MECHANICS

Academician I. I. ARTOBOLEVSKII

LINEARLY ENVELOPING COUPLER CURVES OF MECHANISMS WITH A PRISMATIC PAIR

The present study is devoted to the theory of linearly enveloped coupler curves of mechanisms with three revolute pairs and one prismatic pair.

Consider the slider-crank mechanism \(ABC\) (Fig. 1). In the plane of the connecting rod choose an arbitrary straight line \(u — u\). As was shown earlier \((^1)\), it may be assumed that this straight line passes through point \(B\) at an angle \(\gamma\) to the connecting rod \(BC\). Drop from point \(A\) the perpendiculars \(AF\) and \(AE\) to the straight lines \(u — u\) and \(BC\).

Introduce the notation: \(AB=a\), \(BC=b\), \(AE=p_1\), \(AF=p\), and \(b^2-a^2=m^2\). Then the segment \(p_1\) can be represented as

\[ p_1=\left(b\sin\theta_1\pm\sqrt{b^2\sin^2\theta_1-m^2}\right)\cos\theta_1. \tag{1} \]

The sign of the second term of expression (1) is chosen depending on the assembly conditions of the mechanism.

The equation of the family of straight lines \(u — u\) will be \(x\cos\theta+y\sin\theta=p\), or

\[ x\cos(\theta_1+\gamma)+y\sin(\theta_1+\gamma) = p_1\cos\gamma+\sqrt{a^2-p_1^2}\sin\gamma. \tag{2} \]

Determine the derivative of equation (2) with respect to the angle \(\theta_1\):

\[ y\cos(\theta_1+\gamma)-x\sin(\theta_1+\gamma) = \frac{\partial p_1}{\partial\theta_1}\cos\gamma - \frac{p_1}{\sqrt{a^2-p_1^2}}\frac{\partial p_1}{\partial\theta_1}\sin\gamma. \tag{3} \]

The partial derivative \(\partial p_1/\partial\theta_1=\Psi\) is equal to

\[ \Psi=R+T \tag{4} \]

where \(R=b\cos2\theta_1\) and

\[ T=\frac{(b^2\cos2\theta_1+m^2)\sin\theta_1}{\sqrt{b^2\sin^2\theta_1-m^2}}. \]

From equations (2) and (3), taking equation (4) into account, we obtain the parametric equations for the linearly enveloping coupler curve of the slider-crank mechanism (Fig. 1) in the form \(x=x(\theta_1)\) and \(y=y(\theta_1)\):

\[ \begin{aligned} x={}&[p_1\cos(\theta_1+\gamma)-\Psi\sin(\theta_1+\gamma)]\cos\gamma +\\ &+\left[\sqrt{a^2-p_1^2}\cos(\theta_1+\gamma) +\frac{p_1\Psi}{\sqrt{a^2-p_1^2}}\sin(\theta_1+\gamma)\right]\sin\gamma, \\[6pt] y={}&[p_1\sin(\theta_1+\gamma)+\Psi\cos(\theta_1+\gamma)]\cos\gamma +\\ &+\left[\sqrt{a^2-p_1^2}\sin(\theta_1+\gamma) -\frac{p_1\Psi}{\sqrt{a^2-p_1^2}}\cos(\theta_1+\gamma)\right]\sin\gamma. \end{aligned} \tag{5} \]

The current coordinates \(x_M, y_M\) of the curve linearly enveloping the straight line \(u—u\) are easily determined graphically (Fig. 1). To do this, from the instantaneous center of rotation \(P\) of the connecting rod \(BC\), one must drop the perpendicular \(PM\) to the straight line \(u—u\). The coordinates \(x_{M'}, y_{M'}\) of the curve linearly enveloping the connecting rod \(BC\) are determined if the perpendicular \(PM'\) is dropped to the direction \(BC\).

Fig. 1

The angle \(\alpha\) of rotation of the link \(AB\) is related to the angle \(\theta_1\) by the condition

\[ \cos \theta_1 = \frac{a}{b}\sin \alpha \tag{6} \]

The parametric equations of the curve linearly enveloping the connecting rod \(BC\) will have the form

\[ \begin{aligned} x &= p_1 \cos \theta_1 - \Psi \sin \theta_1,\\ y &= p_1 \sin \theta_1 + \Psi \cos \theta_1, \end{aligned} \tag{7} \]

since for the straight line \(BC\) the angle \(\gamma = 0\).

If the angle \(\gamma = 90^\circ\), then the straight line \(u—u\) passes into the straight line \(u'—u'\) (Fig. 1). The coordinates \(x_{M''}, y_{M''}\) of the curve linearly enveloping the straight line \(u'—u'\) will be found if from the point \(P\) the perpendicular \(PM''\) is dropped to the straight line \(u'—u'\). The parametric equations for this curve will be

\[ x = -\sqrt{a^2 - p_1^2}\sin \theta_1 + \frac{p_1\Psi}{\sqrt{a^2 - p_1^2}}\cos \theta_1, \]

\[ y = \sqrt{a^2 - p_1^2}\cos \theta_1 + \frac{p_1\Psi}{\sqrt{a^2 - p_1^2}}\sin \theta_1. \tag{8} \]

If the lengths of the links of the mechanism satisfy the condition \(a=b\), then the motion of the link \(BC\) will have the character of Cardano motion, in which any straight line belonging to the plane of the connecting rod \(BC\) will have, as a linearly enveloping curve, a straight line or a deformed astroid, with equations for the case \(\gamma = 0\)

\[ x = 2a \sin^3 \theta_1,\qquad y = 2a \cos^3 \theta_1. \tag{9} \]

The loci of points \(E\) and \(F\) (Fig. 1) will be the pole curves of the curves linearly enveloping the straight lines \(u—u\) and \(BC\), if point \(A\) is chosen as the pole of the pole curves. The equations of the pole curve formed by point \(E\) will be

\[ x = p_1 \cos \theta_1,\qquad y = p_1 \sin \theta_1, \tag{10} \]

and, respectively, for point \(F\),

\[ x = -\left(p_1 \cos \gamma + \sqrt{a^2 - p_1^2}\sin \gamma\right)\cos(\theta_1+\gamma); \]

\[ y = \left(p_1 \cos \gamma + \sqrt{a^2 - p_1^2}\sin \gamma\right)\sin(\theta_1+\gamma), \tag{11} \]

where \(p_1\) is determined by equation (1).

For the case when \(a=b\), the equations of the pedals generated by the points \(E\) and \(F\) will have, in polar form, the form

\[ p_1=a\sin 2\theta_1; \tag{12} \]

\[ p_1=a\sin(2\theta_1+\gamma). \tag{13} \]

In this case, for any value of the angle \(\gamma\), the equations of the pedals generated by the points \(E\) and \(F\) will be four-petaled roses.

If, in a four-link mechanism with one translational pair, the driven and driving links enter into revolute pairs with the frame, then we obtain the slider mechanism \(ABC\), shown in Fig. 2, for which it is necessary to consider line-enveloping coupler curves for straight lines rigidly connected with the slider.

Choose in the plane of the slider an arbitrary straight line \(u—u\), passing at an angle \(\gamma\) to the direction \(BC\) of the axis of sliding of the slider. From the point \(A\) drop the perpendiculars \(AF\) and \(AE\) to the straight lines \(u—u\) and \(BC\). Introduce the notation \(AB=a\), \(AC=b\), \(AE=p_1\) and \(AE=p\). The segment \(p_1\) will be equal to

\[ p_1=b\cos\theta_1. \tag{14} \]

Fig. 2

The equation of the family of straight lines will be \(x\cos\theta+y\sin\theta=p\), or

\[ x\cos(\theta_1+\gamma)+y\sin(\theta_1+\gamma) =b\cos\theta\cos\gamma+\sqrt{a^2-b^2\cos^2\theta_1}\,\sin\gamma . \tag{15} \]

The derivative with respect to the angle \(\theta_1\) of equation (15) will be

\[ y\cos(\theta_1+\gamma)-x\sin(\theta_1+\gamma) = \frac{b^2\sin\theta_1\cos\theta_1} {\sqrt{a^2-b^2\cos^2\theta_1}}\sin\gamma -b\sin\theta_1\cos\gamma . \tag{16} \]

From equations (15) and (16) we obtain parametric equations for the line-enveloping curve of the slider mechanism (Fig. 2) in the form \(x=x(\theta_1)\) and \(y=y(\theta_1)\):

\[ x=b\cos^2\gamma+ \left[ \sqrt{a^2-b^2\cos^2\theta_1}\cos(\theta_1+\gamma) - \frac{b^2\sin\theta_1\cos\theta_1} {\sqrt{a^2-b^2\sin^2\theta_1}}\sin(\theta_1+\gamma) \right]\sin\gamma, \]

\[ \tag{17} \]

\[ y=b\sin\gamma\cos\gamma+ \left[ \sqrt{a^2-b^2\cos^2\theta_1}\sin(\theta_1+\gamma) + \frac{b^2\sin\theta_1\cos\theta_1} {\sqrt{a^2-b^2\sin^2\theta_1}}\cos(\theta_1+\gamma) \right]\sin\gamma . \]

The current coordinates \(x_M, y_M\) of the curve line-enveloping the straight line \(u—u\) are easily determined graphically if, from the instantaneous center of rotation \(P\) of the slider, the perpendicular \(PM\) is dropped to the straight line \(u—u\).

The angle \(\alpha\) of rotation of the link \(AB\) is related to the angle \(\theta_1\) by the condition

\[ \operatorname{tg}\theta_1=\frac{b-a\cos\alpha}{a\sin\alpha}. \tag{18} \]

Equations (17) will be the equations of an ellipse or a hyperbola. The latter is not difficult to verify if one takes the angle \(\gamma=90^\circ\) and from equations (17)

exclude the angle $\theta_1$. If $b>a$, then the linearly enveloping curve will be the hyperbola

\[ \frac{x^2}{a^2}-\frac{y^2}{b^2-a^2}=1, \tag{19} \]

and if $b<a$—the ellipse

\[ \frac{x^2}{a^2}+\frac{y^2}{a^2-b^2}=1. \tag{20} \]

As was shown earlier ($^2$), the mechanism $ABC$ (Fig. 2) can always be replaced by the equivalent mechanism $A_0B_0C_0$. The point $A_0$ is the foot of the perpendicular dropped from the point $A$ to the straight line passing through the point $C$ ($C_0$) at the angle $90-\gamma$. The point $B_0$ is the foot of the perpendicular $C_0B_0$, dropped from the point $C$ ($C_0$) to the straight line $u-u$. The straight line $u-u$, if it is regarded as belonging to the mechanism $A_0B_0C_0$, will envelop the same ellipse or hyperbola as in the case when it belongs to the mechanism $ABC$. The point $A_0$ is the center of the ellipse or hyperbola, and the point $C$ ($C_0$) is one of the foci. Thus, the linearly enveloping curves of the slider mechanism will always be ellipses or hyperbolas. If the angle $\gamma=0$, then the point $C$ will be the envelope. If the lengths of the links satisfy the condition $b=a$, then the envelope will also be a point, but lying to the left of the point $A$ at the distance $a$.

The equation of the pedal of the curve linearly enveloping the straight line $u-u$, if the point $A$ is chosen as the pole of the pedal, will have, in polar form, the form

\[ p=b\cos(\theta-\gamma)\cos\gamma+\sqrt{a^2-b^2\cos^2(\theta-\gamma)}\sin\gamma. \tag{21} \]

If the angle $\gamma=90^\circ$, then equation (21) takes the form

\[ p=\sqrt{a^2-b^2\cos^2\theta}. \tag{22} \]

Passing to a rectangular coordinate system, we obtain

\[ (x^2+y^2)^2=a^2x^2\pm b^2y^2. \tag{23} \]

This is a curve of the 4th order, the so-called pedal curve of the type of Vitkov’s curves ($^2$). The upper sign in the equation will correspond to the pedal curve of an ellipse, and the lower sign to the pedal curve of a hyperbola.

Received
15 IV 1961

CITED LITERATURE

$^1$ I. I. Artobolevskii, DAN, 132, No. 1 (1960). $^2$ I. I. Artobolevskii, Theory of Mechanisms for Reproducing Plane Curves, Publishing House of the Academy of Sciences of the USSR, 1959.