A. V. Kvasnikov, Assoc. Prof., STI.

On the Arrangement of Crankshafts in a Five-Cylinder Engine

With the usual arrangement of cylinders, when their number is odd, the three-cylinder design is the most widespread. Among the others, the five-cylinder design has the greatest tendency toward adoption; seven- and nine-cylinder engines are already encountered only in isolated cases. The existence of a five-cylinder engine can be justified chiefly by two reasons. First, when, with a standardized cylinder, a plant obtains different unit powers by varying the number of cylinders, then in the series of cylinder numbers 3, 4, 5, 6 the number 5 is just as logical as all the others (De La Vergne engines). Secondly, in the frequently occurring disputes over the advantages of 4- and 6-cylinder engine installations, the use of a five-cylinder design can undoubtedly, in some cases, provide a proper compromise. The differences between a five-cylinder engine and its nearest neighbors are based almost entirely on the structural features of its shaft.

In the present article these features are examined with respect to their influence on the balance and uniformity of running of the engine.

Basic equations and general conditions for the arrangement of masses

In arranging the masses in a five-cylinder engine, we shall strive to obtain: the most uniform torque, the greatest balance, and the simplest constructive execution of the engine as a whole. We shall assume, as is usually done, that the fluctuations in the magnitude of the torque will be minimal with a uniform alternation of explosions in the engine cylinders. The arrangement of the crankpins satisfying this condition is given in drawing 1 at the left. The angles between neighboring crankpins are equal and are \(72^\circ\). For a two-stroke engine the explosions alternate every \(72^\circ\), and for a four-stroke engine every

\[ \frac{720}{5} = 144^\circ. \]

We shall take this arrangement of the crankpins in the plane of their motion as unchanged for the subsequent discussion. The arrangement of the cylinders along the axis of the shaft, we assume, must be such that, for any specified unbalance, the cylinders and intermediate links, bearings, etc., would either be completely identical, or, if there are differences, identical elements would be placed completely symmetrically with respect to the plane passing through the axis of the middle cylinder perpendicular to the shaft axis. We shall assume that the masses of the individual pistons may also be different, although, of course, by condition they will be placed symmetrically with respect to the middle of the engine. In balancing, we shall first of all try to arrange the masses so that there are no forces and moments of the first order. Since the forces and moments of the first order can nevertheless be balanced by counterweights, we shall also make the check: whether it will not be advantageous to select an arrangement of the crankpins with self-balanced forces and moments of the second order, which are not susceptible to the usual balancing by counterweights.

The general equations for the unbalanced forces and moments arising from the motion of the piston masses are written as follows¹ (if we assume that the crank radii are identical, while the masses reduced to the piston pins may be different):

\[ X = r \omega^{2} \sum A_n \cos n\theta \sum m \cos n\alpha \tag{1} \]

\[ M_y = r \omega^{2} \sum A_n [\cos n\theta \sum m z \cos n\alpha - \sin n\theta \sum m z \sin n\alpha] \tag{2} \]

Here

\(X\) . . . . . free force along the \(X\)-axis, coinciding with the axis of the middle cylinder.

\(M_y\) . . . . free moment rotating the engine about the \(y\)-axis.

\(r\) . . . . . radius of the crank.

\(m\) . . . . . mass reduced to the piston pin.

\(\omega\) . . . . angular velocity of shaft rotation.

\(\theta\) . . . . angle of rotation of the first crank (independent variable).

\(\alpha\) . . . . angle between the first and any other crank.

\(n\) . . . . . number denoting the order of the free force or moment.

\(A_n\) . . . ordinal coefficient, which in the expression for the magnitude of the force of the first order is equal to 1, and for the second order is equal to:

\[ A_2 = \mu + \frac{1}{4}\mu^3 + \frac{5}{128}\mu^5 + \cdots ; \]

where \(\mu\) represents the ratio of the length of the crank radius to the length of the connecting rod.

If one considers only forces and moments of the first and second orders, neglecting the others owing to their small magnitude, then expressions (1) and (2) may be rewritten as:

\[ X = r\omega^2 \cos\theta \sum m\cos\alpha + A_2 r\omega^2 \cos 2\theta \sum m\cos 2\alpha . \tag{1-a} \]

\[ M_y = r\omega^2[\cos\theta \sum m z \cos\alpha - \sin\theta \sum m z \sin\alpha] + \]

\[ + A_2 r\omega^2[\cos 2\theta \sum m z \cos 2\alpha - \sin 2\theta \sum m z \sin 2\alpha] \tag{2-a} \]

The sign \(\sum\) sums the terms over the five cranks. The conditions for self-balancing of the free forces of both orders may be expressed as:

\[ \sum m \sin\alpha = 0, \qquad \sum m \cos\alpha = 0; \tag{3} \]

\[ \sum m \sin 2\alpha = 0, \qquad \sum m \cos 2\alpha = 0; \tag{4} \]

and for the moments:

\[ \sum m z \sin\alpha = 0, \qquad \sum m z \cos\alpha = 0; \tag{5} \]

\[ \sum m z \sin 2\alpha = 0, \qquad \sum m z \cos 2\alpha = 0; \tag{6} \]

The free forces and moments of the first order are absent.

Let us first examine the case when the piston masses are identical.

Then, for the disappearance of the forces of the first order, it is necessary to satisfy, according to (3), the condition:

\[ \sum \sin\alpha = 0, \qquad \sum \cos\alpha = 0; \tag{3-a} \]

For the arrangement of the cranks according to drawing 2, these conditions are fulfilled. With equal angles between neighboring cranks, the order of the greatest existing free force always turns out to be even and, in number, twice the number of cranks. For five cranks we therefore obtain only free forces of the 10th order. In order that the moment of the first order vanish, it is necessary to satisfy the conditions

\[ \sum z \sin\alpha = 0, \qquad \sum z \cos\alpha = 0: \tag{5-a} \]

Let us now form the sum \(\sum z \operatorname{Cos}\alpha\), assuming that the 1st crankpin is located in the middle of the engine and thus \(Z_1=0\).

\[ \sum z \operatorname{Cos}\alpha = Z_2 \operatorname{Cos}72^\circ + Z_3 \operatorname{Cos}144^\circ + Z_4 \operatorname{Cos}216^\circ + Z_5 \operatorname{Cos}288^\circ =0. \]

But

\[ \operatorname{Cos}144^\circ=\operatorname{Cos}216^\circ=-\operatorname{Cos}36^\circ;\qquad \operatorname{Cos}288^\circ=\operatorname{Cos}72^\circ, \]

therefore

\[ \sum z \operatorname{Cos}\alpha =(Z_2+Z_5)\operatorname{Cos}72^\circ -(Z_3+Z_4)\operatorname{Cos}36^\circ =0 \]

or:

\[ (Z_2+Z_5)=(Z_3+Z_4)\frac{\operatorname{Cos}36^\circ}{\operatorname{Cos}72^\circ}; \tag{7} \]

for \(\sum z \operatorname{Sin}\alpha=0\) we obtain,

\[ Z_3-Z_5=(Z_4-Z_2)\frac{\operatorname{Sin}36^\circ}{\operatorname{Sin}72^\circ}. \tag{8} \]

Adding the corresponding parts of equations (7) and (8), and then subtracting, we obtain the following equalities:

\[ Z_2=Z_3+1{,}618Z_4, \]

\[ Z_5=1{,}618Z_3+Z_4. \]

From these equalities we are convinced that the condition of symmetry of the arrangement of the crankpins will be satisfied when the absolute values of \(Z_2\) and \(Z_5\) are equal, i.e., for example, when \((Z_2)=(-Z_5)\), whence it follows that \((Z_4)=(-Z_3)\).

If \(Z_2\) is taken as negative, i.e. placed to the right of the 1st crankpin rather than to the left, as it has just been, then we obtain an arrangement that is the mirror image, or symmetrical to the first, i.e. essentially the very same thing. Putting \(Z_2=1\), we obtain

\[ Z_2=Z_3+1{,}618Z_4;\qquad Z_3=-Z_4+1{,}618Z_2, \]

whence

\[ Z_4=1{,}618Z_2. \tag{9} \]

The arrangement of the crankpins found is given in drawing 1. From the drawing we see that the engine is somewhat longer than in the usual case, when the distance between the axes of the extreme cylinders is equal to \((i-1)a\), where “\(a\)” is the normal distance between cylinders and “\(i\)” is the number of cylinders. From drawing 2 we note that the distance between the extreme cylinders turns out to be greater than the usual by

\[ \frac{5{,}236}{4}=1{,}309=\sim 1{,}31 \]

times.

Let us now find the magnitude of the moment of the 2nd order.

\[ M_2=A_2mr\omega^2 \left[ \operatorname{Cos}2\Theta\sum z\operatorname{Cos}2\alpha -\operatorname{Sin}2\Theta\sum z\operatorname{Sin}2\alpha \right]. \]

The arrangement of the vectors necessary for finding the corresponding sines and cosines is shown in drawing 3:

\[ \sum z \operatorname{Cos}2\alpha=0 \]

\[ \sum z \operatorname{Sin}2\alpha = a\left[ 2{,}618\operatorname{Sin}72^\circ +1{,}618\operatorname{Sin}144^\circ +1{,}618\operatorname{Sin}144^\circ +2{,}618\operatorname{Sin}72^\circ \right] =6{,}88a \]

After this we obtain the following expression for the moment of the second order,

\[ M_2=6{,}88\,A_2amr\omega^2\operatorname{Sin}2\Theta \tag{10} \]

The distances between the cylinders may also be made equal, but in that case it will be necessary to make the piston masses different. To eliminate the free moment of the first order, the conditions (5) must be observed:

\[ \sum m z \operatorname{Sin}\alpha = 0,\qquad \sum m z \operatorname{Cos}\alpha = 0, \tag{5} \]

and, in order to eliminate the free forces of the first order, it is necessary that

\[ \sum m \operatorname{Cos}\alpha = 0,\qquad \sum m \operatorname{Sin}\alpha = 0; \tag{3} \]

carrying out reasoning similar to the preceding, one may obtain the following condition for the self-balancing of the moments of the first order:

\[ m_4 Z_4 = 1.618\,m_2 Z_2, \tag{9-a} \]

where

\[ m_2 Z_2 = -m_5 Z_5 \quad \text{and} \quad m_4 Z_4 = -m_3 Z_3. \]

Since the free forces must also be absent, it is necessary to satisfy the conditions expressed in equalities (3).

The first equality can always be satisfied by choosing the magnitude of \(m_1\).

The second gives:

\[ m_2 - m_5 = (m_4 - m_3)\frac{\operatorname{Sin}36^\circ}{\operatorname{Sin}72^\circ}. \]

With a symmetrical arrangement of the masses, it will evidently be:

\[ m_2 = m_5 \quad \text{and} \quad m_4 = m_3. \]

From the equality \(m_2 Z_2 = -m_5 Z_5\) we find that \(Z_2 = -Z_5\), and assuming that \(Z_2 < Z_4\), we obtain (the case \(Z_2 > Z_4\) is rejected, since it gives considerably worse balancing), in absolute values:

\[ Z_4 = Z_3 = 2Z_2 = 2Z_5. \]

By condition (9-a) we find that \(m_2 = 1.236\,m_4\). The arrangement of the individual masses for this case is shown in drawing 4. The mass of the first piston must be chosen so that \(\sum m \operatorname{Cos}\alpha\) is equal to zero. In drawing 5, on the left, there is given a vector diagram from whose projections the terms \(m\operatorname{Sin}\alpha\) and \(m\operatorname{Cos}\alpha\) are determined. From the drawing it is seen that \(\sum m\operatorname{Sin}\alpha = 0\), and the mass of the 1st piston has no effect on this sum, since \(\operatorname{Sin}\alpha_1 = 0\). From the drawing we conclude that, for \(\sum m\operatorname{Cos}\alpha = 0\), it is necessary to have:

\[ m_1 \operatorname{Cos}0 = 2\operatorname{Cos}36^\circ - 2 \cdot 1.236\operatorname{Cos}72^\circ, \]

or: \(m_1 = 0.854\), if \(m_4 = m_3 = 1\).

Thus, the engine will have no free moments and forces of the first order with equal distances between the cylinders, provided only that (drawing 5, on the right)

\[ m_3 = m_4 = 1.17\,m_1 \quad \text{and} \quad m_2 = m_5 = 1.446\,m_1. \]

Let us now find the magnitudes of the forces and moments of the 2nd order.

In drawing 6 is shown the position of the vectors for determining the terms

\[ m\operatorname{Sin}2\alpha \quad \text{and} \quad m\operatorname{Cos}2\alpha. \]

From the drawing we directly see that

\[ \sum m\operatorname{Sin}2\alpha = 0, \]

\[ \sum m\operatorname{Cos}2\alpha = m_1 + 2 \cdot 1.17\operatorname{Cos}72^\circ - 2 \cdot 1.446\operatorname{Cos}36^\circ = -0.618\,m_1. \]

Thus, we shall have a free force of the second order, varying according to the law

\[ X_2=-0.618 A_2 m r \omega^2 \Cos 2\Theta, \tag{10} \]

where \(m\) is the mass of the small piston; we shall regard this mass as normal; i.e., in the first cylinder the piston will be normal, while in the others the pistons will be made heavier. The magnitude of the normal mass will also be denoted below by the letter \(m\) without accents.

To determine the magnitudes of the moments of the 2nd order, we construct a diagram of the arrangement of the vectors \(mZ\). In Fig. 7 their magnitudes are given. Since \(Z_5\) and \(Z_3\) are negative, we transfer vectors 3 and 5 into the opposite quadrants, where the signs of the projections of the vectors change. From the resulting Fig. 8 we conclude that

\[ \Sigma m z \Cos 2\alpha=0 \quad\text{and}\quad \Sigma m z \Sin 2\alpha=6.15\,a m, \]

after which

\[ M_2=-6.15 A_2 a m r \omega^2 \Sin 2\Theta \tag{11} \]

Free forces and moments of the second order are absent.

We shall first assume that the masses of the individual pistons are identical. In that case, we already know from the preceding that the forces of all orders up to the 10th vanish. In order that the moment of the 2nd order vanish, the conditions must be fulfilled:

\[ \Sigma z \Cos 2\alpha=0 \quad\text{and}\quad \Sigma z \Sin 2\alpha=0. \]

Let us compose these sums:

\[ \Sigma z \Cos 2\alpha = Z_2 \Cos 144^\circ + Z_3 \Cos 288^\circ + Z_4 \Cos 432^\circ + Z_5 \Cos 572^\circ =0; \]

whence

\[ Z_3+Z_4=(Z_2+Z_5)\frac{\Cos 36^\circ}{\Cos 72^\circ} \tag{12} \]

\[ \Sigma z \Sin 2\alpha = Z_2 \Sin 36^\circ - Z_3 \Sin 72^\circ + Z_4 \Sin 72^\circ - Z_5 \Sin 36^\circ =0 \]

and

\[ Z_3-Z_4=(Z_2-Z_5)\frac{\Sin 36^\circ}{\Sin 72^\circ} \tag{13} \]

With a symmetrical arrangement it is evident that, by (12) and (13),

\[ Z_3=-Z_4 \quad\text{and}\quad Z_2=-Z_5, \]

i.e., \(Z_3\) and \(Z_2\) are positive and equal in absolute magnitude respectively to \(Z_4\) and \(Z_5\). From condition (13) we obtain:

\[ Z_3=1.618\,Z_2. \tag{14} \]

The arrangement of the crankpins is presented in Fig. 9.

To find the moments of the first order, we construct a vector diagram, Fig. 10.

From the arrangement of the vectors we see that

\[ \Sigma z \Cos \alpha=0 \]

and

\[ \Sigma Z \Sin \alpha = 2a(1.618\Sin 36^\circ+2.618\Sin 72^\circ) = 6.88\,a. \]

Accordingly, the free moment will vary according to the law:

\[ M_1=6.88\,a m r \omega^2 \Sin\Theta. \tag{15} \]

Assuming that half will be balanced by counterweights or, what is the same thing, transferred into the horizontal plane, we nevertheless obtain that in the vertical and horizontal planes there will act moments reaching a maximum value of \(3.441 a m r \omega^2\).

To this it must be added that the centrifugal forces of the rotating masses of the crank will also create a moment which can be entirely balanced by counterweights. Therefore the counterweights balance the moment:

\[ 3.441 a m r \omega^2 + 6.882 a m_b r \omega^2 = 3.441 a r \omega^2 (m + 2m_b), \]

where \(m_b\) is to be taken as the magnitude of the mass reduced to the crank pin.

Assigning equal distances between the cylinders and using the reasoning given earlier, we obtain the arrangement of the masses and their magnitudes according to Fig. 11. The mass \(m_1\) must be chosen so that the forces of the 2nd order vanish. In Fig. 12 the arrangement of the vectors is given for the forces of the 2nd order when these forces are eliminated. Forces of the first order will exist. The arrangement of the vectors for them is given in Fig. 13. Figure 13 is completely similar to Fig. 6. Therefore here also \(\sum m \cos \alpha = -0.618 m\). That is, the force of the 1st order will have the magnitude

\[ X_1 = -0.618 m r \omega^2 \cos \theta, \tag{16} \]

where \(m\) is the magnitude of the mass of the normal piston.

If half of this force is balanced, then there remains, unbalanced, a free force of the 1st order with a maximum equal to \(0.309 m r \omega^2\). With regard to moments of the 1st order, Fig. 10, given for the arrangement of the vectors, may be considered valid also for our case, with only the difference that the magnitudes of the vectors will be equal to 1.446 and 2.34, i.e. 1.12 times smaller than those indicated in Fig. 10. That is, \(M_1\) will be 1.12 times smaller than that calculated from equation (15)

\[ M_1 = -6.15 a m r \omega^2 \sin \theta. \tag{17} \]

Again, assuming that half of the moment can be balanced by counterweights, we shall have an unbalanced moment \(3.075 a m r \omega^2 \sin \theta\).

The counterweights, as in the preceding case, must also balance the moment produced by the rotating masses \(m_b\). Therefore the moment created by the counterweights is equal to:

\[ 3.075 a m r \omega^2 + 6.15 a m_b r \omega^2 = 3.075 a r \omega^2 [m + 2m_b]. \]

Equal \(m\) and \(a\).

If, in the arrangement according to Fig. 4, the masses are made equal, then the balance will change somewhat, since the forces of the second order will disappear, but moments of the first order will appear. From the arrangement of the vectors (Fig. 14) we see that

\[ \sum z \cos \alpha = 0 \quad \text{and} \quad \sum z \sin \alpha = -0.45a. \]

Therefore

\[ M_1 = 0.45 a m r \omega^2 \sin \theta. \tag{18} \]

The diagram of the vectors \(M_2\) is given in Fig. 15. From it:

\[ \sum z \sin 2\alpha = 4.98a. \]

Whence

\[ M_2 = -4.98 A_2 a m r \omega^2 \sin 2\theta. \tag{19} \]

The total moment will have the magnitude:

\[ M = a m r \omega^2 [0.45 \sin \theta - 4.98 A_2 \sin 2\theta]. \]

And for \(A_2 = 0.225\) \(\left[\mu = -\dfrac{1}{4.5}\right]\)

\[ M = a m r \omega^2 [0.45 \operatorname{Sin}\Theta - 1.12 \operatorname{Sin}2\Theta]. \tag{20} \]

Finding \(M_{\max}\) in the usual way, we ascertain that the moment will reach its greatest absolute values at \(\Theta_1 = 40^\circ 30'\) and \(\Theta_2 = 131^\circ 20'\), where

\[ M'_1 = -0.815 a m r \omega^2 \quad \text{and} \quad M''_1 = 1.449 a m r \omega^2. \]

In Fig. 16 are shown the variations of the moments of the 1st and 2nd orders, as well as their sum.

The masses which may be referred to the crank pin \(m_b\) will also create a free moment, which, however, can be completely balanced by counterweights. If this moment is balanced and the moment of the first order is transferred into the horizontal plane, then the moment created by the counterweights must have the value:

\[ 0.45 a r \omega^2 [m + m_b] \operatorname{Sin}\Theta. \]

Let us now suppose that, with the arrangement according to Fig. 11, the masses will be identical. As in the case just examined, the free forces will be absent and the moments of the 1st and 2nd orders will act.

Determining the moment of the first order in the usual way, we obtain:

\[ M_1 = -4.98 a m r \omega^2 \operatorname{Sin}\Theta. \tag{21} \]

The arrangement of the vectors \(M_2\) is given in Fig. 17, whence we find that

\[ \sum z \operatorname{Cos} 2\alpha = 0 \quad \text{and} \quad \sum z \operatorname{Sin} 2\alpha = 0.45a. \]

Therefore

\[ M_2 = 0.45 A_2 a m r \omega^2 \operatorname{Sin}2\Theta. \tag{22} \]

Analogously to the preceding, we find the required moment from the counterweights, assuming that, because of the large magnitude of the moment of the first order, half of it is transferred into the horizontal plane:

\[ 2.49 a r m \omega^2 + 4.98 a m_b r \omega^2 = 2.49 a r \omega^2 [m + 2m_b]. \]

The total moment will have the following value:

\[ M = a m r \omega^2 [0.45 A_2 \operatorname{Sin}2\Theta - 4.98 \operatorname{Sin}\Theta]. \]

If \(A_2 = 0.225\), then we have the maximum value of the moment at \(\Theta = 92^\circ 25'\), the value of the moment of the second order being negligible, and

\[ M_{\max} = 4.98 a m r \omega^2. \]

Moments absorbed in the frame.

Part of the moments created by the inertia forces acting on the shaft is balanced in the material of the frame. These moments are not so small that they can be disregarded. Their action is especially noticeable in light engines having light aluminum crankcases. Thus, one of the conspicuous shortcomings of an 8-cylinder engine is that, with a great length of the crankcase, its deformations under the action of bending moments from inertia forces either become harmful or compel the crankcase to be made heavier. The latest types of automobile 6-cylinder engines also bear the imprint of a similar influence of the damped moments. This is further promoted by tendencies toward increasing the number of revolutions. The desire to create a rigid crankcase,

if one judges by the latest automobile exhibitions, even leads to constructions that greatly impede access to the moving mechanism and partial repair on site.

Keeping in view only the purpose of comparing various five-cylinder engines with a four-cylinder one, let us determine the magnitudes of the damping moments that depend only on the inertia forces of the masses connected with the piston and that act in the mean section of the crankcase. In Fig. 18 a diagram is given of the arrangement of the cranks in a four-cylinder engine. The moment produced by the masses of pistons 3 and 4 will obviously act on the frame, bending it in the mean section, assuming, of course, that the shaft resistance is not taken into account. The vector diagrams of the first and second orders are shown in Fig. 19.

Carrying out the addition of the vectors and summing the moments, we obtain:

\[ M_4 = (a m r \omega^2)_4 \,[\operatorname{Cos}\Theta + 2 A_2 \operatorname{Cos} 2\Theta]. \]

The subscript 4 denotes affiliation with the four-cylinder engine. Taking \(A_2 = 0.225\), we obtain the maximum value of the bending moment.

\[ M_4 = 1.45\,(a m r \omega^2)_4. \tag{23} \]

Let us now find these moments for all the six cases of the five-cylinder engine examined above.

For the 1st case (see the summary table and Fig. 2) we find the moment of the inertia forces of masses 5 and 3, situated on one side of the middle of the engine. In Fig. 20 the vector diagrams of both orders are presented. For the moments of the 1st order we find that, for masses 3 and 5,

\[ \Sigma z \operatorname{Sin}\alpha = 0 \quad\text{and}\quad \Sigma z \operatorname{Cos}\alpha = -1.618a, \]

therefore

\[ M_1 = -1.618\,a m r \omega^2 \operatorname{Cos}\Theta. \]

The same result can also be reached simply by adding geometrically the vectors 5 and 3. In summing the moments of the 2nd order according to vector diagram II, we conclude that the moments depending on \(\Sigma z \operatorname{Sin} 2\alpha\) have the same sign and are not balanced in the frame. On the contrary, the moments depending in magnitude on the coefficient \(\Sigma z \operatorname{Cos} 2\alpha\) mutually balance. Such a conclusion can also be drawn directly after obtaining equation 10. From Fig. 20, \(\Sigma z \operatorname{Cos}2\alpha = -0.5\,a\), and so the total damping moment will be equal to

\[ M = a m r \omega^2 [1.618 \operatorname{Cos}\Theta + 0.5 A_2 \operatorname{Cos} 2\Theta] \]

and

\[ M_{\max}=1.73\,a m r\omega^2 \tag{24} \]

Proceeding analogously in case II (Fig. 4), we obtain:

\[ M_{\max}=1.54\,a m r\omega^2 \tag{25} \]

Case III gives [see the table and Figs. 14 and 15]

\[ M = a m r\omega^2 [1.309 \operatorname{Cos}\Theta + 0.191 A_2 \operatorname{Cos} 2\Theta], \]

\[ M_{\max}=1.352\,a m r\omega^2. \tag{26} \]

For case IV (Fig. 9)

\[ M_{\max}=0.864\,a m r\omega^2 \tag{27} \]

In case V

\[ M_{\max}=0.772\,a m r\omega^2 \tag{28} \]

In case VI

\[ M_{\max}=0.5\,a m r\omega^2 \tag{29} \]

On the comparison of free forces and moments.

In the summary table given below one can see the comparative characteristic of a 4-cylinder engine and of all the 6 variants, analyzed above, of the arrangement of crankpins and piston masses in a 5-cylinder design. In the table the magnitudes of the free forces are compared, the action of free moments is clarified somewhat conditionally, and the moments from the inertia forces of the masses, referred to the piston pin, which die out in the engine frame, are compared. In making the comparison it is assumed that the engines have:

Identical \(\mu = \dfrac{r}{l}\), i.e., identical ratios of the lengths of the crank radius and the connecting rod.

Identical ratios of the piston stroke to its diameter \(\dfrac{S}{D}\)

Identical mean piston speeds \(C_m\).

Identical indicator diagrams.

Identical powers \(N_i\).

Let us express the quantity \(mr\omega^2\) through these data, identical for the engines being compared. With identical calculated pressures, the volume of the piston crown—and thus its mass as well—may be considered directly proportional to the cube of the piston diameter. For similar pistons, therefore, their masses will be proportional to the cube of the piston diameter, i.e. \(m = BD^3\), where \(B\) is the coefficient of proportionality.

If \(Z\) is the number of cylinders of the engine, \(P_i\) the mean indicated pressure, and \(r\) the crank radius, then the power of a four-stroke engine is expressed as:

\[ N_i = Z \frac{p_i \cdot \pi D^2 \cdot 2r \cdot n}{4 \cdot 2 \cdot 60 \cdot 75} = \frac{p_i C_m}{382} D^2 Z. \]

Then from this

\[ D = \sqrt{\frac{382 N_i}{p_i C_m}} \sqrt{\frac{1}{Z}}. \]

We shall determine the quantity \(r\omega^2\) as follows:

\[ r\omega^2 = \frac{\pi^2 C_m^2}{4r}. \]

After this

\[ mr\omega^2 = \frac{B\pi^2}{2} \cdot \frac{38200 N_i}{p_i} C_m \cdot \frac{D}{S} \cdot \frac{1}{Z} = \frac{C}{Z}. \tag{30} \]

In a 4-cylinder engine with mirror arrangement of the crankpins there is a free force of the 2nd order.

\[ X^1 = 4 A_2 mr\omega^2, \]

or

\[ X^1 = \frac{4 A_2 C}{Z} = A_2 C. \]

In order to be able to approach the comparison of the actions of the unbalanced moments of a 5-cylinder engine and the free force of a 4-cylinder one, we shall determine the arm at which the free force of a 4-cylinder engine will give the same moment as is present in the 5-cylinder one.

In the first combination we have the greatest moment

\[ M_2 = 6{,}88 A_2 a mr\omega^2 = 6{,}88 A_2 \cdot a \frac{C}{5}. \]

Denoting the required lever arm by \(h_1\), we obtain

\[ h_1=\frac{M_2}{X^1}=1.374a. \]

That is, this lever arm is somewhat greater than the distance between adjacent cylinders in a 5-cylinder engine.

For the second combination we have:

\[ \frac{X_2}{X^1}=0.618\,A_2,\qquad \frac{C}{5}:A_2C=0.123. \]

That is, the free force obtained in the 2nd combination is only \(12.3\%\) of the free force of a 4-cylinder engine. For the lever arm of the moment we shall have

\[ h_2=\frac{M_2}{X^1}=1.23a. \]

That is, here the action of the moment will be somewhat less than in the preceding case.

In the third combination we have:

\[ M_{\max}=1.449\,amr\omega^2. \]

Therefore

\[ h_3=\frac{M}{X^1}=\frac{1.449a}{5A_2}=1.29a. \]

In the fourth combination:

\[ h_4=\frac{6.882a}{5A_2}=6.12a. \]

When balancing half the moment, the lever arm nevertheless comes out to \(3.06a\)—considerably greater than in the preceding cases.

Combination V has free forces of the first order

\[ \frac{X_1}{X^1}=\frac{0.618}{5A_2}=0.55. \]

These forces are two times smaller than the free forces of a 4-cylinder engine. For the lever arm we obtain:

\[ h_5=5.47a. \]

In the sixth combination we have only moments, with

\[ h_6=4.43a. \]

The question of the comparative harmfulness of free forces and moments cannot be decided on the basis of data about the engine alone. Besides the properties, i.e. the magnitude and character of the variation, of the forces and moments, the magnitude of the deformations of the parts of the installation in which the engine operates is also affected by the properties of this installation as an elastic system, and by the combination of these properties with the properties of the forces and moments. Therefore the question of the behavior of the installation consists, in general, of three parts.

In the first, the properties of the acting forces are considered; in the second, the character of the free vibrations of the parts of the installation and questions of their strength; and in the third, with the given installation and engine, the actual vibrations and stresses in the parts of the installation are found.

The second and third parts do not enter into the task of the present article, and therefore we shall dwell in a few words only on a pair of particular cases. In Fig. 21

schematically, as a rigid system with center of gravity \(O\), which includes in one case a 4-cylinder engine and in the other a 5-cylinder engine, having a free moment. From the drawing it is clear that if \(H > h\), then the free moment acting on the system with the 4-cylinder engine will be greater. In the first three combinations the arms are so small that, in those cases when the system may be regarded as rigid, the 5-cylinder engine will be better with respect to balancing, since with a 4-cylinder engine the system is acted upon by both a greater moment and also a force.

In Fig. 22 there is shown, approximately to scale, the arrangement of the free forces and moments of an engine in an automobile. In the diagram the automobile is represented as a rigid material beam resting on two spring supports.

Case “a” represents the installation of a 4-cylinder engine, and case “b” is suitable for the first three combinations of a 5-cylinder engine, when the reduced arm differs little in magnitude from the distance between adjacent cylinders. The vibrations of the automobile frame will depend on the characteristics of the elastic supports, on the distribution of the masses along the frame (it is assumed that across the width of the beam the masses are distributed uniformly), on the position of the force between the supports, on the periods of the natural vibrations of the entire system and their relation to the periods of oscillation of the acting forces and moments, and also on the magnitude of the acting forces and moments, which may be judged from the reactions at the supports. With respect to the latter, here too the clear advantage of the 5-cylinder engine is evident.

For comparison of the moments damped in the frame, we form the ratios of the magnitudes of these moments to the moment of a 4-cylinder engine, computed by (23). First we express the quantity \(a mr \omega^2\) through quantities common to all engines. By equality (30)

\[ mr\omega^2 = \frac{C}{Z}. \]

The quantity “\(a\)” for engines close to one another in number of cylinders may, without large error, be taken as directly proportional to the cylinder diameter; but

\[ D = \frac{E}{\sqrt{Z}}. \]

Thus

\[ a mr\omega^2 = \frac{C^1}{\sqrt{Z^3}}. \]

Comparing the moment (24) for the first case with the moment of a 4-cylinder engine, we obtain

\[ \frac{M}{M_4} = \frac{1.73}{1.45} \sqrt{\frac{4^3}{5^3}} = 0.855. \]

Accordingly, for the remaining five cases we obtain the ratios:

\[ 0.76;\ 0.668;\ 0.427;\ 0.382;\ 0.247. \]

Thus, with respect to balancing, the best arrangements are arrangements II and III. The second group with \(X_2 = 0\) is less favorable as regards balancing. If, using the preceding method, one determines the reduced arm of the free moment for an equidistant 3-cylinder engine, then it proves to be equal to \(3.6a\), i.e. smaller than for the 2nd group. The moment balanced in the frame, or crankcase, may be considered especially small in the last two arrangements. This moment is noticeably lower also in the other arrangements in comparison with a 4-cylinder motor, but if one takes into account the smaller height of the crankcase (or frame) of the five-cylinder engine,

then no great advantages are to be expected here from the five-cylinder engine. In this respect, the advantages of the 5-cylinder engine will be more noticeable when compared with a six-cylinder design. A conclusion analogous to the preceding one leads to the following expression for the magnitude of the damping moment in a six-cylinder engine.

\[ M = 1.88\, a m r \omega^2. \]

With respect to moment (23), this will be

\[ \frac{M}{M_a}=\frac{1.88}{1.45}\sqrt{\frac{4^3}{6^3}}=0.705. \]

In designs II and III the moments are somewhat smaller, but the crankcase is certainly stiffer. In the other arrangements, except the first, the difference in favor of the five-cylinder design is considerably more noticeable.

Uniformity of Running

Four- and five-cylinder engines may also be compared with respect to the uniformity of running.

The following assumptions are adopted as the basis of the comparison below:

The indicated power of the engines is 60 PS.

The pistons are taken to be cast iron and aluminum.

The number of revolutions for the 4-cylinder engine is taken as 1200 rpm and 2400 rpm.

The mean speeds of the pistons of the 5-cylinder engines must be the same as for the 4-cylinder engines, i.e. 6.04 m/s and 9.6 m/s.

The reduced weight referred to the piston of the 4-cylinder engine at \(n=1200\) is, for cast iron, \(0.042\ \mathrm{kg/cm^2}\), and for aluminum, \(0.021\ \mathrm{kg/cm^2}\).

The ratio of stroke to diameter is the same for all engines and is equal to 1.2.

For all engines \(\mu\) is the same and equal to 0.25.

The mean indicated pressure, the same for all engines, is taken from one of the existing diagrams as \(5.96\ \mathrm{kg/cm^2}\).

On the basis of the assumptions made, Table I has been compiled.

TABLE I

Engine	4-cylinder	4-cylinder	4-cylinder	4-cylinder	5-cylinder	5-cylinder	5-cylinder	5-cylinder
Piston	Cast iron	Al.	Cast iron	Al.	Cast iron	Al.	Cast iron	Al.
\(D\) mm	126	126	100	100	112.8	112.8	89.5	89.5
\(S\) mm	151	151	120	120	135.4	135.4	107.5	107.5
\(n\) rpm	1200	1200	2400	2400	1340	1340	2680	2680
\(mr\omega^2\) kg	638	319	1015	508	510	255	81.2	40.6

In drawing 23 a constructed diagram of tangential forces is given for the 4-cylinder engine at \(n=1200\), “A” for the case of a cast-iron piston and “B” for an aluminum one. For \(n=2400\), the corresponding diagrams are given in drawing 24. The influence of the piston masses on the course of the lines \(T\) in the 4-cylinder engine is enormous. In the 5-cylinder engine, on the contrary, it is quite negligible. In drawing 25

represented by the curve of tangential forces arising from the inertia forces of the moving pistons in a 5-cylinder engine. The result changes very little also in the case when the pistons are of unequal weight, as occurs in arrangements II and V. A change in the piston material in a 5-cylinder engine likewise has no noticeable effect on the course of the curve \(T\).

The curves of tangential forces for a 5-cylinder engine at \(n = 1340\) and \(n = 2680\) are given in Figs. 26 and 27. The scales along the axes of abscissas and ordinates are the same for all the curves \(T\) shown. To find the mass of the flywheel, it is necessary to find the magnitudes of the excess areas \(F\) in kgmt., and then determine the mass from the relation \(F = M \delta v^2\). Taking in all cases the same degrees of non-uniformity of running \(\delta\) and masses reduced to points having the same circumferential velocities, we obtain Table II.

TABLE II.

Engine	4-cylinder	4-cylinder	4-cylinder	4-cylinder	5-cylinder	5-cylinder
Piston	Cast iron	Al.	Cast iron	Al.	Cast iron or Aluminum	Cast iron or Aluminum
\(n\) rpm	1200	1200	2400	2400	1340	2680
\(F\ \mathrm{mm}^2\)	583	216	1123	419	358	168
\(M\)	100	37	192.6	77	59.7	28.8
\(\Delta M_z\)	100	57.1	206	76	76	40

From the table it is evident that a 5-cylinder engine of the same power as a 4-cylinder one has, in general, a smaller flywheel for one and the same \(\delta\), the difference being very noticeable at high piston speeds.

The curves of the tangential forces \(T\) shown in the drawings represent, on a certain scale, also the curves of variation of the transverse free moments which cause acceleration of the rotating masses, and so forth, and which give rise to the corresponding reaction moment in the engine bearings. Determining the difference between the maximum and minimum moments \(\Delta M_z\), we obtain their relative values, given in the lower row of the table. Here the 5-cylinder engine may be considered better balanced, although for the 4-cylinder engine it is also possible to choose the weight of the pistons very advantageously, so that the transverse free moment becomes small.

With reference to the table in which the principal results obtained above have been brought together, we note that the introduction of self-balancing moments of the 2nd order, even when balancing one half of the moments of the 1st order, brings no advantages with respect to the balancing of the engine. On the contrary, a definitely poorer balance must be noted here. A positive quality of the 2nd group with \(X_2 = 0\) may be considered to be that the moment from the inertia forces, which dies out in the crankcase material, is not large. In case VI, for example, it is 4 times smaller than in a 4-cylinder motor and almost 3 times smaller than in a 6-cylinder one. The second group differs from the first by \(X_1 = 0\), and quite substantially, by the order of firing. The distribution of loads on the bearings, as well as the magnitude of torsional vibrations of the shaft, will depend on this order. The solution of questions concerning torsional vibrations of the shaft and bearing reactions, not considered in the present article, may have a substantial influence on

the choice of one group or the other, if the question of balancing is of secondary importance. If one follows the frequently stated rule of arranging the ignitions in the cylinders so that the sections of shaft between cylinders with adjacent ignitions are as short as possible and as nearly equal to one another as possible, then preference should be given to the first group.

Arrangements I and III with an extended shaft will obviously be worse in both groups, since in these cases the engine is both less well balanced and heavier, and, with the greater shaft length, its torsional vibrations will be more noticeable.

Arrangements II and III may be regarded as almost equivalent in respect to balance, but arrangement III has the important advantage of greater simplicity, since the pistons here are identical.

In addition to the better balancing of the five-cylinder engine (first group) in comparison with the four-cylinder engine, it also has advantages with regard to smoothness of running, especially noticeable at high piston speeds. Its advantages are also indisputable with regard to ease of starting, which will be important, for example, in a marine engine. Among the disadvantages, besides the increase in the number of parts, one may note the more complicated machining of the shaft.

For the article by Ya. V. Ivannikov.

Table of Variants of Crankshaft Arrangement in a 5-Cylinder Engine

Variant	Arrangement of cranks	Free force, 1st order	Free force, 2nd order	Free moment, 1st order	Free moment, 2nd order	Relative free force \[%\]	Reduced arm \(h = M/R\)	Relative closing moment	Distance between cranks	Firing order
—	[[diagram: end view and developed crank arrangement; cylinders marked 1, 2, 3, 4]]	0	\(4{,}42\,mr\omega^2\)	0	0	100	—	100	—	[[diagram: order shown by connected points labeled 1, 2, 3, 4]]
I	[[diagram: end view and developed crank arrangement; cylinders marked 1, 2, 3, 4, 5]]	0	0	0	\(6{,}88\lambda_2amr\omega^2\)	0	\(1{,}37a\)	85	1,31	[[diagram: connected points labeled 1, 2, 3, 4, 5]]
II	[[diagram: end view and developed crank arrangement; cylinders marked 1, 2, 3, 4, 5]]	0	\(0{,}618\lambda_2mr\omega^2\)	0	\(6{,}15\lambda_2amr\omega^2\)	12,3	\(1{,}23a\)	76	1	[[diagram: two points]]
III	[[diagram: end view and developed crank arrangement; cylinders marked 1, 2, 3, 4, 5]]	0	0	\(0{,}45amr\omega^2\)	\(4{,}98\lambda_2amr\omega^2\)	0	\(1{,}29a\)	67	1	[[diagram: two points]]
IV	[[diagram: end view and developed crank arrangement; cylinders marked 1, 2, 3, 4, 5]]	0	0	\(6{,}88amr\omega^2\)	0	0	\(6{,}12a\)	43	1,31	[[diagram: connected points labeled 1, 2, 3, 4, 5]]
V	[[diagram: end view and developed crank arrangement; cylinders marked 1, 2, 3, 4, 5]]	\(0{,}618mr\omega^2\)	0	\(6{,}15amr\omega^2\)	0	55	\(5{,}47a\)	38	1	[[diagram: two points]]
VI	[[diagram: end view and developed crank arrangement; cylinders marked 1, 2, 3, 4, 5]]	0	0	\(4{,}98amr\omega^2\)	\(0{,}45\lambda_2amr\omega^2\)	0	\(4{,}43a\)	25	1	[[diagram: two points]]

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

Fig. 5.

Fig. 6.

Fig. 7.

Fig. 8.

Fig. 9.

Fig. 10.

Fig. 11.

Fig. 12.

Fig. 13.

For the article by Prof. A. V. Krasnikov, “On the Arrangement of Crankshafts in a Five-Cylinder Engine.”

Fig. 14.

Visible labels in the diagram: 1, 2, 3, 4, 5; angle marks \(a\), \(2a\).

Fig. 15.

Visible labels in the diagram: 1, 2, 3, 4, 5; angle marks \(a\), \(2a\).

Fig. 16.

Visible labels in the diagram include: \(0\), \(2\pi\), \(\pi/2\), \(0.45\), \(1.142\,\mathrm{cm}\,w^2\), \(0.982\,K\), \(5a\), \(35a\), \(0.014x\).

Fig. 17.

Visible labels in the diagram: 1, 2, 3, 4, 5; angle marks \(a\), \(2a\), \(116^\circ\), \(39^\circ\).

Fig. 18.

Visible labels in the diagram: 1, 2, 3, 4; \(1.5a\).

Fig. 19.

Visible labels in the diagram: I, II; 3, 4, 5; \(0.5\).

Fig. 20.

Visible labels in the diagram: I, II; 2, 3, 4, 5.

Fig. 21.

Visible labels in the diagram include: \(C\), \(H\), \(x\), \(e\).

Fig. 22.

Visible labels in the diagram include: \(a\), \(b\), \(x\), \(0.014x\), \(0.0286x\).

To the article by Docent A. V. Kvasnikov,
“On the Arrangement of Crankshafts.”

Drawing 23.

Drawing 25.

Drawing 24.

Drawing 26.

Drawing 27.

1. A. Kvasnikov, “Determination of the Unbalance of Piston Machines,” 1923, Izvestiya STI, vol. 48 (II), issue 4. ↩