Prof. S. K. Konnyukhov.

On the Theory of the Wedding Fan

The Wedding fan belongs, by its construction, to the simplest class of machines.

Let us imagine a cylindrical box \(A\) (Fig. 1), inside which passes the axis \(O\), on which the cylinder \(B\) is mounted eccentrically, so that the eccentricity is equal to \(OI\). The pin \(I\) rotates in cylinder \(B\). The axis \(O\) is rigidly connected with cylinder \(B\), so that, if this axis is made to rotate, cylinder \(B\) will roll inside cylinder \(A\). Outside, the axis \(O\) is brought out and rotates in bearings. A pulley is put on this axis, or a handle for rotation is attached. If the line \(OI\) is mentally continued, then at its intersection with the circumference \(A\) one obtains the point of tangency of the circles \(A\) and \(B\). To cylinder \(B\) is attached the plate \(C\), entering the capsule \(D\) and, theoretically speaking, at all times required to touch with its end the inner wall of the capsule \(D\). On the right and left, as shown in Fig. 1, there are arranged channels for the intake of air and its discharge. For the passage of the plate \(C\) into the capsule \(D\), a slot along the generator is made in cylinder \(A\). If the pulley \(B\) is rotated inside cylinder \(A\), then the compressed air will be delivered into the pipe \(F\).

Let us consider the action of this machine from the kinematic point of view.

Let (Fig. 2) \(OB\) be the radius of the large cylinder \(= R\), \(OA\) the radius of the small cylinder \(= r\); \(ABC\) some position of the plate, whose length is usually taken equal to \(R + r\).

If the circumference of radius \(r\) is divided into parts (points \(1, 2, 3, \ldots\)), then it is not difficult to construct by points the transverse section of the capsule, representing in outline an interesting curve having 2 points of inflection and an outgoing point \(B\) (otherwise, an angular point).

In order to find these points, it is necessary first to obtain the equation of the curve. This is what we shall do.

Since in the technical literature we have not happened to encounter a derivation of this equation, it will not be uninteresting to acquaint ourselves with a rather simple method of obtaining this equation. In order to facilitate this derivation as much as possible, we shall prove a theorem well known in the English literature under the name of “Stewart’s theorem.” Suppose we have (Fig. 3) a triangle \(ABC\). Join its vertex \(A\) with some point \(D\) on the base \(BC\). Then drop from point \(A\) to \(BC\) the perpendicular \(AE\).

Now introduce the following notation: \(BC = a\), \(AB = c\), \(AC = b\), \(AD = p\), \(DC = b_1\), \(BD = c_1\).

From the obtuse triangle \(ADC\) and the acute triangle \(ABD\) one may write the following equalities (Fig. 3)

\[ AC^2 = AD^2 + DC^2 + 2\,DC \cdot DE \]

\[ AB^2 = AD^2 + BD^2 - 2\,BD \cdot DE \]

or

\[ b^2 = p^2 + b_1^2 + 2 b_1 \cdot DE \]

\[ c^2 = p^2 + c_1^2 - 2 c_1 \cdot DE \]

Multiplying the first equality by \(c_1\), and the second by \(b_1\), and adding, we then obtain Stewart’s theorem

\[ b^2 c_1 + c^2 b_1 = p^2 a + a b_1 c_1 . \]

We shall apply this theorem to our case.

Referring to drawing 2, take the triangle \(OCA\), in which \(AC\) is taken as the base \(= e\). Further denote \(AB\) by \(Z\), \(BC = e - Z\), and \(OC\) by \(U\). Then we shall have

\[ OA^2 \cdot BC + OC^2 \cdot AB = OB^2 \cdot AC + AC \cdot BC \cdot AB \]

or

\[ r^2(e-Z) + U^2 Z = R^2 e + e \cdot Z \cdot (e-Z). \tag{1} \]

Or else one may take the triangles \(BCE\) and \(ABF\) and likewise find the dependence between \(U\) and \(Z\). But the first method of solving the problem seems to me more elegant.

We shall now pass at once from equation (1) to another, introducing Cartesian coordinates in place of \(U\) and \(Z\).

From the triangle \(BEC\) we have

\[ BC^2 = BE^2 + EC^2 \]

or

\[ (e-Z)^2 = x^2 + (y-R)^2 . \]

Consequently,

\[ Z = e - \sqrt{x^2 + (y-R)^2}. \]

On the other hand, from the triangle \(ODC\) we have

\[ OC^2 = CD^2 + OD^2 \quad \text{or} \quad U^2 = x^2 + y^2 . \]

Substitute these values of \(Z\) and \(U\) into equation (1). Then we obtain the equation of our “fan” curve:

\[ r^2 \sqrt{x^2+(y-R)^2} + (x^2+y^2)\left(e-\sqrt{x^2+(y-R)^2}\right) \]
\[ = R^2 e + e \sqrt{x^2+(y-R)^2}\left(e-\sqrt{x^2+(y-R)^2}\right). \tag{2} \]

If the origin of coordinates is moved to the point \(B\), then

\[ x=x_1,\qquad Y=Y_1+R,\qquad Y_1=Y-R, \]

and the equation assumes a still more simplified form:

\[ r^2\sqrt{x_1^2+y_1^2}+\left[x_1^2+(y_1+R)^2\right]\left(e-\sqrt{x_1^2+y_1^2}\right) \]
\[ =R^2e+e\sqrt{x_1^2+y_1^2}\left(e-\sqrt{x_1^2+y_1^2}\right). \tag{3} \]

And in a polar coordinate system the equation is simplified still more.

Indeed, putting \(x_1=\rho\cos\varphi\) and \(Y_1=\rho\sin\varphi\), we obtain

\[ r^2\rho+\left(\rho^2+2R\rho\sin\varphi+R^2\right)(e-\rho) =R^2e+e\rho(e-\rho) \]

or

\[ \rho\left[\rho^2+2\rho(R\sin\varphi-e)+e^2+R^2-r^2-2Re\sin\varphi\right]=0, \]

and

\[ \rho^2+2\rho(R\sin\varphi-e)+e^2+R^2-r^2-2Re\sin\varphi=0. \]

Solving this equation with respect to \(\rho\), we obtain

\[ \left(\rho-e+R\sin\varphi-\sqrt{r^2-R^2\cos^2\varphi}\right) \left(\rho-e+R\sin\varphi+\sqrt{r^2-R^2\cos^2\varphi}\right)=0. \]

Thus, our curve splits into the following branches:

\[ 1)\ \rho=0;\quad 2)\ \rho=e-R\sin\varphi+\sqrt{r^2-R^2\cos^2\varphi}; \quad 3)\ \rho=e-R\sin\varphi-\sqrt{r^2-R^2\cos^2\varphi}. \]

The first branch represents a point. For the second branch we shall at once find two points. Put \(\varphi=90^\circ\). Then

\[ \rho=e-R+r=2r. \]

Put \(\rho=0\). Then

\[ e-R\sin\varphi+\sqrt{r^2-R^2\cos^2\varphi}=0. \]

Or

\[ (R+r)^2-2(R+r)R\sin\varphi+R^2\sin^2\varphi = r^2-R^2\cos^2\varphi. \]

Expanding the brackets and replacing \(\Cos^2\varphi\) by \(1-\Sin^2\varphi\), we obtain

\[ R^2+2Rr+r^2-2R^2\Sin\varphi-2rR\Sin\varphi+R^2\Sin^2\varphi = r^2-R^2\Sin^2\varphi. \]

This equation gives \(\Sin\varphi=1\), i.e. \(\varphi=90^\circ\). The first point corresponds to the highest point of the curve, and the second to the pole \(B\).

For the third branch, putting \(\varphi=90^\circ\), we obtain

\[ \rho=0. \]

And putting \(\rho=0\), we shall have \(\varphi=90^\circ\). Let us turn to finding the points of inflection of the second curve.

According to the rules of differential calculus, it is necessary to take the second derivative of the equation of the given curve and set this derivative equal to zero.

\[ \rho=e-R\Sin\varphi+\sqrt{r^2-R^2\Cos^2\varphi} \]

\[ \rho^1=-R\Cos\varphi \left( 1+\frac{R\Sin\varphi}{\sqrt{r^2-R\Cos^2\varphi}} \right) \]

\[ \rho''= \frac{ R\Sin\varphi\left(\sqrt{r^2-R^2\Cos^2\varphi}+R\Sin\varphi\right) -R^2\Cos^2\varphi\left(r^2-R^2\Cos^2\varphi+R^2\Sin^2\varphi\right) }{ \sqrt{r^2-R^2\Cos^4\varphi} } =0. \]

In the end, this equation reduces to an equation of the 4th degree, and it is simplest of all to solve it graphically by Lill’s method.

Knowing the first and second derivatives, we shall without difficulty find the radius of curvature of the curve at any point.

\[ \rho_{\mathrm{cr}}= \frac{ \left[ \rho^2+\left(\frac{d\rho}{d\varphi}\right)^2 \right]^{3/2} }{ \rho^2+2\left(\frac{d\rho}{d\varphi}\right)^2-\rho\frac{d^2\rho}{d\varphi^2} }. \]

To find the area bounded by the curve, it would be necessary to take the definite integral

\[ \frac{1}{2}\int_{\varphi_0}^{\varphi}\rho^2\,d\varphi_1, \]

\[ \rho^2= \left(e-R\Sin\varphi+\sqrt{r^2-R\Cos^2\varphi}\right)^2. \]

The quantity of air delivered per second is obviously computed by the formula

\[ Q_{\mathrm{Sec}}= \frac{\pi\,(R^2-r^2)\,L\,n}{60}\,\eta, \]

where \(L\), in meters, is the dimension perpendicular to the plane of the drawing, or the width of the air blower; \(n\) is the number of revolutions per minute, and \(\eta\) is the coefficient of useful action (volumetric).

For a small compression it may be assumed that the process of compression of the air proceeds isothermally, and then the theoretical work of compression, expressed in horsepower, will be

\[ N= \frac{p_1v_1\,\Lg^n\frac{p_2}{p_1}}{75} = 2{,}3026\,\frac{p_1v_1}{75}\,\Lg\frac{p_2}{p_1}. \]

To the article by S. K. Konyukhov, “On the Theory of the Wedding Fan.”

Drawing 3.

Drawing 1.

To the article by S. N. Konyukhov, “On the Theory of the Wedding Fan.”

Drawing 2.