N. P. Shubin

Graphical Method for Determining the Angle of Inclination of the Connecting Rod in a Slider-Crank Mechanism and Its Application to the Calculation of a Flywheel

Tomsk
1928

N. P. Shubin.

Graphical Method for Determining the Angle of Inclination of the Connecting Rod in a Slider-Crank Mechanism and the Application of This Method to the Calculation of a Flywheel

In the graphical calculation of a flywheel it is necessary to know the angle of inclination of the connecting rod to the line of dead centers for various angles of rotation of the crank. Usually the method of intersections is used here. From point K, the position of the crank pin (Fig. 1), with a radius equal to the length of the connecting rod, an intersection is made on the trajectory of the slider and, connecting the resulting point B with the crank pin, the position of the connecting rod is determined.

The inconvenience of this method consists in the fact that, at large scales, one must work with a very large radius of intersection \(l\).

The subject of the present article is the exposition of a method for determining the angle \(\beta\) without using intersections of large radius.

Let us denote:

\[ \begin{aligned} r&\text{ — radius of the crank.}\\ l&\text{ — length of the connecting rod.}\\ \alpha&\text{ — angle of inclination of the crank to the line of dead centers.}\\ \beta&\text{ — angle of inclination of the connecting rod to the line of dead centers.}\\ \frac{r}{l}&=\mu\text{ — ratio of the length of the crank radius to the length of the connecting rod.} \end{aligned} \]

From triangle \(OKB\) (Fig. 1) we find the relation between the angles \(\beta\) and \(\alpha\).

\[ \frac{OK}{\sin \beta}=\frac{BK}{\sin \alpha};\quad \sin \beta=\frac{OK}{BK}\cdot \sin \alpha=\frac{r}{l}\cdot \sin \alpha \]

\[ \sin \beta=\mu\cdot \sin \alpha \ldots (1). \]

The angle of inclination of the connecting rod \(\beta\) is easily found by the following construction. From the center of the main shaft \(O\) we describe a circle with radius equal to the length of the crank \(r\) (Fig. 2). From the center \(O\), along the horizontal diameter to the right (in the direction opposite the slider), we lay off a segment equal to \(\mu^{2}r\), and from point \(O^{1}\), as from a center, we describe a circle of radius \(\mu r\). Further, laying off from point \(O^{1}\) to the left (toward the slider) a segment equal to \(r\), we obtain point \(C\).

For any position of the crank pin \(K\), we find the angle of inclination of the connecting rod as follows: connect point \(K\) with the center of the crank circle \(O\), mark point \(D\)—the intersection of the crank with the auxiliary circle of radius \(\mu r\)—and draw line \(CD\); then the angle \(DCO\) will be equal to the angle \(\beta\).

For the proof, connect point \(D\) with \(O^{1}\) and consider triangle \(ODO^{1}\). From the properties of the sides of a triangle we have:

\[ \frac{OO^{1}}{O^{1}D} = \frac{\sin ODO^{1}}{\sin(180^\circ-\alpha)} = \frac{\sin ODO^{1}}{\sin \alpha} \]

but

\[ OO^1=\mu^2 r \quad \text{and} \quad O^1D=\mu r \]

then

\[ \frac{\mu^2 r}{\mu r}=\frac{\sin ODO^1}{\sin \alpha} \]

therefore

\[ \sin ODO^1=\mu \sin \alpha \ldots\ldots\ldots\ldots (2) \]

Comparing equations 1 and 2, we have

\[ \sin ODO^1=\sin \beta. \]

or

\[ \angle ODO^1=\beta. \]

Let us next consider two triangles: \(\triangle DOO^1\) and \(\triangle CDO^1\).

From \(\triangle CO^1D\) we have

\[ \frac{O^1D}{CO^1}=\frac{\mu r}{r}=\mu \ldots\ldots\ldots (3) \]

From \(\triangle DOO^1\)

\[ \frac{OO^1}{O^1D}=\frac{\mu^2 r}{\mu r}=\mu \ldots\ldots\ldots (4) \]

Equating the left-hand sides of equalities 3 and 4, we obtain

\[ \frac{OO^1}{O^1D}=\frac{O^1D}{CO^1} \]

i.e., two sides of one triangle are proportional to two sides of the other. In addition, both triangles at the vertex \(O^1\) have the common angle \(\alpha-\beta\), included between the proportional sides. Thus \(\triangle DOO^1\) is similar to \(\triangle CDO^1\).

From the similarity of the triangles it follows that

\[ \angle DCO=\beta. \]

Thus, as the angle \(\alpha\) changes from \(0^\circ\) to \(90^\circ\), the angle of inclination of the connecting rod to the line of dead centers will be expressed by the angle \(DCO\).

For crank rotation angles greater than \(90^\circ\), \(\alpha=90^\circ+\gamma\); for example, for the position of the crank pin at point \(K_1\) (Fig. 2), the angle of inclination of the connecting rod \(\beta_1\) is found by joining point \(E\) with point \(O\):

\[ \angle \beta_1=\angle ECO \]

From the triangle \(B_1K_1O\) (Fig. 3) we find the relation between the angles \(\beta_1\) and \(\gamma\):

\[ \frac{OK_1}{B_1K_1} = \frac{\sin \beta_1}{\sin(90^\circ+\gamma)} = \frac{\sin \beta_1}{\cos \gamma}; \qquad \sin \beta_1= \frac{OK_1}{B_1K_1}\cdot \cos \gamma = \frac{r}{l}\cdot \cos \gamma \]

\[ \sin \beta_1=\mu \cos \gamma \ldots\ldots\ldots (5) \]

To prove that the angle \(ECO\) (Fig. 2) is indeed equal to the angle \(\beta_1\), let us consider the triangle \(O^{1}EO\). From the property of the sides of a triangle we have

\[ \frac{OO^{1}}{O^{1}E} = \frac{\sin OEO^{1}}{\sin(90^\circ-\gamma)} = \frac{\sin OEO^{1}}{\cos \gamma} \]

but \(OO^{1}=\mu^{2}r\) and \(O^{1}E=\mu r\);

then

\[ \frac{\mu^{2}r}{\mu r} = \frac{\sin OEO'}{\cos \gamma} \]

therefore

\[ \sin OEO'=\mu\cdot \cos \gamma \ . . . . . . . . . . . . . . . . . . (6) \]

Comparing equations 5 and 6, we have

\[ \sin OEO^{1}=\sin \beta_1, \]

or

\[ \angle OEO^{1}=\beta_1. \]

Let us further prove the similarity of the triangles \(CO^{1}E\) and \(OEO^{1}\). From the triangle \(OEO^{1}\) we have

\[ \frac{OO^{1}}{O^{1}E} = \frac{\mu^{2}r}{\mu r} = \mu \ . . . . . . . . . . . . . . . . (7) \]

From the triangle \(CO^{1}E\)

\[ \frac{O^{1}E}{CO^{1}} = \frac{\mu r}{r} = \mu \ . . . . . . . . . . . . . . . . (8) \]

From 7 and 8 it follows that

\[ \frac{OO^{1}}{O^{1}E} = \frac{O^{1}E}{CO^{1}} \]

Two sides of one triangle are proportional to two sides of the other. Moreover, the angles included between these sides are equal; consequently, these triangles are similar.

From the similarity of the triangles it follows that

\[ \angle ECO=\beta_1 \]

Thus we have proved that the construction of the angle of inclination of the connecting rod by means of the auxiliary circle and the fixed pole \(C\) is valid for any position of the crank pin. On the return stroke all constructions will be analogous.

Application of the graphical method for determining the angle of inclination of the connecting rod to the calculation of a flywheel by the tangential-force diagram.

Let us denote the force acting on the piston at the given position of the crank pin by \(P\).

At point \(B\) (Fig. 1) we resolve the force \(P\) into two components \(S\) and \(N\)

\[ N=P\cdot \operatorname{tg}\beta \quad \text{and} \quad S=\frac{P}{\cos\beta} \]

Let us transfer the force \(S\) along the connecting rod to point \(K\) and resolve it into two components: the tangential force \(T\), acting along the tangent at point \(K\), and the radial force \(Q\), directed along the crank.

\[ Q=S\cos(\alpha+\beta)\qquad T=S\cdot \sin(\alpha+\beta). \]

Substituting the value of \(S\), we obtain

\[ T=P\,\frac{\sin(\alpha+\beta)}{\cos\beta}\ . . . . . . . . . . . . . (9) \]

For the graphical determination of \(T\) (Fig. 4), we lay off, from point \(K\) along the crank and on a definite scale, the force \(P\); from the end of the force \(P\) we draw a vertical line \(MN\) to its intersection with the direction of the connecting rod. The segment \(MN\), on the scale, will be equal to the force \(T\). In this determination of the tangential force, the position of the connecting rod is found by compass arcs with a radius equal to the length of the connecting rod.

As was already noted at the beginning of the article, the use of compass arcs of considerable radius is a substantial drawback of constructions of this kind. With large radii, inaccuracies are possible owing to springing of the legs of the compasses.

Using, in the present case, the method set forth above for determining the angle \(\beta\), we obtain very simply and accurately the magnitude of the tangential force \(T\) without using arcs of large radius.

From the center \(O\) we draw the crank circle of radius \(r\) (Fig. 5); from point \(O^1\) we draw an auxiliary circle of radius \(\mu r\) and mark the position of the pole, point \(C\). We join the crank pin \(K\) to the center of the crank circle \(O\), and the point of intersection of the crank with the auxiliary circle, point \(D\), is joined to the pole \(C\).

To determine the force \(T\), lay off from point \(D\), in the direction of the line \(OK\), the force \(P\) (pressure on the piston), and from the end of the force \(P\) draw a vertical line to its intersection with the line \(CD\).

Then the segment \(LF\) represents the magnitude of the tangential force \(T\) on the same scale on which the force \(P\) was laid off.

For the proof, note that

\[ \angle KOC=\alpha,\qquad \angle DCO=\beta,\qquad \angle LDF=\alpha+\beta \quad\text{and}\quad \angle LFD=90^\circ-\beta. \]

From triangle \(LFD\) we have

\[ \frac{LF}{LD} = \frac{\sin(\alpha+\beta)}{\sin(90^\circ-\beta)} = \frac{\sin(\alpha+\beta)}{\cos\beta}. \]

whence

\[ LF=LD\,\frac{\sin(\alpha+\beta)}{\cos\beta} \]

but \(LD=P\), therefore

\[ LF=P\,\frac{\sin(\alpha+\beta)}{\cos\beta}\ . . . . . . . . . . . . . (10) \]

Comparing equations 9 and 10, we have

\[ LF=T \]

For angles \(\alpha\) greater than \(90^\circ\), \(\alpha=90^\circ+\gamma\), the relation between \(P\) and \(T\) is found from drawing 3.

Let us transfer the force \(S\) along the connecting rod to the point \(K_1\) and resolve it into the forces \(T\) and \(Q\)

\[ \angle B_1 K_1 O = 90^\circ - (\gamma + \beta_1) = \angle K_1 S T \]

from the triangle \(K_1 S T\)

\[ T = S \cdot \sin [90^\circ - (\gamma + \beta_1)] = S \cdot \cos(\gamma + \beta_1) \]

Substituting the value of \(S\), we obtain

\[ T = P \frac{\cos(\gamma + \beta_1)}{\cos \beta_1} \tag{11} \]

To construct \(T\), connect \(K_1\) with the center \(O\) and the point of intersection of the auxiliary circle \(E\) with \(C\) (Fig. 5). From the point \(E\), lay off the force \(P\) along the radius, and from the end of the force \(P\) draw a vertical line to its intersection with the continuation of the line \(EC\).

Then the segment \(MN\) represents the tangential force for the angle of rotation of the crank \(\alpha = 90^\circ + \gamma\), on the same scale on which the force \(P\) was laid off.

For proof, let us note that

\[ \angle NEM = 90^\circ - (\gamma + \beta_1) \quad \text{and} \quad \angle NME = 90^\circ + \beta_1 \]

From the triangle \(EMN\) we have

\[ \frac{MN}{EN} = \frac{\sin [90^\circ - (\gamma + \beta_1)]}{\sin(90^\circ - \beta_1)} = \frac{\cos(\gamma + \beta_1)}{\cos \beta_1} \]

whence

\[ MN = EN \frac{\cos(\gamma + \beta_1)}{\cos \beta_1} \]

but \(EN = P\), therefore

\[ MN = P \frac{\cos(\gamma + \beta_1)}{\cos \beta_1} \tag{12} \]

From equations 11 and 12 it follows that

\[ MN = T \]

Since, in constructing the diagram of working pressures, the Brics method is usually used to determine the position of the piston, it is necessary to show how to apply the method set forth in the present case.

From the center \(O\) of the crank circle (Fig. 6), lay off to the right, at a distance \(\frac{1}{2}\mu r\), the Brics pole \(O''\), and draw an auxiliary circle of radius \(\mu r\) from the center \(O'\), laid off at a distance \(\mu^2 r\) to the right of the Brics pole; from the point \(O'\), lay off \(r\) to the left and mark the constant pole \(C\).

From the Brics pole draw a circle of radius \(r\); divide it into several equal parts (24–32), mark the points of division, and connect them with \(O''\). Connect the points of intersection of the radii with the auxiliary circle with

with the fixed pole \(C\). Laying off from point \(D\), along a radius from center \(O\), the force \(P\), and drawing the vertical line \(LF\) to its intersection with the line \(DC\), we find the segment \(LF\), which will represent the tangential force \(T\).

Indeed, comparing the triangles \(FLD\) (Fig. 6) and \(FLD\) (Fig. 5), we see that, for equal angles of rotation of the crank, they are identical. Consequently, everything set forth above is also valid for Fig. 6.

The drawing shows the construction of the tangential forces for 24 positions of the crank according to the diagram of working pressures. The construction of the characteristic points is indicated by the dotted line.

Using the forces \(T\) found, we then construct the diagram of tangential forces and determine the weight of the flywheel.

Using the construction in Fig. 6, it is also possible to solve the inverse problem easily—to determine the force \(P\) from the tangential force \(T\).

In addition, by the same construction one can determine the velocity of the piston for a given angle of rotation of the crank. Laying off from point \(D\), along a radius from center \(O''\), the segment \(LD\), equal to the velocity of the crank pin at the given instant, we obtain the segment \(LF\), which represents the velocity of the slider.

For small scales of the drawing, it is reasonable, for greater accuracy, to take an auxiliary circle from center \(O'\) of larger radius.

Fig. 7 shows the construction of the tangential conditions for the same diagram of working pressures with a different radius of the auxiliary circle.

Here: \(O\) is the center of the crank circle, \(O''\) is the Brix pole, \(O'\) is the center of the auxiliary circle, and \(C\) is the fixed pole.

The distance from pole \(C\) to center \(O'\) is taken arbitrarily and is equal to \(a\). The radius of the auxiliary circle is equal to \(\mu a\), and the segment \(O''O'\) is equal to \(\mu^{2}a\).

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

Fig. 5.

Drawing 6.

To the article by N. P. Shubin.

To the article by N. P. Shubin.

Drawing 7.