Proceedings of the Tomsk Technological Institute of Emperor Nicholas II

Vol. XXIV. 1911. No. 4.

On the Question of the Coupler Pole

A. V. Ugarov.

Determination of the exact geometric locus of the points of intersection of conjugate chords in a straight-line generating coupler-curve mechanism

In my paper “The Coupler Pole”*) in finding the exact geometric locus of the points of intersection of conjugate chords for a straight-line generating coupler-curve mechanism, I outlined the following path: the desired geometric locus is found as the result of eliminating the variable parameter \(k\) from equations [59] and [69]**).

Since in equation [69] there are terms under the radical sign that contain the parameter, I reduce this equation to rational form by squaring and obtain equation [80] of a very complicated form.

It is obvious that “the elimination of the irrationality turns equation [69] into the product of two equations, of which one represents the line investigated by us, and the other a new line, parallel to the first and separated from it by a variable distance, since only for \(k=\delta\), when \(R_1=R_2=0\), do both equations represent one and the same line.”

“Thus, we see that, with the exception of one particular case, this new line does not pass through the point of intersection of the conjugate chords and, consequently, is entirely superfluous for finding the geometric locus that interests us.”***)

From the quoted excerpt it is clear that, after eliminating the parameter \(k\), we must decompose the equation obtained into a product

*) Proceedings of the Tomsk Technological Institute, 1909, No. 1.
**) See p. 58.
***) Ibid., p. 59.

equations and determine which of them is the one we need.

Equation [80], despite a number of adopted abbreviated notations, turned out to be very complicated, and therefore, in order to simplify the solution of the question, I introduced a certain arbitrary assumption, by means of which I obtained the locus as a hyperbola.*)

It is obvious that such an approximate solution of the question does not give precise indications of the character and properties of the locus, and if some conclusions have turned out to agree with the truth, others may be erroneous.

To ascertain the possibility of the existence of a coupler pole, we must obtain an exact expression for the locus and investigate it.

The path outlined by us, being essentially correct, will lead us, as has already been said, to equation [80]. After substituting into this equation the parameter \(k^2\), determined from equation [59] and expressed by formula [82], we shall obtain “an equation of the 6th degree with respect to \(x\) and \(y\); moreover, this equation will contain all the decreasing powers of the unknowns.”

“Thus, we arrive at the conclusion that finding the exact locus of the points of intersection of conjugate chords for a rectilinear-generating coupler-curve mechanism leads us to the investigation of an equation of the sixth degree with a very large number of terms.”**)

The present article is a continuation and further development of the path outlined.

In order to decompose the equation of the 6th degree, which is to be obtained, into a product of equations of lower degrees, we shall try to simplify the equations of the lines expressed by formulas [59] and [69].

As we know,***) equation [59] represents the polar of the radical center of three circles—the crank circle and two coupler circles; on the other hand, we know that the polar of the radical center always passes through the pole of the coupler radical axis.

Consequently, if we pass to a new rectangular system of coordinates whose axes are parallel to the axes of the former system, and whose origin coincides with the pole of the coupler radical axis, then equation [59], from which subsequently we must determine the value—

*) p. 67, equation [87].
**) See p. 64.
***) Coupler pole—p. 45.

expression for the parameter \(k\), will have a simpler form than the equation of a line passing through the origin.

The coordinates of the pole of the connecting-rod radical axis in the former system are expressed by formulas [61]:

\[ x_p=\frac{r^2}{n},\qquad y_p=-\operatorname{tg}\vartheta\,\frac{r^2}{n}. \]

Thus, in order to pass to the new coordinate system, we have the formulas:

\[ x'=x+\frac{r^2}{n}\quad \text{and}\quad y'=y-\operatorname{tg}\vartheta\,\frac{r^2}{n}, \]

where \(x'\) and \(y'\) refer to the former system, and \(x\) and \(y\) to the new one.

We take the equation of the polar in the form expressed by equation [81]:

\[ (m^2+k^2)x' + (m^2+k^2-2n^2)y'\operatorname{ctg}\vartheta - 2nr^2 = 0. \]

We transform it, relying on the formulas for passing to the new system.

Then we obtain:

\[ (m^2+k^2)x + (m^2+k^2-2n^2)y\operatorname{ctg}\vartheta + (m^2+k^2)\frac{r^2}{n} - \]

\[ -(m^2+k^2-2n^2)\frac{r^2}{n} - 2nr^2 = 0, \]

which, upon reduction, gives:

\[ (m^2+k^2)x + (m^2+k^2-2n^2)y\operatorname{ctg}\vartheta = 0. \tag{1} \]

We then pass to equation [69].

\[ \left\{(m^2+k^2)(n^2+k^2)-4n^2k^2\cos^2\vartheta\right\}x' +\left\{(m^2+k^2)(k^2-n^2)+ \]

\[ +2n^2(n^2+k^2-2k^2\cos^2\vartheta)\right\}y'\operatorname{ctg}\vartheta +2nR_1R_2+ \]

\[ +\frac{n}{2}\left\{4n^2k^2\cos^2\vartheta-(m^2+k^2)^2\right\}=0. \tag{69} \]

Transforming the coordinates, we have the following change of the constant term:

\[ \frac{r^2}{n}\left\{(m^2+k^2)(n^2+k^2)-4n^2k^2\cos^2\vartheta\right\} -\frac{r^2}{n}\left\{(m^2+k^2)(k^2-n^2)+ \]

\[ +2n^2(n^2+k^2-2k^2\cos^2\vartheta)\right\} +\frac{n}{2}\left\{4n^2k^2\cos^2\vartheta-(m^2+k^2)^2\right\}+ \]

\[ +2nR_1R_2. \]

Having collected like terms, we obtain:

\[ 2nr^{2}(m^{2}-n^{2})+\frac{n}{2}\{4n^{2}k^{2}\cos^{2}\vartheta-(m^{2}+k^{2})^{2}\}+2nR_{1}R_{2}. \]

Substituting the expression obtained into equation (69) and clearing the denominator, we have:

\[ \begin{aligned} &2\{(m^{2}+k^{2})(n^{2}+k^{2})-4n^{2}k^{2}\cos^{2}\vartheta\}x +2\{(m^{2}+k^{2})(k^{2}-n^{2}) \\ &\quad +2n^{2}(n^{2}+k^{2}-2k^{2}\cos^{2}\vartheta)\}y\,\operatorname{ctg}\vartheta +4nR_{1}R_{2}+4nr^{2}(m^{2}-n^{2}) \\ &\quad +n\{4n^{2}k^{2}\cos^{2}\vartheta-(m^{2}+k^{2})^{2}\}=0. \end{aligned} \tag{2} \]

In order to simplify equation (2), we multiply equation (1) by \(-2(n^{2}+k^{2})\) and add it to equation (2). Having done this, we obtain a new equation:

\[ \begin{aligned} &-8n^{2}k^{2}x\cos^{2}\vartheta -4n^{2}(m^{2}-2n^{2}-k^{2}+2k^{2}\cos^{2}\vartheta)y\,\operatorname{ctg}\vartheta +4nR_{1}R_{2} \\ &\quad +4nr^{2}(m^{2}-n^{2}) +n\{4n^{2}k^{2}\cos^{2}\vartheta-(m^{2}+k^{2})^{2}\}=0. \end{aligned} \tag{3} \]

Let us now multiply the equation of the polar (1) by \(4n^{2}\) and add it to equation (3). As a result we obtain an equation of the following form:

\[ \begin{aligned} &4n(m^{2}-k^{2}\cos 2\vartheta)x +8nk^{2}y\cos\vartheta\sin\vartheta +4R_{1}R_{2} +4r^{2}(m^{2}-n^{2}) \\ &\quad +4n^{2}k^{2}\cos^{2}\vartheta-(m^{2}+k^{2})^{2}=0. \end{aligned} \tag{4} \]

Thus, instead of equation (2) we have obtained equation (4), which represents a certain straight line from the pencil of straight lines passing through the same point through which the straight lines expressed by equations (1) and (2) also pass.

Consequently, in order to find the geometric locus, we may operate with equations (1) and (4).

For still greater simplification of the subsequent operations, we shall find the point of intersection of the lines (1) and (4).

Let us denote, in abbreviated form, equations (1) and (4) as follows:

\[ Ax+By+C=0, \tag{5} \]

\[ A_{1}x+B_{1}y=0. \tag{6} \]

Then the coordinates of the point of intersection of these lines will be:

\[ x=-\frac{CB_{1}}{AB_{1}-BA_{1}} \quad\text{and}\quad y=\frac{CA_{1}}{AB_{1}-BA_{1}}. \tag{6'} \]

Each of the expressions thus obtained, taken separately, represents a straight line parallel to one of the axes and belonging to our pencil of lines; moreover, the equations of these lines contain only one unknown quantity each.

Let us take the expression for \(y\). We shall have the equation of a straight line parallel to the \(x\)-axis:

\[ (AB_{1}-BA_{1})y=CA_{1}. \]

Substituting for the quantities \(A\), \(B\), \(C\), etc. their values, after some transformations we obtain an equation of the following form:

\[ \begin{aligned} &4n\left\{(m^{2}-k^{2}\cos 2\vartheta)(m^{2}+k^{2}-2n^{2})-2k^{2}(m^{2}+k^{2})\sin^{2}\vartheta\right\}y\,\operatorname{ctg}\vartheta \\ &\quad -(m^{2}+k^{2})\left\{4r^{2}(m^{2}-n^{2})+4n^{2}k^{2}\cos^{2}\vartheta-(m^{2}+k^{2})^{2}\right\} \\ &\quad -4(m^{2}+k^{2})R_{1}R_{2}=0. \end{aligned} \tag{7} \]

Let us simplify the coefficient of \(y\), expanding the small brackets and collecting like terms.

\[ \begin{aligned} &(m^{2}-k^{2}\cos 2\vartheta)(m^{2}+k^{2}-2n^{2})-2k^{2}(m^{2}+k^{2})\sin^{2}\vartheta=\\ &\quad =(m^{2}+k^{2})(m^{2}-k^{2}\cos 2\vartheta-2k^{2}\sin^{2}\vartheta)-2n^{2}(m^{2}-k^{2}\cos 2\vartheta), \end{aligned} \]

which then gives:

\[ \begin{aligned} &(m^{2}+k^{2})(m^{2}-k^{2})-2n^{2}(m^{2}-k^{2}\cos 2\vartheta)=\\ &\quad =(m^{2}+k^{2})(m^{2}-k^{2})-2n^{2}(m^{2}+k^{2}-2k^{2}\cos^{2}\vartheta), \end{aligned} \]

or, finally:

\[ (m^{2}+k^{2})(m^{2}-k^{2}-2n^{2})+4n^{2}k^{2}\cos^{2}\vartheta. \tag{8} \]

Thus equation (7) assumes the form:

\[ \begin{aligned} &4n\left\{(m^{2}+k^{2})(m^{2}-k^{2}-2n^{2})+4n^{2}k^{2}\cos^{2}\vartheta\right\}y\,\operatorname{ctg}\vartheta \\ &\quad -(m^{2}+k^{2})\left\{4r^{2}(m^{2}-n^{2})+4n^{2}k^{2}\cos^{2}\vartheta-(m^{2}+k^{2})^{2}\right\} \\ &\quad -4(m^{2}+k^{2})R_{1}R_{2}=0. \end{aligned} \tag{9} \]

Before freeing equation (9) from irrationality, we transform it, relying on the value of the factor \((m^{2}+k^{2})\), determined from equation (1).

We have:

\[ (m^{2}+k^{2})x+(m^{2}+k^{2}-2n^{2})y\,\operatorname{ctg}\vartheta=0. \tag{1} \]

Whence:

\[ m^{2}+k^{2}=\frac{2n^{2}y\,\operatorname{ctg}\vartheta}{x+y\,\operatorname{ctg}\vartheta}. \tag{10} \]

and

\[ k^{2}=-\frac{m^{2}x-(2n^{2}-m^{2})y\operatorname{ctg}\vartheta}{x+y\operatorname{ctg}\vartheta}. \tag{11} \]

Substituting expressions (10) and (11) into equation (9), we eliminate the parameter \(k\) from all the rational terms of the equation, which gives us the possibility, after reducing like terms, of obtaining further on comparatively less complicated expressions.

Carrying out the indicated operations, after clearing the denominator and making certain cancellations, we obtain a new equation, to which, for convenience, we give the following form:

\[ \begin{vmatrix} x^{3} & -2nm^{2}\cos^{2}\vartheta+\\ +x^{2}y\operatorname{ctg}\vartheta & 2n\left[m^{2}-n^{2}-(3m^{2}-2n^{2})\cos^{2}\vartheta\right]+\\ +xy^{2}\operatorname{ctg}^{2}\vartheta & 2n\left[2m^{2}-3n^{2}-(3m^{2}-4n^{2})\cos^{2}\vartheta\right]+\\ +y^{3}\operatorname{ctg}^{3}\vartheta & -2n(2n^{2}-m^{2})(1-\cos^{2}\vartheta)+\\ +x^{2} & m^{2}n^{2}\cos^{2}\vartheta-r^{2}(m^{2}-n^{2})+\\ +xy\operatorname{ctg}\vartheta & -2(m^{2}-n^{2})(r^{2}-n^{2}\cos^{2}\vartheta)+\\ +y^{2}\operatorname{ctg}^{2}\vartheta & n^{4}-r^{2}(m^{2}-n^{2})-(2n^{2}-m^{2})n^{2}\cos^{2}\vartheta \end{vmatrix} = R_{1}R_{2}(x+y\operatorname{ctg}\vartheta)^{2}. \tag{12} \]

Now we eliminate the irrationality by squaring both sides of equation (12).

Let us write here the values of the irrational quantities \(R_{1}\) and \(R_{2}\):*)

\[ R_{1}= \sqrt{ (n^{2}+k^{2}-2nk\cos\vartheta)r^{2} - \frac{(m^{2}+k^{2}-2mk\cos\vartheta)^{2}}{4} }, \]

\[ R_{2}= \sqrt{ (n^{2}+k^{2}+2nk\cos\vartheta)r^{2} - \frac{(m^{2}+k^{2}+2mk\cos\vartheta)^{2}}{4} }. \]

The product \(R_{1}^{2}R_{2}^{2}\), obviously, is the product of the quantities under the radicals and will have the form:**)

\[ R_{1}^{2}R_{2}^{2} = \left[(n^{2}+k^{2})^{2}-4n^{2}k^{2}\cos^{2}\vartheta\right]r^{4} - \]

\[ -\left[(n^{2}+k^{2}+2nk\cos\vartheta)(m^{2}+k^{2}-2mk\cos\vartheta)^{2}+\right. \]

\[ \left. +(n^{2}+k^{2}-2nk\cos\vartheta)(m^{2}+k^{2}+2mk\cos\vartheta)^{2} \right]\frac{r^{2}}{4} + \]

\[ +\frac{(m^{2}+k^{2}-2mk\cos\vartheta)^{2}(m^{2}+k^{2}+2mk\cos\vartheta)^{2}}{16}. \]

*) p. 61.
**) p. 62.

Substituting into the expression obtained the value of \(k^2\), determined by formula (11), and collecting like terms, we obtain the following:

\[ \begin{aligned} R_1^2 R_2^2 &= \frac{1}{(x+y\operatorname{ctg}\vartheta)^4} \left| \begin{array}{c|l} x^4 & \begin{aligned} &r^4(n^2-m^2)^2+4n^2m^2r^4\cos^2\vartheta \;-\;\\ &\quad -2r^2n^2m^2(m^2-n^2)\cos^2\vartheta \;+\;\\ &\quad +n^4m^4\cos^4\vartheta \end{aligned} \\[1.0em] x^3y\,\operatorname{ctg}\vartheta & \begin{aligned} &4r^4(n^2-m^2)(2n^2-m^2) \;+\;\\ &\quad +8r^4n^2(2m^2-n^2)\cos^2\vartheta \;-\;\\ &\quad -4r^2n^2(n^4+2m^4-4m^2n^2)\cos^2\vartheta \;-\;\\ &\quad -8r^2m^2n^4\cos^2\vartheta \;+\;\\ &\quad +4n^4m^2(m^2-n^2)\cos^4\vartheta \end{aligned} \\[1.0em] x^2y^2\,\operatorname{ctg}^2\vartheta & \begin{aligned} &r^4(3n^2-m^2)(7n^2-5m^2) \;+\;\\ &\quad +r^4(n^2-m^2)(n^2-m^2-24n^2\cos^2\vartheta) \;-\;\\ &\quad -2r^2n^4(n^2-m^2) \;-\;\\ &\quad -4r^2n^2(5n^4+3m^4-9n^2m^2)\cos^2\vartheta \;+\;\\ &\quad +8r^2n^4(2n^2-3m^2)\cos^2\vartheta \;+\;\\ &\quad +4n^4(m^2-n^2)^2\cos^4\vartheta \;+\;\\ &\quad +2m^2n^4\bigl[n^2-(2n^2-m^2)\cos^2\vartheta\bigr]\cos^2\vartheta \end{aligned} \\[1.0em] xy^3\,\operatorname{ctg}^3\vartheta & \begin{aligned} &4r^4(3n^2-m^2)(2n^2-m^2) \;-\;\\ &\quad -8r^4n^2(3n^2-2m^2)\cos^2\vartheta \;-\;\\ &\quad -4r^2n^4(2n^2-m^2) \;-\;\\ &\quad -4r^2n^2(7n^4+2m^4-8n^2m^2)\cos^2\vartheta \;+\;\\ &\quad +8r^2n^4(4n^2-3m^2)\cos^2\vartheta \;+\;\\ &\quad +4n^4(m^2-n^2)\bigl[n^2 \;-\;\\ &\quad\quad -(2n^2-m^2)\cos^2\vartheta\bigr]\cos^2\vartheta \end{aligned} \\[1.0em] y^4\,\operatorname{ctg}^4\vartheta & \begin{aligned} &r^4(3n^2-m^2)^2-4r^4n^2(2n^2-m^2)\cos^2\vartheta \;-\;\\ &\quad -2r^2n^2(3n^2-m^2)\bigl[n^2 \;+\;\\ &\quad\quad +(2n^2-m^2)\cos^2\vartheta\bigr] \;+\;\\ &\quad +8r^2n^4(2n^2-m^2)\cos^2\vartheta \;+\;\\ &\quad +n^4\bigl[n^2-(2n^2-m^2)\cos^2\vartheta\bigr]^2 \end{aligned} \end{array} \right| \tag{13} \end{aligned} \]

Let us return to equation (12).

For brevity, denote the left-hand side of the equation by the letter \(L\). After squaring both sides of the equation we shall have:

\[ L^2=R_1^2R_2^2(x+y\operatorname{ctg}\vartheta)^4 \tag{14} \]

where \(L^2\) is an expression of the following form:

\[ \begin{array}{r|l} x^6 & 4n^2m^4\cos^4\vartheta \\[2mm] x^5y\,\operatorname{ctg}\vartheta & -8n^2m^2\left[m^2-n^2-(3m^2-2n^2)\cos^2\vartheta\right]\cos^2\vartheta \\[2mm] x^4y^2\,\operatorname{ctg}^2\vartheta & 4n^2\left[m^2-n^2-(3m^2-2n^2)\cos^2\vartheta\right]^2 \\ &\quad -8n^2m^2\left[2m^2-3n^2-(3m^2-4n^2)\cos^2\vartheta\right]\cos^2\vartheta \\[2mm] x^3y^3\,\operatorname{ctg}^3\vartheta & 8n^2m^2(2n^2-m^2)(1-\cos^2\vartheta)\cos^2\vartheta \\ &\quad +8n^2\left[m^2-n^2-(3m^2-2n^2)\cos^2\vartheta\right] \\ &\qquad \times\left[2m^2-3n^2-(3m^2-4n^2)\cos^2\vartheta\right] \\[2mm] x^2y^4\,\operatorname{ctg}^4\vartheta & -8n^2(2n^2-m^2)\left[m^2-n^2-(3m^2-2n^2)\cos^2\vartheta\right](1-\cos^2\vartheta) \\ &\quad +4n^2\left[2m^2-3n^2-(3m^2-4n^2)\cos^2\vartheta\right]^2 \\[2mm] xy^5\,\operatorname{ctg}^5\vartheta & -8n^2(2n^2-m^2)(1-\cos^2\vartheta) \\ &\qquad \times\left[2m^2-3n^2-(3m^2-4n^2)\cos^2\vartheta\right] \\[2mm] y^6\,\operatorname{ctg}^6\vartheta & 4n^2(2n^2-m^2)^2(1-\cos^2\vartheta)^2 \\[2mm] x^5 & -4nm^2\left[m^2n^2\cos^2\vartheta-r^2(m^2-n^2)\right]\cos^2\vartheta \\[2mm] x^4y\,\operatorname{ctg}\vartheta & 8nm^2(m^2-n^2)(r^2-n^2\cos^2\vartheta)\cos^2\vartheta \\ &\quad +4n\left[m^2-n^2-(3m^2-2n^2)\cos^2\vartheta\right] \\ &\qquad \times\left[m^2n^2\cos^2\vartheta-r^2(m^2-n^2)\right] \\[2mm] x^3y^2\,\operatorname{ctg}^2\vartheta & -4nm^2\cos^2\vartheta\left[n^4-r^2(m^2-n^2)-(2n^2-m^2)n^2\cos^2\vartheta\right] \\ &\quad -8n\left[m^2-n^2-(3m^2-2n^2)\cos^2\vartheta\right](m^2-n^2)(r^2-n^2\cos^2\vartheta) \\ &\quad +4n\left[m^2n^2\cos^2\vartheta-r^2(m^2-n^2)\right] \\ &\qquad \times\left[2m^2-3n^2-(3m^2-4n^2)\cos^2\vartheta\right] \\[2mm] x^2y^3\,\operatorname{ctg}^3\vartheta & 4n\left[m^2-n^2-(3m^2-2n^2)\cos^2\vartheta\right] \\ &\qquad \times\left[n^4-r^2(m^2-n^2)-(2n^2-m^2)n^2\cos^2\vartheta\right] \\ &\quad -4n(2n^2-m^2)\left[m^2n^2\cos^2\vartheta-r^2(m^2-n^2)\right](1-\cos^2\vartheta) \\ &\quad -8n(m^2-n^2)(r^2-n^2\cos^2\vartheta) \\ &\qquad \times\left[2m^2-3n^2-(3m^2-4n^2)\cos^2\vartheta\right] \\[2mm] xy^4\,\operatorname{ctg}^4\vartheta & 8n(2n^2-m^2)(m^2-n^2)(r^2-n^2\cos^2\vartheta)(1-\cos^2\vartheta) \\ &\quad +4n\left[n^4-r^2(m^2-n^2)-(2n^2-m^2)n^2\cos^2\vartheta\right] \\ &\qquad \times\left[2m^2-3n^2-(3m^2-4n^2)\cos^2\vartheta\right] \\[2mm] y^5\,\operatorname{ctg}^5\vartheta & -4n(2n^2-m^2)\left[n^4-r^2(m^2-n^2)\right. \\ &\qquad\left. -(2n^2-m^2)n^2\cos^2\vartheta\right](1-\cos^2\vartheta) \\[2mm] x^4 & \left[m^2n^2\cos^2\vartheta-r^2(m^2-n^2)\right]^2 \\[2mm] x^3y\,\operatorname{ctg}\vartheta & -4(m^2-n^2)(r^2-n^2\cos^2\vartheta) \\ &\qquad \times\left[m^2n^2\cos^2\vartheta-r^2(m^2-n^2)\right] \\[2mm] x^2y^2\,\operatorname{ctg}^2\vartheta & 4(m^2-n^2)^2(r^2-n^2\cos^2\vartheta)^2 \\ &\quad +2\left[m^2n^2\cos^2\vartheta-r^2(m^2-n^2)\right] \\ &\qquad \times\left[n^4-r^2(m^2-n^2)-(2n^2-m^2)n^2\cos^2\vartheta\right] \\[2mm] xy^3\,\operatorname{ctg}^3\vartheta & -4(m^2-n^2)(r^2-n^2\cos^2\vartheta) \\ &\qquad \times\left[n^4-r^2(m^2-n^2)-(2n^2-m^2)n^2\cos^2\vartheta\right] \\[2mm] y^4\,\operatorname{ctg}^4\vartheta & \left[n^4-r^2(m^2-n^2)-(2n^2-m^2)n^2\cos^2\vartheta\right]^2 \end{array} \tag{15} \]

Substituting into equation (14) the expressions (13) and (15) obtained by us, transferring all terms to one side, carrying out the reduction of like terms, and arranging the coefficients of the unknowns according to the powers of \(\cosinus'a\), we obtain the equation containing the geometric locus sought by us in the following final form:

\[ \left| \begin{array}{l|l} x^6 & n^2m^4\cos^4\vartheta \\[2pt] x^5y\operatorname{ctg}\vartheta & -2n^2m^2(m^2-n^2)\cos^2\vartheta +2n^2m^2(m^2-n^2)\cos^4\vartheta \\[2pt] x^4y^2\operatorname{ctg}^2\vartheta & n^2(m^2-n^2)^2 -2n^2(5m^4-8n^2m^2+2n^4)\cos^2\vartheta \\ & \quad +n^2(15m^4+4n^4-20m^2n^2)\cos^4\vartheta \\[2pt] x^3y^3\operatorname{ctg}^3\vartheta & 2n^2(m^2-n^2)(2m^2-3n^2) -4n^2(5m^4-11m^2n^2 \\ & \quad +5n^4)\cos^2\vartheta +4n^2(5m^4-10m^2n^2+4n^4)\cos^4\vartheta \\[2pt] x^2y^4\operatorname{ctg}^4\vartheta & n^2(6m^4+13n^4-18m^2n^2) +4n^2(14m^2n^2-5m^4 \\ & \quad -9n^4)\cos^2\vartheta +n^2(15m^4+24n^4-40m^2n^2)\cos^4\vartheta \\[2pt] xy^5\operatorname{ctg}^5\vartheta & -2n^2(2n^2-m^2)(2m^2-3n^2) +2n^2(2n^2-m^2)(5m^2 \\ & \quad -7n^2)\cos^2\vartheta -2n^2(2n^2-m^2)(3m^2-4n^2)\cos^4\vartheta \\[2pt] y^6\operatorname{ctg}^6\vartheta & n^2(2n^2-m^2)^2 -2n^2(2n^2-m^2)^2\cos^2\vartheta +n^2(2n^2-m^2)^2\cos^4\vartheta \\[2pt] x^5 & nm^2r^2(m^2-n^2)\cos^2\vartheta\, n^2 -n^3m^4\cos^4\vartheta \\[2pt] x^4y\operatorname{ctg}\vartheta & -nr^2(m^2-n^2)^2 +n(m^2-n^2)(5m^2r^2+m^2n^2-2r^2n^2)\cos^2\vartheta \\ & \quad -m^2n^3(5m^2-4n^2)\cos^4\vartheta \\[2pt] x^3y^2\operatorname{ctg}^2\vartheta & -nr^2(m^2-n^2)(4m^2-5n^2) +2n\bigl[2n^2m^2(m^2-2n^2) \\ & \quad +n^4(n^2+4r^2)+r^2m^2(5m^2-9n^2)\bigr]\cos^2\vartheta \\ & \quad -2n^3(5m^4-8m^2n^2+2n^4)\cos^4\vartheta \\[2pt] x^2y^3\operatorname{ctg}^3\vartheta & n(m^2-n^2)(n^4-6m^2r^2+9r^2n^2) +2n\bigl[3n^2m^2(m^2-3n^2) \\ & \quad +n^4(5n^2+6r^2)+r^2m^2(5m^2-11n^2)\bigr]\cos^2\vartheta \\ & \quad +2n^3(12m^2n^2-5m^4-6n^4)\cos^4\vartheta \\[2pt] xy^4\operatorname{ctg}^4\vartheta & n\bigl[r^2(m^2-n^2)(7n^2-4m^2)+2n^4(m^2-3n^2)\bigr] \\ & \quad -n\bigl[2n^4(8m^2-7n^2)-m^2n^2(4m^2-13r^2) \\ & \quad -r^2(5m^4+8n^4)\bigr]\cos^2\vartheta +n^3(2n^2-m^2)(5m^2-6n^2)\cos^4\vartheta \\[2pt] y^5\operatorname{ctg}^5\vartheta & -n(2n^2-m^2)(n^2-r^2)(n^2+r^2-2n^2\cos^2\vartheta)(1-\cos^2\vartheta) \\[2pt] x^4 & -r^4n^2m^2\cos^2\vartheta \\[2pt] x^3y\operatorname{ctg}\vartheta & r^4n^2(m^2-n^2) -r^2n^2(4r^2m^2-m^2n^2-2r^2n^2)\cos^2\vartheta \\[2pt] x^2y^2\operatorname{ctg}^2\vartheta & r^2n^2\bigl[n^2(n^2-4r^2)-m^2(n^2-3r^2)\bigr] \\ & \quad +r^2n^2\bigl[3m^2(n^2-2r^2)-2n^2(n^2-3r^2)\bigr]\cos^2\vartheta \\[2pt] xy^3\operatorname{ctg}^3\vartheta & r^2n^2\bigl[m^2(3n^2-2m^2)+r^2(3m^2-5n^2)\bigr] \\ & \quad +r^2n^2\bigl[n^2(3m^2-4n^2)+2r^2(3n^2-2m^2)\bigr]\cos^2\vartheta \\[2pt] y^4\operatorname{ctg}^4\vartheta & r^2n^2(n^2-r^2)(2n^2-m^2)(1-\cos^2\vartheta) \end{array} \right|=0 . \tag{16} \]

This is the equation of the 6th degree that we have to decompose into a product of equations of lower degrees, and then determine which of them represents the geometric locus of interest to us.

After several attempts I found that it is most convenient to represent equation (16) in the form of a product of two equations: one equation of the fourth degree and one of the second.

I carry out the decomposition in the following way. Let us take, in general form, two equations with two unknowns, of the orders indicated above.

We shall have:

\[ \begin{gathered} A_1x^4+B_1x^3y+C_1x^2y^2+D_1xy^3+E_1y^4+F_1x^3+G_1x^2y+H_1xy^2+\\ +I_1y^3+K_1x^2+L_1xy+M_1y^2+N_1x+P_1y+Q_1=0, \end{gathered} \tag{17} \]

\[ A_2x^2+B_2xy+C_2y^2+D_2x+E_2y+F_2=0. \; *) \tag{18} \]

If, after multiplying these two equations, we combine like terms and set coefficients before them, then we obtain an equation containing unknown terms of all orders, beginning with the 6th and ending with the zeroth.

Considering equation (16), however, we see that it does not contain unknown terms of order lower than the fourth. Hence we conclude that in a general fourth-degree equation a number of coefficients are equal to zero. Analyzing in a similar way the scheme of coefficients of the equation of the 6th degree obtained by us in general form, we arrive at the conclusion that the coefficients of equation (17) at the unknown terms of order lower than the fourth must be equal to zero.

Thus we obtain that equation (16) can be represented as the result of the product of two equations of the following form:

\[ A_1x^4+B_1x^3y+C_1x^2y^2+D_1xy^3+E_1y^4=0, \tag{19} \]

\[ A_2x^2+B_2xy+C_2y^2+D_2x+E_2y+F_2=0. \tag{20} \]

Multiplication of these two equations by each other gives us an equation of the 6th degree of the following form:

\[ \begin{gathered} \left| \begin{array}{c|c} x^6 & A_1A_2\\ x^5y\operatorname{ctg}\vartheta & A_1B_2+B_1A_2\\ x^4y^2\operatorname{ctg}^2\vartheta & A_1C_2+B_1B_2+C_1A_2\\ x^3y^3\operatorname{ctg}^3\vartheta & B_1C_2+C_1B_2+D_1A_2\\ x^2y^4\operatorname{ctg}^4\vartheta & E_1A_2+D_1B_2+C_1C_2\\ xy^5\operatorname{ctg}^5\vartheta & E_1B_2+D_1C_2\\ y^6\operatorname{ctg}^6\vartheta & E_1C_2 \end{array} \right| + \left| \begin{array}{c|c} x^5 & A_1D_2\\ x^4y\operatorname{ctg}\vartheta & A_1E_2+B_1D_2\\ x^3y^2\operatorname{ctg}^2\vartheta & B_1E_2+C_1D_2\\ x^2y^3\operatorname{ctg}^3\vartheta & C_1E_2+D_1D_2\\ xy^4\operatorname{ctg}^4\vartheta & D_1E_2+E_1D_2\\ y^5\operatorname{ctg}^5\vartheta & E_1E_2 \end{array} \right| +\\[1em] + \left| \begin{array}{c|c} x^4 & A_1F_2\\ x^3y\operatorname{ctg}\vartheta & B_1F_2\\ x^2y^2\operatorname{ctg}^2\vartheta & C_1F_2\\ xy^3\operatorname{ctg}^3\vartheta & D_1F_2\\ y^4\operatorname{ctg}^4\vartheta & E_1F_2 \end{array} \right| =0. \end{gathered} \tag{21} \]

*) With each \(y\) there is a factor \(\operatorname{ctg}\vartheta\) in the corresponding degree, which, for brevity, is omitted here, i.e. included in the coefficients \(B, C, D\), etc.

Considering the patterns of the coefficients of equation (21), we see first of all that the coefficient \(F_2\) of equation (20) is a common factor of the last five terms of equation (21), and since this equation must represent equation (16), we also find \(F_2=-n^2 r^2\). We take the minus sign by considering the coefficient of \(x^4\) in equation (16), in order to obtain the coefficient \(A_1\) of equation (19) with the plus sign.

Having determined \(F_2\), we can thereby determine, from the last five terms of equation (16), all the coefficients of equation (19).

According to what has been said, we obtain that equation (19) has the form:

\[ \begin{aligned} & \left| \begin{array}{c} A_1 x^4\\ B_1 x^3 y\,\operatorname{ctg}\vartheta\\ C_1 x^2 y^2 \operatorname{ctg}^2\vartheta\\ D_1 x y^3 \operatorname{ctg}^3\vartheta\\ E_1 y^4 \operatorname{ctg}^4\vartheta \end{array} \right| = \left| \begin{array}{c} x^4\\ x^3 y\,\operatorname{ctg}\vartheta\\ x^2 y^2 \operatorname{ctg}^2\vartheta\\ x y^3 \operatorname{ctg}^3\vartheta\\ y^4 \operatorname{ctg}^4\vartheta \end{array} \right| \left| \begin{array}{c} r^2 m^2 \cos^2\vartheta\\ r^2(n^2-m^2)+(4r^2m^2-m^2n^2-2r^2n^2)\cos^2\vartheta\\ \bigl[m^2(n^2-3r^2)-n^2(n^2-4r^2)\bigr]-\\ -\bigl[3m^2(n^2-2r^2)-2n^2(n^2-3r^2)\bigr]\cos^2\vartheta\\ \bigl[n^2(2m^2-3n^2)-r^2(3m^2-5n^2)\bigr]-\\ -\bigl[n^2(3m^2-4n^2)+2r^2(3n^2-2m^2)\bigr]\cos^2\vartheta\\ (r^2-n^2)(2n^2-m^2)(1-\cos^2\vartheta) \end{array} \right| =0. \end{aligned} \tag{22} \]

Knowing the coefficients of equations (19) and (22), it is now not difficult to determine also the coefficients of equation (20), relying on equations (21) and (16).

In equation (21) we see that the coefficients of \(x^6\), \(y^6\), \(x^5\), and \(y^5\) have a comparatively simple form, and therefore we shall turn to the same coefficients of equation (16).

Let us turn to \(x^6\).

\[ A_1 A_2 x^6 = n^2 m^4 \cos^4\vartheta. \]

We already know that \(A_1=r^2m^2\cos^2\vartheta\), whence

\[ A_2=\frac{n^2 m^2 \cos^2\vartheta}{r^2}. \]

In order not subsequently to have to deal with coefficients in the form of algebraic fractions, we transform here the value of the coefficient \(A_2\) just found by us.

We know that \(m^2=n^2+r^2-l^2\).*)

On the other hand, we previously had**) the following relation among all the elements of the crank-and-connecting-rod mechanism:

\[ lr=n\sigma\cos\vartheta \]

and

\[ l^2+r^2=n^2+\sigma^2. \]

) See The Coupler Pole, p. 55.
) The Coupler Pole*, p. 49.

On this basis we may write:

\[ m^{2}=r^{2}+n^{2}-l^{2}=2r^{2}-\sigma^{2}. \]

Thus the coefficient \(A_{2}\) will have the form:

\[ A_{2}=\frac{(2r^{2}-\sigma^{2})\,n^{2}\cos^{2}\vartheta}{r^{2}} \]

or

\[ A_{2}=\frac{2r^{2}n^{2}\cos^{2}\vartheta-n^{2}\sigma^{2}\cos^{2}\vartheta}{r^{2}}, \]

which finally gives:

\[ A_{2}=2n^{2}\cos^{2}\vartheta-l^{2}. \tag{23} \]

Passing to \(y^{6}\), we have:

\[ E_{1}C_{2}y^{6} = y^{6}\operatorname{ctg}^{6}\vartheta \left| n^{2}(2n^{2}-m^{2})^{2} - 2n^{2}(2n^{2}-m^{2})^{2}\cos^{2}\vartheta + n^{2}(2n^{2}-m^{2})^{2}\cos^{4}\vartheta. \right. \]

The coefficient of \(y^{6}\operatorname{ctg}^{6}\vartheta\) may be represented in the following form:

\[ E_{1}C_{2}y^{6} = y^{6}\operatorname{ctg}^{6}\vartheta\, n^{2}(2n^{2}-m^{2})^{2}(1-\cos^{2}\vartheta)^{2}. \]

Hence we obtain the value of the coefficient \(C_{2}\):

\[ C_{2} = \frac{n^{2}(2n^{2}-m^{2})^{2}(1-\cos^{2}\vartheta)^{2}}{E_{1}} \]

or, since

\[ E_{1}=(r^{2}-n^{2})(2n^{2}-m^{2})(1-\cos^{2}\vartheta), \]

\[ C_{2} = \frac{n^{2}(2n^{2}-m^{2})}{r^{2}-n^{2}}(1-\cos^{2}\vartheta). \]

We transform this expression so as not to have algebraic fractions.

Relying on the various notations for the quantity \(m^{2}\), we may write:

\[ C_{2} = \frac{n^{2}(2n^{2}-m^{2})}{r^{2}-n^{2}} - \frac{n^{2}(2n^{2}-m^{2})}{r^{2}-n^{2}}\cos^{2}\vartheta \]

or

\[ C_{2} = \frac{n^{2}(2n^{2}-n^{2}-r^{2}+l^{2})}{r^{2}-n^{2}} - \frac{n^{2}(2n^{2}-2r^{2}+\sigma^{2})\cos^{2}\vartheta}{r^{2}-n^{2}}; \]

which gives us:

\[ C_{2} = \frac{ n^{2}(n^{2}-r^{2})+n^{2}l^{2} - 2n^{2}(n^{2}-r^{2})\cos^{2}\vartheta - n^{2}\sigma^{2}\cos^{2}\vartheta }{r^{2}-n^{2}}. \]

Since \(n^2 \sigma^2 \cos^2 \vartheta = r^2 l^2\), we shall have:

\[ C_2 = \frac{ n^2(n^2-r^2)+l^2(n^2-r^2)-2n^2(n^2-r^2)\cos^2\vartheta }{ r^2-n^2 } \]

or, finally:

\[ C_2 = 2n^2\cos^2\vartheta - (n^2+l^2). \tag{24} \]

We pass to the coefficient of \(x^5\) in equation (16):

\[ x^5\left[nm^2r^2(m^2-n^2)\cos^2\vartheta - n^3m^4\cos^4\vartheta\right] = A_1D_2x^5. \]

Since \(A_1 = r^2m^2\cos^2\vartheta\), we bring our expression into a form divisible by this quantity, relying on the value derived earlier,

\[ m^2 = 2r^2-\sigma^2. \]

Whence: \(n^2m^2\cos^2\vartheta = n^2(2r^2-\sigma^2)\cos^2\vartheta\), which, after opening the brackets and substituting, in place of \(n^2\sigma^2\cos^2\vartheta\), the equal quantity \(r^2l^2\), gives us:

\[ n^2m^2\cos^2\vartheta = r^2(2n^2\cos^2\vartheta-l^2). \]

Thus we have:

\[ A_1D_2 = nm^2r^2(m^2-n^2)\cos^2\vartheta - nr^2(2n^2\cos^2\vartheta-l^2)m^2\cos^2\vartheta. \]

Whence:

\[ D_2 = \frac{ m^2r^2n\cos^2\vartheta \left[m^2-n^2+l^2-2n^2\cos^2\vartheta\right] }{ A_1 }, \]

or, since \(m^2=n^2+r^2-l^2\):

\[ D_2 = \frac{ nm^2r^2\cos^2\vartheta\,(r^2-2n^2\cos^2\vartheta) }{ m^2r^2\cos^2\vartheta } = n(r^2-2n^2\cos^2\vartheta). \tag{25} \]

Turning to \(y^5\operatorname{ctg}^5\vartheta\), we have:

\[ y^5\operatorname{ctg}^5\vartheta \left[-n(2n^2-m^2)(n^2-r^2)(n^2+r^2-2n^2\cos^2\vartheta)(1-\cos^2\vartheta)\right] = E_1E_2y^5\operatorname{ctg}^5\vartheta. \]

Since

\[ E_1=(r^2-n^2)(2n^2-m^2)(1-\cos^2\vartheta), \]

it is obvious that

\[ E_2 = n(n^2+r^2-2n^2\cos^2\vartheta). \tag{26} \]

To determine the coefficient \(B_2\), we turn in equation (16) to the term with \(x^5y\operatorname{ctg}\vartheta\).

By equation (21) we know that the scheme of the coefficient of this term is as follows:

\[ x^5y\operatorname{ctg}\vartheta\,[A_1B_2+B_1A_2]. \]

Since \(A_1\), \(A_2\), and \(B_1\) have already been determined by us, finding the value of \(B_2\) no longer presents any difficulty.

Let us form the product of the quantities \(A_2\) and \(B_1\) known to us.

\[ A_2 B_1 = (2 n^2 \cos^2 \vartheta - l^2)\left[r^2(n^2-m^2) + (4r^2m^2 - m^2n^2 - 2r^2n^2)\cos^2\vartheta\right]. \]

To simplify the further operations, transform the coefficient \(B_1\) as follows:

\[ B_1 = r^2(n^2-m^2) + \left[4r^2m^2 - n^2(m^2+2r^2)\right]\cos^2\vartheta, \]

which gives us further:

\[ B_1 = r^2(n^2-m^2) + \left[4r^2m^2 - n^2(4r^2-\sigma^2)\right]\cos^2\vartheta, \]

or,

\[ B_1 = r^2(n^2-m^2) + n^2\sigma^2\cos^2\vartheta + 4r^2(m^2-n^2)\cos^2\vartheta, \]

whence:

\[ B_1 = r^2(n^2-m^2+l^2) + 4r^2(m^2-n^2)\cos^2\vartheta \]

or finally:

\[ B_1 = r^2(2l^2-r^2) + 4r^2(m^2-n^2)\cos^2\vartheta. \]

Thus \(A_2B_1\) assumes the form:

\[ A_2B_1 = (2n^2\cos^2\vartheta-l^2)\left[r^2(2l^2-r^2) + 4r^2(m^2-n^2)\cos^2\vartheta\right]. \]

Multiplying term by term, we obtain:

\[ A_2B_1 = r^2(2n^2\cos^2\vartheta-l^2)(2l^2-r^2) + 4r^2(2n^2\cos^2\vartheta-l^2)(m^2-n^2)\cos^2\vartheta. \]

We found earlier, in deriving the coefficient \(D_2\), that

\[ n^2m^2\cos^2\vartheta = r^2(2n^2\cos^2\vartheta-l^2); \]

therefore the product \(A_2B_1\) may be given the following form:

\[ A_2B_1 = (2l^2-r^2)n^2m^2\cos^2\vartheta + 4(m^2-n^2)m^2n^2\cos^4\vartheta. \]

Subtracting this expression from the coefficient of \(x^5 y \operatorname{ctg}\vartheta\) in equation (16), we obtain:

\[ A_1B_2 = -2n^2m^2(m^2-n^2)\cos^2\vartheta + 2n^2m^2(3m^2-2n^2)\cos^4\vartheta - \]

\[ -(2l^2-r^2)n^2m^2\cos^2\vartheta - 4(m^2-n^2)m^2n^2\cos^4\vartheta, \]

which, after collecting like terms, assumes the form:

\[ A_1B_2 = -r^2n^2m^2\cos^2\vartheta + 2n^2m^4\cos^4\vartheta. \]

Substituting in the second term, instead of \(m^2\), the value \(2r^2-\sigma^2\), we transform our expression as follows:

\[ A_1B_2 = -r^2n^2m^2\cos^2\vartheta + 2n^2m^2(2r^2-\sigma^2)\cos^4\vartheta, \]

or,

\[ A_1B_2 = -r^2n^2m^2\cos^2\vartheta + 4r^2n^2m^2\cos^4\vartheta - 2m^2n^2\sigma^2\cos^2\vartheta\cos^2\vartheta, \]

whence we finally obtain:

\[ A_1 B_2=4 r^2 n^2 m^2 \cos^4\vartheta-r^2 m^2(n^2+2l^2)\cos^2\vartheta . \]

It is now easy to determine the value of the coefficient \(B_2\).

\[ B_2=\frac{4 r^2 n^2 m^2 \cos^4\vartheta-r^2 m^2(n^2+2l^2)\cos^2\vartheta}{A_1}. \]

Substituting here, in place of \(A_1\), the value equal to it, \(A_1=r^2m^2\cos^2\vartheta\), we shall have the value of the coefficient \(B_2\):

\[ B_2=4n^2\cos^2\vartheta-(2l^2+n^2). \tag{27} \]

Thus all the coefficients of equation (20) have been determined.

Next, composing, according to the scheme of equation (21), the coefficients for the remaining powers of \(x\) and \(y\), and equating them to the coefficients of equation (16), we in all cases obtain identities, which also serves as proof of the correctness of our operations.

Thus equation (16), of the sixth degree with respect to \(x\) and \(y\), decomposes into the product of two equations of the second and fourth degrees. We write here both equations found:

\[ \begin{vmatrix} x^4 & r^2m^2\cos^2\vartheta \\ x^3y\ctg\vartheta & r^2(n^2-m^2)+(4r^2m^2-m^2n^2-2r^2n^2)\cos^2\vartheta \\ x^2y^2\ctg^2\vartheta & [m^2(n^2-3r^2)-n^2(n^2-4r^2)] \\ & \quad -[3m^2(n^2-2r^2)-2n^2(n^2-3r^2)]\cos^2\vartheta \\ xy^3\ctg^3\vartheta & [n^2(2m^2-3n^2)-r^2(3m^2-5n^2)] \\ & \quad -[n^2(3m^2-4n^2)+2r^2(3n^2-2m^2)]\cos^2\vartheta \\ y^4\ctg^4\vartheta & (r^2-n^2)(2n^2-m^2)(1-\cos^2\vartheta) \end{vmatrix} =0, \tag{22} \]

\[ \begin{vmatrix} x^2 & 2n^2\cos^2\vartheta-l^2 \\ xy\ctg\vartheta & 4n^2\cos^2\vartheta-(2l^2+n^2) \\ y^2\ctg^2\vartheta & 2n^2\cos^2\vartheta-(l^2+n^2) \\ x & n(r^2-2n^2\cos^2\vartheta) \\ y\ctg\vartheta & n(r^2+n^2-2n^2\cos^2\vartheta) \\ x^0 & -n^2r^2 \end{vmatrix} =0. \tag{28} \]

Now we must determine which of these two equations is the required geometric locus.

For this purpose we shall recall that we have one, found ...

earlier, the point belonging to the geometrical locus; this point is determined by formulas (65)*) having the form:

\[ x=\frac{r^2+n^2-l^2}{2n}\quad \text{and}\quad y=\frac{(l^2-r^2-n^2)^2-4n^2r^2}{2n(l^2-r^2+n^2)\operatorname{ctg}\vartheta}. \]

Since \(r^2+n^2-l^2=m^2\), we give these formulas the following form:

\[ x=\frac{m^2}{2n},\quad y=\frac{m^4-4n^2r^2}{2n(2n^2-m^2)\operatorname{ctg}\vartheta}. \]

These expressions have been found in the previous system of coordinates; we must transform them for the new system.

The transition formulas indicated at the beginning of the present article are as follows:

\[ x'=x+\frac{r^2}{n}\quad \text{and}\quad y'=y-\operatorname{tg}\vartheta\,\frac{r^2}{n}. \]

Consequently, the coordinates of the control point in the new system will be:

\[ x=\frac{m^2}{2n}-\frac{r^2}{n}=\frac{m^2-2r^2}{2n}, \]

\[ y=\frac{m^4-4n^2r^2}{2n(2n^2-m^2)\operatorname{ctg}\vartheta} +\frac{r^2}{n\operatorname{ctg}\vartheta} = \frac{m^4-4n^2r^2+2r^2(2n^2-m^2)} {2n(2n^2-m^2)\operatorname{ctg}\vartheta}. \]

Since \(m^2-2r^2=-\sigma^2\), we obtain:

\[ x=-\frac{\sigma^2}{2n}\quad \text{and}\quad y=-\frac{m^2\sigma^2}{2n(2n^2-m^2)\operatorname{ctg}\vartheta}. \tag{29} \]

Substituting in equation (28) the values of the quantities \(x\) and \(y\), determined by formulas (29), we have:

\[ \frac{\sigma^4}{4n^2}(2n^2\cos^2\vartheta-l^2) + \frac{\sigma^4m^2}{4n^2(2n^2-m^2)} (4n^2\cos^2\vartheta-2l^2-n^2) + \]

\[ + \frac{\sigma^4m^4}{4n^2(2n^2-m^2)^2} (2n^2\cos^2\vartheta-l^2-n^2) -\frac{\sigma^2}{2}(r^2-2n^2\cos^2\vartheta) - \]

\[ -\frac{\sigma^2m^2}{2(2n^2-m^2)} (r^2+n^2-2n^2\cos^2\vartheta)-n^2r^2=0, \]

which, after clearing denominators and collecting like terms—

*) See The Coupler Pole, p. 57.

and substitution, in place of \(m^2\), of the quantity equal to it, \((2r^2-\sigma^2)\), takes the form:

\[ \sigma^2 l^2-2n^2\sigma^2+2r^2\sigma^2-\sigma^4+n^2l^2-r^2l^2-n^4+2n^2r^2-r^4=0. \]

Factoring, we obtain:

\[ -(n^2+\sigma^2)^2+l^2(n^2+\sigma^2)+2r^2(n^2+\sigma^2)-r^2(l^2+r^2)=0. \]

Since \(l^2+r^2=n^2+\sigma^2\), we carry out the reduction, after which we obtain:

\[ -(n^2+\sigma^2)+l^2+2r^2-r^2=0, \]

or

\[ l^2+r^2=n^2+\sigma^2, \]

whence we see that equation (28) is completely satisfied by the coordinates of the control point.

We pass to equation (22).

Substitution of the quantities \(x\) and \(y\) according to formulas (29) gives:

\[ \frac{\sigma^8}{(2n)^4}\cdot r^2m^2\cos^2\vartheta +\frac{\sigma^8}{(2n)^4}\cdot\frac{m^2}{2n^2-m^2} \left[r^2(n^2-m^2)+(4r^2m^2-m^2n^2-2r^2n^2)\cos^2\vartheta\right] \]
\[ +\frac{\sigma^8}{2n^4}\cdot\frac{m^2}{(2n^2-m^2)^2} \left\{m^2(n^2-3r^2)-n^2(n^2-4r^2)\right. \]
\[ \left.-\left[3m^2(n^2-2r^2)-2n^2(n^2-3r^2)\right]\cos^2\vartheta\right\} \]
\[ +\frac{\sigma^8}{2n^4}\cdot\frac{m^6}{(2n^2-m^2)^3} \left\{n^2(2m^2-3n^2)-r^2(3m^2-5n^2)\right. \]
\[ \left.-\left[n^2(3m^2-4n^2)+2r^2(3n^2-2m^2)\right]\cos^2\vartheta\right\} \]
\[ +\frac{\sigma^8}{2n^4}\cdot\frac{m^8}{(2n^2-m^2)^4} (r^2-n^2)(2n^2-m^2)(1-\cos^2\vartheta)=0. \]

Carrying out the reduction and bringing to a common denominator, we obtain:

\[ r^2(2n^2-m^2)^3\cos^2\vartheta +(2n^2-m^2)^2\left[r^2(n^2-m^2)+(4r^2m^2-m^2n^2-2r^2n^2)\cos^2\vartheta\right] \]
\[ +m^2(2n^2-m^2) \left\{m^2(n^2-3r^2)-n^2(n^2-4r^2) -\left[3m^2(n^2-2r^2)\right.\right. \]
\[ \left.\left.-2n^2(n^2-3r^2)\right]\cos^2\vartheta\right\} +m^4\left\{n^2(2m^2-3n^2)-r^2(3m^2-5n^2)\right. \]
\[ \left.-\left[n^2(3m^2-4n^2)+2r^2(3n^2-2m^2)\right]\cos^2\vartheta\right\} +m^6(r^2-n^2)(1-\cos^2\vartheta)=0. \]

Expanding the brackets and doing the reduction of like terms, we obtain as a result:

\[ -2n^6(m^2-2r^2)=0, \]

or

\[ 2n^6 \delta^2 = 0, \]
which is obviously impossible, since \(n\) and \(\delta\) are finite quantities.

Thus equation (22) is not satisfied by the values of the coordinates of the control point and, consequently, does not represent the sought geometric locus; equation (28), as we have seen, has turned into an identity.

In this way we arrive at the conclusion that the geometric locus of the points of intersection of conjugate chords for a curvilinearly generated slider-crank mechanism is expressed by an equation of the second degree.

Let us investigate the geometric meaning of this equation. Let us write it in general form:

\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. \]

The discriminant of this equation has the form:

\[ \Delta = 2(4ACF - AE^2 - CD^2 + BDE - B^2F). \]

We shall determine the value of the discriminant, relying on the following dependence among the coefficients of equation (28):

\[ \begin{aligned} A &= 2n^2\cos^2\vartheta - l^2,\\ B &= (4n^2\cos^2\vartheta - 2l^2 - n^2)\operatorname{ctg}\vartheta = (2A - n^2)\operatorname{ctg}\vartheta,\\ C &= (2n^2\cos^2\vartheta - l^2 - n^2)\operatorname{ctg}^2\vartheta = (A - n^2)\operatorname{ctg}^2\vartheta,\\ D &= n(r^2 - 2n^2\cos^2\vartheta),\\ E &= n(r^2 + n^2 - 2n^2\cos^2\vartheta)\operatorname{ctg}\vartheta = (D + n^3)\operatorname{ctg}\vartheta,\\ F &= -n^2r^2. \end{aligned} \]

Thus, first transforming the discriminant, we have:

\[ \Delta = 2\{ C(4AF - D^2) + E(BD - AE) - B^2F \} \]
or, substituting:

\[ \Delta = 2\{ \operatorname{ctg}^2\vartheta\,(A - n^2)(4AF - D^2) + \operatorname{ctg}\vartheta\,(D + n^3) [\operatorname{ctg}\vartheta\,(2A - n^2)D - \operatorname{ctg}\vartheta\,A(D + n^3)] - \operatorname{ctg}^2\vartheta\,(2A - n^2)^2F \}. \]

Carrying out the operations, we obtain:

\[ \Delta = 2n^4\operatorname{ctg}^2\vartheta\,(-An^2 - Dn - F). \]

Substitute the values of the coefficients \(A\), \(D\), and \(F\).

\[ -(An^2 + Dn + F) = --(2n^4\cos^2\vartheta - n^2l^2 + n^2r^2 - 2n^4\cos^2\vartheta - n^2r^2, \]

or finally:

\[ \Delta=2n^{6}l^{2}\operatorname{ctg}^{2}\vartheta, \]
which is a quantity greater than zero. Thus \(\Delta>0\).

Let us now determine the value of the expression:

\[ H=B^{2}-4AC. \]

\[ H=\operatorname{ctg}^{2}\vartheta\left[(2A-n^{2})^{2}-4A(A-n^{2})\right]. \]

Expanding the brackets, we have:

\[ H=\operatorname{ctg}^{2}\vartheta\left[4A^{2}+n^{4}-4An^{2}-4A^{2}+4An^{2}\right], \]
or
\[ H=n^{4}\operatorname{ctg}^{2}\vartheta>0. \]

Thus equation (28) represents a hyperbola.

Let us now return to the previous system of coordinates, substituting for \(x\) and \(y\) the corresponding quantities:

\[ x=x'-\frac{r^{2}}{n} \quad\text{and}\quad y=y'+\operatorname{tg}\vartheta\,\frac{r^{2}}{n} \quad\text{(see p. 3).} \]

After carrying out all the operations, we obtain the following equation:

\[ \begin{vmatrix} x'^{2} & 2n^{2}\cos^{2}\vartheta-l^{2}\\ x'y'\operatorname{ctg}\vartheta & 4n^{2}\cos^{2}\vartheta-2l^{2}-n^{2}\\ y'^{2}\operatorname{ctg}^{2}\vartheta & 2n^{2}\cos^{2}\vartheta-l^{2}-n^{2}\\ x' & -2n^{3}\cos^{2}\vartheta\\ y'\operatorname{ctg}\vartheta & -n^{3}\cos 2\vartheta \end{vmatrix} \tag{30} \]

Whence we see that the hyperbola we have found passes through the center of the circle of curvature—the origin of the coordinates of the previous system.

We may give equation (30) the following more compact form (omitting the primes on the unknowns, as now unnecessary):

\[ \begin{vmatrix} x^{2} & 2n^{2}\cos^{2}\vartheta\\ xy\operatorname{ctg}\vartheta & n^{2}(4\cos^{2}\vartheta-1)\\ y^{2}\operatorname{ctg}^{2}\vartheta & n^{2}(2\cos^{2}\vartheta-1)\\ x & -2n^{3}\cos^{2}\vartheta\\ y\operatorname{ctg}\vartheta & -n^{3}\cos 2\vartheta \end{vmatrix} = \begin{vmatrix} x^{2} & l^{2}\\ xy\operatorname{ctg}\vartheta & 2l^{2}\\ y^{2}\operatorname{ctg}^{2}\vartheta & l^{2} \end{vmatrix} \]

or, in another way:

\[ \begin{vmatrix} x^{2} & 2\cos\\ xy\operatorname{ctg}\vartheta & 2\cos^{2}\vartheta+\cos 2\vartheta\\ y^{2}\operatorname{ctg}^{2}\vartheta & \cos 2\vartheta\\ x & -2n\cos^{2}\vartheta\\ y\operatorname{ctg}\vartheta & -n\cos 2\vartheta \end{vmatrix} = \frac{l^{2}}{n^{2}}(x+y\operatorname{ctg}\vartheta)^{2}, \]

which further gives:

\[ (x+y\operatorname{ctg}\vartheta-n)(x\cdot 2\cos^{2}\vartheta+y\operatorname{ctg}\vartheta\cos 2\vartheta) = \frac{l^{2}}{n^{2}}(x+y\operatorname{ctg}\vartheta)^{2}, \]

whence we finally have:

\[ (x+y\operatorname{ctg}\vartheta-n)(x\sin 2\vartheta+y\cos 2\vartheta) = \frac{l^{2}(x+y\operatorname{ctg}\vartheta)^{2}}{n^{2}\operatorname{ctg}\vartheta}. \tag{31} \]

In conclusion we note that, substituting into the equation of the locus found by us the coordinates of the point determined by formulas (61) on page 54 of “The Coupler Pole,” we do not obtain an identity.

This shows us that, although the indicated point does satisfy the condition of polarity, it is not the point of intersection of the conjugate chords for the limiting case considered; consequently, all conclusions based on the values of the coordinates of this point were incorrect, as, for example, the numerical conclusion at the end of Chapter Six.