PROCEEDINGS

of the Tomsk Technological Institute

of Emperor Nicholas II

Vol. 18. 1910. No. 2.

II.

A. V. Ugarov.

On the Question of the Graphical Calculation of the Flywheel.

A New Method for Determining Tangential Forces.

1–20.

A New Method for Determining Tangential Forces.

On the question of the graphical calculation of a flywheel.

Piston machines (steam engines and other heat engines), being a mechanism for converting the rectilinear-reciprocating motion of the piston into continuous circular rotation of the main shaft of the machine, perform this conversion under the action on the piston of a variable pressure (of steam or gas).

The transformation of motion is accomplished by means of a connecting-rod-and-crank mechanism.

It is impossible, by means of such a mechanism, to obtain entirely uniform rotation of the main shaft of the machine (during one revolution), as will be shown below.

To reduce the nonuniformity, a flywheel must be installed, which, taking upon itself the excess of living force during the period of accelerated motion, gives back this excess during the period of retardation.

To determine the mass of the flywheel required for the given operating conditions of the machine, it is necessary to obtain an exact picture of the variations of the living force of the flywheel; and this is usually achieved by finding the so-called tangential forces acting on the crank, for any angle of its rotation from the dead positions. Consequently—the magnitude of this angle ranges from \(0^\circ\) to \(180^\circ\).

It is evident that the magnitude of the tangential force depends on the pressure on the piston, and therefore, in order to determine this force, it is necessary to know the position of the piston corresponding to the chosen angle of rotation of the crank. In our consideration we assume a machine of normal dimensions, i.e. one in which the ratio of the length of the crank \(R\) to the length of the connecting rod \(L\) is equal to \(\frac{1}{5}\),

\[ \frac{R}{L}=\frac{1}{5}=\mu . \]

The greater the length of the connecting rod for a given crank, the smaller the value of \(\mu\) will be.

As a first approximation, in the calculations one takes \(\mu=0\) and says that the length of the connecting rod is infinitely large.

The second approximation will be when the value of the quantity \(\mu\) is retained, but one assumes that \(\mu^2=0\), as well as all higher powers of \(\mu\).

The accuracy thereby obtained is quite sufficient for technical purposes.

After what has been said, we pass to the question of determining the position of the piston, or the path traversed by it from the initial (dead-center) position.

For a certain definite angle of rotation of the crank from its dead-center position, the corresponding position of the piston is found by a simple geometrical construction (Fig. 1).

Fig. 1.

If, for the position of the crank pin at the point \(K\), the corresponding position of the piston, or rather of the slide rigidly connected with it, is sought, then from the point \(K\), with a radius equal to the length of the connecting rod \(L\), one intersects the path of the piston and finds the point \(S\).

To save space and for the clarity of the drawing, it is desirable that the positions of the piston in the drawing be in the immediate vicinity of the positions of the crank. This is achieved by shifting the position of the piston in the direction toward the center of the crank circle by the length of the connecting rod; then the dead points of the piston path coincide with the dead positions of the crank, and the required position of the piston \(B\) is obtained if from the point \(S\) we draw an arc with radius equal to \(L\) up to its intersection with the diameter of the crank circle, i.e. the position of the piston is obtained by arc projection with radius \(L\) of the point \(K\) onto the line \(K_0K_0\).

Hence the first approximate construction of the piston position for \(\mu=0\), i.e. for an infinitely long connecting rod, is immediately clear: the arc serving in this case for projecting the point \(K\) has an infinite radius, and the position of the piston is determined as the orthogonal projection of the point \(K\) onto the line of dead points (onto the diameter of the crank circle).

It is clear that, in the approximation under consideration, the motion of the piston is a harmonic oscillation about its mean position.

Analytically, the path \(x\) traversed by the piston from its dead position when the crank is turned through an angle \(\alpha\), for \(\mu=0\), will be expressed as follows:

\[ x=R(1-\cos\alpha). \]

The segment \(K'B\), which in Fig. 1 represents the difference between the true and the approximately found positions of the piston, is called the error of the piston path.

\[ f=K'B=K_0B-K_0K \]

is found from Fig. 2, whence it is seen that

\[ f\cdot(2L-f)=z^2=R^2\sin^2\alpha; \]

or, since \(f\) is very small in comparison with \(2L\), we have:

\[ f\cdot 2L=z^2=R^2\sin^2\alpha; \]

therefore

Fig. 2

\[ f=\frac{R^2}{2L}\sin^2\alpha=\frac{1}{2}\frac{R}{L}R\sin^2\alpha, \]

or, finally,

\[ f=\frac{\mu}{2}R\sin^2\alpha. \tag{1} \]

Thus the true path of the piston is determined as follows:

\[ x=R(1-\cos\alpha)+\frac{\mu}{2}R\sin^2\alpha. \tag{2'} \]

Equation \((2')\) gives us the position of the piston in the second approximation; and this, as we shall see further, more than satisfies the requirements of technical practice.

Since the direction of the arc projection for obtaining the true positions of the piston remains constant (the slider is always located on one side of the machine shaft), it is obvious that, for ob—

...during the motion of the piston (from the right dead center to the left), the errors in circular projection will be subtracted from the piston positions found by orthogonal projection of the crank-pin position, and we obtain the general formula:

\[ x = R(1-\cos \alpha) \pm \frac{\mu}{2} R \sin \alpha. \tag{2} \]

Remark. The greatest error is obtained for \(\alpha = 90^\circ\).

\[ f^{\max} = \frac{\mu}{2} R. \]

Thus, neglecting the magnitude \(f\) in comparison with \(2L\), in determining the error of the piston path, we make the greatest error in the calculation:

\[ \frac{f}{2L} = \frac{\frac{\mu}{2}}{2L} R = \frac{\mu^2}{4}; \]

with accuracy up to this error we also determine the magnitude of the error \(f\).

For \(\mu = \frac{1}{5}\), the greatest error of the calculation will be

\[ \frac{f}{2L} = \frac{1}{5^2 \cdot 4} = 0.01, \]

i.e., in determining the greatest error we obtain the result with an accuracy up to \(0.01\).

Since

\[ f^{\max} = \frac{\mu}{2} R = 0.1R, \]

or \( \frac{1}{20} \) of the entire piston stroke, the inaccuracy of calculation made by us is equal to

\[ 0.01 \frac{1}{20} = 0.0005 \]

of the piston stroke; hence it is clear that the second approximation gives us a very large and quite sufficient degree of accuracy.

For finding the exact positions of the piston, among other methods there exists Müller’s method.

We describe (Fig. 3), from the center \(O\), a circle with radius equal to the length of the crank, and from the same center two more circles with radii equal to the distance of the slide block from the shaft in one and in the other dead-center positions.

A. V. Ugarov.

Obviously, the radius of one circle will be \(L+R\), and of the other \(L-R\); the segment of any radius between these circles represents the full travel of the piston.

If now, from the left dead position of the crank \(K_0\), one draws a circle with radius equal to the length of the connecting rod, then it is obvious that this circle will be tangent to the two circles drawn earlier, and for any angle of rotation of the crank \(\alpha\), the corresponding travel of the piston will be found as the segment \(AB\) of the extended radius of the crank circle.

Drawing 3.

The correctness of this conclusion is clearly seen from the drawing; one need only imagine the crank remaining fixed in the position \(OK_0\), and regard the radius \(OA\) as the line of dead points, rotated with respect to the crank toward the direction of rotation of the machine by the angle \(\alpha\).

Müller’s method gives exact results, but requires drawing circles of large radii.

For an approximate determination of the positions of the piston, Briggs*) gave the following construction (Drawing 4).

The position of the piston corresponding to the angle of rotation of the crank \(\alpha\) is obtained as the orthogonal projection of the point \(K\) onto the diameter of the crank circle, if the vertex of the angle is moved from the center of the circle to the side opposite to the slide by the amount \(f_{\max}=\dfrac{\mu}{2}R\). This is clearly seen from the fact that,

Drawing 4.

) Morskoi Sbornik*, 1890.

that the projection \(K'\) of the point \(K\) is distant from the point \(B\) precisely by the amount of the error corresponding to the angle of rotation of the crank \(\alpha\).

Since \(\dfrac{\mu}{2}R\) is a small quantity, the arc \(K_1K\) may be regarded as a straight line perpendicular to \(OK\), and its length will then be equal to \(f_{\max}\sin\alpha\); \(K'B\), in turn, will be the projection of \(K_1K\); consequently

\[ K'B = K_1K\sin\alpha = f_{\max}\sin\alpha \cdot \sin\alpha = f_{\max}\sin^2\alpha. \]

But \(f_{\max}=\dfrac{\mu}{2}R\), consequently

\[ K'B = \frac{\mu}{2}R\sin^2\alpha, \]

which we already had in formula (1).

Briggs’s method is extremely simple and makes it possible easily to solve the inverse problem, i.e., to find the position of the crank corresponding to a given position of the piston.

The advantage of Briggs’s method over the method of finding the position of the piston by circular projection and Müller’s method consists not only in its simplicity, but also, strange though this may sound with respect to an approximate method, in its accuracy.

Drawing circles of radius equal to the length of the connecting rod on a large scale leads to errors arising from the springing of the legs of the compass; while at small scales we obtain, in general, an inaccurate drawing, precisely because of the smallness of the scale.

Thus the positions of the piston corresponding to any position of the crank can be determined easily and simply by Briggs’s method; as for the corresponding positions of the connecting rod, which are needed when finding the tangential forces, velocities, and accelerations of the links of a crank mechanism, these positions have to be found by the only convenient method—by intersecting, from the given position of the crank pin, the line of travel of the piston with a radius equal to the length of the connecting rod.

The subject of the present article is the exposition of a method for finding tangential forces without having to draw large-radius intersections.

Let us first consider the relation of the forces acting in the crank mechanism.

If, in some definite position of the crank \(K\) (Fig. 5), a certain force \(P\) acts on the piston, then through the connecting rod a force \(S\) is transmitted, obtained by resolving the force \(P\) into two directions: perpendicular to the path of the slider and along the connecting rod.

The first of these forces, \(N\), presses the slider against the guide and does not take part in the rotation of the crank; the second, \(S\), is the force rotating the shaft.

Fig. 5.

If the connecting rod makes an angle \(\beta\) with the direction of motion of the slider, then, evidently,

\[ S=\frac{P}{\cos\beta}, \qquad N=P\tan\beta. \]

Let us determine the angle \(\beta\) as a function of the angle \(\alpha\).

Drop from \(K\) the perpendicular \(KB\); from the triangles \(AKB\) and \(KBO\) we have, respectively,

\[ KB=AK\sin\beta=L\sin\beta, \]

\[ KB=OK\sin\alpha=R\sin\alpha, \]

whence

\[ \sin\beta=\frac{R}{L}\sin\alpha=\mu\sin\alpha. \]

Thus we see that the angle \(\beta=0^\circ\) when \(\alpha=0^\circ\) and \(\alpha=180^\circ\), i.e. in the dead positions of the piston. When \(\alpha=90^\circ\), then \(\beta\) takes its greatest value \(\sin\beta_{\max}=\mu\), which corresponds approximately to the angle \(11^\circ .5\), for \(\mu=\frac{1}{5}\).

Let us return to the force \(S\) (Fig. 5). Being transmitted along the connecting rod, the force \(S\) at the point \(K\) is resolved into two forces: the so-called tangential force \(T=S\sin(\alpha+\beta)\) and the radial force \(Q=S\cos(\alpha+\beta)\). The first of these forces produces the rotation of the machine shaft, i.e. performs work, overcoming useful resistances; the second takes no part in the rotation of the shaft.

If we express the tangential force through the magnitude of the force \(P\) acting on the piston, we obtain

$$ T = S \cdot \sin(\alpha+\beta) = \frac{P}{\cos\beta}\sin(\alpha+\beta) = P \frac{\sin(\alpha+\beta)}{\cos\beta}. \tag{3} $$

Instead of constructing the parallelograms of forces twice, the force \(T\) may be determined by the following construction (Fig. 6).

Fig. 6.

In the position of the crank, from the point \(O\) lay off the force \(P\) as the segment \(OA\); from the point \(A\) draw the line \(AB\) parallel to the direction of the connecting rod; the segment \(OB\), obtained in this way on the vertical through \(O\), will represent the tangential force \(T\).

Indeed, from the triangle \(AOB\) we have:

$$ \frac{OB}{OA} = \frac{\sin(\alpha+\beta)}{\sin(90-\beta)} = \frac{\sin(\alpha+\beta)}{\cos\beta}; $$

whence

$$ OB = P \frac{\sin(\alpha+\beta)}{\cos\beta} = T. $$

Thus we see that, for the graphical determination of the magnitude of the tangential force, it is necessary to know the position of the connecting rod relative to the line of dead centers.

Before proceeding further, let us examine formula (3). From consideration of it it is clear that, even with constant pressure \(P\) on the piston (in reality it is a variable quantity), the tangential force is a variable quantity depending on the angle of rotation of the crank.

Indeed, since \(\sin\beta=\mu\sin\alpha\), then \(\cos\beta=\sqrt{1-\mu^2\sin^2\alpha}\). Then we have:

$$ \frac{\sin(\alpha+\beta)}{\cos\beta} = \frac{\sin\alpha\cos\beta}{\cos\beta} + \frac{\sin\beta\cos\alpha}{\cos\beta}, $$

which, after transformations, gives:

$$ \frac{\sin(\alpha+\beta)}{\cos\beta} = \sin\alpha\left(1+\mu\sqrt{\frac{1-\sin^2\alpha}{1-\mu^2\sin^2\alpha}}\right); $$

thus

\[ T=P\cdot \sin \alpha \left(1+\mu \sqrt{\frac{1-\sin^2\alpha}{1-\mu^2\sin^2\alpha}}\right). \]

If the force \(T\) is a variable quantity, then its moment with respect to the axis of rotation \(O\) is also a variable quantity, and consequently the turning moment on the main shaft is not a constant quantity. It is obvious that this turning moment must at times be greater than the moment of the useful resistances of the machine, referred to the radius of the crank; otherwise the motion of the machine would be impossible.

The least value of the tangential force is \(T_{\min}=0\); this occurs when \(\alpha+\beta=0^\circ\) and \(\alpha+\beta=180^\circ\), since \(\beta=0^\circ\) when \(\alpha=0^\circ\) or \(\alpha=180^\circ\).

For a value of \(\alpha\) different from \(0^\circ\) or \(180^\circ\), \(\beta\) is not equal to \(0^\circ\), and therefore \(\alpha+\beta\) cannot be equal to \(180^\circ\), since \(\alpha\) and \(\beta\) are angles of one and the same triangle.

Hence we see that the tangential force passes through zero at the dead points of the piston. This is also evident because the force \(P\) at the dead points acts entirely along the direction of the radius and gives no component perpendicular to it.

Consequently, if the tangential force is equal to zero not in the dead positions of the mechanism, this can occur only because \(P=0\), i.e., when the force acting on the piston is equal to zero, which indeed happens in steam engines.

The tangential force \(T=P\) when \(\alpha=90^\circ\); indeed, then we shall have

\[ T=P\frac{\sin(\alpha+\beta)}{\cos\beta} =P\frac{\cos\beta}{\cos\beta} =P; \]

therefore, for this position of the crank it is unnecessary to determine the tangential force by construction.

Remark. In addition, \(T=P\) when \(\alpha=90^\circ-2\beta\); then

\[ T=P\frac{\sin(90-2\beta+\beta)}{\cos\beta} =P\frac{\cos\beta}{\cos\beta} =P. \]

Let us determine the angle of rotation of the crank \(\alpha\) corresponding to this value of the force \(T\), assuming \(\mu=\frac{1}{5}\).

If \(\alpha=90^\circ-\beta\), then \(\alpha+\beta=90^\circ-\beta\); consequently

\[ \sin(\alpha+\beta)=\sin(90^\circ-\beta)=\cos\beta; \]

expanding the brackets here,

\[ \sin\alpha\cos\beta+\cos\alpha\sin\beta=\cos\beta \]

and substituting, in place of \(\sin\beta\), the quantity equal to it, \(\mu\cos\alpha\),

\[ \sin\alpha\cos\beta+\mu\cos\alpha\sin\alpha=\cos\beta, \]

\[ \mu\cos\alpha\sin\alpha=\cos\beta(1-\sin\alpha), \]

whence we have

\[ \mu\sin\alpha\sqrt{1-\sin^2\alpha}=(1-\sin\alpha)\sqrt{1-\mu^2\sin^2\alpha}; \]

getting rid of the radicals, we obtain

\[ \mu^2\sin^2\alpha(1-\sin^2\alpha)=(1-\sin\alpha)^2(1-\mu^2\sin^2\alpha); \]

opening the brackets and, for the purpose of reducing like terms, we obtain the quadratic equation

\[ \sin^2\alpha+\frac{1}{2\mu^2}\sin\alpha-\frac{1}{2\mu^2}=0. \]

Substituting here the numerical value of \(\mu\), we have:

\[ \sin^2\alpha+\frac{25}{2}\sin\alpha-\frac{25}{2}=0. \]

Solving this equation, we obtain \(\sin\alpha=0.9307\), which, with accuracy to a minute, corresponds to the angle \(\alpha=68^\circ32'\).

Assuming \(\alpha\) to be approximately \(68^\circ.5\), we can, by means of an ordinary protractor, mark the corresponding point on the crank circle.

For completeness of the investigation it remains still to add that the tangential force \(T\) is equal to the force \(S\) acting along the connecting rod in the case when the crank makes a right angle with the connecting rod.

Indeed, we had earlier

\[ T=S\sin(\alpha+\beta); \]

putting \(\alpha+\beta=90^\circ\), we have \(T=S\).

As we indicated earlier, the force \(S\) is decomposed into two forces, the tangential \(T\) and the radial \(Q\).

Both these forces are expressed as functions of the force \(P\), acting on the piston, by the following formulas:

\[ T=P\frac{\sin(\alpha+\beta)}{\cos\beta};\qquad Q=P\frac{\cos(\alpha+\beta)}{\cos\beta}. \]

It is clear that these formulae, representing the two legs of one and the same triangle, expressed through the hypotenuse \(\dfrac{P}{\cos \beta}\), change their form when the angle of rotation of the crank \(\alpha > 90^\circ\).

Let \(\alpha = 90^\circ+\gamma\), where \(\gamma\) may in turn assume all values between \(0^\circ\) and \(90^\circ\); then

\[ \sin(90+\gamma+\beta)=\cos(\gamma+\beta), \]

\[ \cos(90+\gamma+\beta)=-\sin(\gamma+\beta). \]

Fig. 7.

Since for us only the absolute values \(T\) and \(Q\) are needed, the minus sign has no significance for us; in such a case

\[ T=P\frac{\cos(\gamma+\beta)}{\cos\beta}, \qquad Q=\frac{\sin(\gamma+\beta)}{\cos\beta}, \]

i.e. in the right triangle formed from the tangential, radial, and connecting-rod forces, the tangential and radial forces have changed places with one another (Fig. 7). This circumstance will have to be taken into account in the further exposition, to which we now proceed.

For the graphical determination of the magnitude of the tangential force, as is evident from all that has been set forth, it is necessary to know, in addition to the angle of rotation of the crank from its dead position, also the angle of inclination of the connecting rod to the line of dead points.

The angle of inclination of the connecting rod is easily found by means of the following construction *).

On the diameter of the crank circle (Fig. 8), in the direction from the center toward the crosshead of the machine, lay off a segment equal to \(\mu R\), and from the point \(O'\) thus obtained, as from a center, describe a circle with radius equal to the length of the crank.

) Indicated by me in the article “The Connecting-Rod Pole,” see Proceedings of the Tomsk Technological Institute*, 1909, No. 1.

Then, for any angle of rotation of the crank \(\alpha\), the corresponding angle \(\beta\) will be found as the angle at the vertex of a triangle formed by taking the direction of the crank radius; here the base of this triangle is equal to \(\mu R\), and the closing side connects the point \(O'\) with the point of intersection of the direction of the crank radius with the auxiliary circle.

Fig. 8.

Indeed, denoting the angle at the vertex by the letter \(\varphi\), from the property of the sides of the triangle we have

\[ O'B : O'O = \sin\alpha : \sin\varphi; \]

but \(O'B=R,\ O'O=\mu R\); therefore

\[ O'B : O'O = R : \mu R = 1 : \mu. \]

Consequently

\[ \frac{1}{\mu}=\frac{\sin\alpha}{\sin\varphi}, \]

whence

\[ \sin\varphi=\mu\sin\alpha. \]

But earlier we had a relation between \(\alpha\) and \(\beta\) of the same kind,

\[ \sin\beta=\mu\sin\alpha. \]

Since \(\mu R\) is always less than \(R\), on the basis of the theorem that the smaller angle lies opposite the smaller side of a triangle, we may say that \(\varphi\) cannot be greater than \(90^\circ\), and hence, having \(\sin\varphi=\sin\beta\), we conclude that \(\varphi=\beta\).

For crank rotation angles \(\alpha=90^\circ+\gamma\), we have, respectively,

\[ R : \mu R = \sin(90+\gamma) : \sin\varphi,\quad \sin\beta=\mu\sin(90+\gamma), \]

or

\[ \frac{1}{\mu}=\frac{\cos\gamma}{\sin\varphi},\quad \sin\beta=\mu\cos\gamma, \]

i.e. our construction is also valid for angles greater than \(90^\circ\).

Thus the angle of inclination of the connecting rod to the line of dead points is found very simply.

Consequently, in order to find the tangential force at the point \(A\) (Fig. 9), we construct the angle found and, using one of the constructions known to us, find the force \(T\) from the given force \(P\).

The construction at the point \(A\) of the angle \(\beta\), one of whose sides must be parallel to the line of dead points (the diameter of the crank circle), and the further resolution of the force \(P\) is a rather painstaking task—especially if one takes into account that, in order to obtain an idea of the variations of the force \(T\), it is determined for as large a number as possible of different positions of the crank (for 24–32 positions).

Fig. 9

With the aid of a straightedge and a set square, horizontal and vertical lines are very easily drawn; the usual method of resolving the force \(P\) requires drawing a large number of lines of different inclinations. Therefore we shall try to change the method of resolving the force \(P\) into the forces \(T\) and \(Q\).

Noting that the angle \(BO'K\) (Fig. 9), as an exterior angle with respect to the triangle \(O'KO\), will be equal to \(\alpha+\beta\), we immediately obtain the possibility of simplifying all the constructions.

Let us drop from the point \(K\) a perpendicular to the diameter of the crank circle; then its length \(KB\) will obviously be equal to

\[ KB=KO'\sin(\alpha+\beta). \]

Construct on the segment \(OO'\) a semicircle; then we shall have the point \(C\)—the intersection of the taken radius of the crank \(OA\) with a circle of radius \(\frac{\mu}{2}R\). Connecting the point \(O'\) with the point \(C\), we obtain that \(O'C\) is perpendicular to \(OK\).

Earlier we proved that the angle \(O'KO\) is equal to the angle \(\beta\); consequently, from the right triangle \(O'KC\) we have \(KC=O'K\cos\beta\), whence

\[ O'K = \frac{KC}{\cos \beta}; \]

after substituting this value into the expression for \(KB\), we obtain

\[ KB = KC \frac{\sin(\alpha+\beta)}{\cos \beta}. \]

The formula we have obtained shows that the perpendicular \(BK\), for the given angle of rotation of the crank, represents the magnitude of the tangential force \(T\) on the same scale in which the segment \(KC\) represents the force \(P\) acting on the piston.

This solves the question of finding the tangential force without drawing the corresponding positions of the connecting rod.

Since the vector \(O'K\) (Fig. 9) plays no role in the construction and was introduced by us only to prove the correctness of the stated proposition, there is no need for it in drawing the diagram.

Thus the entire construction is reduced to drawing two circles of radius \(R\) and one of radius \(\mu R\).

To find the true magnitude of the force \(T\) for a given \(P\), similar triangles must be constructed. The simplest way of constructing these triangles will be the following.

Draw, with radius equal to \(KC\) (Fig. 10), the line of dead centers; connect the point \(D\) thus obtained with \(K\), and lay off from \(D\) on \(DK\) a segment \(DE\) equal to \(P\). The perpendicular from \(E\) to the diameter of the crank circle will represent the tangential force \(T\) for the given position of the crank.

Fig. 10.

With the indicated construction we do not even need to drop the perpendicular \(KB\), which simplifies the whole method still further.

For a crank angle close to \(90^\circ\) it may happen that the segment \(KC\) does not intersect the line of dead centers; in such cases we find the magnitude of the force \(T\) by the following construction.

We connect the base of the perpendicular \(KB\) (Fig. 10) with the point \(C\); then laying off from \(C\) along \(CK\) a segment \(CF\) equal to \(P\) and dropping

from \(F\) a perpendicular to \(OK_0\), we have the tangential force \(FG\), as the segment of the perpendicular between the lines \(BC\) and \(KC\).

Up to now we have considered only the tangential force. As for the radial force, on the basis of all that has been said above, its magnitude will be expressed by the segment \(O'B\) on the same scale in which \(KB\) represents the tangential force.

Let us now pass to the consideration of the proposed method as applied to angles of rotation of the crank greater than \(90^\circ\).

Take (Fig. 11) the position of the crank \(OA\) at an angle \(90^\circ+\gamma\) from the left dead position and find for it the angle \(\beta\); then make the resolution of the force \(P\) at the point \(A\); then the tangential force, as we showed earlier, is expressed by the formula:

Fig. 11.

\[ T=P\frac{\cos(\gamma+\beta)}{\cos\beta}. \]

Drop from the point \(K\) a perpendicular to the line of dead points; then the angle \(OKB\) will be equal to \(\gamma\), as is evident from the drawing; and the angle \(O'KB\) will be equal to \(\gamma+\beta\).

Drawing through the point \(O'\) a vertical axis, we obtain the angle \(CO'K\) also equal to \(\gamma+\beta\). In the triangle \(CO'K\), the side

\[ CO'=O'K\cos(\gamma+\beta). \]

Continue the line \(KO\) to its intersection with the circle of radius \(\frac{\mu}{2}R\) at the point \(E\); then, in the right triangle \(O'KE\), the hypotenuse

\[ O'K=\frac{EK}{\cos\beta}. \]

Substituting the value \(O'K\) into the expression for \(CO'\), we have

\[ CO'=EK\frac{\cos(\gamma+\beta)}{\cos\beta}, \]

i.e. the segment of the perpendicular of length \(O'C\) represents the tangential force for the crank angle \(\alpha=90^\circ+\gamma\) on the same scale on which the segment \(EK\) represents the force \(P\) acting on the piston.

Noting that \(KB\) is equal to \(O'C\), as segments of parallel lines between parallel lines, we arrive at the conclusion that also for crank angles greater than \(90^\circ\), the perpendicular to the line of dead centers from the point of intersection of the auxiliary circle with the direction of the crank radius is the measure of the tangential force; i.e. the method set forth is applicable for all angles \(\alpha\) from \(0^\circ\) to \(180^\circ\).

In order, for \(\alpha>90^\circ\), to find the magnitude of the force \(T\), the construction may be simplified still further; we proceed as follows: from the point \(K\), with a radius equal to \(KE\), we cut the line of dead centers; having laid off on the line \(DK\), from the point \(D\), the segment \(DF\), we drop from the point \(F\) the perpendicular \(FG\), which gives the required force \(T\) (Fig. 11).

In this construction we need not draw the lines \(O'K\), \(O'C\), and \(KB\); they were needed only to prove the correctness of the procedure; one has to operate only with the segment of the crank radius, which, incidentally, always intersects the line of dead centers.

Fig. 12.

Let us now pass to the determination of the tangential forces for the reverse motion of the piston, i.e. for \(\alpha>180^\circ\).

We shall first prove that, for crank positions symmetric with respect to the line of dead centers, the angles of inclination of the connecting rod to this line are the same.

Take (Fig. 12) two symmetric positions of the crank \(K\) and \(K'\), to which correspond the crank angles from the left dead position \(180^\circ-\alpha\) and \(180^\circ+\alpha\); then we have

\[ \sin(180^\circ-\alpha)=\sin\alpha,\quad \sin\beta=\mu\sin\alpha; \]

further,

\[ \sin(180^\circ+\alpha)=-\sin\alpha,\quad \sin\beta'=\mu\sin\alpha, \]

i.e. the absolute values of the angles \(\beta\) and \(\beta'\) are the same.

Let us suppose that the force \(P\), acting on the piston, is the same for the two chosen positions of the crank. We shall find the magnitude of the tangential force by the usual construction; for the position of the crank \(K\) we shall have

\[ T=P\frac{\sin(\alpha-\beta)}{\cos\beta}; \]

for the position of the crank \(K'\), respectively,

\[ T'=P\frac{\sin(\alpha-\beta)}{\cos\beta}, \]

i.e. \(T=T'\).

For different \(P\), consequently, we shall have

\[ T:T'=P:P'. \]

whence we conclude that, in order to find the force \(T'\), one may make use of the line \(DK\) already found earlier (Fig. 11).

Fig. 13.

Thus, in order to find the tangential forces during the return stroke of the piston, we may make use of all the constructions made previously, assigning only to the points of division of the crank circle—

of a double numbering, as shown in Drawing 13, which considerably shortens the work *).

To find the magnitude of the tangential force acting on the crank pin, it is necessary to know the pressure \(P\) acting on the piston for the given position of the crank; up to now we have assumed this force to be constant, whereas in reality it is a variable quantity, even passing through a zero value.

The magnitude of the force \(P\) is determined from the so-called diagram of working pressures; this diagram, as is known, is obtained (for steam engines) from transformed indicator diagrams for the forward and return strokes of the piston, taking into account the influence of the masses moving back and forth (the piston, rod, crosshead, and connecting rod).

Knowing the position of the piston corresponding to the adopted position of the crank, we obtain the force \(P\) as the ordinate of the diagram mentioned.

Thus, in order to determine the tangential force, we must know the position of the piston.

If we draw (Drawing 14) from the center \(O\) a circle with radius equal to \(R+\mu R\), then we shall see that this circle will be the outer circle of Müller’s diagram (Drawing 3), provided that \(\mu R\) is equal to half the piston stroke,

\[ \frac{S}{2}; \]

then, for the position of the crank \(A\), the magnitude of the segment \(CB\) represents the path traversed by the piston from the left dead position, while the magnitude of the segment \(BA\) is the path by which the piston has not reached the middle of its stroke; moreover, these paths are represented on the scale in which

Drawing 14.

\[ \frac{S}{2}=R. \]

\(\mu R\) represents half the piston stroke.

Thus, in order to find the true position of the piston, we must lay off from the center \(O\) to the left the segment \(AB\), enlarged on the scale

\[ \frac{R}{\mu R}=\frac{1}{\mu}. \]

*) When drawing the diagram, one should take \(R=10\) or \(20\) cm; in the latter case all the required points are found especially clearly. To take such a radius for the usual method of finding tangential forces is inconvenient, since then, in order to find the positions of the connecting rod, one has to make notches with a radius \(L=5R=100\) cm.

For machines of the normal type \(\mu=\dfrac{1}{5}\); consequently \(\dfrac{1}{\mu}=5\).

In the presence of the so-called proportional compass, the position of the piston is found by a simple measurement of the segment \(AB\).

In the absence of such a compass one may proceed in the following manner: from the center \(O\) (Fig. 14) set off downward along the vertical the segment \(OD=\mu R\); join the point \(D\) by straight lines with the points \(K_0\) and \(K_0\); then, laying off upward from \(D\) the segment \(AB\) (or, respectively, \(NL\) for \(\alpha>90^\circ\)), and drawing the horizontal line \(EF\) and the vertical line \(FG\), we obtain the corresponding position of the piston.

Since we determine the positions of the piston with respect to the middle of its stroke, the circle of radius equal to \(R+\mu R\) becomes superfluous; consequently it is sufficient for us to add the line \(K_0DK_0\) to drawing 13 in order to use it both for finding the true positions of the piston and for determining the tangential forces.

Fig. 15.

One may also use the method of Brix; for this (Fig. 15) we draw a circle of radius \(\dfrac{\mu}{2}R\) concentrically with the crank circle. Then, drawing from the point \(O'\) a second circle of radius \(R\), we obtain, for an arbitrary angle of rotation of the crank \(\alpha\), the triangle \(O'KO''\), completely identical with the triangle

\(O'KO\) of drawing 9. Consequently, all the reasoning given above will also be valid for drawing 15. The positions of the piston will be found by dropping a perpendicular from point \(K'\). Thus one and the same vector \(OK\) will serve to determine the position of the piston and the corresponding tangential force. It is only necessary to add one more eccentric circle with center \(O''\), in order to mark on it equal arcs for a definite number of divisions of the crank circle*).

The question is exhausted. Having found the magnitude of the tangential forces, one usually constructs their diagram, taking as the base the developed crank circle, and determines the mass of the flywheel referred to the crank radius by the known formula

\[ faO = Mv^2\delta, \]

where \(a\) is the scale of the diagram in \(kg/mtr\), \(f\) is the area of the diagram that causes the greatest change in live force, \(M\) is the mass of the flywheel, \(v\) is the circumferential velocity of the crank, and \(\delta\) is the prescribed coefficient of non-uniformity of the machine’s motion.

By the method set forth in the present essay, the inverse problem is also easily solved: from a given constant tangential force, to find the variable force moving the piston—a question encountered in the construction of pumps, air blowers, compressors, and other machines which receive rectilinear reciprocating motion from a continuously rotating driving shaft.

In conclusion, it will not be superfluous to add that by the same method one can also determine the piston velocity \(c_x\) corresponding to a given position of the crank:

As is known, the graphical method for finding the piston velocity is based on the formula:

\[ c_x = v\frac{\sin(\alpha+\beta)}{\cos\beta}; \]

(where \(c_x\) is the variable velocity of the piston, \(v\) is the circumferential velocity of the crank, taken as constant, and \(\alpha\) and \(\beta\) are the same angles that enter into the formula for the tangential force), a formula entirely analogous to the one just given, and therefore everything said about finding tangential forces is applicable also to finding the piston velocity.

Tomsk. January, 1910.

A. Ugarov.

*) In drawing 15 is shown the finding of tangential forces as segments of perpendiculars between two corresponding inclined lines. The whole construction is obtained quickly and easily.