On the Static and Flywheel Moments of the Drive of Connecting Rod-Crank Mechanisms
Abstract Generated abstract
This paper reexamines the reduction of static torque and flywheel moment to the motor shaft in electric drives using connecting rod crank mechanisms, where these quantities vary with crank angle and affect transient calculations and motor sizing. Using force relations, instantaneous centers of rotation, equality of power, and equality of kinetic energy, it derives corrected expressions for the reduced static moment and flywheel moment, including losses in the crank mechanism and transmission. The author argues that earlier formulas by Kunitskii and Popov are inconsistent with the sine theorem and with the basic moment relation at the driven crank. The resulting dependencies are periodic functions of crank angle, with generally unequal amplitudes in the two half periods, including during idle stroke conditions often previously treated as constant.
Full Text
ON THE STATIC AND FLYWHEEL MOMENTS OF THE DRIVE OF CONNECTING-ROD CRANK MECHANISMS
L. I. GANDZHA
It is known that the transient regimes of electric drives of mechanisms with connecting-rod crank transmissions belong to the category of the most difficult to investigate, since such mechanisms determine the dependence of the static and flywheel moments, reduced to the motor shaft, on the path traveled (the angle of rotation of the motor rotor).
Therefore, the correct determination of the functions \(M_{st}\) and \(GD^2\) with respect to the path is of fundamental importance from the point of view of calculations of transient regimes, load diagrams, the drive, and the subsequent determination of the motor power.
In Fig. 1 a diagram is given of a connecting-rod crank mechanism considered by N. P. Kunitskii [1] and borrowed entirely from there. When the crank \(BO\) is turned through \(360^\circ\) through the reducer by motor \(D\), crank \(AO_1\) will perform its function, executing, by means of connecting rod \(AB\), an oscillatory motion within the limits of the angle \(\gamma\).
Fig. 1
If \(Q\) denotes the tangential force on pin \(A\) of crank \(AO_1\), \(Q'\) the force acting along the axis of the connecting rod, and \(M_{st.o1}\) the useful static moment on the working shaft \(O_1\) of the mechanism, then
\[ M_{st.o1}=Q r_1=Q' r_1 \sin(\alpha_1-\beta), \tag{1} \]
whence
\[ Q'=\frac{Q}{\sin(\alpha_1-\beta)}. \tag{2} \]
Transferring the force \(Q'\) along the axis of the connecting rod to pin \(B\) of crank \(BO\) makes it possible to determine the tangential force at pin \(B\):
\[ T = Q' \sin(\alpha-\beta) \tag{3} \]
or, substituting (2) into (3),
\[ T = Q \frac{\sin(\alpha-\beta)}{\sin(\alpha_1-\beta)} . \tag{4} \]
The tangential force \(T\) can be determined in another way. If the directions of the cranks \(BO\) and \(AO_1\) are extended until they intersect at point \(C\), then the latter is the instantaneous center of rotation of the system. Then the circumferential velocities \(v_A\) and \(v_B\) of pins \(A\) and \(B\) of the cranks can be found from the expressions:
\[ \begin{aligned} v_A &= \omega \cdot AC \\ v_B &= \omega \cdot BC \end{aligned} \Biggr\}, \tag{5} \]
where \(\omega\) is the instantaneous angular velocity of the system.
Neglecting friction losses in the links of the connecting-rod crank mechanism, we obtain the equality of the powers developed by the forces \(Q\) and \(T\)
\[ Q \cdot v_A = T \cdot v_B ; \tag{6} \]
from (5) we have:
\[ \frac{v_A}{v_B} = \frac{AC}{BC} ; \tag{7} \]
moreover, from (6) we have:
\[ T = Q \frac{v_A}{v_B} . \tag{8} \]
Then, on the basis of (7) and (8), we obtain
\[ T = Q \frac{AC}{BC}, \tag{9} \]
and on the basis of (4) and (9)
\[ \frac{AC}{BC} = \frac{\sin(\alpha-\beta)}{\sin(\alpha_1-\beta)} . \tag{10} \]
Turning to triangle \(ABC\), we find
\[ \sin \angle BAC = \sin \left[180^\circ - (\alpha_1-\beta)\right] = \sin(\alpha_1-\beta). \tag{11} \]
Comparing (10) and (11), we see that equality (10) represents the familiar sine theorem as applied to triangle \(ABC\). Thus, equality (10) confirms the correctness of the derivations obtained.
The static moment reduced to axis \(O\) will be equal to
\[ M_{\mathrm{st}\,o} = \frac{Tr}{\eta_{\sh}}, \tag{12} \]
where \(\eta_{\sh}\) is the efficiency of the connecting-rod crank mechanism.
Taking (4) into account, we obtain
\[ M_{\mathrm{st}\,o} = \frac{Q \cdot r_1}{\eta_{\sh}} \frac{\sin(\alpha-\beta)}{\sin(\alpha_1-\beta)} . \tag{13} \]
The static moment, reduced to the motor shaft, will be equal to
\[ M_{st}=\frac{Q\cdot r}{\eta_{\sh}\eta_{r}\cdot i}\, \frac{\sin(\alpha-\beta)}{\sin(\alpha_1-\beta)}, \tag{14} \]
where \(\eta_{r}\) is the efficiency and \(i\) is the transmission ratio of the reducer.
Let \((GD^2)_{\mathrm{red},\,o_1}\) denote the flywheel moment of all parts of the mechanism driven relative to the connecting rod, reduced to the shaft \(O_1\).
Then the weight \(G_A\) of the system, reduced to pin \(A\), will be:
\[ G_A=\frac{(GD^2)_{\mathrm{red},\,o_1}}{(2r_1)^2}. \tag{15} \]
This weight may be reduced to pin \(B\) on the basis of equality of live forces, if the losses in the links of the connecting-rod crank mechanism are neglected. Then
\[ \frac{G_A v_A^2}{2g}=\frac{G_B v_B^2}{2g}, \tag{16} \]
where \(g\) is the acceleration due to gravity.
The weight \(G_B\) of the system, reduced to pin \(B_1\), will be equal to
\[ G_B=G_A\,\frac{v_A^2}{v_B^2} =G_A\,\frac{(AC)^2}{(BC)^2}, \]
or, taking (10) into account,
\[ G_B=G_A\,\frac{\sin^2(\alpha-\beta)}{\sin^2(\alpha_1-\beta)}. \tag{17} \]
The flywheel moment of the mechanism, reduced to the axis \(O\), will be equal to
\[ (GD^2)_{\mathrm{red},\,o}=G_B(2r)^2. \tag{18} \]
Substituting (17) into (18) and taking into account the losses in the links of the connecting-rod crank mechanism, we obtain
\[ (GD^2)_{\mathrm{red},\,o} =4r^2\,\frac{G_A}{\eta_{\sh}}\, \frac{\sin^2(\alpha-\beta)}{\sin^2(\alpha_1-\beta)}. \tag{19} \]
Denoting by \((GD^2)_o\) the total flywheel moment of the reducer, coupling, brake pulley, and the armature (or rotor) of the motor, we obtain the flywheel moment reduced to the motor shaft
\[ (GD^2)_{\mathrm{red}} =(GD^2)_o+ 4r^2\,\frac{G_A}{\eta_{r}\cdot\eta_{\sh}\cdot i}\, \frac{\sin^2(\alpha-\beta)}{\sin^2(\alpha_1-\beta)}. \tag{20} \]
This case was considered by N. P. Kunitskii [1] and Prof. V. K. Popov [2]. The indicated authors arrived at completely incorrect results, obtaining
\[ M_{st}=\frac{G\cdot r}{\eta\cdot i}\, \sin(\alpha_1-\beta)\cdot\sin(\alpha-\beta), \]
\[ (GD^2)_{\mathrm{red}} =(GD^2)_o+ 4r^2\,\frac{G_A}{i^2}\, \sin(\alpha_1-\beta)\sin(\alpha-\beta), \]
\[ Q'=Q\cdot\sin(\alpha_1-\beta)\quad\text{and} \]
\[ \frac{ac}{bc}=\sin(\alpha_1-\beta)\cdot\sin(\alpha-\beta), \]
where \(\eta=\eta_{\sh}\cdot\eta_{r}\).
The latter of the conclusions cited by N. P. Kunitskii and Prof. V. K. Popov contradicts the sine theorem as applied to triangle \(ABC\), which is not difficult to establish by comparing it with expression (10).
If one takes into account the third equality obtained by the above-mentioned authors, then, as follows from Fig. 1, we obtain:
\[ Q' r_1 \cdot \sin(\alpha_1-\beta)=Qr_1, \]
or
\[ Q\cdot \sin(\alpha_1-\beta)\cdot r_1\cdot \sin(\alpha_1-\beta)=Qr_1, \]
or
\[ \sin^2(\alpha_1-\beta)=1, \]
which is absurd.
The reason for the incorrect results arrived at by the above-mentioned authors in [1], [2] is that in their reasoning they did not proceed from equality (1).
In addition, in the expression for \((GD^2)_{pr}\) they did not take into account the losses in the kinematic chain from the output shaft \(O_1\) to the motor shaft.
It follows from expressions (14) and (20) that \(M_{st}\) and \((GD^2)_{pr}\) are periodic functions of the angle \(\alpha\).
If we assume that \(M_{st.O_1}=\text{const}\), which may occur, for example, during the idle stroke of the mechanism, then
\[ M_{st.O_1}=Qr_1=\text{const} \]
and
\[ Q=\text{const}. \]
During the idle stroke we shall have
\[ Q_{xx}=\frac{M_{xxO}}{r_1} \]
and then, according to (14),
\[ M_{xx\,pr}=\frac{Q_{xx}\cdot r}{\eta_{\sh}\eta_p\cdot i}\, \frac{\sin(\alpha-\beta)}{\sin(\alpha_1-\beta)} = M_{xx}\frac{\sin(\alpha-\beta)}{\sin(\alpha_1-\beta)}, \tag{21} \]
where
\[ M_{xx}=\frac{Q_{xx}\cdot r}{\eta_{\sh}\eta_p\cdot i}=\text{const}. \]
In Fig. 2 the dependence \(M_{xx\,pr}=f(\alpha)\) according to (21) is represented graphically.
Fig. 2
It follows from it that the static moment reduced to the motor shaft is, in the general case, a periodic function; moreover, the amplitude values in the positive and negative half-periods are not the same.
This curve refutes the analogous curve given by the above-mentioned authors, where, in the idle-stroke portion, the static moment is represented as a constant quantity.
In the same Fig. 2 the curve \((GD^2)_{pr}=f(\alpha)\) is shown according to expression (20).
This curve also differs from the analogous curve given in [1] and [2], and shows that the amplitudes of the curve \((GD^2)_{pr}\) in the first and second half-periods are not the same.
Conclusions
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In the present article an error is revealed in the derivation of the dependencies \(M_{\mathrm{st.\,pr.}} = f(\alpha)\) and \((GD^2)_{\mathrm{pr.}} = f_1(\alpha)\) for the common kinematic drive scheme according to Fig. 1.
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Correct formulas are given for calculating the indicated dependencies, taking into account losses in the links of the connecting-rod crank mechanism.
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It is shown that the amplitude values of both \(M_{\mathrm{st.\,pr.}}\) and \((GD^2)_{\mathrm{pr.}}\) in the first and second half-periods are, in the general case, different.
Literature
- N. P. Kunitskii—Electrical equipment of rolling and heat-treatment shops, 1938.
- V. K. Popov—Fundamentals of Electric Drive, 1945.