Page:Scientific Memoirs, Vol. 1 (1837).djvu/639

in terms of the diameter of the cylindrical armature, and the thickness of the wire and its silky envelope. We have however the semi-diameter of a convolution, if half the thickness of the iron cylinders is ${}=q$,

for the first		series	${}=q+{\frac {b+\beta }{2}}$
——	second	———	${}=q+{\frac {3}{2}}(b+\beta )$
——	third	———	${}=q+{\frac {5}{2}}(b+\beta )$
——	nth	———	${}=q+{\frac {2n-1}{2}}(b+\beta )$

whence we have the length of a convolution or

$c$	for the first		series	${}={\big (}2q+(b+\beta ){\big )}\pi$
$c$	——	second	———	${}={\big (}2q+3(b+\beta ){\big )}\pi$
$c$	——	third	———	${}={\big (}2q+5(b+\beta ){\big )}\pi$
$c$	——	nth	———	${}={\big (}2q+(2n-1)(b+\beta ){\big )}\pi .$

If we substitute the first value of $c$ in the equation for $m$, we obtain

$p_{1}={\frac {\alpha b^{2}f}{\alpha \pi {\big (}2q+(b+\beta ){\big )}+(b+\beta )m}}.$

The electromotive power is for two series of convolutions with regard to the above law, No. 2,

${}=2{\frac {\alpha }{b+\beta }}\cdot f;$

but the resistance is equal to that of the two series of convolutions, together with the piece $m$, therefore

${}={\frac {{\frac {\alpha }{b+\beta }}(2q+b+\beta )\pi +{\frac {\alpha }{b+\beta }}(2q+3(b+\beta ))\pi +m}{b^{2}}}$

${}={\frac {\alpha \pi (4q+4(b+\beta ))+m(b+\beta )}{b^{2}(b+\beta )}},$

consequently the power of the current for two series, or

${\begin{aligned}p_{2}&={\frac {2\cdot {\frac {\alpha }{b+\beta }}\cdot f}{\frac {\alpha \pi (4q+4(b+\beta ))+m(b+\beta )}{b^{2}(b+\beta )}}}\\&={\frac {2\alpha b^{2}f}{\alpha \pi (4q+4(b+\beta ))+m(b+\beta )}}.\end{aligned}}$