of the quantity \(k\).

Substituting in place of \(x\) and \(y\) in equation [59] the quantities determined by formulas [61], we have:

\[ (l^{2}-r^{2}-n^{2}-k^{2})\frac{r^{2}}{n} -\operatorname{ctg}\vartheta\,(l^{2}-r^{2}+n^{2}-k^{2})\operatorname{tg}\vartheta\,\frac{r^{2}}{n} +2nr^{2}=0. \]

It is easy to see that we have before us the identity

\[ -2n^{2}r^{2}+2n^{2}r^{2}=0. \]

The quantity \(k\) in equation [59] varies within the limits from \(0\) to \(c\). For all intermediate values of the quantity \(k\), equation [59] represents the polar of the radical center of a system of three circles.

Let us now see what line this equation represents for the limiting values of the quantity \(k\).

For \(k=5\) equation [59] assumes the form

\[ \tag{62} (l^{2}-r^{2}-n^{2}-\sigma^{2})x+\operatorname{ctg}\vartheta\,(l^{2}-r^{2}+n^{2}-\sigma^{2})y+2nr^{2}=0. \]

In deriving the conditions for the possibility of existence of a rectilinear-generating crank-and-connecting-rod mechanism, we obtained the dependence between the elements of the mechanism expressed by formula [21]:

\[ lr=n\sigma\cos\vartheta \qquad \text{and} \]

\[ l^{2}+r^{2}-2lr=n^{2}+\sigma^{2}-2n\sigma\cos\vartheta. \]

The second of these formulas, on the basis of the first, gives:

\[ l^{2}+r^{2}=n^{2}+\sigma^{2} \]

and

\[ n^{2}-r^{2}=l^{2}-\sigma^{2}. \]

Substituting these values into the coefficients of \(x\) and \(y\) of equation [62], we obtain:

\[ \tag{62 bis} -2r^{2}x+2\operatorname{ctg}\vartheta\,(l^{2}-\sigma^{2})y+2nr^{2}=0. \]

Dividing the equation by 2 and multiplying it by \(\sin\vartheta\), we give the resulting equation the following form:

\[ -r^{2}\sin\vartheta\cdot x+\cos\vartheta\,(l^{2}-\sigma^{2})y+nr^{2}\sin\vartheta=0. \]

From formula [21] we have:

\[ \cos\vartheta=\frac{lr}{n\sigma}. \]

Substitute this value into our equation:

\[ -r^{2}\sin\vartheta\cdot x+\frac{lr}{n\sigma}(l^{2}-\sigma^{2})y+nr^{2}\sin\vartheta=0. \]

Clearing the denominator:

\[ -r^{2}n\sigma\sin\vartheta\cdot x+lr(l^{2}-\sigma^{2})y+n^{2}r^{2}\sigma\sin\vartheta=0. \]

Divide by \(rn\):

\[ -r\sigma\sin\vartheta\cdot x+\frac{l}{n}(l^{2}-\sigma^{2})y+nr\sigma\sin\vartheta=0. \]

In the coefficient of \(y\) we substitute for \((l^{2}-\sigma^{2})\) the equal quantity \((n^{2}-r^{2})\) (see above).

Then this coefficient takes the form:

\[ \frac{l(n^{2}-r^{2})}{n}. \]

Expanding the parentheses, we obtain:

\[ \frac{ln^{2}-lr^{2}}{n}=ln-\frac{lr}{n}\,r. \]

But

\[ lr=n\sigma\cos\vartheta . \]

Whence

\[ ln-\frac{lr}{n}r=ln-r\sigma\cos\vartheta . \]

Substituting the found value of the coefficient of \(y\) into the equation under consideration, we obtain the final form of formula [62]:

\[ -r\sigma\sin\vartheta\,x+(ln-r\sigma\cos\vartheta)y+rn\sigma\sin\vartheta=0. \]

This equation, as we already know, represents the chord of the dead points and was obtained by us earlier (see formula [47]).

This was to be expected, since for \(k=\sigma\) the coupler-crank radical axes pass into tangent lines, and we know that the line joining the points of tangency serves as the polar for the point of intersection of the tangents. The line joining the points of tangency is precisely the chord of the dead points considered by us earlier.

Let us now see into what formula [59] turns for another limiting value of the quantity \(k\), namely for \(k=0\).

Substituting this value of \(k\) into equation [59], we obtain:

\[ \tag{63} (l^2-r^2-n^2)x+\operatorname{ctg}\vartheta\,(l^2-r^2+n^2)y+2nr^2=0. \]

We know that for \(k=0\) both coupler circles coincide; in other words, in the case under consideration we have before us a system of two circles: one coupler circle and one crank circle.

It is clear that the radical axis of these two circles will represent the merger of the two coupler-crank radical axes.

Indeed: for \(k=0\), \(\alpha=0\) and \(\beta=0\), and therefore the equations of the radical axes, expressed by formulas [28] and [38], represent one and the same straight line

\[ 2nx-(r^2-l^2+n^2)=0, \]

which, as we already know (see formula [48]), is the mean chord.

Thus, the two radical axes in the case under consideration merge into one line, which represents the limit of the inscribed quadrilateral having as its vertices the system of conjugate points.

Let us return to equation [63].

We know that it, being a particular case of formula [59], must be satisfied by the coordinates of the pole of the coupler radical axis, i.e. must represent a line passing through the mentioned pole.

In other words, equation [63] must represent the polar of some point lying on the coupler radical axis.

It is easy to see that this point has coordinates of the following form:

\[ \tag{64} x=\frac{r^2+n^2-l^2}{2n},\qquad y=-\operatorname{ctg}\vartheta\,\frac{(l^2-r^2+n^2)}{2n}. \]

Let us see what point this is.

For this purpose let us find the point of intersection of the mean chord with the connecting-rod radical axis.

The equation of the connecting-rod radical axis, as we derived earlier, has the form:

\[ \tag{60} x\cos\vartheta-y\sin\vartheta-n\cos\vartheta=0. \]

The equation of the mean chord:

\[ \tag{48} 2nx-(r^2+n^2-l^2)=0. \]

Solving these two equations simultaneously, we obtain that the coordinates of the point of intersection of the lines expressed by these equations will have the form:

\[ x=\frac{r^2+n^2-l^2}{2n} \quad\text{and}\quad y=-\operatorname{ctg}\vartheta\,\frac{(l^2-r^2+n^2)}{2n}. \]

Thus we come to the conclusion that equation [63] represents the polar of the point of intersection of the mean chord with the connecting-rod radical axis.

On the other hand, we know that the mean chord for the case under consideration represents the coalescence of two connecting-rod-crank radical axes.

We further know that the point of intersection of such axes, being one of the vertices of the polar triangle for all possible values of the quantity \(k\), moves along the connecting-rod radical axis.

Thus, we may assert that the point found is a vertex of the polar triangle for the limiting case under consideration.

Further, it is known to us that the second vertex of this triangle is the point of intersection of the diagonals of the inscribed quadrilateral, which, as we have just seen, for the case being considered degenerates into a straight line expressed by the equation of the mean chord.

Consequently, in order to find the limiting position of the point of intersection of the diagonals (the second vertex of the polar triangle), we must find the point of intersection of the mean chord with the limiting polar expressed by equation [63].

Let us do this.

The equation of the mean chord:

\[ \tag{48} 2nx-(r^2+n^2-l^2)=0. \]

The equation of the limiting polar:

\[ (l^2-r^2-n^2)x+\operatorname{ctg}\vartheta(l^2-r^2+n^2)y+2nr^2=0. \]

Solving these equations jointly, we find the coordinates of the point of intersection of the lines expressed by these equations:

\[ \tag{65} \begin{cases} \displaystyle x=\frac{r^2+n^2-l^2}{2n},\\[1.2em] \displaystyle y=\frac{(l^2-r^2-n^2)^2-4n^2r^2}{2n\operatorname{ctg}\vartheta(l^2-r^2+n^2)}. \end{cases} \]

It is obvious that the point found belongs to the required geometric locus.

Let us now find the point of intersection of the limiting polar with the axis of abscissas.

For this, as we know, in the equation of the polar one must set \(y\) equal to zero.

Doing this, we obtain:

\[ (l^2-r^2-n^2)x+2nr^2=0. \]

Whence

\[ \tag{66} x=\frac{2nr^2}{r^2+n^2-l^2}. \]

Let us write the equation of the polar of this point.

It will have the form:

\[ \frac{2nr^2}{r^2+n^2-l^2}x-r^2=0. \]

Clearing the denominator and reducing the equation by \(r^2\), we obtain:

\[ 2nx-(r^2+n^2-l^2)=0. \]

This equation represents the middle chord, which, as we saw earlier, passes through two vertices of the polar triangle and, consequently, is the polar of the third vertex.

Thus, the point determined by formula [66] is the third vertex of the polar triangle for the limiting case under consideration.

Relying on the preceding reasoning, let us return to the chord of dead points, which represents the second limiting polar for the case when \(k=\sigma\).

On substituting the value \(k=\sigma\) into equation [59], we obtained equation [62], which, after certain transformations, took the form [62 bis]

\[ -2r^2x+2\operatorname{ctg}\vartheta(l^2-\sigma^2)y+2nr^2=0. \]

Divide the whole equation by \(-2n\).

Then we have:

\[ \frac{r^{2}x}{n}-\operatorname{ctg}\vartheta\,\frac{(l^{2}-\sigma^{2})}{n}\,y-r^{2}=0. \]

This equation represents the polar of a point whose coordinates have the form:

\[ \tag{67} \left\{ \begin{aligned} x&=\frac{r^{2}}{n},\\[4pt] y&=-\operatorname{ctg}\vartheta\,\frac{(l^{2}-\sigma^{2})}{n}. \end{aligned} \right. \]

Let us take formulas [58], which give the magnitudes of the coordinates of the point of intersection of the coupler-crank radical axes in the general case:

\[ \tag{58} \left\{ \begin{aligned} x&=-\frac{l^{2}-r^{2}-n^{2}-k^{2}}{2n} =\frac{r^{2}+n^{2}+k^{2}-l^{2}}{2n},\\[6pt] y&=-\operatorname{ctg}\vartheta\,\frac{l^{2}-r^{2}+n^{2}-k^{2}}{2n}. \end{aligned} \right. \]

Substituting \(k=\sigma\) into these formulas, we have:

\[ x=\frac{r^{2}+n^{2}-l^{2}+\sigma^{2}}{2n} \quad\text{and}\quad y=-\operatorname{ctg}\vartheta\,\frac{l^{2}-r^{2}+n^{2}-\sigma^{2}}{2n}. \]

We had earlier, on the basis of formula [21]:

\[ l^{2}+r^{2}=n^{2}+\sigma^{2}. \]

Substituting in the formulas under consideration, instead of \((n^{2}+\sigma^{2})\), the equal quantity \((l^{2}+r^{2})\), and instead of \((n^{2}-r^{2})\), the equal quantity \((l^{2}-\sigma^{2})\), we have:

\[ \tag{67} \left\{ \begin{aligned} x&=\frac{r^{2}-l^{2}+l^{2}+r^{2}}{2n}=\frac{r^{2}}{n},\\[6pt] y&=-\operatorname{ctg}\vartheta\,\frac{l^{2}-\sigma^{2}+l^{2}-\sigma^{2}}{2n} =-\operatorname{ctg}\vartheta\,\frac{(l^{2}-\sigma^{2})}{n}. \end{aligned} \right. \]

Thus, the point determined by formulas [67] is one of the vertices of the polar triangle for the limiting case under consideration.

We had that the polar of the point determined by formulas [67] represents the chord of the dead points.

On the other hand, we know that the point of intersection of the diagonals of the inscribed quadrilateral, being the second vertex of the polar triangle, must lie on the chord of the dead points and serve as the pole of the opposite side of the polar triangle.

Further, we know that the chord of the dead points passes through the pole of the coupler radical axis; this pole, however, lies on the perpen-

...the perpendicular dropped from the center of the curvature circle (the origin of coordinates) onto the coupler radial axis.

Consequently, the chord of the dead points intersects the aforesaid perpendicular at the pole.

Let us verify whether this pole is not the point of intersection of the diagonals of the inscribed quadrilateral for the limiting case under consideration, i.e. the second vertex of the polar triangle.

If such an assumption is correct, then, obviously, the following must hold: the line joining the point determined by formulas [67] with the pole of the coupler radial axis must be the polar of the third vertex of the polar triangle and, consequently, the polar of the point of intersection of the chord of the dead points with the coupler radial axis; and since, as we know, both these lines pass through the point \(M\)—the midpoint of the path of the slider—then the line under consideration must be the polar of the point \(M\).

Let us prove this.

The general equation of the line passing through two points has the form:

\[ (y_1-y_2)x-(x_1-x_2)y+(x_1y_2-y_1x_2)=0. \]

Here, let \(x_1\) and \(y_1\) be the coordinates of the pole, and \(x_2,y_2\) the coordinates of the point determined by formulas [67].

Consequently, we have:

\[ x_1=\frac{r^2}{n}, \qquad y_1=-\operatorname{tg}\vartheta\,\frac{r^2}{n}, \]

\[ x_2=\frac{r^2}{n}, \qquad y_2=-\operatorname{ctg}\vartheta\,\frac{(l^2-\sigma^2)}{n}. \]

Forming the coefficients of the equation, we have:

\[ y_1-y_2=\operatorname{ctg}\vartheta\,\frac{(l^2-\sigma^2)}{n} -\operatorname{tg}\vartheta\,\frac{r^2}{n} = \frac{(l^2-\sigma^2)-\operatorname{tg}^2\vartheta\,r^2}{n\cdot \operatorname{tg}\vartheta}, \]

\[ x_1-x_2=\frac{r^2}{n}-\frac{r^2}{n}=0, \]

\[ x_1y_2-y_1x_2 = \operatorname{tg}\vartheta\,\frac{r^2}{n}\cdot\frac{r^2}{n} - \operatorname{ctg}\vartheta\,\frac{(l^2-\sigma^2)}{n}\cdot\frac{r^2}{n} = \]

\[ = -\frac{r^2}{n}\cdot \frac{(l^2-\sigma^2)-\operatorname{tg}^2\vartheta r^2}{n\operatorname{tg}\vartheta}. \]

Consequently, the equation being formed will have the form:

\[ \frac{(l^2-\sigma^2)-tg^2\vartheta r^2}{n\cdot tg\vartheta}\,x - \frac{r^2}{n}\cdot \frac{(l^2-\sigma^2)-tg^2\vartheta r^2}{n\cdot tg\vartheta} =0, \]

which, upon reduction and clearing the denominator, gives:

\[ \tag{68} nx-r^2=0. \]

It is obvious that we have before us the equation of the polar of the point \(M\), lying on the axis of abscissas at a distance \(n\) from the origin of coordinates. Thus we are convinced that our supposition agrees with the truth.

From this we draw the consequence: the point of intersection of the diagonals of the inscribed quadrilateral for the second limiting case is the pole of the connecting-rod radical axis; consequently, its coordinates will be:

\[ \tag{61} x=\frac{r^{2}}{n}\quad \text{and}\quad y=-\operatorname{tg}\vartheta\,\frac{r^{2}}{n}. \]

It is clear that this point, like the point determined by formulas [65], belongs to the required geometric locus.

The two points found, as we shall see further on, will prove extremely useful to us in finding the geometric locus of the points of intersection of conjugate chords.

From all that has been set forth above we conclude that, for all values of the quantity \(k\), the line determined by equation [59] acquires geometric properties that in no way contradict the conditions of the problem under consideration.

Let us now turn to the line determined by equation [56].

\[ \bigl[(n-\alpha)\beta \xi_{1}N_{2}+(n+\alpha)\beta \xi_{2}N_{1}\bigr]x+ \bigl[\xi_{2}(n+\alpha)^{2}N_{1}-\xi_{1}(n-\alpha)^{2}N_{2}\bigr]y+ \]

\[ +2n\beta\bigl[R_{1}R_{2}-\xi_{1}\xi_{2}(n^{2}-\alpha^{2})\bigr]=0. \]

This line, as belonging to the pencil of lines passing through the point of intersection of the conjugate chords determined for the general case by equations [53] and [54], must obviously also be satisfied by particular values of the quantity \(k\).

In other words, for \(k=\sigma\) and \(k=0\) this line must pass through the already found points of the geometric locus; that is, for particular values of the quantity \(k\), equation [56], after substituting into it the coordinates of these points, must become an identity.

Let us carry out this verification.

First of all, we transform equation [56], substituting into it the values of the quantities \(\alpha,\beta,\xi_{1},\xi_{2},N_{1}\), and \(N_{2}\).

We have:

\[ N_{1}=(n-\alpha)^{2}+\beta^{2} =n^{2}+\alpha^{2}-2n\alpha+\beta^{2} =n^{2}+k^{2}\cos^{2}\vartheta-2nk\cos\vartheta+ \]

\[ +k^{2}\sin^{2}\vartheta =n^{2}+k^{2}-2nk\cos\vartheta, \]

\[ N_{2}=(n+\alpha)^{2}+\beta^{2} =n^{2}+k^{2}+2nk\cos\vartheta. \]

Further:

\[ \xi_{1} = \frac{r^{2}-l^{2}+(n-\alpha)^{2}+\beta^{2}}{2(n-\alpha)} = \frac{r^{2}-l^{2}+n^{2}+k^{2}-2nk\cos\vartheta}{2(n-k\cos\vartheta)}, \]

\[ \xi_2=\frac{r^2-l^2+(n+\alpha)^2+\beta^2}{2(n+\alpha)} =\frac{r^2-l^2+n^2+k^2+2nk\cos\vartheta}{2(n+k\cos\vartheta)}. \]

Let us take up the coefficient of \(x\) in equation [56]; for brevity let us denote it by \(A\).

Substituting into \(A\) the values of the quantities entering into it, we have:

\[ A=\beta\bigl[(n-\alpha)N_2\xi_1+(n+\alpha)N_1\xi_2\bigr]= \]

\[ =k\sin\vartheta\left[ \frac{r^2-l^2+n^2+k^2-2nk\cos\vartheta}{2} (n^2+k^2+2nk\cos\vartheta)+ \right. \]

\[ \left. +\frac{r^2-l^2+n^2+k^2+2nk\cos\vartheta}{2} (n^2+k^2-2nk\cos\vartheta) \right]. \]

Let us, for brevity, denote the quantity \((r^2-l^2+n^2)\) by \(m^2\). Then, expanding the brackets and, for bringing like terms together, we finally obtain:

\[ A=k\cdot\sin\vartheta\bigl[(m^2+k^2)(n^2+k^2)-4n^2k^2\cos^2\vartheta\bigr]. \]

Let us pass to the coefficient of \(y\), denoting it by \(B\).

\[ B=\xi_2(n+\alpha)^2N_1-\xi_1(n-\alpha)^2N_2= \]

\[ =\frac{(m^2+k^2+2nk\cos\vartheta)(n+k\cos\vartheta)(n^2+k^2-2nk\cos\vartheta)}{2}- \]

\[ -\frac{(m^2+k^2-2nk\cos\vartheta)(n-k\cos\vartheta)(n^2+k^2+2nk\cos\vartheta)}{2}. \]

Expanding the brackets and, for bringing like terms together, we obtain:

\[ B=k\cos\vartheta\bigl[(m^2+k^2)(k^2-n^2)+2n^2(n^2+k^2-2k^2\cos^2\vartheta)\bigr]. \]

Denoting by the letter \(C\) the last term, after substitution we obtain:

\[ C=2n\beta\bigl[R_1R_2-\xi_1\xi_2(n^2-\alpha^2)\bigr]= \]

\[ =2nk\cdot\sin\beta\, \frac{4R_1R_2-(m^2+k^2)^2+4n^2k^2\cos^2\vartheta}{4}. \]

Dividing the whole equation by \(k\cdot\sin\vartheta\), we have:

\[ \tag{69} \bigl[(m^2+k^2)(n^2+k^2)-4n^2k^2\cos^2\vartheta\bigr]x+ \]

\[ +\operatorname{ctg}\vartheta\bigl[(m^2+k^2)(k^2-n^2)+2n^2(m^2+k^2-2k^2\cos^2\vartheta)\bigr]y+2nR_1R_2+ \]

\[ +\frac{n}{2}\bigl[4n^2k^2\cos^2\vartheta-(m^2+k^2)^2\bigr]=0. \]

Let us investigate this equation for \(k=\sigma\) and \(k=0\).

We had earlier (see p. 31) that for \(k=\sigma\), \(R_1=R_2=0\).

Thus, substituting in equation [69], in place of \(k\), the value \(\sigma\), we obtain:

\[ \bigl[(m^2+\sigma^2)(n^2+\sigma^2)-4n^2\sigma^2\cos^2\vartheta\bigr]x+ \]

\[ +\operatorname{ctg}\vartheta\bigl[(m^2+\sigma^2)(\sigma^2-n^2)+2n^2(n^2+\sigma^2-2\sigma^2\cos^2\vartheta)\bigr]y+ \]

\[ +2n^3\sigma^2\cos^2\vartheta-\frac{n(m^2+\sigma^2)^2}{2}=0. \]

Since for \(k=\sigma\) the point of intersection of the conjugate chords will have coordinates determined by formulas [61], we shall substitute these quantities for \(x\) and \(y\) in the equation obtained:

\[ \tag{61} \left\{ \begin{aligned} x&=\frac{r^2}{n},\\ y&=-\operatorname{tg}\vartheta\,\frac{r^2}{n}. \end{aligned} \right. \]

Then we have:

\[ \bigl[(m^2+\sigma^2)(n^2+\sigma^2)-4n^2\sigma^2\cos^2\vartheta\bigr]\frac{r^2}{n}- \]

\[ -\bigl[(m^2+\sigma^2)(\sigma^2-n^2)+2n^2(n^2+\sigma^2-2\sigma^2\cos^2\vartheta)\bigr]\frac{r^2}{n}+ \]

\[ +2n^3\sigma^2\cos^2\vartheta-\frac{n}{2}(m^2+\sigma^2)^2=0. \]

Clearing the denominator and bringing like terms together, we obtain:

\[ 2r^2\bigl[(m^2+\sigma^2)(n^2+\sigma^2)-4n^2\sigma^2\cos^2\vartheta-(m^2+\sigma^2)(\sigma^2-n^2)- \]

\[ -2n^2(n^2+\sigma^2-2\sigma^2\cos^2\vartheta)\bigr]+4n^4\sigma^2\cos^2\vartheta-n^2(m^2+\sigma^2)^2=0. \]

Carrying out certain transformations in the large brackets, we have:

\[ 2r^2\bigl[2n^2(m^2+\sigma^2)-2n^2(n^2+\sigma^2)\bigr]+4n^4\sigma^2\cos^2\vartheta-n^2(m^2+\sigma^2)^2=0, \]

which further gives:

\[ 4n^2r^2(m^2-n^2)+4n^4\sigma^2\cos^2\vartheta-n^2(m^2+\sigma^2)^2=0. \]

Dividing by \(n^2\) and noting that, for \(m^2=r^2-l^2+n^2\), \((m^2-n^2)=(r^2-l^2)\), we have:

\[ 4r^2(r^2-l^2)+4n^2\sigma^2\cos^2\vartheta-(r^2-l^2+n^2+\sigma^2)^2=0. \]

We had earlier (see formula 21) that

\[ n\sigma\cos\vartheta=lr \quad\text{and}\quad (n^2+\sigma^2)=(r^2+l^2). \]

Substituting these quantities into the result obtained, we have:

\[ 4r^2(r^2-l^2)+4r^2l^2-(r^2-l^2+r^2+l^2)^2=0, \]

which gives finally the identity:

\[ 4r^4-4r^4=0. \]

Let now, in equation [69], \(k=0\).

For \(k=0\), \(\alpha=\beta=0\), and consequently,

\[ R_1=R_2=\sqrt{\,n^2\left[r^2-\left(\frac{r^2-l^2+n^2}{2n}\right)^2\right]\,}. \]

Thus equation [69] assumes the form:

\[ m^2n^2x+\operatorname{ctg}\vartheta\,(2n^4-m^2n^2)y+2n^3r^2 -\frac{2n^3(r^2-l^2+n^2)^2}{4n^2} -\frac{nm^4}{2}=0. \]

For \(k=0\) the point of intersection of the conjugate chords is determined by formulas [65]:

\[ \tag{65} \left\{ \begin{aligned} x&=\frac{r^2+n^2-l^2}{2n}=\frac{m^2}{2n},\\[4pt] y&=\frac{(l^2-r^2-n^2)^2-4n^2r^2}{2n\operatorname{ctg}\vartheta\,(l^2-r^2+n^2)}. \end{aligned} \right. \]

Substituting these quantities into the equation found, we have:

\[ \frac{m^4n^2}{2n} +\frac{(2n^4-m^2n^2)\bigl[(l^2-r^2-n^2)^2-4n^2r^2\bigr]} {2n(l^2-r^2+n^2)} +2n^3r^2-\frac{nm^4}{2}-\frac{nm^4}{2}=0. \]

Observing that \((l^2-r^2-n^2)=-m^2\) and \((l^2-r^2+n^2)=2n^2-m^2\), we can give the equation under consideration the following form:

\[ \frac{m^4n^2}{2n} +\frac{n^2(2n^2-m^2)(m^4-4n^2r^2)} {2n(2n^2-m^2)} +2n^3r^2-\frac{nm^4}{2}-\frac{nm^4}{2}=0, \]

which, after cancellation and collection of like terms, turns into the identity:

\[ -2n^2r^2+2n^2r^2=0. \]

Thus we see that the line expressed by equation [56], derived from the general equations of the conjugate chords, preserves its geometrical meaning also for the special cases, and consequently we have full grounds to introduce it into our analysis in determining the geometrical locus.

CHAPTER SIX.

Determination of the geometric locus of the points of intersection of conjugate chords by an approximate method. Numerical example.

We are faced with the question: to find the geometric locus of the points of intersection of the lines expressed by equations [56] and [59].

Let us write out these equations here, taking, instead of equation [56], its transformed form—equation [69].

Thus, we have:

\[ [59]\qquad (l^{2}-r^{2}-n^{2}-k^{2})x+\operatorname{ctg}\vartheta\,(l^{2}-r^{2}+n^{2}-k^{2})y+2nr^{2}=0, \]

\[ [69]\qquad \bigl[(m^{2}+k^{2})(n^{2}+k^{2})-4n^{2}k^{2}\cos^{2}\vartheta\bigr]x+ \]

\[ +\operatorname{ctg}\vartheta\bigl[(m^{2}+k^{2})(k^{2}-n^{2})+2n^{2}(n^{2}+k^{2}-2k^{2}\cos^{2}\vartheta)\bigr]y+ \]

\[ +2nR_{1}R_{2}+\frac{n}{2}\bigl[4n^{2}k^{2}\cos^{2}\vartheta-(m^{2}+n^{2})^{2}\bigr]=0. \]

Each of these equations represents a certain line, changing its position depending on the change of the quantity \(k\).

In order to find the geometric locus of the points of intersection of these lines, we must eliminate \(k\) from both equations.

We shall do this as follows: determine \(k\) from equation [59], insert the value obtained into equation [69]; the result of the substitution will be an equation defining the desired geometric locus.

If we look at equation [69], we shall see that it contains the quantity \(R_{1}R_{2}\)—the product of two irrational quantities; further, from formula [43] it is clear that the quantity \(k\) of interest to us enters under both radical signs.

Thus, in order to seek the geometric locus, we must first of all free ourselves from the irrationality in equation [69].

We shall free ourselves from this irrationality by isolating the irrational quantity and squaring the entire equation.

Before carrying out the indicated algebraic operations, let us determine their geometrical meaning.

Provisionally we may write equation [69] in the following form:

\[ Ax+By+C=0, \]

where

\[ C=C_{1}+2nR_{1}R_{2}. \]

Thus, we have:

\[ Ax+By+C_{1}+2nR_{1}R_{2}=0. \]

We eliminate the irrational quantity:

\[ Ax+By+C_1=-2nR_1R_2. \]

Squaring the equation, we then have:

\[ (Ax+By+C_1)^2=4n^2R_1^2R_2^2. \]

We move all terms to the left-hand side:

\[ (Ax+By+C_1)^2-4n^2R_1^2R_2^2=0. \]

We factor:

\[ (Ax+By+C_1+2nR_1R_2)(Ax+By+C_1-2nR_1R_2)=0, \]

or

\[ (Ax+By+C)(Ax+By+C_1-2nR_1R_2)=0. \]

From this we see that the elimination of the irrationality transforms equation [69] into the product of two equations, one of which represents the line we are investigating, while the other is a new line, parallel to the first and at a variable distance from it, since only for \(k=0\), when \(R_1=R_2=0\), do both equations represent one and the same line.

Thus we see that, with the exception of one special case, this new line does not pass through the point of intersection of the conjugate chords and, consequently, is entirely superfluous for finding the geometrical locus that interests us.

Having noted this, we proceed to perform the algebraic operations. We eliminate the irrational quantity in equation [69]; then we have:

\[ [(m^2+k^2)(n^2+k^2)-4n^2k^2\cos^2\vartheta]x+ \]

\[ +\operatorname{ctg}\vartheta\,[(m^2+k^2)(k^2-n^2)+2n^2(n^2+k^2-2k^2\cos^2\vartheta)]y+ \]

\[ +\left[2n^3k^2\cos^2\vartheta-\frac{n(m^2+k^2)^2}{2}\right] =-2nR_1R_2. \]

On squaring, we may write the result conventionally in the following form:

\[ A^2x^2+B^2y^2+C_1^2+2ABxy+2AC_1x+2BC_1y=4n^2R_1^2R_2^2. \]

Let us determine the coefficients of this equation.

\[ A^2=[(m^2+k^2)(n^2+k^2)-4n^2k^2\cos^2\vartheta]^2. \]

We expand the small brackets inside the large ones.

\[ A^2=[k^4+k^2(n^2+m^2-4n^2\cos^2\vartheta)+m^2n^2]^2. \]

Denoting provisionally by the letter \(p^2\) the quantity \((n^2+m^2-4n^2\cos^2\vartheta)\), we have:

\[ A^2=[k^4+k^2p^2+m^2n^2]^2. \]

Squaring, we obtain:

\[ [70]\qquad A^2=k^8+2k^6p^2+k^4(p^4+2m^2n^2)+2k^2p^2m^2n^2+m^4n^4. \]

We now determine the coefficient \(B^2\).

\[ B^2=\operatorname{ctg}^2\vartheta\bigl[(m^2+k^2)(k^2-n^2)+2n^2(n^2+k^2-2k^2\cos^2\vartheta)\bigr]^2. \]

Expanding the inner brackets and collecting the quantities \(k\) in descending powers:

\[ B^2=\operatorname{ctg}^2\vartheta\,[k^4+k^2(n^2+m^2-4n^2\cos^2\vartheta)+n^2(2n^2-m^2)]^2. \]

Denoting \((2n^2-m^2)\) by \(c^2\), we have:

\[ B^2=\operatorname{ctg}^2\vartheta\,[k^4+k^2p^2+n^2c^2]^2. \]

Squaring, we obtain:

\[ [71]\qquad B^2=\operatorname{ctg}^2\vartheta\,[k^8+2k^6p^2+k^4(p^4+2n^2c^2)+2k^2p^2n^2c^2+n^4c^4]. \]

We pass to the coefficient \(C_1^2\).

\[ C_1^2=\left[2n^3k^2\cos^2\vartheta-\frac{n(m^2+k^2)^2}{2}\right]^2 =\frac{n^2}{4}\left[4n^2k^2\cos^2\vartheta-(m^2+k^2)^2\right]^2. \]

Expanding the inner brackets, we have:

\[ C_1^2=\frac{n^2}{4}\left[4n^2k^2\cos^2\vartheta-m^4-k^4-2m^2k^2\right]^2 \]

\[ =\frac{n^2}{4}\left[-\{k^4+2k^2(m^2-2n^2\cos^2\vartheta)+m^4\}\right]^2. \]

Denoting \((m^2-2n^2\cos^2\vartheta)\) by \(a^2\), we have:

\[ C_1^2=\frac{n^2}{4}\left[-(k^4+2k^2a^2+m^4)\right]^2. \]

Squaring, we obtain:

\[ [72]\qquad C_1^2=\frac{n^2}{4}\left[k^8+4k^6a^2+2k^4(m^4+2a^4)+4k^2a^2m^4+m^8\right]. \]

We determine the coefficient \(A.B\).

\[ A.B=\operatorname{ctg}\vartheta[(m^2+k^2)(n^2+k^2)-4n^2k^2\cos^2\vartheta]\bigl[(m^2+k^2)(k^2-n^2)+ \]

\[ +2n^2(n^2+k^2-2k^2\cos^2\vartheta)\bigr]. \]

As before, we shall have:

\[ A.B=\operatorname{ctg}\vartheta\,[k^4+k^2p^2+m^2n^2]\,[k^4+k^2p^2+n^2c^2]. \]

Multiplying and collecting the quantities \(k\), we obtain:

\[ A.B=\operatorname{ctg}\vartheta\left[k^8+2k^6p^2+k^4(m^2n^2+p^4+n^2c^2)+k^2p^2n^2(m^2+c^2)+m^2n^4c^2\right]. \]

Since \(c^2=2n^2-m^2\), then \(c^2+m^2=2n^2\), and consequently,

\[ [73]\qquad A.B=\operatorname{ctg}\vartheta\left[k^8+2k^6p^2+k^4(p^4+2n^4)+2k^2p^2n^4+m^2n^4c^2\right]. \]

We pass to the coefficient \(A.C_1\).

\[ A.C_1=\left[(m^2+k^2)(n^2+k^2)-4n^2k^2\cos^2\vartheta\right] \left[2n^3k^2\cos^2\vartheta-\frac{n(m^2+k^2)^2}{2}\right]. \]

Taking \(\dfrac{n}{2}\) outside the brackets from the second factor, we have:

\[ A.C_1=-\frac{n}{2}\left[k^4+k^2p^2+m^2n^2\right]\left[k^4+2k^2a^2+m^4\right]. \]

Multiplying, we obtain:

\[ [74]\qquad A.C_1=-\frac{n}{2}\left[ k^8+k^6(p^2+2a^2)+k^4(m^2n^2+2p^2a^2+m^4) +k^2m^2(2n^2a^2+p^2m^2)+m^6n^2 \right]. \]

We shall now determine the last coefficient \(B.C_1\). In accordance with the former designations, we shall have:

\[ B.C_1=\operatorname{ctg}\vartheta\left[k^4+k^2p^2+n^2c^2\right] \left[4n^2k^2\cos^2\vartheta-(m^2+k^2)^2\right]\frac{n}{2}. \]

Taking the minus sign outside the brackets from the third factor, we have:

\[ B.C_1=-\frac{n.\operatorname{ctg}\vartheta}{2} \left[k^4+k^2p^2+n^2c^2\right] \left[k+2k^2a^2+m^4\right]. \]

Performing the multiplication and arranging according to powers of the quantity \(k\), we obtain:

\[ [75]\qquad B.C_1=-\frac{n.\operatorname{ctg}\vartheta}{2} \left[ k^8+k^6(p^2+2a^2)+k^4(n^2c^2+2p^2a^2+m^4) +k^2(2a^2n^2c^2+m^4p^2)+m^4n^2c^2 \right]. \]

We pass to the term \(4n^2R_1^2R_2^2\).

From formulae [43], after substitution, we obtain:

\[ R_1=\sqrt{(n^2+k^2-2nk\cos\vartheta)r^2-\frac{(m^2+k^2-2nk\cos\vartheta)^2}{4}} \quad \text{and} \]

\[ R_2=\sqrt{(n^2+k^2+2nk\cos\vartheta)r^2-\frac{(m^2+k^2+2nk\cos\vartheta)^2}{4}}. \]

Consequently, \(R_1^2 R_2^2\) will be equal to the product of the radical quantities.

Thus, we may write:

\[ R_1^2 R_2^2 = \left[(n^2+k^2)^2-4n^2k^2\cos^2\vartheta\right]r^4 - \]

\[ -\frac{r^2}{4} \left[ (n^2+k^2+2nk\cos\vartheta)(m^2+k^2-2nk\cos\vartheta)^2 + \right. \]

\[ \left. +(n^2+k^2-2nk\cos\vartheta)(m^2+k^2+2nk\cos\vartheta)^2 \right] + \]

\[ + \frac{(m^2+k^2-2nk\cos\vartheta)^2(m^2+k^2+2nk\cos\vartheta)^2}{16}. \]

Expanding the parentheses in the first term:

\[ [76]\qquad \left[n^4+k^4+2n^2k^2-4n^2k^2\cos^2\vartheta\right]r^4 = \left[k^4+2n^2k^2(1-2\cos^2\vartheta)+n^4\right]r^4. \]

Expanding the parentheses in the second term and collecting like terms, we find that the second term takes the form:

\[ -\frac{r^2}{4} \left[ 2(n^2+k^2)\left[(m^2+k^2)^2+4n^2k^2\cos^2\vartheta\right] -16(m^2+k^2)n^2k^2\cos^2\vartheta \right]. \]

Dividing by \(2\) and expanding the parentheses further, we obtain:

\[ -\frac{r^2}{2} \left[ (n^2+k^2)(m^4+k^4+2m^2k^2+4n^2k^2\cos^2\vartheta) -8(m^2+k^2)n^2k^2\cos^2\vartheta \right]. \]

Carrying out the multiplication and taking the quantities \(k\) outside the parentheses, we shall have:

\[ -\frac{r^2}{2} \left[ (n^2+k^2)\left[m^4+k^4+2k^2(m^2+2n^2\cos^2\vartheta)\right] - \right. \]

\[ \left. -8m^2n^2k^2\cos^2\vartheta -8n^2k^4\cos^2\vartheta \right] = \]

\[ = -\frac{r^2}{2} \left[ k^6+k^4(n^2+2m^2-4n^2\cos^2\vartheta) +k^2\left[n^2(2n^2+m^2-8m^2\cos^2\vartheta)+4n^4\cos^2\vartheta\right] +n^2m^4 \right]. \]

Or, introducing here the quantity \(p^2=n^2+m^2-4n^2\cos^2\vartheta\), we finally obtain:

\[ [77]\qquad -\frac{r^2}{2} \left[ k^6+k^4(p^2+m^2) +k^2\left[m^2(2p^2-m^2)+4n^4\cos^2\vartheta\right] +n^2m^4 \right]. \]

Now let us turn to the last term:

\[ \frac{(m^2+k^2-2nk\cos\vartheta)^2(m^2+k^2+2nk\cos\vartheta)^2}{16}. \]

Coupler Pole

We shall perform all the operations successively:

\[ \frac{(m^2+k^2-2nk\cos\vartheta)^2(m^2+k^2+2nk\cos\vartheta)^2}{16} = \frac{\left[(m^2+k^2)^2-4n^2k^2\cos^2\vartheta\right]^2}{16} = \]

\[ = \frac{(m^2+k^2)^4+16n^4k^4\cos^4\vartheta -8(m^2+k^2)n^2k^2\cos^2\vartheta}{16}. \]

Raising to the corresponding powers and collecting the quantities \(k\), we shall have:

\[ \frac{ k^8+4k^6(m^2-2n^2\cos^2\vartheta) +2k^4(3m^4-8n^2m^2\cos^2\vartheta+8n^4\cos^4\vartheta) +4k^2m^4(m^2-2n^2\cos^2\vartheta)+m^8 }{16}. \]

We previously denoted the quantity \((m^2-2n^2\cos^2\vartheta)\) by the letter \(a^2\); therefore we shall finally have the expression for the last term:

\[ [78]\qquad \frac{ k^8+4k^6a^2+2k^4(m^4+2a^4)+4k^2m^4a^2+m^8 }{16}. \]

Now let us write our equation, using all the quantities found from [70] to [78].

Then we shall have:

\[ [79]\qquad \left\{ \begin{aligned} &\bigl[k^8+2k^6p^2+k^4(p^4+2m^2n^2)+2k^2p^2m^2n^2+m^4n^4\bigr]x^2+\\ &\quad+\bigl[k^8+2k^6p^2+k^4(p^4+2n^2c^2)+2k^2p^2n^2c^2+n^4c^4\bigr]\operatorname{ctg}^2\vartheta\,y^2+\\ &\quad+2\bigl[k^8+2k^6p^2+k^4(p^4+2n^4)+2k^2p^2n^4+m^2n^4c^2\bigr]\operatorname{ctg}\vartheta\,xy-\\ &\quad-2\cdot\frac{n}{2}\bigl[k^8+k^6(p^2+2a^2)+k^4(m^2n^2+2p^2a^2+m^4)+\\ &\qquad\qquad\qquad\qquad\qquad\qquad +k^2m^2(2n^2a^2+p^2m^2)+m^6n^2\bigr]x-\\ &\quad-2\cdot\frac{n}{2}\bigl[k^8+k^6(p^2+2a^2)+k^4(c^2n^2+2p^2a^2+m^4)+\\ &\qquad\qquad\qquad\qquad\qquad\qquad +k^2(2a^2n^2c^2+m^4p^2)+m^4n^2c^2\bigr]\operatorname{ctg}\vartheta\,y+\\ &\quad+\frac{n^2}{4}\bigl[k^8+4k^6a^2+2k^4(m^4+2a^4)+4k^2a^2m^4+m^8\bigr] \end{aligned} \right. \]

\[ \begin{aligned} ={}&4n^2r^4\bigl[k^4+2n^2k^2(1-2\cos^2\vartheta)+n^4\bigr]-\\ &-4n^2\cdot\frac{r^2}{4} \bigl[ k^6+k^4(p^2+m^2)+k^2\{m^2(2p^2-m^2)+4n^4\cos^2\vartheta\}+n^2m^4 \bigr]+\\ &+\frac{4n^2\bigl[k^8+4k^6a^2+2k^4(m^4+2a^4)+4k^2m^4a^2+m^8\bigr]}{16}. \end{aligned} \]

We free equation [79] from the denominator and transfer all terms to the left-hand side; then, expanding the brackets and arranging

arranging the equation in descending powers of the quantity \(k\), we give it the following form:

\[ \tag{80} \left\{ \begin{aligned} &+k^8\ \bigl|\, x^2+\operatorname{ctg}^2\vartheta\, y^2+2\operatorname{ctg}\vartheta\, xy-nx-n\operatorname{ctg}\vartheta\, y+ \\[2mm] &+k^6\ \bigl|\, 2p^2x^2+2p^2\operatorname{ctg}^2\vartheta\, y^2+4p^2\operatorname{ctg}\vartheta\, xy -n(p^2+2a^2)x- \\[1mm] &\hspace{35mm}{}-n(p^2+2a^2)\operatorname{ctg}\vartheta\, y+2n^2r^2+ \\[2mm] &+k^4\ \bigl|\,(p^4+2m^2n^2)x^2+(p^4+2n^2c^2)\operatorname{ctg}^2\vartheta\, y^2+ \\[1mm] &\hspace{35mm}{}+2(p^4+2n^4)\operatorname{ctg}\vartheta\, xy -n[m^2n^2+2p^2a^2+m^4]x- \\[1mm] &\hspace{35mm}{}-n[n^2c^2+2p^2a^2+n^4]\operatorname{ctg}\vartheta\, y +2n^2r^2(m^2+p^2-2r)+ \\[2mm] &+k^2\ \bigl|\, 2m^2n^2p^2x^2+2n^2c^2p^2\operatorname{ctg}^2\vartheta\, y^2 +4p^2n^4\operatorname{ctg}\vartheta\, xy- \\[1mm] &\hspace{35mm}{}-m^2n[2n^2a^2+m^2p^2]x -n[2a^2n^2c^2+m^4p^2]\operatorname{ctg}\vartheta\, y+ \\[1mm] &\hspace{35mm}{}+2n^2r^2\bigl[m^2(2p^2-m^2)+4n^4\cos^2\vartheta -4n^2r^2(1-2\cos^2\vartheta)\bigr]+ \\[2mm] &+k^0\ \bigl|\, m^4n^4x^2+n^4c^4\operatorname{ctg}^2\vartheta\, y^2 +2m^2n^4c^2\operatorname{ctg}\vartheta\, xy-m^6n^3x- \\[1mm] &\hspace{35mm}{}-m^4n^3c^2\operatorname{ctg}\vartheta\, y +2r^2n^4(m^4-2r^2n^2) \end{aligned} \right|=0. \]

To determine the geometric locus, we must in equation [80] substitute, in place of \(k^n\), the quantity determined from equation [59]. In the notation adopted by us, this equation assumes the following form:

\[ \tag{81} (m^2+k^2)x+\operatorname{ctg}\vartheta\,(m^2+k^2-2n^2)y-2nr^2=0. \]

We determine from equation [81] the value of the quantity \(k^2\):

\[ \tag{82} k^2=\frac{2nr^2-m^2x-\operatorname{ctg}\vartheta\,(m^2-2n^2)y}{x+\operatorname{ctg}\vartheta\, y}. \]

Considering formula [82], we arrive at the conclusion that the expression \(k^8\) will contain the quantities \(x\) and \(y\) to the fourth degree and, consequently, after substituting all the \(k\)’s in equation [80], we shall obtain an equation of the 6th degree with respect to \(x\) and \(y\); moreover, this equation will also contain all the decreasing degrees of the unknowns.

Thus, we arrive at the conclusion that finding the exact geometric locus of the points of intersection of the conjugate chords for the rectilinear-derivative crank-and-connecting-rod mechanism leads us to the investigation of an equation of the sixth degree with a very large number of terms.

Leaving the further development of the proposed path for a subsequent time, let us now introduce into our problem a certain arbitrary assumption that considerably simplifies our investigation.

As will be shown below by a numerical example, the path described by the point of intersection of the conjugate chords is extremely small; it is traversed almost along a straight line, and the errors arising from the assumptions we have made lie far beyond the limits of accuracy of ordinary drawing instruments.

For every value of the quantity \(k\) within the limits between \(0\) and \(c\), we obtain, in the general case, a quadrilateral inscribed in the centrode circle, two of whose sides are the crank-rod radical axes expressed by equations [28] and [38].

The point of intersection of the diagonals of this quadrilateral, which are the conjugate chords, will lie inside the quadrilateral between the radical axes and, consequently, inside the angle formed by lines [28] and [38].

If we join this point with the vertex of the angle, we obtain a line belonging to the pencil passing through the radical center, since the vertex of the angle, as was shown above, is the radical center whose coordinates are expressed by formulas [58].

Let us turn to equations [28] and [38].

These equations have the form:

\[ \begin{aligned} [28]\qquad &2(n-\alpha)x+2\beta y-(n-\alpha)^2-\beta^2-r^2+l^2=0,\\ [38]\qquad &2(n+\alpha)x-2\beta y-(n+\alpha)^2-\beta^2-r^2+l^2=0. \end{aligned} \]

Expanding the brackets, we then obtain:

\[ 2nx-2\alpha x+2\beta y-n^2-\alpha^2+2n\alpha-\beta^2-r^2+l^2=0, \]

\[ 2nx+2\alpha x-2\beta y-n^2-\alpha^2-2n\alpha-\beta^2-r^2+l^2=0. \]

The result of termwise addition and subtraction of these two equations is two new ones, defining two lines belonging to the same pencil.

Carrying out the indicated operation, we obtain:

\[ 4nx-2(n^2+\alpha^2+\beta^2+r^2-l^2)=0, \]

\[ 4\alpha x-4\beta y-4n\alpha=0. \]

Making reductions and substituting for \(\alpha\) and \(\beta\) their values in terms of \(k\), we have:

\[ [83]\qquad 2nx-(n^2+r^2-l^2+k^2)=0, \]

\[ [60]\qquad x\cos\vartheta-y\sin\vartheta-n\cos\vartheta=0. \]

Considering the equations obtained, we see that one of them represents the equation previously found by us for the crank-rod radical axis (see p. 46), while the other represents a straight line perpendicular to the axis of abscissas, i.e. to the base line of our mechanism.

By the line expressed by equation [83], we shall replace the line, unknown to us, passing through the radical center and the point of intersection of the conjugate chords.

Such a replacement, being arbitrary, in the final result gives us an insignificant error, since the chosen line will always lie between the radical axes and, consequently, the locus obtained by means of it will differ little in its properties from the true one.

The admissibility of such a replacement is, moreover, confirmed by investigating equation [83] for the limiting values of \(k\).

Indeed, for \(k=0\), we obtain:

\[ \tag{84} 2nx-\left(n^{2}+r^{2}-l^{2}\right)=0 . \]

This equation, as we know, represents the mean chord (see [48]).

Let \(k=\sigma\); then equation [83] takes the form:

\[ 2nx-\left(n^{2}+r^{2}-l^{2}+\sigma^{2}\right)=0 . \]

We had earlier, from formula [21], that \(n^{2}+\sigma^{2}=l^{2}+r^{2}\); consequently, for \(k=\sigma\), equation [83] assumes the final form:

\[ \tag{85} nx-r^{2}=0 . \]

Equations [84] and [85] show us that these lines pass through points whose abscissas are respectively equal to

\[ \tag{86} x_{0}=\frac{n^{2}+r^{2}-l^{2}}{2n},\qquad x_{\sigma}=\frac{r^{2}}{n}. \]

Since our task now consists in finding, for each given \(k\), the point of intersection of the line expressed by equation [83] with the polar expressed by formula [59]—consequently, in other words, for the limiting values of the quantity \(k\), to find on the corresponding polars the points whose abscissas would be expressed by formulas [86].

As is not difficult to see, these points are the points previously found by us, determined by formulas [61] and [67], which, as has already been proved, belong to the sought locus.

Thus, we may say that the introduction into our analysis of the line expressed by formula [83] does not contradict the essence of the matter, and for the limiting values of the quantity \(k\) gives complete agreement with the truth.

Let us solve jointly equations [81] and [83] with respect to the quantity \(k\).

\[ \tag{83} 2nx-\left(n^{2}+r^{2}-l^{2}+k^{2}\right)=2nx-\left(m^{2}+k^{2}\right)=0, \]

[81].

\[ (m^{2}+k^{2})x+\operatorname{ctg}\vartheta(m^{2}+k^{2}-2n^{2})y-2nr^{2}=0. \]

From [83] we have:

\[ (m^{2}+k^{2})=2nx. \]

Substituting this quantity in [81], we then obtain:

\[ 2nx^{2}+\operatorname{ctg}\vartheta(2nx-2n^{2})y-2nr^{2}=0, \]

which, after reduction, gives:

\[ \tag{87} x^{2}+xy\cdot\operatorname{ctg}\vartheta-ny\operatorname{ctg}\vartheta-r^{2}=0. \]

Equation [87] represents (under the assumption made) the required geometrical locus.

Since we have before us, evidently, a curve of the second order, in order to determine its kind we must form the discriminant from the coefficients of the given equation.

If we have before us a general equation of a curve of the second order of the following form:

\[ Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0, \]

then, in order to determine the geometrical kind of the curve, it is necessary to examine two algebraic expressions composed of the coefficients of the given equation:

\[ H=B^{2}-4AC, \]

\[ \Delta=2(4ACF-AE^{2}-CD^{2}+BDE-B^{2}F). \]

Turning to equation [87], we see that it represents the general equation of a curve of the second order with coefficients:

\[ A=1,\quad B=\operatorname{ctg}\vartheta,\quad C=0,\quad D=0,\quad E=-n\operatorname{ctg}\vartheta,\quad F=-r^{2}. \]

Consequently,

\[ H=B^{2}-4AC=\operatorname{ctg}^{2}\vartheta>0, \]

\[ \Delta=2(-AE^{2}-B^{2}F)=-2(n^{2}\operatorname{ctg}^{2}\vartheta+r^{2}\operatorname{ctg}^{2}\vartheta)=-2\operatorname{ctg}^{2}\vartheta(n^{2}+r^{2}). \]

Since the quantity \(H=\operatorname{ctg}^{2}\vartheta\) is always greater than zero for \(\vartheta\) lying between \(0^\circ\) and \(90^\circ\), and only between these limits is the existence of a rectilinearly generating slider-crank mechanism possible—and \(\Delta\) is not equal to zero, we conclude that equation [87] represents a hyperbola.

Having thus obtained an approximate notion of the character of the geometrical locus of interest to us, we shall not investigate equation [87], since it does not correspond to reality.

For the solution of the question we have posed: “is it possible, without great error, to admit in a rectilinearly generating curv—

in the crank mechanism the existence of a coupler pole”—we shall choose the following path.

In deriving equation [60] we showed that for each value of \(k\) we obtain, as the result of the intersection of the radical axes, the corresponding radical center.

We further showed that this radical center moves along the coupler radical axis, expressed by equation [60].

Let us determine the character of the motion of the radical center.

Formulas [58] give the coordinates of the points of intersection of the radical axes for an indeterminate \(k\).

We write them here:

\[ [58]\quad \left\{ \begin{aligned} x&=-\,\frac{l^{2}-r^{2}-n^{2}-k^{2}}{2n} =\frac{m^{2}+k^{2}}{2n},\\[4pt] y&=-\operatorname{ctg}\vartheta\, \frac{l^{2}-r^{2}+n^{2}-k^{2}}{2n} =-\operatorname{ctg}\vartheta\, \frac{2n^{2}-m^{2}-k^{2}}{2n}. \end{aligned} \right. \]

At the greatest value of the quantity \(k\), namely at \(k=\sigma\), the abscissa of the radical center takes the form:

\[ x=\frac{m^{2}+\sigma^{2}}{2n} =\frac{n^{2}+r^{2}-l^{2}+\sigma^{2}}{2n} =\frac{r^{2}}{n}. \]

This expression is known to us from formulas [67] and [61].

When the quantity \(k\) changes from \(\sigma\) to \(0\), i.e., as the slider moves away from the dead position, we have that the abscissa of the radical center decreases together with the decrease of \(k\).

Indeed, for the general case we have:

\[ x=\frac{m^{2}+k^{2}}{2n} =\frac{m^{2}}{2n}+\frac{k^{2}}{2n}. \]

Thus, we arrive at the conclusion that, as the slider moves away from its dead positions, the radical center moves along the coupler radical axis, receding from the midpoint of the slider path.

At the lower limiting value \(k=0\), the abscissa of the radical center takes the form:

\[ x=\frac{m^{2}}{2n}. \]

We had this expression in formula [64].

On the basis of what has just been said, we draw the following conclusion: as the slider moves from the dead points toward the middle, the polar of the radical center, expressed by equation [59], rotates about the pole of the coupler radical axis in the clockwise direction, for the adopted arrangement of the drawing.

We showed earlier that, for limiting values of the magnitude \(k\), the coordinates of the points of intersection of conjugate chords are expressed by formulas [65] and [67].

Let us determine the relative position of these points with respect to the coordinate axes.

Let (see Drawing X*) these points be, respectively, \(Q\) and \(P\).*

The abscissa of the point \(Q\) is

\[ x_q=\frac{r^2+n^2-l^2}{2n}. \]

The abscissa of the point \(P\) has the form: \(x_p=\dfrac{r^2}{n}\). We shall prove that \(x_q<x_p\).

We write the inequality:

\[ \frac{r^2+n^2-l^2}{2n}<\frac{r^2}{n}. \]

Clearing the inequality of denominators:

\[ r^2+n^2-l^2<2r^2. \]

Isolating \(n^2\), we then have:

\[ n^2<r^2+l^2. \]

Hence we conclude that the inequality is valid, since we proved much earlier that

\[ r^2+l^2=n^2+\sigma^2. \]

Thus we arrive at the conclusion that the point \(Q\) lies to the left of the point \(P\); in other words, the curve defining the locus must, when the pole rotates clockwise, pass inside the angle \(EPA\), formed by the limiting positions of the pole.

We can also verify the correctness of what has been said in the following way.

Suppose that the curve defining the locus of the points of intersection of conjugate chords, beginning (for \(k=\sigma\)) at the point \(P\), passes inside the angle \(BPF\).

Since this curve must arrive (for \(k=0\)) at the point \(Q\), it is obvious that it must again pass through the point \(P\) in order to be situated inside the angle \(EPA\).

Since we have proved that the radical center—one of the vertices of the polar triangle—when \(k\) varies from \(\sigma\) to \(0\), continuously moves away from the midpoint of the slider’s path, it is clear that the double passage of the locus curve through the point \(P\) must correspond to two different positions of the radical center; consequently, we must conclude that the point \(P\) is twice a point of intersection of conjugate chords.

* Drawing X is made without regard to scale, in order to obtain as large an angle \(EPA\) as possible.
A. U.

In other words, under our assumption of the existence of a curve inside the angle \(BPF\), three lines must intersect at the point \(P\): the chord of the dead points and two associated chords.

Let us see whether this is possible.

We must prove that the three lines expressed by equations [47], [53], and [54] intersect at one point.

It is necessary, therefore, that the determinant composed of the coefficients of these equations be equal to zero.

We form the determinant, relying on the conventional notation of equations [53] and [54]:

\[ \left| \begin{array}{ccc} A+a, & B+b, & C+c \\ A-a, & B-b, & C-c \\ -r\sigma\sin\vartheta, & (ln-r\sigma\cos\vartheta), & rn\sigma\cos\vartheta \end{array} \right| =\Delta. \]

For arbitrary \(k\), i.e. for \(k\) less than \(\sigma\), this determinant is not equal to 0; when \(k\) is equal to \(\sigma\), equations [47], [53], and [54] express one and the same straight line.

Thus, our assumption of the existence of a curve inside the angle \(BPF\) contradicts the truth.

For each value of \(k\) we obtain, respectively, a quadrilateral inscribed in the crank circle, the points of intersection of whose diagonals and opposite sides form a polar triangle. Two of the vertices of this triangle, namely, the radical center and the point of intersection of the associated chords, we have already considered.

Let us turn to the third vertex.

To obtain the coordinates of this vertex, we must (see drawing \(V\)) write the equations of the lines passing through the points \((x'_1,y'_1)\), \((x_2,y_2)\) and \((x_1,y_1)\), \((x'_2,y'_2)\), and then solve these two equations jointly.

It is easy to see that as a result we shall obtain complicated equations, analogous to equations [53] and [54], with irrational quantities entering into them and, consequently, requiring very complicated manipulations.

Therefore we shall proceed by another path.

We shall prove that for \(k\) not equal to 0 and \(\sigma\) (for these values we have already found the third vertex of the polar triangle) the coordinates of the third vertex have a finite value, i.e. are not infinitely large.

In order for the abscissa \(x\) to be infinitely large, it is necessary that the sides of the inscribed quadrilateral be parallel to one another.

Let us have (see Drawing X) two sides \(HJ\) and \(KL\) parallel to one another.

The two other sides, being radical axes, intersect at the point \(G\) on the line \(MS\). The diameter of the circle perpendicular to the chords bisects them and, obviously, passes through the point \(G\).

Thus, for an infinitely large value of the abscissa \(x\), it is necessary that the line joining the point of intersection of the radical axes with the center of the curvilinear circle make equal angles with the radical axes. Let us see whether this is possible.

The point of intersection of the radical axes for an arbitrary \(k\) is expressed by formulas [58]:

\[ x=\frac{m^2+k^2}{2n},\qquad y=-\operatorname{ctg}\vartheta\,\frac{2n^2-m^2-k^2}{2n}. \]

The equation of the line joining this point with the center will have the following form:

\[ \operatorname{ctg}\vartheta\,\frac{2n^2-m^2-k^2}{2n}\,x -\frac{(m^2+k^2)}{2n}\,y=0, \]

which, upon reduction and clearing the denominator, gives:

\[ [88]\qquad (2n^2-m^2-k^2)\cos\vartheta\,x-(m^2+k^2)\sin\vartheta\,y=0. \]

Or, conventionally,

\[ Ax+By=0. \]

From formulas [28] and [38] we write here, in a slightly transformed form, the equations of the radical axes:

\[ [89]\qquad (n-k\cos\vartheta)x+k\sin\vartheta\,y -\frac{n^2+r^2-l^2+k^2-2nk\cos\vartheta}{2}=0, \]

\[ [90]\qquad (n+k\cos\vartheta)x-k\sin\vartheta\,y -\frac{n^2+r^2-l^2+k^2+2nk\cos\vartheta}{2}=0. \]

The tangent of the angle \(\varphi_1\) between the lines [88] and [89] will be expressed as follows:

\[ [91]\qquad \operatorname{tg}\varphi_1= \frac{Ak\sin\vartheta-B(n-k\cos\vartheta)} {A(n-k\cos\vartheta)+Ak\sin\vartheta}. \]

The tangent of the angle \(\varphi_2\) between the lines [88] and [90], respectively, will have the following form:

\[ [92]\qquad \operatorname{tg}\varphi_2= \frac{-Ak\sin\vartheta-B(n+k\cos\vartheta)} {A(n+k\cos\vartheta)-Ak\sin\vartheta}. \]

Since we have assumed that the angles \(\varphi_1\) and \(\varphi_2\) are equal to each other, consequently there must be the equality:

\[ \frac{Ak\sin\vartheta-B(n-k\cos\vartheta)} {A(n-k\cos\vartheta)+Ak\sin\vartheta} = \frac{-Ak\sin\vartheta-B(n+k\cos\vartheta)} {A(n+k\cos\vartheta)-Ak\sin\vartheta}. \]

Subtracting one from each part of the proportion obtained, we shall then have:

\[ \frac{-(n-k\cos\vartheta)(A+B)} {A(n-k\cos\vartheta)+Ak\sin\vartheta} = \frac{-(n+k\cos\vartheta)(A+B)} {A(n+k\cos\vartheta)-Ak\sin\vartheta}. \]

Reducing the result obtained by \(-\dfrac{A+B}{A}\), we have:

\[ \frac{n-k\cos\vartheta} {n-k(\cos\vartheta-\sin\vartheta)} = \frac{n+k\cos\vartheta} {n+k(\cos\vartheta-\sin\vartheta)}. \]

Taking the products of the means and extremes of the proportion:

\[ (n-k\cos\vartheta)[n+k(\cos\vartheta-\sin\vartheta)] = (n+k\cos\vartheta)[n-k(\cos\vartheta-\sin\vartheta)]. \]

Expanding the brackets and doing the reduction of like terms, we obtain:

\[ n^2-kn\cos\vartheta+nk(\cos\vartheta-\sin\vartheta)-k^2\cos\vartheta(\cos\vartheta-\sin\vartheta) = \]

\[ = n^2+kn\cos\vartheta-nk(\cos\vartheta-\sin\vartheta)-k^2\cos\vartheta(\cos\vartheta-\sin\vartheta), \]

or,

\[ 2kn\cos\vartheta-2kn(\cos\vartheta-\sin\vartheta)=0. \]

Whence: \(\sin\vartheta=0\), or the angle \(\vartheta=90^\circ\).

In deriving the conditions for the possibility of existence of a rectilinearly generated connecting-rod mechanism, we proved that the angle \(\vartheta\) cannot be equal to \(90^\circ\).

Thus we have proved that the abscissa of the third vertex of the polar triangle cannot assume infinitely large values.

Reasoning in the same way, we shall prove that the ordinate of the third vertex also cannot be an infinitely large quantity.

On the basis of this conclusion we are convinced that the geometric locus of the intersections of conjugate chords can be only a closed curve, and we were right not to examine equation [87], which represents a hyperbola.

To decide the question whether it is possible to admit the existence of a connecting-rod pole in the mechanism under consideration, we shall nevertheless suppose that the third vertex of the polar triangle lies on the \(x\)-axis and has receded to an infinitely great distance, so that its polar is the \(y\)-axis. In making such an assumption, we consciously increase the error to our disadvantage, i.e. we increase the length of the segment of the curve inside the angle \(EPA\).

Let us assume that the point of the curve \(T\) farthest from the point \(P\) will lie midway between the points \(t_1\) and \(t_2\). Let us find its ordinate, since its abscissa is equal to zero.

For this, let us determine the ordinates of the points of intersection of the limiting polars \(EF\) and \(AB\) with the \(y\)-axis.

As we know, the general equation of the polar, expressed by formula [59], gives, for \(k=0\) and \(h=\sigma\), the following two equations:

\[ [47]\qquad -r\sigma\sin\vartheta\,x+(ln-r\sigma\cos\vartheta)y+rn\sigma\sin\vartheta=0, \]

\[ [63]\qquad (l^2-r^2-n^2)x+\operatorname{ctg}\vartheta(l^2-r^2+n^2)y+2nr^2=0. \]

We have already shown earlier that equation [47] may be written in the form of equation [62 bis]:

\[ -2r^2x+2\operatorname{ctg}\vartheta(l^2-\sigma^2)y+2nr^2=0. \]

We find the point of intersection of this line with the axis of ordinates; for this we set \(x\) equal to zero.

Then we obtain:

\[ 2\operatorname{ctg}\vartheta(l^2-\sigma^2)y_1+2nr^2=0. \]

Whence:

\[ y_1=-\frac{nr^2}{l^2-\sigma^2}\operatorname{tg}\vartheta. \]

In the same way we determine the point of intersection of line [63] with the axis of ordinates,

\[ y_2=-\frac{2nr^2}{l^2-r^2+n^2}\operatorname{tg}\vartheta. \]

The ordinate of the point of the curve \(T\), according to what has been adopted, will be equal to the half-sum of the ordinates found:

\[ y=\frac{y_1+y_2}{2} =-\operatorname{tg}\vartheta\frac{nr^2}{2} \left(\frac{1}{l^2-\sigma^2}+\frac{2}{l^2-r^2+n^2}\right). \]

We transform this expression:

\[ y=-\operatorname{tg}\vartheta\cdot\frac{nr^2}{2}\cdot \frac{l^2-r^2+n^2+2l^2-2\sigma^2} {(l^2-\sigma^2)(l^2-r^2+n^2)} = \]

\[ =-\operatorname{tg}\vartheta\cdot\frac{nr^2}{2}\cdot \frac{3l^2-2\sigma^2+n^2-r^2} {(l^2-\sigma^2)(l^2-r^2+n^2)}. \]

We have had earlier the relation:

\[ n^2+\sigma^2=l^2+r^2. \]

Whence:

\[ n^2-r^2=l^2-\sigma^2 \quad\text{and}\quad l^2-r^2+n^2=2l^2-\sigma^2. \]

Consequently, we shall obtain:

\[ y=-\operatorname{tg}\vartheta\cdot\frac{nr^2}{2} \frac{(4l^2-2\sigma^2)} {(l^2-\sigma^2)(2l^2-\sigma^2)}, \]

which, after reduction, gives finally:

\[ \tag{93} y=-\frac{nr^2}{l^2-\sigma^2}\operatorname{tg}\vartheta . \]

Thus, we have at our disposal three points of the curve of the locus of intersections of conjugate chords. The coordinates of these points are expressed by formulas [61], [65], and [93]:

\[ \tag{65} \begin{cases} x_1=\dfrac{r^2+n^2-l^2}{2n},\\[6pt] y_1=\dfrac{(l^2-r^2-n^2)^2-4n^2r^2}{2n\,\operatorname{ctg}\vartheta\,(l^2-r^2+n^2)}. \end{cases} \]

\[ \tag{61} \begin{cases} x_2=\dfrac{r^2}{n},\\[6pt] y_2=-\operatorname{tg}\vartheta\,\dfrac{r^2}{n}. \end{cases} \]

\[ \tag{93} \begin{cases} x_3=0,\\[6pt] y_3=-\dfrac{nr^2}{l^2-\sigma^2}\operatorname{tg}\vartheta . \end{cases} \]

We shall now show that, for normal ratios of the dimensions of a rectilinear-generating crank mechanism, the distances between these three points are extremely small, so that, in tracing, these three points merge into a segment of a straight line of insignificant length.

For this purpose let us turn to a numerical example.

Let us have a certain slide mechanism in which, as is often allowed, the extreme points of the slide—the places of attachment of the eccentric rods—move along straight lines (see drawing XI). Dividing the slide mentally by a line passing through the center of the shaft \(O\) and the midpoint of the slide \(B\) into two symmetrical parts, we obtain, evidently, two identical rectilinear-generating crank mechanisms.

Let us take the radius of the eccentric \(OA=r\) as unity; the distance from the center of the shaft to the midpoint of the slide we shall take equal, as is often the case, to \(20r\), and the length of half the slide to \(2.5r\).

Let us determine the remaining dimensions characterizing our generating mechanism.

We shall find the length of the segment \(OM\)—the base line.

From the right triangle \(OMB\) we have:

\[ \overline{OM}^{\,2}=n^2=\overline{OB}^{\,2}+\overline{BM}^{\,2}, \]

or,

\[ n^{2}=20^{2}+2.5^{2}, \]

\[ =400+6.25, \]

\[ =406.25, \]

whence:

\[ n=\sqrt{406.25}= \]

\[ =20.15. \]

We determine the angle \(\vartheta=\angle NMO=\angle MOB\).

\[ \cos\vartheta=\frac{OB}{OM}=\frac{20}{20.15}=0.99255. \]

To this value of the cosine there corresponds an angle of \(7^\circ\).
From the tables we find \(\operatorname{tg}\vartheta=0.12278\).
Then

\[ \operatorname{tg}^{2}\vartheta=0.015129 \sim 0.015. \]

We had the following relation between the elements of the derived mechanism:

\[ l\cdot r=n\cdot \sigma \cos\vartheta \quad \text{and} \quad l^{2}-\sigma^{2}=n^{2}-r^{2}. \]

Here \(l\) is the length of the eccentric rod, \(\sigma\) is the length of the semipath of the upper end of the slide.

We substitute the values known to us into the first formula:

\[ l\cdot 1=20.15\cdot 0.99255\,\sigma. \]

Carrying out the multiplication, we obtain:

\[ l=19.9998825\,\sigma \sim 20\sigma. \]

Substituting this value of \(l\) into the second formula, we have:

\[ l^{2}-\sigma^{2}=n^{2}-r^{2}, \]

\[ 400\sigma^{2}-\sigma^{2}=(20.15)^{2}-1, \]

or,

\[ 399\sigma^{2}=405.25. \]

Whence:

\[ \sigma^{2}=\frac{405.25}{399}, \]

\[ =1.0154, \]

and

\[ \sigma=\sqrt{1.0154}, \]

\[ =1.008. \]

Now we can find the length of the eccentric rod:

\[ l=20\delta= \]

\[ =20{,}16. \]

Thus, the elements of our mechanism are as follows:

Radius of the eccentric . . . . . . . . . \(r=1,\)

Length of the eccentric rod . . . . . . \(l=20{,}16,\)

Length of the main rocker . . . . . . . \(n=20{,}15,\)

Length of the path of the end of the slide . . . \(2\sigma=2{,}016.\)

Angle \(\vartheta\) . . . . . . . . . . \(=7^\circ,\)

\(\operatorname{tg}^{2}\vartheta\) . . . . . . . . . \(=0{,}015.\)

Let us now determine the magnitude of the angle \(EPA\) (see Drawing X) between the limiting positions of the polar.

For the limiting values of the quantity \(k\) we determined the coordinates of the third vertices of the corresponding polar triangles, and thereby found that both of these vertices lie on the \(x\)-axis.

The coordinates of the point \(M\) will evidently be

\[ x_m=n, \]

\[ y_m=0. \]

The coordinates of the point \(N\)—by formula [66]:

\[ x_n=\frac{2nr^2}{r^2+n^2-l^2}, \]

\[ y_n=0. \]

The coordinates of the point \(P\)—by formula [61]:

\[ x_p=\frac{r^2}{n}, \]

\[ y_p=-\operatorname{tg}\vartheta \frac{r^2}{n}. \]

To determine the cosine of the angle \(MPN\), we need to know the magnitudes of the sides of triangle \(PMN\).

Let us determine them, knowing the coordinates of the vertices of the triangle.

\[ \overline{PM}^{\,2}=(x_m-x_p)^2+(y_m-y_p)^2, \]

\[ \tag{94} \overline{PM}^{\,2} =\left(n-\frac{r^{2}}{n}\right)^{2} +\left(\operatorname{tg}\vartheta\,\frac{r^{2}}{n}\right)^{2}, \]

\[ =\frac{n^{4}-2n^{2}r^{2}+r^{4}\left(1+\operatorname{tg}^{2}\vartheta\right)}{n^{2}}. \]

We pass to the second side.

\[ \overline{PN}^{\,2}=(x_n-x_p)^2+(y_n-y_p)^2, \]

\[ \overline{PN}^{\,2} =\left(\frac{2nr^{2}}{m^{2}}-\frac{r^{2}}{n}\right)^{2} +\left(\operatorname{tg}\vartheta\,\frac{r^{2}}{n}\right)^{2}, \]

\[ =\left(\frac{2n^{2}r^{2}-m^{2}r^{2}}{m^{2}n}\right)^{2} +\operatorname{tg}^{2}\vartheta\,\frac{r^{4}}{n^{2}}. \]

Taking \(\dfrac{r^{4}}{n^{2}}\) outside the brackets,

\[ \tag{95} \overline{PN}^{\,2} =\frac{r^{4}}{n^{2}} \left[ \left(\frac{2n^{2}-m^{2}}{m^{2}}\right)^{2} +\operatorname{tg}^{2}\vartheta \right], \]

\[ =\frac{r^{4}}{n}\cdot \frac{4n^{2}(n^{2}-m^{2})+m^{4}\left(1+\operatorname{tg}^{2}\vartheta\right)}{m^{4}}. \]

The length of the third side \(MN\) is easily determined by the difference of the abscissae \(x_n\) and \(x_m\),

\[ \overline{MN}=x_n-x_m=\frac{2nr^{2}}{m^{2}}-n= \]

\[ =\frac{n(2r^{2}-m^{2})}{m^{2}}. \]

Whence:

\[ \tag{96} \overline{MN}^{\,2}=\frac{n^{2}(2r^{2}-m^{2})}{m^{4}}. \]

Calling \(\angle MPE\) by the letter \(\alpha\), we have:

\[ \cos\alpha= \frac{\overline{PM}^{\,2}+\overline{PN}^{\,2}-\overline{MN}^{\,2}} {2\,\overline{PM}\cdot\overline{PN}}. \]

Substituting into this formula the expressions [94], [95] and [96], we obtain for the cosine the following quantity:

\[ \cos\alpha= \frac{ \dfrac{n^{4}-2n^{2}r^{2}+r^{4}(1+\operatorname{tg}^{2}\vartheta)}{n^{2}} + \dfrac{4r^{4}n^{2}(n^{2}-m^{2})+r^{4}m^{4}(1+\operatorname{tg}^{2}\vartheta)}{m^{4}n^{2}} - \dfrac{n^{2}(2r^{2}-m^{2})^{2}}{m^{4}} } { 2 \sqrt{\dfrac{n^{4}-2n^{2}r^{2}+r^{4}(1+\operatorname{tg}^{2}\vartheta)}{n^{2}}} \sqrt{\dfrac{4r^{4}n^{2}(n^{2}-m^{2})+r^{4}n^{4}(1+\operatorname{tg}^{2}\vartheta)}{m^{4}n^{2}}} }. \]

Reducing to a common denominator and canceling like terms, we give the preceding expression the following form:

\[ [97]\quad \cos \alpha = \frac{ r^2 m^2(1+\operatorname{tg}^2 \vartheta)+n^2(2n^2-2r^2-m^2) }{ \sqrt{\,n^4-2n^2r^2+r^4(1+\operatorname{tg}^2 \vartheta)\,}\, \sqrt{\,4n^2(n^2-m^2)+m^4(1+\operatorname{tg}^2 \vartheta)\,} }. \]

Let us substitute their numerical values for the letters. We denoted by the letter \(m^2\) the quantity \(n^2+r^2-l^2\). Let us determine its value.

\[ n^2=(20,15)^2=406,0225, \]

\[ r^2=(1)^2=1,0000, \]

\[ l^2=(20,16)^2=406,4250. \]

Consequently, \(m^2=0,5975\).

Whence

\[ m^4=0,35700625 \]

and

\[ 2n^2-2r^2-m^2=809,4475. \]

Let us determine the first radical of the denominator.

\[ \sqrt{\,n^4-2n^2r^2+r^4(1+\operatorname{tg}^2\vartheta)\,} = \sqrt{\,n^2(n^2-2r^2)+r^4(1+\operatorname{tg}^2\vartheta)\,}. \]

Substituting the numerical values:

\[ n^2-2r^2=404,0225, \]

\[ n^2(n^2-2r^2)=164042,22550625 \quad\text{and} \]

\[ r^4(1+\operatorname{tg}^2\vartheta)=1,015. \]

Thus we obtain the radical:

\[ \sqrt{164043,24050625}=405,0225. \]

Let us take up the second radical of the denominator. Substituting the numerical values:

\[ n^2-m^2=405,4250; \]

\[ 4n^2(n^2-m^2)=658446,68825, \]

\[ m^4(1+\operatorname{tg}^2\vartheta)=0,3623613. \]

Whence the whole radical:

\[ \sqrt{\,4n^2(n^2-m^2)+m^4(1+\operatorname{tg}^2\vartheta)\,} = \sqrt{658447,0506113}. \]

Extracting the root of the number, we obtain:

\[ \sqrt{\,4n^2(n^2-m^2)+m^4(1+\operatorname{tg}^2\vartheta)\,} = 811,44881. \]

For the numerical value of \(cosinus\) we have the fraction:

\[ \cos \alpha = \frac{0{,}5975 \cdot 1{,}015 + 406{,}0225 \cdot 809{,}4475} {405{,}0225 \cdot 811{,}4481}. \]

Whence we finally obtain:

\[ \cos \alpha = 0{,}9999981, \]

i.e., a quantity differing extremely little from unity, and, consequently, the angle \(\alpha\) is almost equal to zero.

If we determine \(\alpha\) with an accuracy to seconds, we obtain:

\[ \angle MPN = \alpha = 0^\circ 6'20'', \]

i.e., about one tenth of a degree.

Thus we have the right to say that the angle \(MPN\), in which the curve of the locus of the points of intersection of the conjugate chords is contained, may practically be taken as equal to zero and, consequently, the curve turns into a straight line.

Let us now determine the magnitude of the segment of the rectilinear path of the points of intersection of the conjugate chords.

The initial point of this path has coordinates expressed by formulas [61]; the final point (under the known assumption) is determined by formula [93].

Thus, the whole path will be determined as the distance between these two points.

Denoting this distance by the letter \(d\), we obtain the following expression:

\[ d = \sqrt{ \left(\frac{r^2}{n}\right)^2 + \left( -\operatorname{tg}\vartheta \frac{r^2}{n} + \operatorname{tg}\vartheta \frac{nr^2}{n^2-r^2} \right)^2 }. \]

We perform the transformation.

\[ d = \sqrt{ \frac{r^4}{n^2} + \operatorname{tg}^2\vartheta \left( \frac{r^2 n}{n^2-r^2} - \frac{r^2}{n} \right)^2 } = \]

\[ = \sqrt{ \frac{r^4}{n^2} + \frac{r^4}{n^2} \operatorname{tg}^2\vartheta \cdot \frac{r^4}{(n^2-r^2)^2} }. \]

Taking \(\frac{r^2}{n}\) out from under the radical sign, we obtain:

\[ d = \frac{r^2}{n} \sqrt{ 1+ \operatorname{tg}^2\vartheta \frac{r^4}{(n^2-r^2)^2} }. \]

We substitute here the numerical values. Then

\[ d = \frac{1}{20{,}15} \sqrt{ 1+ \frac{0{,}015}{(405{,}0225)^2} }. \]

In view of the exceedingly small numerical value of the fraction in the radius, we discard it and finally obtain

\[ d=\frac{1}{20.15}=0.0496 \sim 0.05 . \]

Thus, under the assumption we have made, the point of intersection of the conjugate chords moves rectilinearly along the chord of the dead points and describes a path of about five hundredths of the radius.

Consequently, if the radius of the eccentric is equal to \(100^{m}/m\), the path traversed by this point is equal to \(5^{m}/m\).

Let us suppose that our assumption, by which we obtained formula [93], greatly exaggerates the result just indicated.

Let us assume that the path under consideration is bounded only by the points determined by formulas [61] and [65].

Let us determine the length of the segment, and in doing so simplify our calculation, assuming that the chord of the dead points is parallel to the fundamental line, i.e. that the ordinates of formulas [61] and [65] are equal to one another.

Then the required segment is equal to the difference of the abscissas:

\[ \begin{aligned} d &= x_2 - x_1 = \\ &= \frac{r^2}{n} - \frac{r^2+n^2-l^2}{2n} = \\ &= \frac{r^2-n^2+l^2}{2n}. \end{aligned} \]

Since the numerical values of \(n\) and \(l\) differ very little from one another, we shall assume that \(l^2-n^2\) is equal to zero.

Then

\[ d=\frac{r^2}{2n}=\frac{1}{40.3}\sim 0.025 . \]

Thus, even with a rough calculation, we find that the path traversed by the point of intersection of the conjugate chords along the chord of the dead points, for a radius of \(100^{m}/m\), is equal to \(2.5^{m}/m\).

Consequently, we are not entitled to admit the existence of a coupler pole in a rectilinear-generating crank mechanism.

On the other hand, direct drawing, even on a large scale, shows us that all the conjugate chords intersect approximately at the point of intersection between the chord of the dead points and the mean chord. (See drawing V.)

This may occur because, as the slider moves away from its dead position even by small distances, the point of intersection of the conjugate chords quickly reaches the point of intersection of the mean chord with the chord of the dead points, and remains near this place until the slider arrives at the middle of its path.

CHAPTER SEVEN

The curvilinear-sliding crank mechanism. Conditions for the possibility of its existence. Properties of conjugate chords.

Let us take a slider mechanism with a rectilinear slide, oscillating about the point \(M\) (Plate XII).

The line \(OM\), connecting the center of rotation of the main shaft with the center of oscillation of the slide, divides the slide into two curvilinear-sliding crank mechanisms.

Let us undertake the study of mechanisms of this kind.

In the first chapter we mentioned that the base line of a connecting-rod crank mechanism is the line connecting the center of rotation of the machine shaft with the midpoint of the path of the slider; then we indicated that the angle of inclination of the tangent at the midpoint of the curvilinear path of the slider characterizes the given mechanism.

Suppose that we have (see Plate XIII) a curvilinear-sliding crank mechanism. The path of the slider follows the arc \(NML\) of a circle of radius \(O_1L=\rho\); to this path there corresponds the central angle

\(NO_1L=\varphi\); then the chord corresponding to this angle is \(NL=2\sigma=2\rho\sin\dfrac{\varphi}{2}\), whence

\[ [98]\qquad \sigma=\rho\sin\dfrac{\varphi}{2}. \]

The versed sine of the arc \(NML\)—the line \(MP=h\)—is determined in terms of the radius \(\rho\) and the angle \(\varphi\) as follows:

\[ [99]\qquad h=\rho\left(1-\cos\dfrac{\varphi}{2}\right). \]

The base line of the given mechanism is the line \(MO=\sqrt{OO_1^{\,2}+O_1M^{\,2}}\).

Let us determine the magnitude of the crank radius and the length of the connecting rod.

The dead points of the path of the slider are the points \(N\) and \(L\). Their distances from the center \(O\) of the main shaft determine, as we already know from the first chapter, the magnitudes of the crank and the connecting rod.

Thus we arrive at the conclusion that, for the magnitudes of the crank radius and the length of the connecting rod, the curvilinearity of the slider path, when the positions of the dead points are unchanged, has no significance whatever; in other words, the position of the dead points on the crank circle depends exclusively on the position of the dead points of the slider path. Hence we derive the consequence: the chord of the dead points of the crank circle remains unchanged when the dead points of the slider path are unchanged.

Leaving the same dead points \(N\) and \(L\), let us give the radius of the arc \(NML\)—\(\rho\)—an infinitely large value. Then we obtain a straight-line generating mechanism with the slider path \(NPL\), where \(P\) is the midpoint of the slider path and the line \(OP\) is the principal line of the resulting mechanism. This line \(OP=n_1\), in distinction from the line \(OM=n\), we shall call the second principal line of the mechanism under consideration.

For the straight-line generating mechanism we have proved that the chord of the dead points, when prolonged, passes through the midpoint of the slider path.

On this basis we may say that in a curvilinear-generating crank mechanism the chord of the dead points of the crank circle meets the second principal line at the midpoint of the chord subtending the dead points of the curved path of the slider.

Having determined the radius of the crank and the length of the connecting rod as the half-difference and half-sum of the distances \(ON\) and \(OL\), we have before us a completely defined curvilinear-generating mechanism.

By notching from the midpoint \(M\) of the curved path of the slider a crank circle with radius equal to the length of the connecting rod, we obtain the mean chord \(CD\).

It is clear that the mean chord is perpendicular to the first principal line.

From all that has been said we conclude that the curvilinear-generating mechanism is completely determined by two principal lines, by the angle \(\vartheta\) between the tangent at the midpoint of the slider path and the first principal line, and, finally, by the radius \(\rho\) of the circular path of the slider.

Denoting by \(\vartheta_1\) the angle between the second principal line and the chord subtending the dead points of the slider path, on the basis of formulas [24] we may write the conditions for the possible existence of a curvilinear-generating connecting-rod-and-crank mechanism:

\[ [100]\qquad \left\{ \begin{aligned} l \cdot r &= n_1 \sigma \cos \vartheta_1,\\ l-r &\ge n_1 \sin \vartheta_1,\\ \rho \sin \frac{\varphi}{2} &= \sigma \le n_1 \cos \vartheta_1,\\ \rho &> r. \end{aligned} \right. \]

If we take on the curve \(NML\) two points \(m\) and \(m_1\), symmetric with respect to the point \(M\)—the midpoint of the slider path—and from these points notch the crank circle with a radius equal to the length of the connecting rod, then we obtain two conjugate chords \(HG\) and \(EF\), which intersect in the drawing at the point \(\lambda\)—at the point of intersection of the mean chord

…and the chords of the dead points. In other words, we have before us the coupler pole.

Let us show that the point \(\lambda\) is only the visible coupler pole and that, mathematically speaking, the point of intersection of the conjugate chords describes a certain path.

In the first chapter we indicated that the system of conjugate points forms a quadrilateral inscribed in the crank circle.

The intersection of all conjugate chords at one point means that at this point the diagonals of all inscribed quadrilaterals intersect. We know that the point of intersection of the diagonals of an inscribed quadrilateral is one of the vertices of the corresponding polar triangle.

The property of polar triangles is as follows: each vertex is the pole of the opposite side.

Thus, if all conjugate chords intersect at one point—and, consequently, one of the vertices of the polar triangles is fixed—then it is necessary that the other two vertices of these triangles, in all their changes, move along a straight line which is the polar of the fixed vertex.

The second vertex of the polar triangle in the mechanism under consideration will represent—analogously to the rectilinear-generating mechanism considered—the radical center of the system of three circles: the crank circle and the two coupler circles.

Since the centers of the coupler circles in our investigation are always taken symmetrically with respect to the midpoint of the path of the slider, we conclude that the coupler radical axis for all coupler circles will be one and the same; in other words, the radical center of the system of three circles—the second vertex of the polar triangle—will, for all displacements of the slider, move along a straight line.

Consequently, if we prove that, for some position of the slider, the third vertex of the polar triangle does not lie on this straight line, our assertion that all conjugate chords intersect at one point is incorrect.

Let us take (see drawing XIV) the chord \(NL\), which subtends the dead points of the path of the slider, as the rectilinear-generating mechanism. For such a mechanism, when \(k=\sigma\), we have already found that the third vertex of the polar triangle will be the point \(P\), lying on the line \(HM\)—the coupler radical axis; the corresponding radical center will be at the point \(G\).

When \(k=0\), i.e. for the midpoint of the path of the slider, the inscribed quadrilateral becomes the middle chord, as we already know, perpendicular to the first principal line.

The first vertex of the polar triangle for the case under consideration will be situated at the point \(E\)—the point of intersection of the mean chord \(CB\) with the polar of the corresponding radical center \(F\)—the second vertex of the polar triangle, which obviously lies on the line \(HM\).

The pole of the mean chord \(CD\)—the third vertex of the polar triangle—must lie on the line \(OM\), perpendicular to \(CD\) and passing through the center \(O\).

On the other hand, we know that, for all conjugate chords to be able to intersect at one point, it is necessary that the third vertex of the polar triangle also lie on the line \(HM\).

Thus, the pole of the mean chord must be situated simultaneously on the line \(OM\) and on the line \(HM\); in other words, it must coincide with the point \(M\), which is at a distance \(n\) from the center of the crank circle.

Let us show that this assertion does not correspond to reality.

Let the \(x\)-axis coincide with the line \(OM\), and let the \(y\)-axis be perpendicular to this line and pass through the center of the crank circle.

Let us find the equation of the mean chord. It will obviously be the equation of the radical axis for the case when the slider coincides with the point \(M\).

We write the equations of the circles of the crank and of the connecting rod, taking the crank radius equal to \(r\), the length of the connecting rod equal to \(l\), and the length of the segment of the first base line equal to \(n\).

Then we shall have:

\[ x^{2}+y^{2}-r^{2}=0; \]

\[ (x-n)^{2}+y^{2}-l^{2}=0. \]

Subtracting one equation from the other, we obtain the equation of the radical axis—the mean chord—in the following form:

\[ 2nx-(r^{2}+n^{2}-l^{2})=0. \]

This equation, as representing the mean chord—the limit of the inscribed quadrilateral when \(k=0\)—must represent the polar of the third vertex of the polar triangle for the case under consideration. It is easy to see that the pole of this straight line will have the coordinates:

\[ x=\frac{2nr^{2}}{r^{2}+n^{2}-l^{2}} \quad\text{and}\quad y=0. \]

In order that the third vertex of the polar triangle coincide with the point \(M\), whose coordinates are \(x=n\) and \(y=0\), the following condition must be fulfilled:

\[ \frac{2r^{2}}{r^{2}+n^{2}-l^{2}}=1, \]

which in turn gives:

\[ n^2 = r^2 + l^2. \]

On the basis of what was set forth much earlier, relying on formulas [100], we may write the following:

\[ r^2 + l^2 = n_1^2 + \sigma^2. \]

Whence we have:

\[ n^2 = n_1^2 + \sigma^2. \]

The geometrical meaning of the expression obtained is as follows: the square of the hypotenuse is equal to the sum of the squares of the legs.

In other words: the second principal line is perpendicular to the chord connecting the dead positions of the slider, i.e. \(\vartheta_1 = 90^\circ\), and this, as we know, is impossible.

Thus, the conjugate chords of a curvilinear-generating mechanism do not intersect at a single point, and a mathematical coupler pole does not exist in the mechanism under consideration.

Proceeding analogously to what we did in the investigation of the rectilinear-generating slider-crank mechanism, we may prove that the angle between the limiting positions of the polar radical centers, i.e. the angle within which the geometric locus of the points of intersection of the conjugate chords of a curvilinear-generating mechanism is contained, is so small that we have the right to assert that the limiting poles coincide with one another.

It follows from this that the point of intersection of the conjugate chords moves along a straight line (the chord of the dead points), describing a certain path.

The segment of the chord of the dead points along which the point of intersection of the conjugate chords moves we shall call the coupler polar segment *).

CHAPTER EIGHT.

General conclusions.

The consideration of slider-crank mechanisms of all three kinds allows us to draw the following conclusions:

*) This term is not particularly successful, but we shall nevertheless retain it, relying on the fact that, for a certain mutual relation of the parts of derived slider-crank mechanisms, this segment becomes so small that it may be taken as a visible (not mathematical) point—and then this segment becomes a coupler pole.

A. U.

a) to every crank mechanism in a crank circle there corresponds one definite chord of dead points;

b) to every crank mechanism there corresponds one definite mean chord;

c) the intersection of two definite chords—the mean chord and the chord of dead points—gives, for each mechanism, one definite point: the coupler pole of the mechanism;

d) the position of the coupler pole within the crank circle depends exclusively on the dimensions and mutual arrangement of the parts of the crank mechanism;

e) a change in the dimensions and mutual arrangement of the parts of the crank mechanism, entailing a change in its kinematic properties, at the same time also changes the position of the coupler pole, and consequently,

f) the position of the coupler pole fully determines the kinematic properties of the corresponding crank mechanism;

g) since the properties and conditions of steam distribution of a steam engine are determined by the kinematic properties of the primary and secondary crank mechanisms, the study of the properties of the corresponding coupler poles entails the explanation of the phenomena occurring in the steam-distributing organs during their displacements.

CHAPTER NINE.

Properties of the coupler pole of the primary crank mechanism.

Leaving aside the consideration of derived crank mechanisms*), we shall show here certain properties of the coupler pole in the primary connecting-rod-and-crank mechanism.

The fundamental proposition: all conjugate chords intersect at one point, lying on the line of dead points at a distance

\[ \xi = \frac{1}{2n} \]

from the center of the crank circle in the direction toward the slider.

Here \(n\) is an abstract number indicating how many times the length of the connecting rod is greater than the radius of the crank. Consequently, when \(n\) is known, the position of the coupler pole is completely exactly known to us.

*) This question will be set forth in detail in the work I am preparing for print: “Investigation of slotted mechanisms by the method of the coupler pole.”

A. U.

Let us suppose that we know (see drawing II) two conjugate points \((x_1, y_1)\) and \((x_2, y_2)\), found by graphical construction or analytically.

Let these points be the points \(B\) and \(E\) in drawing XV. Since in the basic crank mechanism all conjugate chords intersect one another on the line of dead points \(MM_1\), we easily find, from the chord \(BE\), the chord \(AE\) conjugate to it.

When the crank is in point \(B\), the piston will move away from the left dead point \(M\), assuming an infinitely long connecting rod, by the distance \(MH\).

If we wish to determine the true position of the piston, taking into account the finite length of the connecting rod, then, following Schorch’s method, we must draw through point \(B\) a circle of radius equal to the length of the connecting rod, and then obtain point \(J\), which indicates the magnitude of the error \(HJ\) that has occurred from the assumption of an infinitely great length of the connecting rod.

Let us see whether this quantity \(HJ\) has any relation to the point of intersection \(\lambda\) of the conjugate chords for the given position of the crank at point \(B\).

Since the points \(A\) and \(B\) are symmetric with respect to the line of dead points \(MM_1\), the line \(AB\) is perpendicular to the line \(MM_1\); on the same grounds the line \(EF\) is parallel to the line \(AB\).

The lines \(AF\) and \(BE\) intersect, as conjugate chords, at the point \(\lambda\).

Let us connect the points \(B\) and \(J\) by a straight line and extend this line to its intersection with the crank circle at point \(C\). In a similar way we shall find the point \(D\), which represents the true position of the piston corresponding to the position of the crank at point \(F\).

We shall show that the lines \(BC\) and \(FG\) may be regarded as parallel to one another.

Let us find the coordinates of point \(B\).

Let the origin of coordinates coincide with the center of the crank circle (see drawing II).

Then the equation of this circle will be:

\[ x^2 + y^2 - r^2 = 0. \]

For the position of the slider, removed from the middle of its path by the amount \(k\), we shall have, respectively:

\[ [x - (n + k)]^2 + y^2 - n^2 = 0. \]

Solving these equations simultaneously, we obtain for point \(B\) (drawing XV) the following coordinate values:

\[ x_b = \frac{k^2 + 2nk + r^2}{2(n+k)}, \qquad y_b = \frac{\sqrt{(r^2-k^2)[(2n+k)^2-r^2]}}{2(n+k)}. \]

The coordinates of the point \(J\) will have the following form:

\[ x_j=k;\qquad y_j=0. \]

Let us now write the equation of the line \(BC\), passing through the points \(B\) and \(J\).
As is not difficult to see, the equation of this line will have the following form:

\[ \sqrt{(r^2-k^2)[(2n+k)^2-r^2]}\cdot x-(r^2-k^2)\cdot y-k\sqrt{(r^2-k^2)[(2n+k)^2-r^2]}=0. \]

The equation of the line \(FG\) is obtained from the equation of the line \(BC\), if in it we put, instead of \(+k\), the quantity \((-k)\).
Thus, we obtain:

\[ \sqrt{(r^2-k^2)[(2n-k)^2-r^2]}\cdot x-(r^2-k^2)\cdot y+k\sqrt{(r^2-k^2)[(2n-k)^2-r^2]}=0. \]

Let us find the angle between the lines \(BC\) and \(FG\).
If we have two lines expressed by the equations:

\[ Ax+By+C=0\quad \text{and} \]

\[ A_1x+B_1y+C_1=0, \]

then the tangent of the angle between these lines will have the following form:

\[ \operatorname{tg}\varphi=\frac{AB_1-BA_1}{AA_1+AB_1}. \]

In the equations expressing the lines \(BC\) and \(FG\), the coefficients of \(y\) are the same; consequently, we have:

\[ \operatorname{tg}\varphi=\frac{B\cdot(A-A_1)}{A\cdot(A_1+B_1)}. \]

Substituting into this expression the values of the quantities \(A\), \(B\), \(A_1\), and \(B_1\), we then obtain:

\[ \operatorname{tg}\varphi= -\frac{(r^2-k^2)\left\{\sqrt{(r^2-k^2)[(2n+k)^2-r^2]}-\sqrt{(r^2-k^2)[(2n-k)^2-r^2]}\right\}} {\sqrt{(r^2-k^2)[(2n+k)^2-r^2]}\cdot \left\{\sqrt{(r^2-k^2)[(2n-k)^2-r^2]}-(r^2-k^2)\right\}}, \]

which, after reduction by \(\sqrt{r^2-k^2}\), gives us:

\[ \operatorname{tg}\varphi= -\frac{(r^2-k^2)\left\{\sqrt{(2n+k)^2-r^2}-\sqrt{(2n-k)^2-r^2}\right\}} {\sqrt{(2n+k)^2-r^2}\left\{\sqrt{(r^2-k^2)[(2n-k)^2-r^2]}-(r^2-k^2)\right\}}. \]

For a machine of normal type, the length of the connecting rod is equal to five times the length of the crank radius: if the crank radius is equal to 1, then \(n=5\).
The quantity \(k\) may take values from zero to unity.

Compute, for various values of \(k\), the magnitude \(\operatorname{tg}\varphi\).

Determining separately the values of the terms of the algebraic fraction that expresses \(\operatorname{tg}\varphi\), we can compile the following table:

\(r=1,\ n=5\)	0.8	0.6	0.4	0.2
\(k=\)	0.8	0.6	0.4	0.2
\(1-k^2=\)	0.36	0.64	0.84	0.96
\(\sqrt{1-k^2}=\)	0.6	0.8	0.917	0.98
\(2n+k=\)	10.8	10.6	10.4	10.2
\((2n+k)^2-r^2=\)	115.64	111.36	107.16	103.04
\(\sqrt{(2n+k)^2-r^2}=\)	10.75	10.55	10.35	10.15
\(2n-k=\)	9.2	9.4	9.6	9.8
\((2n-k)^2-r^2=\)	83.64	87.36	91.16	95.04
\(\sqrt{(2n-k)^2-r^2}=\)	9.145	9.346	9.55	9.75
\(\operatorname{tg}\varphi=\)	0.010	0.010	0.008	0.005

Examining this table, we see that \(\operatorname{tg}\varphi\) is measured by a very small quantity; restricting ourselves to two figures, we have: \(\operatorname{tg}\varphi=0.01\), which corresponds to an angle of \(0^\circ34'\).

On the basis of what has been set forth, we conclude that the lines \(BC\) and \(FG\) may be regarded as parallel, and, consequently, the line \(CF\) passes through the center \(O\) of the crank circle.

This leads us to an important conclusion, namely: if, for some arbitrary position of the crank pin at the point \(B\) on the crank circle, we know in some way the point of intersection of the conjugate chords for the given case, then, having found the point \(A\), symmetric to \(B\), we easily find the points \(F\) and \(E\), conjugate to the two named points; having then found the point \(F\) and joined it by a straight line with the center of the circle, we find the point \(C\). The straight line \(CB\) intersects the line of dead points \(MM_1\) at the point \(J\), showing the true position of the piston for a finite length of connecting rod.

Thus, we see that there exists a definite geometrical relation between the point of intersection of the conjugate chords and the position of the piston.

Let us examine this relation further.

Join point \(B\) with the center \(O\), and extend this straight line until it intersects the circle at point \(G\). It is clear that the line \(FG\) will be parallel to the line \(BC\), as chords resting on a diameter. Consequently, the angle \(EFG\) will be equal to the angle \(ABC\), as angles with parallel sides. From this we conclude that the angle \(ABC\) is equal to the angle \(EBG\), as being measured by the same arc \(EG\).

The property found enables us to formulate the relation between the true position of the piston and the point of intersection of the conjugate chords in the following way:

For an arbitrary position of the crank on the circle, the angle formed by the line connecting the position of the crank with the corresponding true position of the piston, and by the perpendicular to the line of dead points, is equal to the angle formed with the radius of the crank by the line connecting the crank pin with the point of intersection of the corresponding conjugate chords.

In the second chapter we showed that we make an error of about one thousandth part of the radius—for the ratio of the length of the connecting rod to the length of the crank equal to 5—if we assume that in the basic mechanism there exists a connecting-rod pole.

We may neglect this error for circles with a diameter up to \(500^{m}/_{m}\) *), which circle is extremely large for investigating the steam distribution of a steam engine being designed or studied.

If, in investigations of steam distributions, the influence of the finite length of the eccentric rods is also neglected, then in no case should the influence of the finite length of the connecting rod be neglected.

Schorch’s method, the most convenient and most frequently used in such cases, carries within itself the limits of the diameter of the crank circle.

Indeed, with a connecting-rod length five times the crank radius, for a crank circle with a diameter of \(500^{m}/_{m}\), it is necessary, by Schorch’s method, to draw circles with a radius of \(1250^{m}/_{m}\), which is a great inconvenience, since it is quite impossible to observe simultaneously both ends of compasses opened to such a distance.

*) For such a diameter, the error allowed by us is equal to one quarter of a millimeter.

A. U.

The method of the coupler pole makes it possible to take into account the finite length of the connecting rod by comparatively simple methods.

Let the position of the crank be given at the point \(B\) (see drawing XV) and the position of the coupler pole \(\lambda\).

From the point \(B\) we draw three lines: the diameter \(OB\), the vector \(\lambda B\), and the perpendicular to the line of dead points; we continue all these lines until they intersect the circle.

The position of the point \(J\)—the true*) distance of the piston from the dead position—can be determined in several ways.

The first method, the most accurate, since it is based on measuring and setting off large arcs: we measure with dividers the arc \(EG\) and set it off from the point \(A\) to the right (in the direction of rotation of the machine indicated in the drawing); we obtain the point \(C\).

The straight line \(BC\) intersects the line of dead points at the required point.

The second method is less accurate:

We draw the three lines \(BH\), \(BO\), and \(B\lambda\), without extending them to their intersections with the circle. Then, with a radius equal to \(BH\), we notch the line \(B\lambda\), obtaining the point \(L\). At the point \(L\) we erect a perpendicular to \(B\lambda\). It is obvious that this perpendicular is equal in magnitude to the segment \(HJ\). Setting off from the point \(H\) to the right the segment \(LN\), we find the required position of the piston.

Since the angle \(\lambda BO\) is in general small**), then, instead of the segment of the perpendicular \(LN\), we may take the length of the chord \(LP\), which differs from it only extremely little. On this basis there is an even simpler, though less accurate, third method for determining the position of the piston for a given position of the connecting rod:

We draw the three lines: \(BH\), \(B\lambda\), and \(BO\); with a radius equal to \(BH\) we notch the sides of the angle \(\lambda BO\)—the resulting chord is set off to the right from the point \(H\).

The true position of the piston may also be found by a fourth, extremely simple, method.

If, in a manner similar to that set forth on p. 88, we find the magnitude of the variable angle \(ABC\) (see drawing XV), we shall see that this angle varies extremely little, namely:

*) By the “true” distance is here meant the path of the piston, determined graphically with greater or lesser accuracy.
A. U.

**) With a fivefold ratio of the length of the connecting rod to the crank, this angle, as is not difficult to see, will vary from zero (at the dead position of the crank) to its maximum value of about \(6^\circ\) (at the middle position of the slide), and then will again decrease to zero (for the other dead position).
A. U.

$r=1,\quad n=5$	$r=1,\quad n=5$	$r=1,\quad n=5$	$r=1,\quad n=5$	$r=1,\quad n=5$
$k=$	0.8	0.6	0.4	0.2
$\operatorname{tg}\varphi_1=$	0.056	0.075	0.088	0.096

and we, without making a significant error, may take $\operatorname{tg}\varphi_1=0.1$.

On this basis, the true positions of the piston are found by the intersection of lines parallel to the line $BC$ and drawn from the corresponding positions of the crank, with the diameter of the crank circle. The direction of these lines is most simply found from the position of the crank corresponding to the mean position of the piston. (See Drawing II.)

Let us solve the inverse problem.

Given the position of the piston $C$ (see Drawing XVI), find the corresponding true position of the crank.

This problem can be solved by the same four methods, of varying accuracy.

Here we shall solve it only by the most accurate one. Through the point $C$ we draw a perpendicular $AB$ to the line of dead centers. From the point $B$ thus found we draw the diameter $BG$ and the vector $B\lambda$ extended to the point $H$. Setting off the arc $HG$ to the left from the point $A$, we find the line $FB$, giving the point $D$—the foot of the perpendicular dropped onto the line of dead centers from the required position of the crank.

Thus the problem is solved.

The proposed method, which reduces the finding of the mutual positions of the piston and the crank to the ruler and simple compass, cannot be considered especially complicated, all the more so because, in the study of steam distributions, one usually confines oneself to finding only the principal points of the diagrams.

We shall now find analytically the dependence between the stroke of the piston, the angle of rotation of the crank, and the distance of the connecting-rod pole from the center of the crank circle.

Let us denote the length $O\lambda$ by the letter $\mu$ (see Drawing XV). From the point $\lambda$ let fall a perpendicular to the position of the crank $OB$, corresponding to a rotation of the crank from the left dead position through the angle $\omega$.

Having made this construction, we obtain two similar right triangles $HBJ$ and $\lambda BK$. On this basis we may write the following proportion:

\[ \frac{HJ}{BH}=\frac{\lambda K}{BK}. \]

Whence we have:

\[ HJ=\frac{BH\cdot \lambda K}{BK}. \]

The distance of the piston from the left dead point, corresponding to the angle of rotation of the crank, will be expressed by the segment \(MJ=MH+HJ\).

If the radius of the crank circle is taken equal to unity, then we can express each of the segments under consideration as follows:

\[ BH=r\sin\omega=\sin\omega, \]

\[ \lambda K=\mu\sin\omega, \]

\[ BK=r-\mu\cos\omega=1-\mu\cos\omega, \]

\[ MH=r(1-\cos\omega)=1-\cos\omega. \]

Thus we obtain:

\[ HJ=\frac{BH\cdot \lambda K}{BK} =\frac{\mu\sin^2\omega}{1-\mu\cos\omega}. \]

Denoting the path traversed by the piston by the letter \(x\), on the basis of what has been said we write the following formula:

\[ MJ=MH+HJ= \]

\[ x=1-\cos\omega+\frac{\mu\sin^2\omega}{1-\mu\cos\omega}. \]

Carrying out the algebraic operations:

\[ x=\frac{(1-\cos\omega)(1-\mu\cos\omega)+\mu\sin^2\omega}{1-\mu\cos\omega}. \]

Expanding the brackets:

\[ x=\frac{1-\cos\omega-\mu\cos\omega+\mu\cos^2\omega+\mu\sin^2\omega}{1-\mu\cos\omega}. \]

Or finally:

\[ [101]\qquad x=\frac{(1+\mu)(1-\cos\omega)}{1-\mu\cos\omega}. \]

Formula \([101]\), expressing the desired analytical relation between the piston stroke, the angle of rotation of the crank, and the polar distance, shows us that the path traversed by the piston is a fourth proportional to the quantities found.

From this we obtain still another method of determining the true position of the piston. From the center \(O\) of the crank circle (see drawing XVII), with a radius equal to the magnitude \(\mu\), draw a concentric circle.

Parallel to the line of dead points \(AA_1\), at a distance from it equal to \(1+\mu\), draw the line \(MN\). For an arbitrary position of the crank \(B\), the corresponding stroke of the piston is found as follows: through the point of intersection of the crank radius with the circle of radius \(\mu\), draw a line perpendicular to \(AA_1\), and continue it to its intersection with the line \(MN\) at the point \(E\). Join the point \(E\) by a straight line with the point \(A\). This straight line will intersect the perpendicular from the point \(B\) to the line \(AA_1\) at the point \(R\). The segment \(CR\), by its length, will represent the required stroke of the piston.

This is evident from the following.

The right triangles \(ARC\) and \(AEF\) are similar to one another. Therefore:

\[ \frac{RC}{AC}=\frac{EF}{AF}, \]

whence

\[ RC=\frac{AC\cdot EF}{AF}. \]

But

\[ AC=1-\cos\omega, \]

\[ EF=1+\mu, \]

\[ AF=AO-FO= \]

\[ =1-\mu\cos\omega. \]

Therefore:

\[ RC=\frac{(1+\mu)(1-\cos\omega)}{1-\mu\cos\omega}=x. \]

For an angle of rotation of the crank greater than \(90^\circ\), for example for the point \(D\), by the same construction we find:

\[ x=LJ. \]

The last-described method of graphically determining the path traversed by the piston is the most accurate, since in it there is no need to measure angles (very small ones), and, moreover, all the points necessary for the construction are obtained on the drawing as sharply defined.

Using formula [101], we shall determine the velocity of the piston for a finite length of connecting rod. Let us denote the velocity of the piston by the letter \(c\); the angular velocity

\[ w=\frac{v}{r}, \]

where \(v\), the velocity of the crank, is considered a constant quantity equal to

\[ \frac{\pi\cdot r\cdot n}{30}, \]

depending on the number of revolutions of the machine \(n\) per minute.

The angular velocity is \(w=\dfrac{d\omega}{dt}\).

The velocity of the piston will be found as the first derivative of the path with respect to time.

Consequently:

\[ c=\frac{dx}{dt}\quad \text{or}\quad c=w\frac{dx}{d\omega}=v\frac{dx}{d\omega}, \]

since we take the radius of the crank to be equal to unity.

We take the derivative of the piston path from formula [101]

\[ \frac{dx}{d\omega} =(1+\mu)\cdot \frac{\sin\omega(1-\mu\cos\omega)-\mu\sin\omega(1-\cos\omega)} {(1-\mu\cos\omega)^2}. \]

Expanding the brackets in the numerator, we have:

\[ \frac{dx}{d\omega} =(1+\mu)\cdot \frac{\sin\omega-\mu\sin\omega\cos\omega-\mu\sin\omega+\mu\sin\omega\cos\omega} {(1-\mu\cos\omega)^2}, \]

which, after cancellation and factoring, gives:

\[ \frac{dx}{d\omega} =(1-\mu^2)\frac{\sin\omega}{(1-\mu\cos\omega)^2}. \]

Thus, the velocity of the piston, for a finite length of the connecting rod, is ultimately expressed by the following formula:

\[ \tag{102} c=v(1-\mu^2)\frac{\sin\omega}{(1-\mu\cos\omega)^2}. \]

Let us now find the angle of rotation of the crank corresponding to the maximum velocity of the piston.

For this we must take the derivative of the quantity \(c\) in formula [102] and set it equal to zero.

In other words, we must take the second derivative of formula (101), i.e., determine the acceleration of the piston.

For the position of the crank corresponding to zero acceleration, we shall evidently have the maximum velocity.

Denoting the acceleration of the piston by the letter \(p\), we obtain the following formula:

\[ p=\frac{d^2x}{dt^2}=\frac{dc}{dt}. \]

Since \(w=v=\dfrac{d\omega}{dt}\), we write:

\[ p=v^2\frac{d^2x}{d\omega^2}. \]

The second derivative of the path with respect to time is obtained from the first derivative.

We had earlier:

\[ \frac{dx}{d\omega}=(1-\mu^2)\frac{\sin\omega}{(1-\mu\cos\omega)^2}. \]

Whence:

\[ \frac{d^2x}{d\omega^2}=(1-\mu^2)\frac{\cos\omega(1-\mu\cos\omega)^2-2(1-\mu\cos\omega)\mu\sin^2\omega}{(1-\mu\cos\omega)^4}. \]

Carrying out the reduction, we obtain:

\[ \frac{d^2x}{d\omega^2}=(1-\mu^2)\frac{\cos\omega(1-\mu\cos\omega)-2\mu\sin^2\omega}{(1-\mu\cos\omega)^3}. \]

We transform the numerator of the resulting fraction, expanding the brackets and expressing sinus through cosinus:

\[ \frac{d^2x}{d\omega^2}=(1-\mu^2)\frac{\cos\omega-\mu\cos^2\omega-2\mu+2\mu\cos^2\omega}{(1-\mu\cos\omega)^3}, \]

which gives us:

\[ [103]\qquad \frac{d^2x}{d\omega^2}=(1-\mu^2)\frac{\mu\cos^2\omega+\cos\omega-2\mu}{(1-\mu\cos\omega)^3}. \]

The acceleration of the piston will be expressed by the following formula:

\[ [104]\qquad p=v^2(1-\mu^2)\frac{\mu\cos^2\omega+\cos\omega-2\mu}{(1-\mu\cos\omega)^3}. \]

To solve the question of the maximum velocity, we set the numerator of formula [103] equal to zero. Then we have the following equation:

\[ \mu\cos^2\omega+\cos\omega-2\mu=0. \]

Dividing by \(\mu\), we have:

\[ \cos^2\omega+\frac{1}{\mu}\cos\omega-2=0. \]

Whence:

\[ \cos\omega=-\frac{1}{2\mu}\pm\sqrt{\frac{1}{4\mu^2}+2}, \]

or

\[ \cos\omega=\frac{-1\pm\sqrt{1+8\mu^2}}{2\mu}. \]

Since \(\mu\), in relation to the radius of the crank, represents a proper fraction, and since, on the other hand, the absolute value

since the cosinus cannot be greater than unity, from the two roots obtained for the equation we retain only one.

Then we have that the acceleration is equal to zero when the crank is turned through an angle whose cosinus is determined by the following expression:

\[ [105]\qquad \cos \omega=\frac{-1+\sqrt{1+8\mu^{2}}}{2\mu}. \]

Let us have a machine in which the ratio of the length of the connecting rod to the radius of the crank is equal to 5, i.e. \(n=5\).

Then

\[ \mu=\frac{1}{2n}=0.1. \]

Let us determine the magnitude of the cosinus of the angle at which the acceleration is equal to zero.

Then we shall have:

\[ \cos \omega=\frac{-1+\sqrt{1+0.08}}{0.2}= \]

\[ =\frac{-1+1.03923}{0.2}= \]

\[ =0.19615, \]

which corresponds, with an accuracy to minutes, to an angle of \(78^\circ 41'\).

The path traversed by the piston, corresponding to this angle of rotation of the crank, we determine by formula [101].

Substituting into it the quantities known to us:

\[ x=\frac{(1+0.1)(1-0.19615)}{1-0.1\cdot 0.19615}. \]

Carrying out the operations on the numbers, we obtain:

\[ x=0.902. \]

By this amount the piston will have moved away from its left dead position.

Let us determine at what angle the connecting rod is situated with respect to the crank at the maximum velocity of the piston.

For this we must know the distance of the piston from the main shaft. Obviously, at the left dead position the piston is distant from the shaft by a distance equal to the sum of the lengths of the connecting rod and the radius of the crank, i.e. \((n+r)\).

For our case this distance is

\[ s=(n+r)= \]

\[ =(5+1)=6. \]

The distance of the piston from the shaft for the position of the crank under consideration will be obtained as follows:

\[ a=s-x \]

or

\[ a=6-0.902= \]

\[ =5.098. \]

Thus, we know the three sides of the triangle: \(n\), \(r\), and \(a\). Denoting by the letter \(\alpha\) the angle between the connecting rod and the crank, we obtain for the cosine of this angle the following expression:

\[ \cos \alpha=\frac{n^2+r^2-a^2}{2nr}. \]

Substituting here the numerical values:

\[ \cos \alpha=\frac{5^2+1^2-(5.098)^2}{2\cdot 5\cdot 1}. \]

The numerical calculations give:

\[ \cos \alpha=\frac{26-25.98960}{10}= \]

\[ =0.00104. \]

Determining from tables the angle corresponding to the found value of the cosine, to an accuracy of one second, we obtain

\[ \alpha=89^\circ 56'20''. \]

Thus, the result found fully confirms the accepted assumption that the piston reaches its greatest velocity at the moment when the connecting rod forms a right angle with the crank.

We found that, for this position of the piston, the cosine of the angle of rotation of the crank is equal to \(0.19615\).

If we take, in round numbers, the value of the cosine equal to \(0.20\), then we obtain the following relation, of interest to us, between the position of the crank at the maximum velocity of the piston and the polar distance of the crank mechanism under consideration:

\[ \cos \alpha=0.20=2\cdot 0.1=2\mu. \]

Hence we draw the conclusion that the perpendicular erected to the line of dead centers at a distance from the center equal to twice the polar distance intersects the crank circle at the point corresponding to the maximum velocity of the piston, i.e. at the point for which the acceleration of the piston is equal to zero.

Knowing this point, we can easily, by means of a construction already known to us, determine on the diameter of the crank circle, which represents the line of dead points of the piston stroke, the point at which the so-called curve of inertia forces intersects the line of the piston stroke.

The point removed from the center of the crank circle in the direction toward the crosshead by twice the pole distance may be useful, in constructing diagrams on a large scale, not only for the curve of inertia forces, but also for constructing the so-called diagram of tangential forces.

For constructing the diagram of tangential forces it is necessary, as is known, to resolve the force acting at each given instant on the piston into radial and tangential forces, which is possible only under the condition of two given quantities: the magnitude and the direction of the force to be resolved.

We obtain the magnitude of the force from the diagram of working pressures and inertia forces.

We can draw these diagrams on the same large scale as the valve diagrams.

As for the direction of the force, it coincides with the direction of the connecting rod, the position of which is found by construction, by cutting the line of the piston stroke from the corresponding points of the crank circle with a radius equal to the length of the connecting rod on a definite scale.

This construction also accounts for the comparatively very small scale adopted in drawing the diagrams of the projected machine.

With the aid of the point \(\lambda_1\) (see drawing XVIII), removed from the center of the circle by twice the pole distance, we obtain the possibility, by a comparatively simple construction, of determining for each given position of the crank the corresponding angle of inclination of the connecting rod to the line of dead points, and consequently of determining the direction of the force acting along the connecting rod.

For an arbitrary position of the crank, suppose at point \(B\), we have from the right triangle \(ABC\):

\[ BC = AB \cdot \sin \beta, \]

where \(AB\) is the length of the connecting rod, and \(\beta\) is the angle made by the connecting rod with the line of the piston stroke.

On the other hand, from the right triangle \(OBC\) we have:

\[ BC = OB \sin \omega, \]

where \(OB\) is the radius and \(\omega\) is the angle of rotation of the crank.

On the basis of these two formulas we have:

\[ AB \cdot \sin \beta = OB \sin \omega. \]

Or

\[ \frac{AB}{OB}=\frac{\sin\omega}{\sin\beta}. \]

But the ratio

\[ \frac{AB}{OB}=\frac{n}{1}. \]

Consequently:

\[ \frac{\sin\omega}{\sin\beta}=n. \]

On the other hand, we have:

\[ \mu=\frac{1}{2n}. \]

Whence:

\[ 2\mu=\frac{1}{n}. \]

Or

\[ n=\frac{1}{2\mu}. \]

Thus, we have the proportion:

\[ \frac{1}{2\mu}=\frac{\sin\omega}{\sin\beta}. \]

Since in every triangle the sides are proportional to the sines of the opposite angles, in order to determine the angle \(\beta\) we must have a triangle with variable angle \(\omega\), lying opposite a side equal to unity, i.e. the radius of the crank.

On the basis of what has been said, the graphical method for determining the angle \(\beta\) reduces to the following.

On the line of dead points of the crank circle, in the direction toward the slider, we lay off from the center a segment \(O\lambda_1\), equal to \(2\mu\); from the point \(\lambda_1\), as from a center, we describe a circle with radius equal to the length of the crank. This circle will obviously be the geometric locus of the vertices of triangles for which the base is equal to \(2\mu\), and the angle adjacent to the base will be equal to the angle of rotation of the crank.

Thus, in order to find the angle formed by the connecting rod with the line of dead points for the position of the crank at the point \(B\), we proceed as follows.

We prolong the radius \(OB\) to its intersection at the point \(D\) with the auxiliary circle, and then join the point \(D\) with the point \(\lambda_1\). The angle \(\lambda_1DO\) will be the required one.

Indeed, from the triangle \(\lambda_1DO\) we have

\[ \frac{\lambda_1D}{\lambda_1O}=\frac{\sin BO\lambda_1}{\sin \lambda_1DO}. \]

But \(\lambda_1D=1\) by construction,

\[ \lambda_1O=2\mu, \]

\[ \angle BO\lambda=\omega. \]

Therefore:

\[ \frac{\sin \lambda_1 DO}{\sin BO\lambda_1} = \frac{\sin \lambda_1 DO}{\sin \omega} = \frac{1}{2\mu} = n, \]

i.e.

\[ \sin \lambda_1 DO=\sin \beta . \]

The drawing shows three positions of the crank, and the corresponding constructions have been made for them.

To obtain the required direction of the connecting rod, one must construct, for the corresponding position of the crank, the angle found, in such a way that one of its sides is parallel to the line of direction of the piston stroke.

Let us see whether the connecting-rod pole found by us is connected with any already known properties of the steam engine.

Let us have (see drawing XIX) the crank circle. The point \(B\) represents an arbitrary position of the crank.

We find, by any of the methods known to us, the point \(E\), representing the corresponding true position of the piston.

At the point \(E\) we erect a perpendicular to the direction of the piston stroke and extend it to the point \(C\)—its intersection with the crank circle. Through the point \(C\) we draw two lines: the central line \(CO\), and the line parallel to the given position of the crank, \(C\lambda_1\).

Since, by what was proved earlier, the line \(CO\) is parallel to the line \(B\lambda\), we have before us two similar triangles, \(\lambda OB\) and \(CO\lambda_1\). We therefore write the proportion:

\[ \frac{B\lambda}{CO}=\frac{\lambda O}{\lambda_1 O}. \]

Whence:

\[ \lambda_1 O=\frac{CO\cdot \lambda O}{B\lambda}. \]

But \(CO\) is the radius of the crank, equal to unity;

\(\lambda O\) is the pole distance, equal to \(\mu\);

\(B\lambda\) is a vector of variable magnitude, depending on the angle of rotation of the crank.

Denoting this vector by the letter \(\rho\), we have:

\[ \rho^2=1+\mu^2-2\mu\cos\omega . \]

At the left dead position of the crank,

\[ \omega=0^\circ \quad \text{and} \quad \cos\omega=1, \]

therefore:

\[ \rho_0^2=1+\mu^2-2\mu=(1-\mu)^2 . \]

Whence:

\[ \rho_0 = 1 - \mu . \]

At the right dead position, \(\omega = 180^\circ\) and \(\cos \omega = -1\).

Consequently:

\[ \rho_{180}^{2} = 1 + \mu^{2} + 2\mu = (1 + \mu)^2 . \]

Whence:

\[ \rho_{180} = 1 + \mu . \]

Thus, for a connecting rod equal to five times the radius, we have the minimum and maximum magnitudes of the vectors in the following form:

\[ \rho_{\min} = 1 - \mu = 1 - 0.1 = 0.9 . \]

\[ \rho_{\max} = 1 + \mu = 1 + 0.1 = 1.1 . \]

Thus, we see that the segment \(O\lambda_1\) changes its magnitude within the limits:

\[ \frac{\mu}{0.9} \quad \text{and} \quad \frac{\mu}{1.1}. \]

Or, since \(\mu = 0.1\), we have:

\[ O\lambda_1 = \left( \frac{1}{9} - \frac{1}{11} \right). \]

The mean value of \(O\lambda_1\) will, obviously, be very close to \(\mu\).

If we take the segment \(O\lambda_1 = \mu\), then we obtain the point first indicated by the mechanical engineer F. A. Briks in his article: “Improvement in the Distribution of Steam in Steam Engines”*).

Indeed, we obtained earlier that \(\mu = \dfrac{1}{2n}\).

Here \(1\) is the crank radius, and \(n\) is an abstract number indicating the ratio of the length of the connecting rod to the radius.

Denoting the crank radius by the letter \(R\) and the length of the connecting rod by the letter \(L\), we obtain for \(n\) the following value:

\[ n = \frac{L}{R}. \]

Whence:

\[ \mu = \frac{R}{2n} = \frac{R^2}{2L}. \]

) Morskoi Sbornik*—1890, Nos. 1 and 2.

This is precisely the magnitude of the segment laid as the basis of the well-known, in terms of the accuracy of the results attainable by it, diagram of F. A. Briks.

CHAPTER TEN.

Conclusion.

From all that has been set forth above, it is clear that every crank mechanism is characterized by its corresponding coupler pole.

Turning to the basic mechanism, as the one examined in greatest detail in the present article, we see that the coupler pole can be associated with all the known properties of this mechanism.

Since the graphical method of calculation for steam engines, especially with multiple expansion of steam, is at present generally accepted, and since the assessment of the qualities and dimensions of the machine being built depends on the accuracy with which the diagrams included in the design are executed, it is clear that the larger the scale on which the diagrams are made, the more accurate the results obtained will be.

The principal diagrams in design are: volumetric indicator diagrams and, as derivatives of these latter, diagrams of working pressures together with the forces of inertia; further—diagrams of tangential forces and, finally, diagrams of steam distribution.

The method of the coupler pole, making it possible to determine the true position of the piston for any angle of rotation of the crank, makes it possible to obtain volumetric diagrams on a large scale.

Thereby we obtain the possibility of carrying out the construction of all the remaining diagrams likewise in large dimensions.

The curves of expansion and compression of steam, obtained in indicator diagrams by the graphical method, prove under such conditions to be more accurate.

In steam-distribution diagrams we can attain very accurate results by taking into account the influence of the finite length of the rods of the coupler and eccentric mechanisms according to the method proposed by F. A. Briks, or else by separately constructing the travels of the piston and the slide valve, relying on the known properties of the coupler pole.

By the latter method it becomes possible to draw with extreme accuracy the so-called elliptical diagrams of the motion of slide valves.

In short, the application of the coupler pole allows us to increase to a considerable degree the accuracy of the graphical investigation of steam engines.

Crank and valve steam-distribution mechanisms can be reduced graphically to eccentric mechanisms, and therefore, a priori, one may say that the method of the coupler pole may also prove useful in designing the above-mentioned steam-distribution mechanisms, making it possible to take into account the influence of the finite length of the rods of the entire steam-distribution mechanism.

I hope to turn to a more detailed study of the question touched upon in these last lines in the near future.

INDEX

of certain articles in the journal literature in which, in one way or another, the influence of the finite length of the connecting rod and eccentric rod is taken into account:

Zeitschr. d. Ver. d. Ing.

1860—p. 25.
Hertzer—Ein neues Diagramm für Schiebersteuerungen mit sehr kurzer Lenkerstange.

1. Schorch—Kolben- und Schieberdiagramme.

1878—p. 445.
A. Seemann—Zur Theorie der Schiebersteuerungen.

1880—p. 513.
A. Hollenberg—Graphische Darstellung der Schieberbewegung bei Dampfmaschinen.

1883—p. 136.
L. Pinzger—Zur Construction der Beschleunigungscurve des Kreuzkopfes eines Kurbelmechanismus.

1890—p. 1320.
Kirsch—Ueber die graphische Bestimmung der Kolbenbeschleunigung.

1891—p. 129.
F. Vaes—Ueber die graphische Bestimmung der Kolbenbeschleunigung.

1894—p. 297.
K. Reinhardt—Schieberdiagramm für die Meyersche Steuerung.

1894—p. 770.
L. Janse—Ueber Schieberdiagramme.

1896—p. 904.
R. Land—Der Geschwindigkeits- und Beschleunigungsplan für Mechanismen.

1897—p. 431.
F. A. Brix—Das bizentrische polare Exzenterschieberdiagramm.

1908—p. 141.
L. Bandiss—Beitrag zur Ausmittlung des Kulissenantriebes bei der Heusinger-Steuerung.

Dingler’s Polytechnisches Journal.

1858–59 pp. 241, 314, 315.
H. Fuhst—Anwendung des Zeuner’schen Diagrammes auf Steuerungen mit kurzen Exzenterstangen.

1876—p. 289.
V. Sirk—Ueber das Fehlerglied der einfachen Schiebersteuerung.

1876—p. 283.
V. Thallmayer—Construction des Fehlergliedes bei der einfachen Schiebersteuerung.

1877—p. 137.
— Schieberdiagrammograph.

1881—p. 161.

Müller-Melchiors—W. D. Mark’s Construction of the Error Link in Zeuner’s Slide-Valve Diagram.
1881—p. 249.

Brandt—A New Construction of Zeuner’s Slide-Valve Diagrams.
1906—p. 451.

Goldberger—Exact Construction of Slide-Valve Diagrams.
Proceedings of the Society for the Promotion of Industry.
1908.

W. Hartmann—The Acceleration of Rolling Motion.
Maschinen-Constructeur.
1870—p. 231.

Neubert—Slide-Valve Gear Diagrams.
1872—p. 230.

— The Error Link of Zeuner’s Theory of Slide-Valve Gears.
1888—p. 22.

G. Hoefer—On the Construction of Zeuner’s Error Link.
Annales Industrielles.
1886—p. 17.

M. Demoulin—Sinusoidal Diagram of Steam Distribution.
1887—p. 305.

I. Claeys—Diagram Giving the Simultaneous Positions of the Piston and the Slide Valve.
Le Génie Civil.
1889—p. 365.

I. Claeys—Empirical Plots Relating to the Positions and Velocities of the Piston of Steam Engines.
Portefeuille économique des machines.
1890—p. 39.

M. Dubost—A Means of Taking Strict Account of the Obliquity of Connecting Rods in Distribution Diagrams.
American Machinist.
1897—No. 10.

Zeuner and Bilgram slide-valve diagrams.
Engineering.
1893—p. 418.

W. Dalby—Harmonic Valve Diagram.
Naval Collection.
1890—No. 1.

F. A. Brix—Improvement in Steam Distribution by a Simple Slide Valve.
1893—No. 5.

— Double-center slide-valve diagram.

Drawing I.

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

Fig. 5.

Fig. 6.

Fig. 7.

Fig. 8.

Fig. 9.

Fig. 10.

Drawing III.

To the article by A. V. Sharov
“The Coupler Pole.”

Plate 1

Fig. II

Fig. III

Fig. IV

Chromolith. by Eng. [[unclear: name]], Kharkov.

Drawing V.

Drawing VI.

Drawing VII.

Drawing VIII.

Drawing IX.

Plate 2

To the article by A. V. Zhdanov, “The Coupler Pole.”

Fig. V

Visible labels and notation include: \(A\), \(B\), \(M\), \(N\), \(P(\alpha,\beta)\), \(Q(\alpha_1,\beta_1)\), \(X\), \(Y\), \(k\), \(k'\), \(d_1\), \(d_2\), \(d\), \(p\), \(\theta\), \(n\).

Fig. VII

Visible labels and notation include: \(O\), \(a\), \(b\), \(c\), \(A\), \(B\), \(M\), \(N\), \(\alpha\), \(\theta\), \(n\), \(m\), \(k\), \(t\), \(l\).

Fig. VIII

Visible labels and notation include: \(O\), \(D\), \(C\), \(A\), \(B\), \(M\), \(n\), \(l\), \(\theta\).

Fig. IX

Visible labels and notation include: \(O\), \(D\), \(C\), \(A\), \(B\), \(M\), \(n\), \(l\), \(\theta\), \(X\).

Fig. X

Visible labels and notation include: \(O\), \(a\), \(b\), \(c\), \(d\), \(e\), \(f\), \(g\), \(i\), \(j\), \(k\), \(m\), \(n\), \(r\), \(x\), \(y\), \(z\), \(A\), \(B\), \(M\), \(N\), \(\alpha\), \(\theta\), \(X\), \(Y\).

Plate XII.

Plate XIII.

Plate XV.

Plate XVII.

Plate XVIII.

To the article by A. V. Bulgakov, “The Coupler Pole.”
Plate 3.

Drawing XIII.

Drawing XIV.

Drawing XV.

Drawing XVI.

Drawing XVII.

Drawing XVIII.

Drawing XIX.

Chromolithography by Zamenhof, St. Petersburg.