A. V. Kvasnikov.

DETERMINATION OF THE IMBALANCE OF RECIPROCATING MACHINES.

PREFACE.

The present article is a supplement, in methodological [[unclear: form/part]], to the author’s work, the origin of which is evident from the following extract from the preface to a manuscript kept in the Laboratory of Heat Engines at the Tomsk Technological Institute.

“...Below the author gives general expressions for determining the unbalanced forces and moments arising in the most widespread types of multi-cylinder reciprocating machines. These expressions were obtained by him as early as the beginning of 1919, when, on the basis of a resolution of the Mechanical Faculty of the Tomsk Technological Institute, he had to develop, on the assignment of Prof. V. G. Karnenko, the topic ‘Balancing of masses in internal-combustion engines according to V. Arshaulov and O. Kölsch.’ These expressions were later taken by him as the basis for the systematic presentation and supplementation of a considerable number of particular cases [[unclear: examined]] in the first part of Prof. G. Karnenko’s work, ‘Balancing of masses in internal-combustion machines.’ However, the author, who at that time did not set himself the task of developing the equations of free forces and moments, considered the question unfinished and requiring separate consideration. In carrying out his solution, the author also makes a small addition, consisting in the determination of the free transverse moment caused by the motion of the connecting rod. Taking into account the nonuniformity of rotation of the shaft, and also the application of expressions of the complete dependence of the free forces and moments on the angle of rotation of the crank, gives the expressions a complicated external form, which is valuable, however, in that it gives an understanding of the complete picture of the forces arising. Practice requires simplicity and only sufficient accuracy; therefore, further on, examples indicate how, after a number of possible simplifications, the expressions acquire a form convenient for operations with them. 30 August 1920. Tomsk, Gertsenovskaya 31.”

A. V. Kvasnikov.

January 1923.

Conditions and notation

If there are no reservations, the center of gravity of the reciprocating part of the machine is assumed to coincide with the piston axis; the center of gravity of the piston is assumed to coincide with the axis of the cylinder, and the direction of its motion passes through the axis of the shaft. The arrangement of the cranks is symmetrical with respect to their straight lines [[unclear: end of sentence obscured]].

The following letter designations are most often used:

$\theta$ . . . . . . . . angle of rotation of the crank [[unclear: first]].

$\varphi$ . . . . . . . . [[unclear: angle of rotation]] of the [[unclear: counterweight]].

$\beta$ . . . . . . . . angle of deflection of the connecting rod.

$\omega$ . . . . . . . . angular velocity of the shaft.

$\xi$ . . . . . . . . angular acceleration $\dfrac{d\omega}{dt}$.

$r$ . . . . . . . . radius of the crank.

$l$ . . . . . . . . length of the connecting rod.

$\mu$ . . . . . . . . $\dfrac{r}{l}$.

$s_a$ . . . . . . . . distance of the center of gravity of the connecting rod from the axis [[unclear: continuation obscured]].

$s_b$ . . . . . . . . [[unclear: distance of the center of gravity of the connecting rod from the axis]] [[unclear]].

$m$ . . . . . . . . mass of the piston and [[unclear: of the parts]] connected [[unclear]].

$m_l$ . . . . . . . . mass of the connecting rod.

$m_r$ . . . . . . . . mass of the crank, referred to [[unclear: the radius]] of its [[unclear: center of gravity]]; its true static moment of the masses of the crank with respect to the shaft axis.

$\left.\begin{array}{c}
X\
Y\
Z_0
\end{array}\right}$ Components of the free forces.

$\left.\begin{array}{c}
M_x\
M_y\
M_z
\end{array}\right}$ Components of the free moments.

$\mathcal{M}$ . . . . . . . . Static moment.

$J$ . . . . . . . . Moment of inertia.

The signs of the free forces and moments are adopted according to the scheme given in Fig. 1 for a single mechanism, and in Fig. 2 for their combination.

General concepts

In a piston machine we have two groups of bodies: the first group consists of parts performing periodic, definite motions relative to the second group, which must be motionless on the foundation. We shall call the latter the frame. The periodicity of the motion of the masses serves as a sign of the presence of relative accelerations of bodies of the 1st and 2nd groups, and thus as a sign of the existence of forces of interaction between them.

Taking fixed coordinate axes connected with the frame of the machine, we obtain the equations of motion of an individual mass in the general form:

\[ x=\sum P_n \operatorname{sn}(At+p)_n,\qquad y=\sum Q_n \operatorname{sn}(At+q)_n,\qquad z=\sum R_n \operatorname{sn}(At+r)_n, \]

where $n=0,1,2,3,\ldots$.

for the inertia forces

\[ X=-m\frac{d^{2}x}{dt^{2}}, \quad \text{etc., we find} \]

\[ X=m\sum N_n \operatorname{Cs} n(At+p), \quad Y=m\sum S_n \operatorname{Cs} n(At+q), \quad Z=m\sum T_n \operatorname{Cs} n(At+r). \]

For the frame, considered separately, these forces will be external. In considering the question whether the frame will tend to move translationally, we form the sums

\[ X_0=\sum X,\quad Y_0=\sum Y,\quad Z_0=\sum Z, \]

or

\[ X_0=\sum_{m_1}^{m'} m\sum N_n \operatorname{Cs} n(At+p), \]

\[ Y_0=\sum_{m_1}^{m'} m\sum S_n \operatorname{Cs} n(At+q), \]

\[ Z_0=\sum_{m_1}^{m'} m\sum T_n \operatorname{Cs} n(At+q). \]

In practice, summation over \(n\) is often reduced to the addition of only a few terms, since the coefficients \(N\), \(S\), and \(T\) may decrease rapidly.

From these sums the magnitudes of the components of the unbalanced, or free, force are determined. Since this force does not necessarily pass through the center of gravity of the frame, unbalanced, or free, moments also appear; these should be determined from the formulas

\[ M_x=\sum (Yz-Zy), \quad M_y=\sum (Zx-Xz), \quad M_z=\sum (Xy-Yx). \]

Since the frame is usually firmly connected with the upper part of the foundation, and the latter, like the surrounding ground, does not remain absolutely immobile, then in general the frame, under the action of free forces and moments, acquires certain motions, depending chiefly on the nature of the elastic properties of the foundation. Thus the free forces and moments, in the general case, die away, doing work in the masses constituting the foundation and the soil beneath it.

The moving masses of greatest magnitude in a piston machine are the masses of the parts of the crank mechanism, and usually in determining the imbalance only these are taken into account.

As a first approximation, without taking exactly into account the influence of the connecting rod, all the masses may be divided into those moving rectilinearly back and forth, assigned to the piston pin, and those moving in a circle, conventionally assigned to the crank pin. The action of these masses is easy to trace from Fig. 3, which represents the arrangement of the moving masses in a vertical type of multi-cylinder engine. As can be seen, the moving masses exist only along two axes, \(x\) and \(y\). Therefore it is evident that \(Z_0=0\), and only the free forces \(X_0\) and \(Y_0\) can have a finite value. The origin of coordinates is taken as passing through the center of gravity of the frame, and the moments of the inertia forces of the individual masses with respect to the center of gravity give, in sum, the free moments rotating the frame about the axes \(x\), \(y\), and \(z\). From the drawing it is not difficult to see that \(M_y\) is produced by the masses assigned both to the piston and to the crank pin. \(M_x\) and \(M_z\) are obtained only at the expense of the rotating masses, since the pistons are constantly in the plane \(y=z\). If the action of the connecting rod is not taken into account, then it may be considered that \(M_z\) exists as a consequence of the presence of angular accelerations of the shaft and of the masses rigidly connected with it. \(M_y\) and \(M_z\) are often called longitudinal moments, and \(M_x\) transverse.

1. Free forces and moments in the connecting-rod mechanism

Connecting rod with \(\beta = 0\).

Let us first consider the simplest case, when

\[ \beta = \mathrm{const} = 0, \]

and let us also assume that

\[ \omega = \mathrm{constans}. \]

The forces acting on the frame as a result of the accelerations of the crank masses are determined most simply. For an elementary mass of the crank, situated at a distance \(\rho\) from the axis of rotation, the inertia force has the magnitude

\[ \Delta m \cdot \rho \omega^2. \]

Summing them over the whole crank, we obtain the centrifugal force of inertia of the entire crank:

\[ R_1 = \sum \Delta m \rho \omega^2 = \omega^2 \sum \Delta m \rho = \mathfrak{M}\omega^2, \]

and, for \(\mathfrak{M} = m_r r\),

the whole action of the crank may be replaced by some concentrated mass located at the crank radius, of mass \(m_r\). As a result, when determining the unbalance, we must consider the motion of a piston having mass

\[ m = m_k + m_l + m_r, \]

where \(m_k\) and \(m_l\) are situated at its ends. The equations of motion of the connecting rod as a plane system will be

\[ \sum X = \sum m \frac{d^2 x}{dt^2}, \qquad \sum Y = \sum m \frac{d^2 y}{dt^2}, \]

\[ \sum (Xy - Yx) = \sum m \left( y \frac{d^2 x}{dt^2} - x \frac{d^2 y}{dt^2} \right). \]

It is obvious that the forces acting on the supports are determined, in magnitude, by the right-hand sides of these equations taken with the opposite sign. These will be the free forces. Let us determine them.

For any material element of the connecting rod \(s\), cut out by two planes perpendicular to the axis \(AB\), its coordinates of the center of gravity are

\[ x = r \cos \theta + b, \qquad y = r \sin \theta, \tag{2} \]

whence the velocity of \(\Delta m_i\), assumed concentrated in it, is

\[ x' = - r \omega \sin \theta, \qquad y' = r \omega \cos \theta, \]

and the acceleration is

\[ x'' = - r \omega^2 \cos \theta, \qquad y'' = - r \omega^2 \sin \theta. \]

Thus, for the inertia forces of this elementary mass \(\Delta m_i\), we shall have the expressions

\[ \Delta X = \Delta m \, r \omega^2 \cos \theta, \qquad \Delta Y = \Delta m \, r \omega^2 \sin \theta. \]

To determine the free forces, we sum these expressions over the entire connecting rod from \(A\) to \(B\):

\[ X_0 = \sum \Delta X = r \omega^2 \cos \theta \sum \Delta m, \qquad Y_0 = \sum \Delta Y = r \omega^2 \sin \theta \sum \Delta m. \]

therefore,

\[ \begin{aligned} X_\theta&=mr\omega^2\cos\theta,\\ Y_\theta&=mr\omega^2\sin\theta,\\ m&=m_i+m_l+m_r, \end{aligned} \tag{3} \]

From expressions (3) it follows that the resultant free force \(R\) is constant in magnitude and has the direction of the crank radius,

\[ R=mr\omega^2. \tag{4} \]

For the sum of the [[unclear: free]] moments about the \(z\)-axis we obtain

\[ \sum \Delta M_z = \sum \Delta m r\omega^2 \cos\theta \, r\sin\theta - mr\omega^2\sin\theta \, C s\theta, \]

\[ =mr^2\omega^2\sin\theta\, C s\theta + r\omega^2\sin\theta\sum b\cdot \Delta m; \]

but \(\sum b\cdot \Delta m\) is the static moment of the masses of the connecting rod relative to the point [[unclear: \(C\), the crank pin]]; therefore

\[ \sum b\,\Delta m = mS_b = \mathcal{M} b; \]

determining further \(M_z\) from its expression

\[ M_z=\sum \Delta X\,Y-\sum \Delta Y\,x, \]

as a result we obtain

\[ M_z=-\mathcal{M} b r\omega^2\sin\theta. \tag{5} \]

Hence, if \(\theta=\dfrac{\pi}{2}\) and \(\theta=\dfrac{3\pi}{2}\), we obtain, in one revolution, two absolute maxima

\[ M_z=-\mathcal{M} b r\omega^2. \]

In such a mechanism both the force and the moment can easily be eliminated by the action of balancing weights. In fact, \(R\) (4) can be balanced by adding a mass \(m_1\) situated symmetrically to the mass \(m\), diametrically opposite to it; here the condition that determines it is only

\[ m_1 r_1 = mr. \]

When \(M_z=0\), obviously one of the factors on the right-hand side of equation (5) is equal to zero. In practice this will be when

\[ \mathcal{M}b=0 \quad \text{or} \quad S_b=0; \]

Thus, in order that no free moment exist, the mass of the connecting rod must be arranged along its axis so that its center of gravity coincides with the crank pin. Fig. 5 schematically illustrates such balancing.

Ordinary crank mechanism.

Let us have, with the previous notation, the mechanism shown in Fig. 6. We assume the rotation of the shaft to be nonuniform, as occurs in machines. The coordinates of the connecting-rod element will be

\[ \begin{aligned} x&=r\cos\theta+b\cos\beta,\\ y&=a\sin\beta=a+\sin\theta, \end{aligned} \tag{6} \]

where

\[ \mu=\frac{r}{l}=\frac{\sin\beta}{\sin\theta}. \]

Excluding the angle \(\beta\) from the expression for \(x\), we obtain

\[ x = r\cos\theta + b\sqrt{1-\mu^2\sin^2\theta} = r\cos\theta + b\left( 1-\frac{1}{2}\mu^2\sin^2\theta -\frac{1}{8}\mu^4\sin^4\theta -\frac{1}{16}\mu^6\sin^6\theta -\ldots \right). \]

but has

\[ \operatorname{Sin}^{2}\theta=\frac{1}{2}(1-\operatorname{Cs}2\theta), \]

\[ \operatorname{Sin}^{4}\theta=\frac{1}{8}(\operatorname{Cs}4\theta-4\operatorname{Cs}2\theta+3), \]

\[ \operatorname{Sin}^{6}\theta=-\frac{1}{32}(\operatorname{Cs}6\theta-6\operatorname{Cs}4\theta+15\operatorname{Cs}2\theta-10), \]

etc., then

\[ \begin{aligned} x &=r\operatorname{Cs}\theta+b\Bigg[ \left(1-\frac{1}{4}\mu^{2}-\frac{3}{64}\mu^{4}-\frac{5}{256}\mu^{6}-\ldots\right) \\ &\qquad+\operatorname{Cs}2\theta\left(\frac{1}{4}\mu^{2}+\frac{15}{512}\mu^{6}+\ldots\right) -\operatorname{Cs}4\theta\left(\frac{1}{64}\mu^{4}+\frac{3}{256}\mu^{6}+\ldots\right) \\ &\qquad+\operatorname{Cs}6\theta\left(\frac{1}{512}\mu^{6}+\ldots\right)+\ldots \Bigg], \end{aligned} \]

or, more simply,

\[ x=r\operatorname{Cs}\theta+b\left(C_{0}+C_{2}\operatorname{Cs}2\theta+C_{4}\operatorname{Cs}4\theta+\ldots C_{n}\operatorname{Cs}n\theta+\ldots\right). \tag{7} \]

We next determine the acceleration of the element \(\theta\).

\[ x'=\frac{dx}{dt}=\frac{d\theta}{dt}\frac{dx}{d\theta}=\omega\frac{dx}{d\theta}. \]

But since

\[ \frac{d\omega}{dt}=\varepsilon, \]

is the angular acceleration, we shall finally obtain for the acceleration of the element the expression

\[ x''=\omega^{2}\frac{d^{2}x}{d\theta^{2}}+\varepsilon\frac{dx}{d\theta}. \]

After substitution we obtain:

\[ x'=-r\omega\operatorname{Sin}\theta-b\omega(2C_{2}\operatorname{Sin}2\theta+4C_{4}\operatorname{Sin}4\theta+\ldots nC_{n}\operatorname{Sin}n\theta+\ldots), \]

and

\[ \begin{aligned} x''={}&-r\omega^{2}\operatorname{Cs}\theta -b\omega^{2}(4C_{2}\operatorname{Cs}2\theta+16C_{4}\operatorname{Cs}4\theta+\ldots n^{2}C_{n}\operatorname{Cs}n\theta+\ldots) \\ &-r\varepsilon\operatorname{Sin}\theta -b\varepsilon(2C_{2}\operatorname{Sin}2\theta+4C_{4}\operatorname{Sin}4\theta+\ldots nC_{n}\operatorname{Sin}n\theta+\ldots). \end{aligned} \tag{8} \]

Similarly, for the \(y\)-axis we obtain:

\[ y=a+\omega\operatorname{Cs}\theta, \]

\[ y''=-r\omega^{2}\operatorname{Sin}\theta+a\omega\varepsilon\operatorname{Cs}\theta. \]

The elementary forces of inertia are expressed as

\[ \Delta X=-\Delta m\,x'', \qquad \Delta Y=-\Delta m\,y'', \]

or, after substitution,

\[ \left\{ \begin{aligned} \Delta X={}&\Delta m\,r\omega^{2}\operatorname{Cs}\theta +\Delta m\,b\omega^{2}(4C_{2}\operatorname{Cs}2\theta+\ldots) \\ &+\Delta m\,r\varepsilon\operatorname{Sin}\theta +\Delta m\,b\varepsilon(2C_{2}\operatorname{Sin}2\theta+\ldots),\\ \Delta Y={}&\Delta m\,a r\omega^{2}\operatorname{Sin}\theta -\Delta m\,a r\varepsilon\operatorname{Cs}\theta. \end{aligned} \right. \tag{10} \]

For the entire mass of the connecting rod, situated from \(A\) to \(B\),

\[ \begin{aligned} X_{0} &=\sum \Delta X =r\omega^{2}\operatorname{Cs}\theta\sum\Delta m +\omega^{2}(4C_{2}\operatorname{Cs}2\theta+\ldots n^{2}C_{n}\operatorname{Cs}n\theta+\ldots)\sum\Delta m\,b \\ &\quad+r\varepsilon\operatorname{Sin}\theta\sum\Delta m +\varepsilon(2C_{2}\operatorname{Sin}2\theta+\ldots nC_{n}\operatorname{Sin}n\theta+\ldots)\sum\Delta m\,b, \end{aligned} \]

since

\[ \sum\Delta m=m \quad\text{and}\quad \sum\Delta m\,b=m\delta=\mathfrak{M}_{cb}. \]

\[ X_0=m r\omega^2 \operatorname{Cs}\theta +mS_b\omega^2(4C_2\operatorname{Cs}2\theta+\cdots) +mr\xi\operatorname{Sin}\theta +mS_a\xi(2C_2\operatorname{Sin}2\theta+\cdots), \]

likewise, for \(Y_0\) we obtain

\[ Y_0=\sum \Delta Y =m_b\omega^2\operatorname{Sin}\theta\sum \Delta ma -m_b\xi\operatorname{Cs}\theta\sum \Delta ma, \]

or

\[ Y_0=mS_b\omega^2\operatorname{Sin}\theta-mS_a\xi\operatorname{Cs}\theta; \]

let us present \(X_0\) and \(Y_0\) in a more convenient form. Put

\[ C_n=\mu B_n. \]

Then, taking the terms outside the brackets in the expression for \(X_0\), we shall have:

\[ \begin{aligned} X_0={}&\left(m+\frac{mS_b}{l}\right)r\omega^2\operatorname{Cs}\theta +\frac{mS_b}{l}\left(4B_2\operatorname{Cs}2\theta+\cdots+\right)\\ &+\left(\frac{mS_a}{l}+\frac{mS_b}{l}\right)r\xi\operatorname{Sin}\theta +\frac{mS_b}{l}r\xi\left(2B_2\operatorname{Sin}2\theta+\cdots\right); \end{aligned} \]

let us call the coefficients at \(\operatorname{Cs}n\theta\) by \(A_n\), and introduce the notation

\[ m_a=\frac{mS_b}{l}, \qquad m_k=\frac{mS_a}{l}. \tag{11} \]

Then we finally obtain:

\[ \begin{aligned} X_0={}&m_b r\omega^2\operatorname{Cs}\theta +m_a r\omega^2\left(\operatorname{Cs}\theta +A_2\operatorname{Cs}2\theta +A_4\operatorname{Cs}4\theta+\cdots A_n\operatorname{Cs}n\theta+\cdots\right)\\ &+m_b r\xi\operatorname{Sin}\theta +m_a r\xi\left(\operatorname{Sin}\theta +\frac{1}{2}A_2\operatorname{Sin}2\theta+\cdots +\frac{1}{n}A_n\operatorname{Sin}n\theta+\cdots\right); \end{aligned} \tag{12} \]

also

\[ Y_0=m_b r\omega^2\operatorname{Sin}\theta -m_k r\xi\operatorname{Cs}\theta. \tag{13} \]

The factor \(f(\theta)\) is determined by the character of the variation of the indicator pressures, of the useful resistances, and by the magnitude and arrangement of the moving masses. The terms not containing \(\xi\) give the values of components independent of the form of the indicator diagram.

From the equations it is further evident that, in determining the free forces in the crank mechanism, one may consider only one link of its mechanism, the mass of which (see the case with \(\beta=0\))

\[ m=m_k+m_l+m_r. \]

This mass may be considered as distributed over its ends \(A\) and \(B\) according to equalities (11). From the mass \(m_b\), concentrated at the crank pin, centrifugal and tangential inertia forces are obtained; in the whole mechanism they are free, and their resultant is easily obtained in the form

\[ R=m_b r\sqrt{\omega^4+\xi^2}. \tag{14} \]

The mass \(m_a\) creates a free force directed along one straight line. In magnitude it may be regarded as composed of an infinite series of forces of the 1st, 2nd, 4th, etc., even orders. It is easy to see that the terms determining the magnitudes of the forces of the separate orders have an entirely identical structure. The force of any \(n\)-th order may be represented as projected onto the direction of motion \(A\) of the centrifugal force of some mass rotating on radius \(r\) with angular velocity \(n\omega\). Its magnitude will be, for example,

\[ A_n m_a r\omega^2\operatorname{Cs}n\theta =\frac{A_n m_a}{n^2}\,r\,(n\omega)^2\operatorname{Cs}n\theta =m_n r\omega_a^2\operatorname{Cs}n\theta. \]

that is, for an independent force this is the same as

\[ A_n \frac{m_a}{n}\, r \sin n\theta = \frac{A_n m_a}{n^2}\, r\, \mu \sin n\theta [[\unclear]] \]

It may be assumed that the mass is unchanged in its magnitude. Then

\[ A_n m_a r \omega^2 \cos n\theta = m_a \frac{A_n r}{n^2}(n\omega)^2 \cos n\theta = m_a [[\unclear]] \]

Therefore the inertia forces of rectilinearly moving masses should be represented in the general form as follows:

\[ X_0=\sum m_a r_n \omega_n^2 \cos n\theta+\sum m_a b_n \sin n\theta . \]

Here

\[ r_n=\frac{A_n r}{n^2}, \qquad \omega_n=n\omega, \]

where for the force of the first order \(A_n=1\), and for the rest according to Table 1. Usually the forces of the first and second orders are of practical importance, more rarely those of the fourth. The forces of the remaining orders are so insignificant that, in balancing, they may be disregarded and are determined only for the characterization of the machine being studied. From drawing 7 one can see the relative magnitudes of the forces of the 2nd, 4th, and 6th orders for \(\mu=0.25\). The force of the 4th order is roughly 10 times smaller than the force of the 2nd order; consequently, in ordinary practice it is not taken into account. The coefficients \(A\) are expressed by the following equalities:

\[ \begin{aligned} A_2&=\left(\mu+\frac{1}{4}\mu^3+\frac{5}{128}\mu^5+\cdots\right),\\ A_4&=-\left(\frac{1}{4}\mu^3+\frac{3}{16}\mu^5+\cdots\right),\\ A_6&=\left(\frac{9}{128}\mu^5+\cdots\right). \end{aligned} \tag{16} \]

For the values most commonly found in piston machines, the coefficients are given in Table 1.

Table 1

\(\mu\)	\(\dfrac{1}{4}\)	\(\dfrac{1}{4.5}\)	\(\dfrac{1}{5}\)	\(\dfrac{1}{6}\)
\(A_2\)	0.2540	0.2250	0.2020	0.1678
\(A_4\)	−0.0041	−0.0028	−0.0020	−0.0012
\(A_6\)	0.0001	0.0000	0.0000	0.0000

From it one can see that, relative to the greatest free force of the first order, the maximum free force of the 2nd order is approximately from 17% to 25%, depending on the value of \(\mu\). Correspondingly, for the force of the 4th order we have limits of 0.1%—0.4%, and for the 6th—0.00%—0.01%.

The magnitudes of these forces are also affected by the quantities \(r\), \(m_a\), and \(\omega\). With respect to \(m_a\), and chiefly \(\omega\), there arises the question of taking the forces of higher orders as real.

On the Balancing of Free Forces

By applying counterweights, the free forces arising from the action of the mass \(m_b\) may be balanced. If a counterweight of mass \(m_1\) is placed on the crank \(r_1\), taking

\[ m_c r_1 = m_b r_1, \]

and choosing the corresponding \(r_1\) and \(r_1\), we obtain two equal and oppositely directed centrifugal forces, and also tangential forces; the remaining free forces of the masses \(m_a\) act along a constant straight line. Obviously, by introducing an additional mass \(m_2\), one can partly or completely balance the force of the first order \(m_a r \omega^2 \operatorname{Cs}\theta\), but in so doing this additional mass gives a free component along the \(Y\)-axis equal in magnitude to all the values of the balanced one. Arranging the masses along the connecting rod in such a way that its center of gravity lies at \(A\) (Fig. 8), i.e., putting in equations (11)

\[ S_b = 1,\quad S_a = 0, \]

we obtain from (12) and (13)

\[ X_0 = m r \omega^2 \left\{ A_n \operatorname{Cs} n\theta + m r \xi \sum_{n} \frac{1}{n} A_n \operatorname{Sin} n\theta \right\}, \qquad Y_0 = 0 . \tag{17} \]

Thus, with the indicated arrangement of the masses along the axis of the connecting rod, it is possible to obtain a free force acting only in one direction.

In order, as far as possible, to balance the free forces, it is advantageous to arrange the masses in such a way that \(S_b\) decreases; in doing so \(m_a\) decreases, while the action of the increasing \(m_b\) can be annulled by a counterweight. In the limiting case,

\[ S_b = 0,\quad S_a = 1; \]

then equations (12) and (13) give

\[ \begin{aligned} X_0 &= m r \omega^2 \operatorname{Cs}\theta + m r \xi \operatorname{Sin}\theta,\\ Y_0 &= m r \omega^2 \operatorname{Sin}\theta - m r \xi \operatorname{Cs}\theta . \end{aligned} \tag{19} \]

According to the preceding, these forces are balanced without remainder by a counterweight. Such a case of perfect balancing of the free forces is shown schematically in Fig. 9.

This balancing, however, is impractical, since the additional mass \(m_2\) is either too large or stands too far from the center of the crank, which is inconvenient from the design standpoint. Moreover, as will be seen below, the free transverse moment \(M_z\) increases, owing to the increase in the moment of inertia of the masses of the connecting rod relative to its center of gravity.

Transverse Moment \(M_z\)

The direction of the inertia forces of the piston always passes through the axis of the moments \(z\) (see Fig. 6); therefore it may be stated in advance that these forces will not give free moments. The same is true for the centripetal forces.

forces of the masses of the mechanism. With an uneven motion, the resulting moment of the forces of the masses rigidly connected with the shaft will be expressed as

\[ M_z = J_z \,\ddot{\theta}; \]

where \(J_z\) is the polar moment of inertia of the rotating masses.

The masses of the connecting rod also possess a free moment; consequently, it is necessary to represent the state as two parts: one—the rotation of the connecting rod about its center of gravity, and the other—the translational motion of the latter; moreover, the direction of motion of the latter does not always pass through [[unclear]]. Let us determine the moment of the forces of inertia of the connecting rod \(M'_z\).

\[ M'_z=\Sigma(\Delta X\,y-\Delta Y\,x), \]

for an arbitrary element \(e\) (Fig. 6), the forces of inertia are determined by equations (10). Substituting their values into the expression for \(M'_z\), we obtain, grouping the terms in pairs, which we shall combine below:

\[ \begin{aligned} M'_z={}& \Sigma \Delta m\,r\omega^2 \Cs\theta \cdot a\mu\Sin\theta -\Sigma \Delta m\,a\mu\omega^2\Sin\theta \cdot r\Cs\theta \\ &+\Sigma \Delta m\,b\omega^2(4C_2\Cs2\theta+\cdots)\,a\mu\Sin\theta \\ &-\Sigma \Delta m\,a\mu\omega^2\Sin\theta \cdot b(C_0+\cdots+C_n\Cs n\theta+\cdots) \\ &+\Sigma \Delta m\,b\varepsilon(2C_2\Sin2\theta+\cdots)\,a\mu\Sin\theta \\ &+\Sigma \Delta m\,a\mu\varepsilon\Cs\theta \cdot b(C_0+\cdots+C_n\Cs n\theta+\cdots) \\ &+\Sigma \Delta m\,r\varepsilon\Sin\theta\; [[unclear]] \end{aligned} \tag{19} \]

The first pair [[unclear]]:

\[ M'_1=-\mu\omega^2\Sin\theta\,[ -C_0+3C_2\Cs2\theta+\cdots +(n^2-1)C_n\Cs n\theta+\cdots]\,\Sigma ab\,\Delta m . \]

Introduce the notation:

\[ \Sigma ab\,\Delta m=H; \]

then

\[ M'_1=-H\mu\omega^2\Sin\theta\,[C_0+3C_2\Cs2\theta+\cdots -(n^2-1)C_n\Cs n\theta+\cdots]. \tag{21} \]

Expression (21) can be given another form. Take \(\mu\Sin\theta\) outside the brackets; then we obtain a series again with variable part of the form \(\Sin\theta\Cs n\theta\). By the known trigonometric identity

\[ \Sin\alpha\Cs\beta=\frac12\Sin(\alpha+\beta)+\frac12\Sin(\alpha-\beta) \]

one may write

\[ 2\Sin\theta\Cs n\theta=\Sin(n+1)\theta-\Sin(n-1)\theta. \]

Applying this identity to the terms in brackets and combining sines of equal orders, we finally obtain, instead of (21):

\[ M'_1=-H\omega^2(D_1\Sin\theta+D_3\Sin3\theta+D_5\Sin5\theta+\cdots+D_k\Sin k\theta) \tag{22} \]

or

\[ M'_1=-H\omega^2\Sigma D_k\Sin k\theta, \tag{22A} \]

where \(k=1,3,5,\ldots(2n+1)\). The values of \(D_k\) are determined from the equalities

\[ \begin{aligned} D_1&=\left(\mu+\frac18\mu^3+\frac{3}{64}\mu^5+\frac{25}{1024}\mu^7+\cdots\right),\\ D_3&=-\left(\frac38\mu^3+\frac{27}{128}\mu^5+\frac{185}{1024}\mu^7+\cdots\right),\\ D_5&=\left(\frac{15}{128}\mu^5+\frac{125}{1024}\mu^7+\cdots\right),\\ D_7&=-\left(\frac{35}{1024}\mu^7+\cdots\right). \end{aligned} \tag{23} \]

denoting by \(D_i\) the values given in Table II.

The coefficient \(H\) is determined as follows:

\[ H=\sum ab\,\Delta m=\sum a(l-a)\Delta m = l\sum a\,\Delta m-\sum a^2\,\Delta m . \]

Denoting the static moment of the mass of the connecting rod with respect to point \(A\) by the letter \(\mathcal{M}_a\), and the moment of inertia with respect to the same point by \(J_a\), we may write

\[ H=\mathcal{M}_a l-J_a ; \tag{24} \]

in exactly the same way we obtain

\[ H=\mathcal{M}_b l-J_b . \]

If \(J_s\) is the moment of inertia of the connecting rod with respect to its center of gravity,

\[ J_a=J_s+m_1 S_a^2 . \]

Then, instead of (24), we may obtain

\[ H=\mathcal{M}_a l-J_s-m_1 S_a^2 =(S_a+S_b)m_1S_a-J_s-m_1S_a^2, \]

and further—

\[ H=m_1S_aS_b-J_s , \tag{25} \]

or also

\[ H=m_1S_aS_b-m_1k^2 , \tag{26} \]

where \(k\) is the radius of gyration.

Now expression (22) may be written as

\[ M=\left(J_s-m_1S_aS_b\right)\omega^2\sum D\sin k\varphi . \tag{27} \]

According to this formula it is easy to compile a table of moments for an arbitrary connecting rod, taking only those quantities which are contained in the parentheses as constant.

If, in a given mechanism, there is a whole number of repetitions of these in one revolution of the crank, then both here and for free forces the frequencies or moments of any order are expressed by twice the number denoting the order.

Table II

\(k\)	\(\dfrac{1}{4}\)	\(\dfrac{1}{4.5}\)	\(\dfrac{1}{5}\)	\(\dfrac{1}{6}\)
\(D_1\)	0.2826	0.2236	0.2010	0.1678
\(D_3\)	−0.0060	−0.0042	−0.0032	−0.0017
\(D_5\)	0.0001	0.0001	0.0000	0.0000
\(D_7\)	−0.0000	−0.0000	−0.0000	−0.0000

Composing the following terms of series (10):

\[ \frac{M_{II}}{H\varepsilon} = \Bigl[ \sin\theta\,(2C_2\sin 2\theta+\cdots+nC_n\sin n\theta+\cdots) +\cos\theta\,(C_0+\cdots+ C_n\cos n\theta+\cdots) \Bigr]\sum ab\,\Delta m, \]

Introducing into the corresponding brackets \(\sin\theta\), \(\cos\theta\) by the identities,

\[ 2\sin\theta\sin n\theta=-\cos(n+1)\theta+\cos(n-1)\theta, \]

\[ 2\cos\theta\cos n\theta=\cos(n+1)\theta+\cos(n-1)\theta. \]

Then, after adding the coefficients at cosines of equal angles, we obtain for \(M_{II}\) the expression:

\[ M_{II} = H\varepsilon \left( \frac{1}{1}D_1\cos\theta + \frac{1}{3}D_3\cos3\theta + \frac{1}{5}D_5\cos5\theta +\cdots \right), \tag{28} \]

or

\[ M_{II} = H\varepsilon \sum \frac{1}{k}D_k\cos k\theta . \]

Finally, the last pair of terms of equation (19) gives

\[ M_{III}=\ [[unclear: expression]] \]

or, with

\[ \frac{m_b S_a}{1}=m_b, \]

\[ M_{III}=m_b r^2\varepsilon . \tag{30} \]

Now, instead of (19), one may write:

\[ M^*=M_I+M_{II}+M_{III}, \]

and, after substitution:

\[ M_z^* = -H\omega^2\sum D_k\sin k\theta + H\varepsilon\sum\frac{1}{k}D_k\cos k\theta + m_b r^2\varepsilon . \]

This will be the final expression for the free moment due to the reciprocating motion of the connecting rod.

The total free transverse moment will have the value

\[ M_z=M_z^*+M_z', \]

or

\[ M_z = -H\omega^2\sum D_k\sin k\theta + H\varepsilon\sum\frac{1}{k}D_k\cos k\theta + m_b r^2\varepsilon + J_z\varepsilon; \]

combining the last two terms, we obtain

\[ M_z = -H\omega^2\sum D_k\sin k\theta + H\varepsilon\sum\frac{1}{k}D_k\cos k\theta + J_z\varepsilon, \tag{32} \]

where \(J_z\) is the polar moment of the rotating masses, including \(m_b\), concentrated at the crank pin. Expression (32) is final for the free transverse moment. The greatest value in it is usually possessed by the last term \(J_z\varepsilon\). In the first two terms of this expression, moments only of the first order—and more rarely of the third—can have practical significance. From Table I it is easy to see that, from the maximum of the first-order moment, the moments of the third and fifth orders amount, respectively, to: \(2\), \(4\%\)—\(1\%\) and \(0.04\%\)—\(0.00\%\). In drawing 10 is shown the course of the free moments of the 1st and 3rd orders at \(\omega=\mathrm{const.}\)

On the balancing of \(M_z\)

As is known, in reciprocating machines \(\xi\) is usually not equal to zero; therefore, except for the case considered above, \(m_b r^2 \xi\) in equation (31) will be a substantial quantity. Unlike \(M_I\), which has an entirely different law of variation, it can be balanced under the condition

\[ H=0. \]

By equality (24) this condition is expressed as

\[ \mathcal{M}_a l - J_a = 0, \]

which gives

\[ l=\frac{J_a}{\mathcal{M}_a}. \tag{33} \]

In other words, the condition states that the length of the connecting rod must be equal to its reduced length as a physical pendulum with point of suspension \(A\) (also \(B\)). By equality (26), the condition \(H=0\) gives:

\[ S_a S_b = k^2, \]

\[ (1-S_a)S_a = S_a - S_a^2 = k^2, \]

and hence, for the position of the center of gravity of the balanced connecting rod,

\[ S_a=\frac{1}{2}\pm\sqrt{\frac{1}{4}-k^2}. \tag{34} \]

The remaining part of the unbalanced moment \(m_b r^2 \xi\) is less characteristic in its manifestations than the first. In magnitude it represents a very small fraction of the analogous moment \(M_z = J_z \xi\), which varies according to the same law.

As an example of the condition \(H=0\), let us consider a connecting rod with a uniformly distributed mass (Fig. 11). With \(AC=1\) and \(AB=L\),

\[ \mathcal{M}_a = m\frac{L}{2} \quad\text{and}\quad J_a=\frac{1}{3}mL^2, \]

therefore, by (33),

\[ l=\frac{J_a}{\mathcal{M}_a}=\frac{2}{3}L, \]

and by (34),

\[ S_a=\frac{1}{2}\pm\frac{1}{4},\quad S_a'=\frac{3}{4},\quad S_a''=\frac{1}{4}. \]

This means that balancing is effected when the center of gravity is located at \(G_1\) and \(G_2\), respectively, with additional masses placed along \(BC\) and along \(AC\).

In normally designed connecting rods we have

\[ m S_a S_b > J_s \quad \text{and} \quad S_b < S_a, \]

therefore, in order to reduce \(H\), and together with it also \(M_I+M_{II}\), it is necessary, by properly arranging the masses, to bring the center of gravity of the connecting rod closer to the crankpin. Complete balancing of the slider-crank mechanism, even in the absence of angular acceleration, by means of additional masses placed along the axes of the connecting rod and crank, cannot be achieved. For the disappearance of the free forces the condition is \(S_b=0\) (see drawing 9), i.e. the conditions for complete balancing would be:

\[ k^2=S_a S_b;\quad S_b=0, \]

whence

\[ k=0,\quad \text{or}\quad J_s=0, \]

which is not possible.

It is assumed that the center of the shaft has been displaced and its coordinates with respect to the newly chosen coordinate system are taken as constant and equal to \(\lambda\) and \(\eta\). Thus the equation of moments relative to the present position of the center of gravity at \(O\) is

Let us denote the coordinates of the center of the shaft \(Z\), relative to the new system having its origin at \(O\), by \(\lambda\) and \(\eta\). Then

\[ M_o=\sum\left[\Delta X(y+\lambda)-\Delta Y(x-\eta)\right], \]

whence

\[ M_o=\sum(\Delta Xy-\Delta Yx)+\sum(\Delta X\lambda+\Delta Y\eta) =M'_z+M_A, \]

where \(M'_z\) has already been found earlier (31). We determine \(M_A\):

\[ M_A=\sum \Delta X\lambda+\sum \Delta Y\eta =\lambda\sum\Delta X+\eta\sum\Delta Y. \]

But

\[ \sum\Delta X=X_0,\quad \sum\Delta Y=Y_0, \]

therefore

\[ M_A=X_0\lambda+Y_0\eta. \]

Using (12) and (18), we obtain

\[ \begin{aligned} M_A={}&\lambda m_a r\omega^2\Cos\theta [[unclear: damaged part of formula]] +\frac{1}{2}A_2\Sin 2\theta+\cdots\Big)\\ &+\lambda m_b r\omega^2\Cos\theta+\lambda m_b r\varepsilon\Sin\theta\\ &+\eta m_b r\omega^2\Sin\theta-\eta m_b r\varepsilon\Cos\theta . \end{aligned} \]

From Fig. 13 it is clear that, if

\[ \varphi=\operatorname{arctg}\frac{\lambda}{\eta} \quad\text{and}\quad \rho=\sqrt{\lambda^2+\eta^2}, \tag{35} \]

then

\[ \lambda\Sin\theta-\eta\Cos\theta=-\rho\Cos(\theta+\varphi), \]

\[ \lambda\Cos\theta+\eta\Sin\theta=\rho\Sin(\theta+\varphi). \]

Therefore the preceding expression for \(M_A\) is converted to

\[ M_A=\lambda m_a r\omega^2\sum A_n\Cos n\theta +\lambda m_b r\varepsilon\sum \frac{1}{n}A_n\Sin n\theta \tag{36} \]

\[ +\rho m_b r\omega^2\Sin(\theta+\varphi) -\rho m_b r\varepsilon\Cos(\theta+\varphi). \]

Considering the action of the crank masses separately, we obtain an additional moment due to them in the form

\[ M'_A=\rho m_r r\omega^2\Sin(\theta+\varphi) -\rho m_r r\varepsilon\Cos(\theta+\varphi), \]

where

\[ m_r r=[[unclear: formula]] \]

the static moment of the mass of the crank relative to \(Z\). Thus expression (36) may be represented by replacing everywhere \(m_b\) by \(m_b+m_r\), or, bringing it to equality (11), by

\[ m=m_k+m_l+m_r. \]

In the case most often encountered, when \(\lambda=0\) and, consequently, \(\varphi=0,\ \rho=\eta\),

\[ M_A=\eta m_b r\omega^2\Sin\theta-\eta m_b r\varepsilon\Cos\theta. \tag{37} \]

This expression is suitable for most vertical engines, which are found [[unclear: continuation cut off]].

Joint Action of Several Mechanisms.

Free forces.

Let us take, as the general case of the mutual arrangement of crank mechanisms, the case shown in Fig. 14. By means of various simplifications one may obtain from it the majority of mechanisms of modern multi-cylinder reciprocating machines. We shall denote the angle of inclination of the cylinders to the \(X\)-axis by the letter \(\gamma\), and the angle between the first and an arbitrary crank by \(\alpha\).

We shall regard the angle of deflection of the first crank \(\theta\) as an independent variable for the entire mechanism. We shall denote the angle of deflection of an arbitrary crank from the axis of its cylinder by \(\psi\).

With an oblique arrangement of the coordinates, the inertia forces of a separate mechanism will give, along the axes, the components (Fig. 15):

\[ \begin{aligned} X &= X_n \operatorname{Cos}\gamma - Y_n \operatorname{Sin}\gamma,\\ Y &= X_n \operatorname{Sin}\gamma + Y_n \operatorname{Cos}\gamma, \end{aligned} \tag{37} \]

where for \(X_n\) and \(Y_n\) we have the expressions (12) and (13) already obtained earlier.

Obviously, the resultant components will be:

\[ \begin{aligned} X_0 &= \sum X = \sum X_n \operatorname{Cos}\gamma - \sum Y_n \operatorname{Sin}\gamma,\\ Y_0 &= \sum Y = \sum X_n \operatorname{Sin}\gamma + \sum Y_n \operatorname{Cos}\gamma . \end{aligned} \tag{38} \]

Substituting \(X_n\) and \(Y_n\) from expressions (12) and (13), and introducing the current angle of an arbitrary crank \(\psi\), we obtain:

\[ \begin{aligned} X_0 ={}& \omega^2 \sum^{n}\left\{ m_a r A_n \operatorname{Cos} n\psi \operatorname{Cos}\gamma + \xi \sum^{n}\left\{\frac{1}{n} m_a r A_n \operatorname{Sin} n\psi \operatorname{Cos}\gamma \right.\right.\\ &\quad + \omega^2 \sum m_b r \operatorname{Cos}\psi \operatorname{Cos}\gamma + \xi \sum m_b r \operatorname{Sin}\psi \operatorname{Cos}\gamma\\ &\quad \left.{}- \omega^2 \sum m_b r \operatorname{Sin}\psi \operatorname{Sin}\gamma + \xi \sum m_b r \operatorname{Cos}\psi \operatorname{Sin}\gamma, \\[4pt] Y_0 ={}& \omega^2 \sum^{n}\left\{ m_a r A_n \operatorname{Cos} n\psi \operatorname{Sin}\gamma + \xi \sum^{n}\left\{\frac{1}{n} m_a r A_n \operatorname{Sin} n\psi \operatorname{Sin}\gamma \right.\right.\\ &\quad + \omega^2 \sum m_b r \operatorname{Cos}\psi \operatorname{Sin}\gamma + \xi \sum m_b r \operatorname{Sin}\psi \operatorname{Sin}\gamma\\ &\quad \left.{}+ \omega^2 \sum m_b r \operatorname{Sin}\psi \operatorname{Cos}\gamma - \xi \sum m_b r \operatorname{Cos}\psi \operatorname{Cos}\gamma . \end{aligned} \tag{39} \]

Here and below, the sign \(\sum^n\) sums terms of different orders for the inertia forces, and the sign \(\sum\) means that the sum under the sign is taken over the mechanisms.

From Fig. 14 it is evident that the relation is

\[ \psi = \beta - \alpha + \eta - \gamma . \tag{40} \]

Let the angle between the first and an arbitrary cylinder be \(\beta\), and also

\[ \alpha - \beta = \varphi . \tag{41} \]

Then from (40)

\[ \psi = \theta + \varphi . \tag{42} \]

Introducing this value of \(\psi\) into equation (39), let us first consider the terms with the factor \(m_a\):

\[ \begin{aligned} X_1 ={}& \omega^2 \sum^{n}\left\{ m_a r A_n \operatorname{Cos} n(\theta+\varphi)\operatorname{Cos}\gamma + \xi \sum^{n}\left\{\frac{1}{n}m_a r A_n \operatorname{Sin} n(\theta+\varphi)\operatorname{Cos}\gamma,\right.\right.\\ Y_1 ={}& \omega^2 \sum^{n}\left\{ m_a r A_n \operatorname{Cos} n(\theta+\varphi)\operatorname{Sin}\gamma + \xi \sum^{n}\left\{\frac{1}{n}m_a r A_n \operatorname{Sin} n(\theta+\varphi)\operatorname{Sin}\gamma. \end{aligned} \tag{43} \]

*) In an arbitrary mechanism, \(\varphi\) is the actual angle of its crank relative to the axis of its cylinder (the angle at \(\theta = 0\)); see Fig. 16.

These expressions give the positions of the equivalent masses referred to the end of the crank driving the machine. In this form they are convenient to use when there are no simple formulas for the positions for individual crank mechanisms.

We shall further take \(n\theta\) as the common variable for all [[unclear: mechanisms]].

\[ \begin{aligned} X_i={}& \sum_1^n \omega^2\left( \cos n\theta \sum m_a r A_n \cos n\varphi \cos\gamma -\sin n\theta \sum m_a r A_n \sin n\varphi \cos\gamma \right) \\ &+\sum_1^n \xi\left( \cos n\theta \sum_1^n m_a r A_n \sin n\varphi \cos\gamma +\sin n\theta \sum_1^n m_a r A_n \cos n\varphi \cos\gamma \right), \\ Y_i={}& \sum_1^n \omega^2\left( \cos n\theta \sum m_a r A_n \cos n\varphi \sin\gamma -\sin n\theta \sum m_a r A_n \sin n\varphi \sin\gamma \right) \\ &+\sum_1^n \xi\left( \cos n\theta \sum_1^n m_a r A_n \sin n\varphi \sin\gamma +\sin n\theta \sum_1^n m_a r A_n \cos n\varphi \sin\gamma \right). \end{aligned} \tag{44} \]

For the masses \(m_b\), from expr. (39), it is easy to obtain:

\[ \left. \begin{aligned} X_{\Pi} &= \omega^2 \sum m_b r \cos(\varphi+\theta) +\xi \sum m_b r \sin(\varphi+\theta),\\ Y_{\Pi} &= \omega^2 \sum m_b r \sin(\varphi+\theta) -\xi \sum m_b r \cos(\varphi+\theta). \end{aligned} \right\} \tag{45} \]

Introducing:

\[ \varphi_1=\theta+\varphi \tag{46} \]

and moving the preceding \(\theta\) outside the summation sign, we obtain:

\[ \left. \begin{aligned} X_{\Pi}={}& \omega^2\left(\cos\theta\sum m_b r\cos\varphi_1 -\sin\theta\sum m_b r\sin\varphi_1\right) \\ &+\xi\left(\cos\theta\sum m_b r\sin\varphi_1 +\sin\theta\sum m_b r\cos\varphi_1\right),\\ Y_{\Pi}={}& \omega^2\left(\cos\theta\sum m_b r\sin\varphi_1 +\sin\theta\sum m_b r\cos\varphi_1\right) \\ &-\xi\left(\cos\theta\sum m_b r\cos\varphi_1 -\sin\theta\sum m_b r\sin\varphi_1\right). \end{aligned} \right\} \tag{47} \]

The resultants for the free forces will obviously be:

\[ \begin{aligned} X_0 &= X_i+X_{\Pi},\\ Y_0 &= Y_i+Y_{\Pi}. \end{aligned} \]

From expressions (45) it is seen that \(X_{\Pi}\) and \(Y_{\Pi}\) represent the free projections of the centrifugal and tangential forces of inertia of the masses \(m_b\). Therefore the resultant of these forces can be balanced by a counterweight. To do this, in addition to the magnitude of the resultant, it is necessary to know its position for some fixed state of the mechanism. Let us first find \(R_{\Pi}\) for the constructed forces. We introduce the notation:

\[ \sum m_b r\cos\varphi_1=R\cos\delta,\qquad \sum m_b r\sin\varphi_1=R\sin\delta, \tag{48} \]

thus

\[ R=\sqrt{\left(\sum m_b r\cos\varphi_1\right)^2+ \left(\sum m_b r\sin\varphi_1\right)^2}. \tag{49} \]

Then expressions (47) are rewritten as:

\[ \left. \begin{aligned} X_{\Pi} &= R\omega^2\cos(\theta+\delta)+R\xi\sin(\theta+\delta),\\ Y_{\Pi} &= R\omega^2\sin(\theta+\delta)-R\xi\cos(\theta+\delta). \end{aligned} \right\} \tag{50} \]

Hence the centrifugal force of inertia is

\[ R_{\Pi}=R\omega^2. \]

Setting here \(\theta=0\), we determine \(\delta\) from (50) by the equality

\[ \tg\delta=\frac{\Sin\delta}{\Cs\delta}=\tg\delta =\frac{\sum m_b r\Sin\varphi_1}{\sum m_b r\Cs\varphi_1}. \tag{52} \]

The quantity \(B\), determined by (49), is obviously constant for any \(\theta_1\), i.e. also for \(\theta_1=0\). Therefore expression (49) may be replaced by:

\[ B=\sqrt{\left(\sum m_b r\Cs\alpha\right)^2+\left(\sum m_b r\Sin\alpha\right)^2}. \tag{49A} \]

Similarly, for the tangential forces of inertia one may obtain

\[ R'_{\mathrm{II}}=B\xi . \tag{53} \]

Their direction is, obviously, always perpendicular to \(R'_{\mathrm{I}}\).

A counterweight whose mass is

\[ m_1=\frac{B}{r_1} \]

and whose radius \(r_1\) is directed at the angle

\[ \delta_1=\delta+\pi \]

(see drawing 17) will balance them.

But even without counterweights the free forces of the separate \(m_b\) can balance one another. Obviously this will be when \(R=0\), or, by expression (49A), when

\[ \begin{aligned} \sum m_b r\Cs\alpha&=0,\\ \sum m_b r\Sin\alpha&=0. \end{aligned} \tag{54} \]

Returning to expression (44), we divide it into parts independent of \(\xi\), introducing the following notation:

\[ \begin{aligned} \sum m_a r A_n \Cs n\beta \Cs n\xi &=B_n\sum m_a r A_n \Cs n\beta \Sin n\xi = b_n\Sin n\xi,\\ \sum m_a r A_n \Sin n\beta \Cs n\xi &=C_n\sum m_a r A_n \Sin n\beta \Sin n\xi = c_n\Sin n\xi, \end{aligned} \tag{55} \]

\[ \begin{aligned} X'_{\mathrm{I}}&=\sum' E_n\omega^2\Cs n(\theta+\lambda),\\ Y'_{\mathrm{I}}&=\sum' G_n\omega^2\Cs n(\theta+\tau); \end{aligned} \tag{56} \]

after this, the final construction of the expression determining the magnitude of the free forces for \(\omega=\mathrm{Const}\) may be given as:

\[ \begin{aligned} X_0&=\omega^2\sum' E_n\Cs n(\theta+\lambda)+B\omega^2\Cs(\theta+\delta),\\ Y_0&=\omega^2\sum' G_n\Cs n(\theta+\tau)+B\omega^2\Sin(\theta+\delta), \end{aligned} \tag{57} \]

where, of course, \(B,\delta,E_n,\lambda,G_n,\tau\) are determined from relations (48), (49), (55).

Considering the forces of the different orders separately, it is not difficult to ascertain that the resultant of each of them is the radius vector of an ellipse, the magnitude and position of whose principal axes are determined by the values of the indicated constants.

Let us dwell somewhat on the force of the first order, the most important in its practical significance. Its projections will be:

\[ X'=E\omega^2\Cs(\theta+\lambda),\qquad Y'=G\omega^2\Cs(\theta+\tau), \]

determining the magnitude \(R'\) as:

\[ R'=\omega^2\sqrt{E^2\Cs^2(\theta+\lambda)+G^2\Cs^2(\theta+\tau)}, \]

one can find analytically both \(R'_{\max}\) and its position relative to the coordinate axes, i.e. the position of the major axis of the ellipse. It is simpler, however, to find these graphically, taking the data of drawing 17, where the major axis is drawn in two hatched diameters.

In particular cases \(R'\) may be obtained acting along one straight line.

This will be for [[unclear: condition]] the root

\[ R'=\omega^2\sqrt{E^2+G^2}\operatorname{Cs}(\theta-\lambda), \]

and the direction of action is determined by the relation

\[ \frac{Y'}{X'}=\frac{G}{E}=\text{const.} \]

A more interesting case is that in which

\[ \tau=(2k+1)\frac{\pi}{2}+\lambda \quad \text{and} \quad E=G; \]

then

\[ X'=E\omega^2\operatorname{Cs}(\theta+\lambda),\quad Y'=\pm E\omega^2\operatorname{Sin}(\theta+\lambda) \quad \text{and} \quad R'=E\omega^2, \]

i.e. the resultant is constant in magnitude and rotates with angular velocity \(\omega\) in the direction of rotation of the shaft or else opposite to it.

In this case the free force of the first order may be balanced by a counterweight when the rotation of the shaft and of the vector \(R'\) agree.

For mutual balancing of the free forces of any order it is evidently necessary that

\[ \begin{aligned} &\sum m_a r A_n \operatorname{Cs} n\varphi \operatorname{Cs}\gamma=0,\quad \sum m_a r A_n \operatorname{Sin} n\varphi \operatorname{Cs}\gamma=0,\\ &\sum m_a r A_n \operatorname{Cs} n\varphi \operatorname{Sin}\gamma=0,\quad \sum m_a r A_n \operatorname{Sin} n\varphi \operatorname{Sin}\gamma=0. \end{aligned} \tag{68} \]

Using the identities

\[ \begin{aligned} &2\operatorname{Cs} n\varphi \operatorname{Cs}\gamma =\operatorname{Cs}(n\varphi-\gamma)+\operatorname{Cs}(n\varphi+\gamma),\\ &2\operatorname{Sin} n\varphi \operatorname{Sin}\gamma =\operatorname{Cs}(n\varphi-\gamma)-\operatorname{Cs}(n\varphi+\gamma),\\ &2\operatorname{Sin}\alpha \operatorname{Sin}\beta =\operatorname{Sin}(\alpha+\beta)+\operatorname{Sin}(\alpha-\beta), \end{aligned} \]

instead of conditions (68) we obtain:

\[ \begin{aligned} &\sum m_a r A_n \operatorname{Cs}(n\varphi+\gamma)=0;\quad \sum m_a r A_n \operatorname{Sin}(n\varphi+\gamma)=0,\\ &\sum m_a r A_n \operatorname{Cs}(n\varphi-\gamma)=0;\quad \sum m_a r A_n \operatorname{Sin}(n\varphi-\gamma)=0, \end{aligned} \]

and, combining them again,

\[ \begin{aligned} &\sum m_a r A_n \operatorname{Cs}(n\varphi\pm\gamma)=0,\\ &\sum m_a r A_n \operatorname{Sin}(n\varphi\pm\gamma)=0. \end{aligned} \tag{69} \]

Longitudinal moments.

For the moments about the axes \(x\) and \(y\) we have (see Fig. [[unclear: figure number]])

\[ M_x=\sum(Yz-Zy), \]

\[ M_y=\sum(Zx-Xz). \]

Since \(Z=0\), then

\[ M_x=\sum Yz \quad \text{and} \quad M_y=-\sum Xz. \]

Using equations (37), we further obtain

\[ M_x=\sum X_n z\operatorname{Cs}\gamma+\sum Y_n z\operatorname{Sin}\gamma; \]

\[ -M_y=\sum X_n z\operatorname{Cs}\gamma-\sum Y_n z\operatorname{Sin}\gamma. \]

These expressions for the free moments differ from the vectors (68) in that under the summation sign there is still the factor \(z\); therefore the derivations can be carried out in exactly the same way as for the free forces. Performing them, we obtain (for \(\omega=\Omega\)):

\[ \left. \begin{aligned} M_x &= \sum^n \omega^2\bigl(\cos n\theta \sum m_a r z A_n \cos n\varphi \sin\gamma -\sin n\theta \sum m_a r z A_n \sin n\varphi \sin\gamma\bigr) \\ &\quad +\omega^2\bigl(\cos\theta \sum m_b r z \sin\varphi_1 +\sin\theta \sum m_b r z \cos\varphi_1\bigr),\\[4pt] -M_y &= \sum^n \omega^2\bigl(\cos n\theta \sum m_a r z A_n \cos n\varphi \cos\gamma -\sin n\theta \sum m_a r z A_n \sin n\varphi \cos\gamma\bigr) \\ &\quad +\omega^2\bigl(\cos\theta \sum m_b r z \cos\varphi_1 -\sin\theta \sum m_b r z \sin\varphi_1\bigr). \end{aligned} \right\} \tag{61} \]

In abbreviated form we have

\[ \left. \begin{aligned} M_x &= \omega^2 \sum^n S_n \cos n(\theta+\eta) +K\omega^2\sin(\theta+\varepsilon),\\ -M_y &= \omega^2 \sum^n T_n \cos n(\theta+\mu) +K\omega^2\cos(\theta+\varepsilon), \end{aligned} \right\} \tag{62} \]

where \(S_n,T_n,K,\eta,\mu,\varepsilon\) are determined from the equalities

\[ \begin{gathered} \sum m_a r z A_n \cos n\varphi \sin\gamma = S_n\cos\eta,\qquad \sum m_a r z A_n \sin n\varphi \sin\gamma = S_n\sin\eta,\\ \sum m_a r z A_n \cos n\varphi \cos\gamma = T_n\cos\mu,\qquad \sum m_a r z A_n \sin n\varphi \cos\gamma = T_n\sin\mu,\\ \sum m_b r z \sin\varphi_1 = K\sin\varepsilon,\qquad \sum m_b r z \cos\varphi_1 = K\cos\varepsilon. \end{gathered} \]

The values of \(\varphi\) and \(\varphi_1\) are determined from equalities (41) and (46).

Analogously to equalities (60), for the annihilation of the resultant moment of the free forces of the entire mechanism we shall have

\[ \left. \begin{aligned} \sum m_b r z \sin\varphi_1 &= 0,\\ \sum m_b r z \cos\varphi_1 &= 0, \end{aligned} \right\} \quad\text{or}\quad K=0, \tag{63} \]

i.e., it is necessary to place the rotating masses so that \(K=0\), for which the equalities must be satisfied:

\[ \left. \begin{aligned} \sum m_b r z \sin\alpha &= 0,\\ \sum m_b r z \cos\alpha &= 0. \end{aligned} \right\} \tag{64} \]

Transverse moment

By formula (31), for an individual element we have

\[ M_\tau=-\omega^2\gamma H\sin k\varphi +\varepsilon H\left\{\frac{1}{k}D_k\cos k\varphi+m_b r^2\right\}. \]

The algebraic summation of the quantities of the moments of the individual elements gives the magnitude of the transverse moment for the entire mechanism, i.e.,

\[ M_\tau=\sum M_\tau, \]

or

\[ M_\tau=-\omega^2\sum^k \gamma H D_k\sin k\varphi +\varepsilon\sum^k \gamma \frac{H}{k}D_k\cos k\varphi +\varepsilon\sum m_b r^2. \tag{65} \]

Taking the variable \(\theta\) outside the sign \(\sum\). Then

\[ \begin{aligned} M_\tau &=-\omega^2\sum^k\gamma\bigl(\cos k\theta\sum H D_k\sin k\alpha +\sin k\theta\sum H D_k\cos k\alpha\bigr)\\ &\quad+\varepsilon\sum^k\gamma\bigl((\cos k\theta)\sum \frac{H}{k}D_k\cos k\alpha -\sin k\theta\sum \frac{H}{k}D_k\sin k\alpha\bigr) +\varepsilon\sum m_b r^2. \end{aligned} \tag{66} \]

To obtain the final result, it is necessary to add to the moment just found the moment of the tangential inertia forces of the masses connected with the shaft,

\[ M_{\tau}=-J'_z \xi . \]

That is,

\[ M_z=M_t+M'_{\tau}. \]

Introducing the notation

\[ \sum H D_k \Sin k\varphi=P_k \Sin k\varepsilon,\qquad \sum H D_k \Cos k\varphi=P_k \Cos k\varepsilon, \tag{67} \]

and putting

\[ J_z+\sum m_k l^2=J_z, \]

we write the expression for \(M_z\) as follows:

\[ M_z=-\omega^2 \sum^{k} P_k \Sin k(\theta+\varepsilon) +\xi \sum^{k-1} k P_k \Cos k(\theta+\varepsilon)+J_z \xi . \tag{68} \]

Under absolute uniformity we shall have the moment

\[ M_{\varphi}=-\omega^2 \sum^{k} P_k \Sin k(\theta+\varepsilon). \tag{69} \]

In order that it be balanced in the mechanism, it is obviously necessary to attain fulfillment of the conditions:

\[ \sum H D_k \Sin k\varphi=0, \qquad \sum H D_k \Cos k\varphi=0. \tag{70} \]

III. Particular Case

Unbalanced forces and moments in internal-combustion engines.

As is known, in internal-combustion engines the individual crank mechanisms are identical. Therefore, in the expressions obtained earlier for the free forces and moments, one may take constant \(m_i, F_i, H_i\), while in terms of the same order.

In the formulas given below we shall neglect, in determining the free forces and longitudinal moments, quantities \(\xi\). Because of the small magnitude of \(\xi\), the forces and moments depending on it are very small in comparison with those determined for \(\omega=c\). Moreover, most of them do not affect the greatest values of the free forces obtained for \(\omega=c\); they only change the character of the course of the curve \(X_0\) and the relation between them. The reason is the circumstance that the curves of the principal free forces are cosinusoids, while the curves \(X_{\xi}\) retain the general character of sinusoids.*)

Thus, from (44) and (47),

\[ \begin{aligned} X_0={}&m_a r\omega^2 \sum^{n} A_n\left(\Cos n\theta \sum \Cos n\varphi -\Sin n\theta \sum \Sin n\varphi\right) \\ &+m_b r\omega^2\left(\Cos\theta \sum \Cos\varphi_1 -\Sin\theta \sum \Sin\varphi_1\right), \\[4pt] Y_0={}&m_a r\omega^2 \sum^{n} A_n\left(\Cos n\theta \sum \Cos n\varphi \Sin\gamma +\Sin n\theta \sum \Sin n\varphi \Sin\gamma\right) \\ &+m_b r\omega^2\left(\Cos\theta \sum \Sin\varphi_1 +\Sin\theta \sum \Cos\varphi_1\right). \end{aligned} \tag{72} \]

\[ \left. \begin{aligned} &\\[-1.5em] &\\ \end{aligned} \right\} \]

*) Balancing of masses of double-acting engines according to V. Arshalov and O. Kolsch [[unclear: remainder of footnote]].

\[ \left. \begin{aligned} X_0 &= m_a r\omega^2 \sum_{1}^{n} E_n \,\mathrm{Cs}\,n(\theta+\psi)+m_b r\omega^2 B\,\mathrm{Cs}(\theta+\delta),\\ Y_0 &= m_a r\omega^2 \sum_{1}^{n} G_n \,\mathrm{Sin}\,(\theta+\gamma)+m_b r\omega^2 B\,\mathrm{Sin}(\theta+\delta), \end{aligned} \right\} \tag{72} \]

where here:

\[ \sum A_n \mathrm{Cs}\,n\varphi\,\mathrm{Cs}\,\gamma=E_n\mathrm{Cs}\,n\lambda,\quad \sum A_n \mathrm{Sin}\,n\varphi\,\mathrm{Cs}\,\gamma=E_n\mathrm{Sin}\,n\lambda, \]

\[ \sum A_n \mathrm{Cs}\,n\varphi\,\mathrm{Sin}\,\gamma=G_n\mathrm{Sin}\,n\eta,\quad \sum A_n \mathrm{Sin}\,n\varphi\,\mathrm{Sin}\,\gamma=G_n\mathrm{Sin}\,n\eta, \]

\[ \sum \mathrm{Cs}\,\delta=B\mathrm{Cs}\,\delta,\quad \sum \mathrm{Sin}\,\delta=B\mathrm{Sin}\,\delta. \]

For the conditions of mutual balancing we have, instead of (54) and (60),

\[ \left. \begin{aligned} &\mathrm{Cs}\,\alpha=0,\quad \sum \mathrm{Sin}\,\alpha=0,\\ &\sum \mathrm{Cs}(n\varphi+\gamma)=0,\quad \sum \mathrm{Sin}(n\varphi+\gamma)=0. \end{aligned} \right\} \tag{73} \]

Finally, for the longitudinal moments we obtain:

\[ \left. \begin{aligned} M_x &= m_a r\omega^2 \sum_{1}^{n} A_n \left(\mathrm{Cs}\,n\theta\sum z\,\mathrm{Cs}\,n\varphi\,\mathrm{Sin}\,\gamma -\mathrm{Sin}\,n\theta\sum z\,\mathrm{Sin}\,n\varphi\,\mathrm{Sin}\,\gamma\right)\\ &\quad +m_b r\omega^2\left(\mathrm{Cs}\,\theta\sum z\,\mathrm{Sin}\,\varphi_1 +\mathrm{Sin}\,\theta\sum z\,\mathrm{Cs}\,\varphi_1\right),\\[4pt] -M_y &= m_a r\omega^2 \sum_{1}^{n} A_n \left(\mathrm{Cs}\,n\theta\sum z\,\mathrm{Cs}\,n\varphi\,\mathrm{Cs}\,\gamma -\mathrm{Sin}\,n\theta\sum z\,\mathrm{Sin}\,n\varphi\,\mathrm{Cs}\,\gamma\right)\\ &\quad +m_b r\omega^2\left(\mathrm{Cs}\,\theta\sum z\,\mathrm{Cs}\,\varphi_1 -\mathrm{Sin}\,\theta\sum z\,\mathrm{Sin}\,\varphi_1\right). \end{aligned} \right\} \tag{74} \]

or, by (62),

\[ \left. \begin{aligned} M_x &= m_a r\omega^2 \sum_{1}^{n} G_n\,\mathrm{Cs}\,n(\theta+\eta) +m_b r\omega^2 K\,\mathrm{Sin}(\theta+\varepsilon),\\ -M_y &= m_a r\omega^2 \sum_{1}^{n} E_n\,\mathrm{Cs}\,n(\theta+\lambda) +m_b r\omega^2 K\,\mathrm{Cs}(\theta+\varepsilon), \end{aligned} \right\} \tag{75} \]

where the designations \(G_n,\ E_n\), and \(K\) are determined, with the corresponding changes, as above. For the condition of mutual balancing we have:

\[ \left. \begin{aligned} &\sum z\,\mathrm{Sin}\,\alpha=0,\quad \sum z\,\mathrm{Cs}\,\alpha=0,\\ &\sum z\,\mathrm{Cs}(n\varphi+\gamma)=0,\quad \sum z\,\mathrm{Sin}(n\varphi+\gamma)=0. \end{aligned} \right\} \tag{76} \]

Finally, for the transverse moment we shall have the expression

\[ M_z=-H\omega^2 \sum_{1}^{k} D_k \left(\mathrm{Cs}\,k\theta\sum \mathrm{Sin}\,k\varphi +\mathrm{Sin}\,k\theta\sum \mathrm{Cs}\,k\varphi\right), \tag{77} \]

or

\[ M_z=-H\omega^2 \sum_{1}^{k} P_k\,\mathrm{Sin}\,k(\theta+\rho_k), \tag{78} \]

where

\[ \sum D_k \mathrm{Sin}\,k\varphi=P_k\mathrm{Sin}\,k\rho_k,\quad \sum D_k \mathrm{Cs}\,k\varphi=P_k\mathrm{Cs}\,k\rho_k. \]

Hence, for mutual balancing of the connecting rods it is necessary to satisfy:

\[ \sum \mathrm{Sin}\,k\varphi=0,\quad \sum \mathrm{Cs}\,k\varphi=0. \tag{79} \]

When considering the most common types of internal-combustion engines, one may observe that they have a plane of symmetry \(X-Z\) (Fig. 14). In virtue of this symmetry, to each cylinder making an angle \(\varphi\) with the vertical there corresponds a cylinder with angle \(-\varphi\); all their crank angles are, respectively, \(+n\varphi\) and \(-n\varphi\). Referring to equation (74), we conclude that in such a case they have

\[ \sum \mathrm{Sin}\,n\varphi\,\mathrm{Cs}\,\gamma=0,\quad \sum \mathrm{Cs}\,n\varphi\,\mathrm{Sin}\,\gamma=0. \]

The usual symmetry of the arrangement of the cranks gives

\[ \sum \mathrm{Sin}\,\varphi_1=0,\quad \sum \mathrm{Sin}\,k\varphi=0. \]

After this, the angle $\psi$ and (74) pass into

\[ \begin{aligned} X_0&=m_a r\omega^2\sum_1^k A_n \operatorname{Cs} n\theta \sum \operatorname{Cs} n\varphi\,\operatorname{Cs}\varphi +m_b r\omega^2 \operatorname{Cs}\theta \sum \operatorname{Cs}\varphi,\\ Y_0&=m_a r\omega^2\sum_1^k A_n \operatorname{Sin} n\theta \sum \operatorname{Sin} n\varphi\,\operatorname{Sin}\varphi +m_b r\omega^2 \operatorname{Sin}\theta \sum \operatorname{Sin}\varphi,\\ M_z&=-H\omega^2\sum_1^k D_n \operatorname{Sin} k\theta \sum \operatorname{Cs} k\varphi . \end{aligned} \tag{71a} \]

In multicrank engines with an even number of cranks, the angles between the cranks are equal. Therefore, for them

\[ \sum \operatorname{Cs}\alpha=0,\qquad \sum \operatorname{Sin}\alpha=0, \]

and, consequently, the masses $m_b$ balance one another and, for the same reason, $\sum \operatorname{Cs}\varphi=0$, so that there are no transverse moments of the first order.

The simplifications of expressions (74) are easily carried out in particular cases. We have an important case when the cylinders and cranks are arranged symmetrically with respect to the planes $X-Z$ and $X-Y$. Conditions (76) are then completely fulfilled, and the engine is entirely free of free moments.

Let us now consider the special case of the arrangement of the cylinders.

Uniform arrangement

With this arrangement of the cylinders we have one general notation. The angles between adjacent cylinders are usually equal. According to the diagram we have

\[ \varphi-\alpha-\beta=-\beta=-\gamma, \qquad \gamma_1-\gamma_i=0, \]

and so, in expressions (74a),

\[ \sum \operatorname{Cs}\varphi_1=\sum \operatorname{Cs}0=i, \]

where $i$ is the number of cylinders; i.e., instead of (71) we have here

\[ \begin{aligned} X_0&=m_a r\omega^2 \sum_1^k A_n \operatorname{Cs} n\theta \sum \operatorname{Cs} n\varphi\,\operatorname{Cs}\varphi +m_b r\omega^2 \operatorname{Cs}\theta,\\ Y_0&=m_a r\omega^2 \sum_1^k A_n \operatorname{Sin} n\theta \sum \operatorname{Sin} n\varphi\,\operatorname{Sin}\varphi +m_b r\omega^2 \operatorname{Sin}\theta . \end{aligned} \tag{80} \]

With a number of cylinders greater than 3, further simplifications are possible. Namely, in such a case there always exist forces of the first order which, however, can be balanced without remainder. Forces of higher orders appear only for a certain number of cylinders, and the lowest order is expressed by the number of cylinders $i$. In fact:

\[ \sum \operatorname{Cs} n\gamma\,\operatorname{Cs}\gamma =\frac12\sum \operatorname{Cs}(n+1)\gamma +\frac12\sum \operatorname{Cs}(n-1)\gamma, \]

\[ \sum \operatorname{Sin} n\gamma\,\operatorname{Sin}\gamma =-\frac12\sum \operatorname{Cs}(n+1)\gamma +\frac12\sum \operatorname{Cs}(n-1)\gamma . \]

For $n=1$

\[ \sum \operatorname{Cs}^2\gamma =\frac12\sum \operatorname{Cs}2\gamma+\frac12, \]

but for $i>3$, as will be indicated below, $\sum \operatorname{Cs}2\gamma=0$. Thus

\[ \sum \operatorname{Cs}^2\gamma=\frac12, \qquad \text{also}\qquad \sum \operatorname{Sin}^2\gamma=\frac12 . \]

Then, for the free forces, there remain, for the time being, only those of the first order; we obtain

\[ X_0=(0.5m_1+m_0)\,r\omega^2\cos\theta, \tag{81} \]

\[ Y_0=(0.5m_1+m_0)\,r\omega^2\sin\theta . \]

The geometrical sum is represented as

\[ R=(0.5m_1+m_0)\,r\omega^2 . \tag{82} \]

This force can easily be finally balanced by the centrifugal force of a counterweight.

The forces of the higher orders will exist depending on whether there exists one of the sums

\[ \sum \cos 3\gamma,\quad \sum \cos 5\gamma,\quad \sum \cos 7\gamma,\ldots \]

For determining them we shall use the identity

\[ \cos\alpha+\cos2\alpha+\cos3\alpha+\cdots+\cos i\alpha = \frac{\sin\frac{i\alpha}{2}}{\sin\frac{\alpha}{2}}\, \cos\frac{(i+1)\alpha}{2}. \]

In our case \(\alpha=k\gamma_1\), where \(k\) is an odd number. Thus the separate sums will be equal to

\[ \frac{\sin k\frac{i\gamma_1}{2}}{\sin k\frac{\gamma_1}{2}}\, \cos\frac{(i+1)}{2}k\gamma_1. \]

[[unclear: one line obscured by damage]]

\[ \sum \cos k\gamma = \frac{\sin k\frac{i\gamma_1}{2}}{\sin k\frac{\gamma_1}{2}}\, \cos\frac{(i+1)}{2}k\gamma_1 . \]

This sum is always equal to zero if \(k\) is not a multiple of \(i\), since in the latter case

\[ \sum \cos k\gamma = \frac{0}{0}\, \cos\frac{(i+1)}{2}k\gamma_1 . \]

This circumstance reveals to us that when \(k\) is a multiple of \(i\), the angles between the cylinders, multiplied by \(k\), are equal to a whole number of \(2\pi\), and this means they will be \(k\)-fold [[unclear: crank angles]]:

\[ \sum \cos k\gamma=i . \]

Thus, for 5-cylinder engines we have free unbalanced forces only of the 15th, 25th, etc., orders. Owing to their insignificance, they may be neglected. Since the number of odd numbers, etc., cannot be a multiple of \(i\) even, star engines with an even number of cylinders (double-acting) have no free forces of the higher orders except the first.

There are no longitudinal moments in star engines, since all the axes act in one plane. Considering the crankshaft of the first cylinder and also the center of gravity of the engine, we find

\[ M_z=0,\qquad M_y=0. \]

In the case when a star engine has two cranks at an angle \(\varphi\), that is, when it is composed of two engines, we have

remaining double moments vanish only when the number of cylinders is more than three, since each of the terms has no free forces owing to equation [[unclear: number]].

When determining the free transverse [[unclear]], we note that \(\sum Cs\, k/l\) is not equal to zero when

\[ \frac{k}{l}=\text{an integer}, \]

therefore, for an even number of cylinders the connecting rods [[unclear: are arranged]].

For an odd number, the lowest-order moment will be

\[ M_z=-iD_1H\omega^2\sin i\theta . \]

V-shaped arrangement.

Engines with a V-shaped arrangement of cylinders usually consist of several elements made up of pairs of cylinders with a common crank, as is shown in Fig. 20. We shall measure the angles from the \(x-x\) axis; then we shall have:

\[ \varphi_1=\gamma,\quad \varphi_2=\pi+\gamma,\quad \varphi_3=\theta-\gamma,\quad \varphi_4=\pi+\theta-\gamma . \]

Restricting ourselves [[unclear: to the case]] and disregarding the influence of the masses \(m_b\), [[unclear: rotating]] [[unclear: together with the counterweight]], from formulas (72), for one element we obtain:

\[ \begin{aligned} X_0&=2m_a r\omega^2\left(Cs^2\gamma\,Cs\theta+A_2\,Cs\,2\gamma\,Cs^2\theta\right),\\ Y_0&=2m_a r\omega^2\left(\Sin^2\gamma\,\Sin\theta+A_2\,\Sin2\gamma\,\Sin\gamma\,\Sin^2\theta\right). \end{aligned} \]

It is clear from this that both forces of the first and second orders are represented by ellipses whose principal axes coincide with the coordinate axes.

In the most usual case, when the angle between the cylinders is equal to \(\dfrac{\pi}{2}\), i.e. \(\gamma=\dfrac{\pi}{4}\), we obtain that

\[ Cs^2\gamma=\Sin^2\gamma=\frac{1}{2},\quad Cs\,2\gamma=0,\quad \Sin2\gamma=1, \]

and

\[ \begin{aligned} X_0&=m_a r\omega^2 Cs\theta,\\ Y_0&=m_a r\omega^2\Sin\theta+A_2\Sin2\theta . \end{aligned} \]

That is, the forces of the first order are again completely balanced; only the forces of the 2nd order remain, acting along the \(y\)-axis.

In another particular case, when the cylinders are opposite one another, i.e. \(\gamma=90^\circ\),

\[ X_0=0, \]

\[ Y_0=2m_a r\omega^2\Sin\theta . \]

Combining 2 elements according to Fig. 20, we obtain the usual 4-cylinder engine, for which

\[ \begin{aligned} X_0&=4m_a r\omega^2 A_2 Cs\,2\gamma\,Cs^2\theta=E\,Cs\,2\theta,\\ Y_0&=4m_a r\omega^2 A_2 \Sin2\gamma\,\Sin\gamma\,\Sin2\theta=F\,\Sin2\theta . \end{aligned} \]

At \(\gamma=45^\circ\) we again obtain a free force only along the \(x\)-axis. A 4-cylinder engine is obviously equivalent to an 8-cylinder engine, which, with the same weight of the moving parts, gives forces of the second order twice as large.

In a V-shaped arrangement, the engine elements are arranged along the axis in the same way as, for example, in the case with 4 cylinders; the planes [[unclear: of the cylinders]] are arranged through one plane of symmetry. Under this condition (68) [[unclear: yields]] ...

are compensated completely, and thus the free longitudinal moments are balanced.

Determining \(M_z\) for a two-cylinder engine from Eq. (77a), we have

\[ M_z=-2H\omega^2\left(D_1\operatorname{Cs}\gamma\sin\theta +D_3\operatorname{Cs}3\gamma\sin3\theta+\cdots\right), \]

and for a four-cylinder engine

\[ M_z=0, \]

which will remain true for all engines derived from it.

Normal arrangement

By a normal arrangement we shall mean one in which the cylinders are placed on one side of the shaft, their axes and [[unclear: word]] lying in one plane (Fig. 21). In this case

\[ \xi=0,\quad \lambda=0,\quad \mu=\pi. \]

From the expressions (71a) we obtain, taking into account that, when the angles between adjacent cranks are equal,

\[ \sum \operatorname{Cs} n\alpha=0, \]

\[ \left. \begin{aligned} X_0&=m_a r\omega^2\sum_{1}^{p} A_n\,\operatorname{Cs} n\beta\,\operatorname{Cs} n\theta,\\ Y_0&=0. \end{aligned} \right\} \tag{84} \]

Since \(\sum \operatorname{Cs} n\alpha\) will be nonzero when \(n/i\) is an integer, then for an even number of cylinders the free force of the lowest order is determined as

\[ X_0=i\,m_a r\omega^2 A_1\operatorname{Cs} i\theta, \tag{85} \]

and for an odd number,

\[ X_0=i\,m_a r\omega^2 A_{2i}\operatorname{Cs} 2i\theta. \tag{86} \]

In determining the transverse free moment from (77a), we obtain for an even number of cylinders \(i\)

\[ M_z=0, \]

and for an odd \(i\)

\[ M_z=-H i\omega^2 D_i\sin i\theta, \]

if only the moment of the lowest order is considered.

LITERATURE

J. Wittenberg, “Ermittlung des Massendruckes der hin—und her gehenden Teile der Dampfmaschinen,” Z. d. V. d. I., 1896.

W. Schlippe, “Analysis of the inertia forces,” Engineering, 1897.

H. Lorenz, Dynamik der Kurbelgetriebe, Leipzig, 1901.

H. Schubert, Theory of the balancing of inertia forces by Schlick’s method, translated by Matison, St. Petersburg, 1902.

A. Sharp, Balancing of Engines, London, 1907.

O. Kölsch, Gleichgang und Massenkräfte bei Fahr—und Flugzeugmaschinen, Berlin, 1911.

V. Arshaulov, “On the balancing of inertia forces in Diesel engines,” Neftyanoe delo, 1912.

Radzig, “Report on balancing,” Bulletin of the Society of Technologists, 1913.

V. Karpienko, Balancing of the masses of internal-combustion engines, 1920, manuscript in the library of the Tomsk Technological Institute.

CONTENTS

Introduction

	Page
Preface
Conditions and notation	1
General concepts	2

I. Free forces and moments in the crank mechanism

	Page
Connecting rod with $\beta = 0$	4
Ordinary crank mechanism	6
On the balancing of free forces	9
Transverse moment	9
On the balancing of $M$[[unclear: moment designation]]	16
Influence of the position of the machine’s center of gravity on the magnitude of [[unclear: remainder of line]]	14

II. Combined action of several mechanisms

	Page
Free forces	15
Longitudinal moments	18
Transverse moment	[[unclear: page number obscured]]

III. Particular forms

	Page
Unbalanced forces and moments in internal-combustion engines	20
Star-shaped arrangement	[[unclear: page number obscured]]
V-shaped arrangement	24
Normal arrangement	25
Literature	25

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4. Determination of the actual direction [[unclear: remaining words of the vertical caption]].

Fig. 5.

Fig. 6.

Fig. 7.

Fig. 8.

Fig. 14.

Visible diagram annotations: \(x\), \(y\), \(z\), \(z_2\), \(y_1\), \(r_2\), \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(\alpha\), \(\varphi\), \(1\), \(2\), \(3\), \(I\), \(II\), \(III\), \(IV\), \(V\).

Fig. 15.

Visible diagram annotations: \(x_n\), \(y_n\), \(V\), \(VI\).

Visible annotations in the upper diagrams and curves: \(x\), \(y\), \(z\), \(O\), \(O_1\), \(O_2\), \(r_1\), \(r_2\), \(r_1\varphi_2\), \(r_2\varphi_2\), \(a = 1\), \(I\), \(II\), \(III\), \(IV\), \(V\), \(VI\), \(b_n\), \(b_m\), \(b_1\), \(b_2\), \(h\), \(m\), \(m_1\), \(m_2\), \(\rho_2\), \(y_1\), \(y_2\), \(z_1\), \(z_2\), \(x_2\), \(r_2\).

Fig. 18.

Fig. 16.

Fig. 17.

Fig. 19.

Fig. 20.

Fig. 21.