Selection of a Rational Gear Ratio for a Reducer in Mechanisms with a Trapezoidal Velocity Profile

I. P. Trofimuk

(Presented by the scientific seminar of the Electromechanical Faculty)

It is known that the time required to traverse a prescribed displacement path depends on the ratio of the kinetic-energy reserves of the motor and of the working machine, which is determined by the gear ratio of the reducer. Therefore, in designing drives that are optimal with respect to response speed, the attention of researchers is drawn to the problem of determining the gear ratio which, all other conditions being equal, provides maximum speed of response for a given motor and mechanism.

The choice of gear ratio for a triangular velocity profile is known. The question of choosing the gear ratio in the case when the motor reaches steady-state speed is more difficult. This question has been discussed in the literature; however, the proposed methods are rather laborious and of little use for engineering calculations.

Below a new, very simple approximate method is given for determining the optimal gear ratio of a reducer for this case of motion.

The notation adopted is as follows:

$M_\lambda$ — starting or braking torque of the motor;

$M_n$ — rated torque of the motor;

$M_p$ — resisting torque on the shaft of the working machine;

$M_c$ — resisting torque referred to the motor shaft;

$J_o$ — moment of inertia of the motor and coupling;

$J_p$ — moment of inertia of the working machine with respect to its own shaft;

$\omega_o$ — rated speed of the motor (angular);

$\omega_p$ — rated speed of the working machine (angular);

$i$ — gear ratio of the reducer;

$t_j$ — time of the transient starting–braking processes;

$t_o$ — time of operation at rated speed;

$\alpha$ — path (angle of rotation) of the shaft of the working machine in radians;

$\alpha_j$ — path traversed during the transient processes;

\(x_0\) is the distance traveled during operation at nominal speed.

The remaining notation is introduced in the text.

Suppose that a distance \(\alpha\) is specified, the processing time of which determines the speed of response of the mechanism.

Let us assume that we have selected the optimal gear ratio for a triangular velocity diagram. But the specified distance \(\alpha\) of the working machine is such that the motor operates for some time \(t_0\) at nominal speed. A similar process is shown by curve 1 (Fig. 1). The technical and design data of many mechanisms are such that they allow the maximum speed to be increased within larger or smaller limits, and the limit of this increase is not precisely established. If the gear ratio is reduced somewhat, then, owing to an increase in the average speed, the entire process will be completed faster; see curve 2 (Fig. 1). However, with a further reduction of the gear ratio, the slow course of the transient starting-braking processes leads to a significant increase in the time required to process the specified distance, despite the higher steady-state speed.

Fig. 1

Let us determine the time required to process the specified distance:

\[ t_{\Sigma}=t_j+t_0, \tag{1} \]

where

\[ t_0=\frac{\alpha_0}{\omega_p}. \]

Since

\[ \alpha_0=\alpha-\alpha_j \quad \text{and} \quad \alpha_j=\frac{\omega_p t_j}{2}, \]

then:

\[ t_{\Sigma}=\frac{\alpha}{\omega_p}+\frac{t_j}{2}, \tag{2} \]

The time of the starting-braking transient processes is determined from the basic equation of drive motion by the known formula:

\[ t_j=\frac{2\omega_0 J_0}{M_\lambda}\cdot \frac{i^2+I}{i^2-m^2}, \quad \text{where} \quad I=\frac{J_p}{J_0}, \quad m=\frac{M_p}{M_\lambda}; \tag{3} \]

Then:

\[ t_{\Sigma}=\frac{\alpha\cdot i}{\omega_0} + \frac{\omega_0 J_0}{M_\lambda}\cdot \frac{i^2+I}{i^2-m^2}. \tag{4} \]

To determine the optimal gear ratio, we differentiate (4) with respect to \(i\) and set the derivative equal to zero; then we obtain:

\[ \frac{dt_{\Sigma}}{di} = \frac{\omega_0 J_0}{M_\lambda}\cdot \frac{2i(i^2-m^2)-(i^2+I)2i}{(i^2-m^2)^2} + \frac{\alpha}{\omega_0} =0 \tag{5} \]

Transform (5) into the form:

\[ A \cdot \frac{1}{(i^2-m^2)^2}=1, \tag{6} \]

where

\[ A=\frac{2\omega_0^2 J_n (m^2+1)}{aM_\lambda}. \tag{7} \]

Taking \(i^4\) out in the denominator and introducing the quantity \(\mu=\frac{m}{i}\), we then have:

\[ \frac{A}{i^4(1-\mu^2)^2}=1 \tag{8} \]

where

\[ \mu=\frac{m}{i}. \tag{8a} \]

Equation (8) makes it possible to find an approximate graphical solution easily. Write (8) in the form:

\[ i_{\mathrm{opt}}=K\sqrt[3]{A} \tag{9} \]

where

\[ K=\sqrt[3]{\frac{1}{(1-\mu^2)^2}} . \tag{9a} \]

The value \(K\) depends only slightly on the gear ratio.

Thus, if we take \(M_\lambda=2M_n\) and \(M_c=(0\div1.2)M_n\),

then

\[ \mu=\frac{M_c}{M_\lambda}=0\div0.6 \quad\text{and}\quad K=1\div1.35. \]

Introducing the notation \(i_0=\sqrt[3]{A}\), we obtain

\[ i_{\mathrm{opt}}=Ki_0=(1\div1.35)i_0; \]

to solve (8), we rewrite (8a) and (9a) in the form:

\[ \mu=f(\mu_0,K) \tag{10} \]

where

\[ \mu_0=\frac{m}{i_0}. \tag{10a} \]

\[ K=\varphi(\mu) \tag{11} \]

Fig. 2

The function \(K=\varphi(\mu)\) depends only on the current value \(\mu\) and is represented by a single curve of the form shown in Fig. 2. The function \(f(\mu_0,K)\) represents a family of branches of hyperbolas, specific to each individual case. Each of these branches is completely determined by the value \(\mu_0\):

\[ \mu_0=\frac{m}{i_0}=\frac{m}{\sqrt[3]{A}} \tag{10a} \]

The form of the function \(f(\mu_0,K)\) is shown in Fig. 3. From expressions (9) and (10) it is evident that the value \(i_{\mathrm{opt}}\) can be found by solving (10) and (11) jointly, that is, from the coordinates \(\mu_1\) and \(K_1\) of the point of intersection of the curves \(\varphi(\mu)\) and \(f(\mu_0;K)\), as shown in Fig. 4. Indeed, having found \(K_1\), we obtain \(i_{\mathrm{opt}}=K_1 i_0\) with an accuracy determined by the line thickness and the scale of the graph. The matter is simplified by the fact that it is sufficient to have the family of curves \(f(\mu_0K)\) constructed once. Such a family is given in Fig. 5 for intervals \(\Delta K=0.02\), that is,

Fig. 3

Fig. 4

From Fig. 5 one can determine \(i_{\mathrm{opt}}\) with an accuracy up to \(\dfrac{\Delta K}{2}=1\%\). Greater accuracy is not needed in this case, since near \(i=i_{\mathrm{opt}}\) the curve \(t_2(i)\) is very flat.

Fig. 5

Let us represent the quantity \(A\) in a form convenient for engineering calculation; for this purpose we pass from \(\omega_0\) to \(n_{\mathrm{n}}\)—the nominal speed of the motor—

in rpm, from $\alpha$ to $S$—the number of revolutions of the shaft of the working machine, from $J_0$ to $GD_0^2$—the flywheel moment of inertia in kg·m²; then:

\[ A=\frac{n_{\mathrm{н}}^{2}\cdot GD_0^{2}}{11320\,M_{\lambda}\cdot S}\cdot (m^{2}+1) \tag{12} \]

Thus, the process of determining the optimum gear ratio of the reducer for given $\alpha$, $M_p$, and $J_p$ reduces to the following simple operations:

1. A series of motors is selected by known methods.

2. For each motor, the quantity $A$ is determined from (12).

3. The quantities are determined

\[ i_0=\sqrt[3]{A} \]

1. The quantities are determined

\[ \mu_0=\frac{M_p}{M_{\lambda}\cdot i_0} \]

1. Along the abscissa axis of Fig. 5 the value $\mu_0$ is plotted, and a curve parallel to the curves of the family $f(\mu,K)$ is drawn until it intersects the curve $\varphi(K)$, as shown in Fig. 4.

The coordinates of the intersection of the curves are determined.

1. The optimum gear ratios $i_{\mathrm{opt}}=K i_0$ are determined for each motor.

2. From (4) the operating time is determined. One may also use the following formula, which is more convenient for engineering calculation:

\[ t_{\Sigma}=\frac{60\cdot S}{n_{\mathrm{н}}}+\frac{n_{\mathrm{н}}GD_0^{2}}{375\,M_{\lambda}}\cdot\frac{i^{2}+1}{i^{2}-m^{2}}. \tag{13} \]

1. The times are compared and the motor providing the greatest speed of operation is selected.

The simplicity and convenience of applying the method described allow us to recommend it for engineering calculations of the optimum gear ratio.