D. G. Stanko-

Vector Flywheel Diagram.

In designing engine machines that have a connecting-rod crank mechanism, a question arises connected with the uniformity of the machine’s motion. For uniform motion of the machine it is necessary that the driving forces imparting rotation to the shaft of the machine remain at all times equal to the forces of resistance.

Owing to the structural arrangement of the connecting-rod crank mechanism, we cannot obtain a constant tangential force developed by the driving forces on the crank pin.

If we assume that the forces of useful resistance, in steady motion, are constant in magnitude, then we are faced with questions of excess and deficiency of the work of the driving forces during one revolution of the shaft. In order to obtain uniform motion of the machine, it is necessary to install an energy accumulator. The latter must absorb the excess energy of the driving forces and give up this accumulated energy during the period of deficiency of the work of the driving forces.

The energy accumulator is made in the form of a flywheel mounted on the shaft. The relation between the greatest excess or deficient work and the mass of the flywheel (through the moment of inertia of the flywheel) may be expressed as follows¹:

\[ A = J\omega_{\mathrm{cp}}^{2}\delta, \tag{1} \]

where:

\(A\) — the greatest excess or deficient work of the driving forces,
\(J\) — the moment of inertia of the rotating masses,
\(\omega_{\mathrm{cp}}\) — the mean angular velocity,
\(\delta\) — the degree of nonuniformity.

Starting from the first equation, we can determine the dimensions of the flywheel. The determination of the excess or deficient work is connected with the construction of the flywheel diagram.

The flywheel diagram is constructed from the diagram of tangential forces, and the latter is connected with the diagram of working pressures.

The diagram of working pressures is a function of the piston position and of the pressure difference on the right and left sides of the piston. The working pressure is determined from indicator diagrams.

Let us dwell somewhat more fully on the determination of the tangential force. Denote by:

\(P\) — the working pressure,
\(S\) — the force acting along the connecting rod,
\(N\) — the normal component perpendicular to the line of travel of the piston,
\(T\) — the tangential force in the direction perpendicular to the crank radius,
\(R_k\) — the component along the crank radius.

The determination of the tangential force is reduced to the addition and subtraction of force vectors.

The vector notation is represented as follows (Fig. 1):

\[ P = S + \to N \tag{2} \]

\[ S = T + \to R_k \tag{3} \]

From a comparison of (2) and (3) we have the value of the tangential force equal to

\[ T = P \to N \to R_k \tag{4} \]

Addition and subtraction of vectors are conceived geometrically.

In Fig. 1 (a), the tangential force for the connecting-rod-and-crank mechanism has been found; in adding the vectors the parallelogram rule was used.

The crank circle is divided into 8 equal parts. In the general case, for each position of the piston the value of the working pressure \(P\) is taken arbitrarily.

Usually the value of the working pressure is taken from the diagram of working pressures.

For positions 2, 3, 5, 7, the decomposition of the force \(P\) has been carried out directly at the end of the crank. The whole process of construction is clear from the drawing.

Determining the tangential forces by the parallelogram rule for each position of the connecting-rod-and-crank mechanism is somewhat cumbersome. Several methods have been proposed for determining the tangential force.

The initial data, among the methods known to the author, are given by the equation,

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

where \(\varphi\) is the angle between the crank radius and the path of the slider,
\(\beta\) is the angle between the direction of the connecting rod and the path of the slider.
(The remaining designations in the equation are as before.)

The diagram of tangential forces was obtained, as it were, by an artificial method. Trigonometric properties of the triangle were used. Then the flywheel diagram was constructed separately. In essence, determination of the tangential force, as was said above, is nothing other than the geometrical addition of force vectors.

The author proposes a new method of determining the tangential force, by constructing a vector flywheel diagram.

Along the way, the properties of the vector flywheel diagram are revealed, and the problem is solved by bringing the positions of the connecting rod to the pole.

In construction, the vector flywheel diagram gives, superposed, the diagram of tangential forces and the flywheel diagram, i.e., the intermediate construction of the diagram of tangential forces is avoided.

To an even greater degree, the advantage of applying the vector flywheel diagram appears in multicylinder engines\(^3\).

We proceed to the exposition of the method of constructing the vector flywheel diagram.

We choose [Fig. 1 (b)] an arbitrary pole \(O\).

From the pole \(O\), as center, we draw a circle with an arbitrary radius, in accordance with the scale of forces and taking into account the diagrams that are placed on the drawing.

\(^1\) \(+\to\) plus—addition of vectors.
\(\to\) minus—subtraction of vectors.
\(^2\) The method of Prof. Ugarov, N. P. Shubin, F. Retscher, and others.
\(^3\) The author proposes to illuminate this question in a special work.

We divide the circumference into an equal number of parts¹. It is better to choose the number of divisions so that the divisions fit into a right angle an integral number of times. Through the division points of the circumference and the center \(O\) we draw radii. For each position of the crank-and-connecting-rod mechanism we construct the force plan².

In constructing the force plan, we shall have the required tangential forces directed along radii from the center \(O\). The radii along which the tangential forces are directed form a figure similar to the positions of the crank, but rotated by \(90^\circ\) in the direction of rotation [compare Figs. 1(a) and 1(b)]. This property follows from the fact that the tangential force is directed perpendicular to the radius of the crank. We proceed to determine the tangential force.

In the force plan all working pressures will be laid off, to some scale³, along the line \(2—6\), passing through the pole \(O\) [Fig. 1(b)].

Let us agree to call the position of the connecting rod “highest” when, during the piston stroke toward the crankshaft, the radius of the crank forms a right angle with the line of the piston stroke, and the position of the connecting rod “lowest” when, during the piston stroke away from the crankshaft, the radius of the crank likewise forms a right angle with the line of the piston stroke. Then it may be said that the working pressures are laid off on the crank radii corresponding to the “highest” and “lowest” positions of the connecting rod, if the similarity of the tangential forces in the force plan is taken into account.

Let us trace how the tangential force is found.

For the first position we have the geometrical sum of vectors [Fig. 1(b)]

\[ P_1 = S_1 + \to N_1 \]

\[ S_1 = T_1 + \to R_{k_1}. \]

The directions of the acting forces were discussed earlier.

Since our construction satisfies the vector equations, then

\[ T_1 = P_1 \to N_1 \to R_{k_1}. \]

For positions 2 and 4 (the highest and lowest positions of the connecting rod) we have:

\[ T_2 = P_2; \qquad N_2 = R_{k_2} \]

\[ T_4 = P_4; \qquad N_4 = R_{k_4} \]

i.e. the working pressures are equal in magnitude and direction to the tangential forces, and in these positions the lines of action of \(P\) and \(T\) coincide. The components \(N\) and \(R_k\) are equal in magnitude, but are directed in opposite directions. In the dead positions \(0—4\) the components of the tangential force are equal to zero.

Analogously to the first position analyzed above, the tangential forces are constructed and found for all the remaining positions of the crank-and-connecting-rod mechanism.

From the entire construction the following follows: in order to find the tangential force it is necessary to lay off the working pressure from the pole \(O\) on the radii-vectors of the tangential force corresponding to the highest and lowest positions of the connecting rod; from the end of the vector of working pressure draw a perpendicular until it intersects the corresponding direction of the force \(S\) (in other words, the direction of the connecting rod), and, finally, from the obtained point of intersection (on the direction of the connecting rod) drop a perpendicular until it intersects with

corresponding to the direction of the tangential force. Then the magnitude of the tangential force is determined by the segment between the point of intersection of the dropped perpendicular and the pole \(O\). The direction of the tangential force will always be from the pole. The hodograph of the tangential force, shown by the dotted line, gives us the vector flywheel diagram. On the force plane the pencil of connecting-rod directions, otherwise the force \(S\), splits into two, for the forward and return stroke of the piston. We shall consider the forward stroke to be when the piston moves in the direction toward the crankshaft, and the return stroke—away from the crankshaft. If in Fig. 1 (b) the motion of the connecting-rod-and-crank mechanism with radius \(r\) is represented, then the semicircles of radius \(r\), with their convexities, will be directed from the straight line 6–2 toward the point 0–8.

The construction just described is somewhat difficult, since special marking of the positions of the tangential force is necessary, and it is necessary to know the direction of the connecting-rod lines.

The posed problem of determining the tangential force can be solved much more simply.

We superpose the lines of direction of the tangential forces with the radii of the crank.

For the superposition it is necessary to rotate Fig. 1 (b) through \(90^\circ\) in the direction opposite to the rotation of the crank, together with all the constructions made. In Fig. 1 (c) the force plane is shown in the rotated position, and the marking of the crank radii [Fig. 1 (a)] and the direction of the tangential forces [Fig. 1 (c)] coincide.

The pencil of connecting-rod directions in the superposed position in Fig. 1 (c) is directed toward the slider (piston), with respect to Fig. 1 (a). The arrows \(x\) and \(y\) show the direction of rotation along the circumference of radius \(r\). It is not difficult to remember that the direction of rotation of the radius \(r\) is from the pole \(O\), which is also shown in Fig. 1 (b) (c) by the arrows \(x\) and \(y\).

A somewhat more complicated question is that of determining the direction of the connecting rod, since the pole of the force plane is taken as the initial position for the connecting rod.

To solve this problem we proceed as follows: we draw the straight line \(O\ 0{,}6\), with an arbitrary radius \(l\) from the center \(O\) [Fig. 2 (a)] we cut off an arc. At a distance \(r = l\), we draw a straight line parallel to \(O\ 0{,}6\), until it intersects the arc of radius \(l\). From the point of intersection 3 we drop a perpendicular to the straight line \(O\ 0{,}6\), and find the center \(O_1\). From the center \(O_1\) we cut off a circle of radius \(r\).

To determine the direction for any position of the crank, it is necessary, from a point taken on the circumference of radius \(r\), to draw a straight line parallel to the horizontal \(O\ 0{,}6\), until it intersects the arc of radius \(l\). Then the desired direction of the connecting rod is determined by the straight line drawn through the pole \(O\) and the point of intersection on the arc of radius \(l\).

To prove the correctness of bringing the connecting rod to the pole, let us single out some position of the connecting-rod-and-crank mechanism [Fig. 2 (b)] and drop perpendiculars from the points \(k\) and \(x\) to the line of travel of the piston. We have two right triangles, equal to one another, i.e.

\[ \triangle Okl = \triangle zxy. \]

The equal triangles are translated progressively in the direction of the trajectory of the slider; consequently, the straight lines \(Ok\) and \(zx\) are parallel to one another; otherwise, the superposed direction of the connecting rod is parallel to the actual direction of the connecting rod, which was what had to be proved.

Summarizing all that has been set out above, let us note the basic properties of the force plane in the superposed position.

1. In the force plane the tangential forces are directed from the pole along the corresponding directions of the crank radius. To determine the actual-

Fig. 1.

Fig. 2

[a] [b]

Fig. 3.

the actual direction of the tangential force, it is necessary on the force plane to rotate the tangential force by \(90^\circ\) in the direction of rotation.

1. The working pressures are laid off on the crank radii corresponding to the lowest and highest positions of the connecting rod.

2. The bundle of directions of the connecting rod on the force plane is divided into two—for the forward and return strokes—and is distributed toward the crosshead from the crank radii corresponding to the lowest and highest positions of the connecting rod.

3. The marking of the crank circle of radius \(r\), when the connecting rod is referred to the pole, begins from the points lying nearer to the pole; moreover, the circle of radius \(r\) is divided into two semicircles—for the forward stroke, on the crank radius corresponding to the highest position of the connecting rod, and for the return stroke, on the crank radius corresponding to the lowest position of the connecting rod.

4. In determining the tangential force, from the end of the vector of the working pressure we erect a perpendicular until it intersects the corresponding direction of the connecting rod of the circle of radius \(r\), and from the point of intersection we drop a perpendicular until it intersects the corresponding crank radius.

5. The tangential force always emerges from the pole. To determine the actual direction of the tangential force, it is necessary to rotate the latter by \(90^\circ\) in the direction of rotation.

6. The tangential forces are equal to zero in the dead positions and when the working pressures are equal to zero.

7. With negative working pressures, when the motion of the piston does not coincide with the direction of the acting force, the determination of the tangential force is analogous to item 5, with the exception of the direction of the connecting rod and of the crank radius, since the latter are directed to the opposite side.

Let us illustrate the proposed method by an example. Suppose that we have the indicator diagram shown in Fig. 3(a), with mean indicator pressure \(p_i\) and piston stroke \(H = 2R\), where \(R\) is the crank radius.

In Fig. 3(a) we construct the diagram of working pressures (the dotted closed curve). The working pressure at each moment is the algebraic sum of the pressures on the right and left sides of the piston\(^1\). The piston stroke of the diagram of working pressures is taken on the same scale as for the indicator diagram.

Taking into account the inertia of the reciprocating parts [Fig. 3(b)], we construct the curve of inertial pressures by Tolle’s method. We reduce the diagram of working pressures to a line parallel to the piston stroke \(24—12\), while algebraically adding the working pressures to the inertia forces.

The reduced diagram of working pressures is shown by the solid curve. From the center \(O\) we scribe a circle of crank radius \(R\).

Taking into account the influence of the finite length of the connecting rod, we draw from the Briggs pole \(O\) a dividing circle of arbitrary radius; in our case the crank radius is taken. To find the Briggs pole \(O\), it is necessary, from the center \(O\), in the direction opposite to the crosshead, to lay off the segment \(\frac{1}{2}\mu R\).\(^2\)

\(^1\) For the method of constructing indicator diagrams and diagrams of working pressures, see the corresponding courses on steam engines, e.g., Zhiritskii, Steam Engines.

\(^2\) More precisely, the segment is equal to

\[ \frac{R}{\mu}\left[1-\sqrt{1-\mu^2}\right]. \]

The circle drawn from the Bricks pole \(O\) is divided into 24 equal parts, and radii are drawn to their intersections with the circle of center \(O\).

The numbered points of intersection on the circle of center \(O\) will link the piston positions with the crank positions. On the crank radii corresponding to the lowest and highest positions of the connecting rod, we reduce the position of the connecting rod to the pole \(O\) by the method set forth above.

In Fig. 3(b) we have the marking of the points of the diagram of working pressures, the directions of the connecting rod, and the directions of the tangential forces.

To find the tangential force, it is necessary to transfer the working pressure along the guide to the corresponding position of the connecting rod. From the point of intersection, erect a perpendicular until it intersects the corresponding direction of the tangential force.

The construction process is shown in Fig. 3(b) for positions 2 and 14.

Dropping the perpendicular from the point of intersection on the connecting rod may be done with a template made of stiff paper.

In finding the tangential force corresponding to negative pressure, it is necessary to continue the direction of the connecting rod from the pole \(O\), and to transfer the working pressure onto the extension of the connecting rod. From the obtained point of intersection we erect a perpendicular until it intersects the direction of the tangential force. The direction of the tangential force under negative pressures does not coincide with the position of the crank, but will be turned by \(180^\circ\) in the direction of rotation.

The construction of the tangential force, for negative working pressure, is shown in Fig. 3(b) for position 11.

Having determined all the tangential forces, we join the ends of the vectors along the outline and obtain the hodograph of the radius vector of the tangential force.

From the center \(O\) we draw a circle with radius \(T_{\mathrm{cp}}\), corresponding to the mean tangential force. The expression for the work over one revolution through the mean indicated pressure is expressed as follows:

\[ W_1 = 4R\,p_i . \tag{5} \]

The expression for the work through the mean tangential force will be

\[ W_2 = 2\pi R\,T_{\mathrm{cp}} . \tag{6} \]

By the law of conservation of energy, neglecting losses, \(W_1 = W_2\);

then

\[ 4R\,p_i = 2\pi R\,T_{\mathrm{cp}} . \tag{7} \]

Hence

\[ T_{\mathrm{cp}} = \frac{2p_i}{\pi}. \tag{8} \]

With equal scales of forces in the diagrams: indicator, working-pressure, and flywheel diagrams, the number of millimeters corresponding to \(p_i\) must be multiplied by 2 and divided by \(\pi\); then in millimeters we obtain the radius \(T_{\mathrm{cp}}\). The intersection of the circle of radius \(T_{\mathrm{cp}}\) with the hodograph of the vector of the tangential force gives us the crank rotation angles corresponding to excess and deficiency of tangential force during one revolution. The excesses and deficiencies of the tangential forces formed by the hodographs are hatched.

Within the angle \(\angle \alpha_1 = \angle 1_1' O 7_1'\) — excess \(+f_1'\)
″ ″ \(\angle \alpha_2 = \angle 7_1' O 13_1'\) — deficiency \(-f_2'\)
″ ″ \(\angle \alpha_3 = \angle 13_1' O 19_1'\) — excess \(+f_3'\)
″ ″ \(\angle \alpha_4 = \angle 19_1' O 1_1'\) — deficiency \(-f_4'\)

The negative tangential forces obtained from the corresponding negative working pressures, as was said earlier, do not lie on the posi-

positions of the crank radius, but turned by \(180^\circ\) in the direction of rotation. At the areas \(-f_2'\) and \(-f_4'\), the negative tangential forces are reduced to the corresponding positions of the cranks. On the area \(-f_2'\) there is a segment (dotted), obtained from the negative tangential forces laid off from the circumference of radius \(T_{cp}\) within the angle \(\angle 9_1' O 12\); the same holds for the area \(-f_4'\), but within the angle \(\angle 22_1' O 24\). The determination of the greatest excess or deficient work during one revolution is reduced to the following:

1. Denoting by \(t_f\) the variable vector, and by \(t_{cp}\) the mean vector of the areas \(f_1'\), \(f_2'\), \(f_3'\), and \(f_4'\), we have

\[ \left. \begin{aligned} +\, t_{cp_1} &= \frac{1}{n_1}\sum t_{f_1},\\ -\, t_{cp_2} &= \frac{1}{n_2}\sum t_{f_2},\\ +\, t_{cp_3} &= \frac{1}{n_3}\sum t_{f_3},\\ -\, t_{cp_4} &= \frac{1}{n_4}\sum t_{f_4}. \end{aligned} \right\} \tag{9} \]

In the equations, the value \(n\) is the number of variable vectors \(t_f\) taken. The excess or deficient work is expressed as follows:

\[ \left. \begin{aligned} +\, A_1 &= \alpha_1 R t_{cp_1} F,\\ -\, A_2 &= \alpha_2 R t_{cp_2} F,\\ +\, A_3 &= \alpha_3 R t_{cp_3} F,\\ -\, A_4 &= \alpha_4 R t_{cp_4} F. \end{aligned} \right\} \tag{10} \]

The values \(\alpha_1\), \(\alpha_2\), \(\alpha_3\), and \(\alpha_4\) are in radians, \(F\) is the piston area in \(\text{cm}^2\), \(R\) is in meters; then the calculated work will be in kilogram-meters.

Knowing, then, the greatest of the excess or deficient areas of work, we determine the size of the flywheel in the usual manner.

1. For greater accuracy, the number of divisions should be not less than 16, preferably 24–32 parts. ↩↩

2. Analogously to the kinematic construction of the plan of velocities and accelerations. ↩

3. The scale in constructing the force plan in Fig. 1(b) is taken to be the same as in determining the force in Fig. 1(a). ↩