Page:On Governors.pdf/5

of this pressure is taken off by a spring which acts on the centrifugal piece. The force acting on $\scriptstyle \ B$ to turn it round is therefore

$F'\left[{\frac {dx}{dt}}\right]-C';$

and if we remember that the velocity varies within very narrow limits, we may write the expression

$F\left({\frac {dx}{dt}}-V_{1}\right);$

where $\scriptstyle \ F$ is a new constant, and $\scriptstyle \ V_{1}$ is the lowest limit of velocity within which the governor will act.

Since this force necessarily acts on $\scriptstyle \ B$ in the positive direction, and since it is necessary that the break should be taken off as well as put on, a weight $\scriptstyle \ W$ is applied to $\scriptstyle \ B$ tending to turn it in the negative direction; and, for a reason to be afterwards explained, this weight is made to hang in a viscous liquid, so as to bring it to rest quickly.

The equation of motion of $\scriptstyle \ B$ may then be written

(6)	$B{\frac {d^{2}x}{dt^{2}}}=F\left({\frac {dx}{dt}}-V_{1}\right)-Y{\frac {dy}{dt}}-W,$

where $\scriptstyle \ Y$ is a coefficient depending on the viscosity of the liquid and on other resistances varying with the velocity, and $\scriptstyle \ W$ is the constant weight. Integrating this equation with respect to $\scriptstyle \ t$, we find

(7)	$B{\frac {dy}{dt}}=F\left(x-V_{1}t\right)-Yy-Wt$

If $\scriptstyle \ B$ has come to rest, we have

(8)	$x=\left(V_{1}+{\frac {W}{F}}\right)t+{\frac {Y}{F}}y$

or the position of the machine is affected by that of the governor, but the final velocity is constant, and

(9)	$V_{1}+{\frac {W}{F}}=V_{,}$

where $\scriptstyle \ V_{1}$ is the normal velocity.

The equation of motion of the machine itself is

(10)	$M{\frac {d^{2}x}{dt^{2}}}=P-R-F\left({\frac {dx}{dt}}-V_{1}\right)-Gy$

This must be combined with equation (7) to determine the motion of the whole apparatus. The solution is of the form

(11)	$x=A_{1}e^{n_{1}t}+A_{2}e^{n_{2}t}+A_{3}e^{n_{3}t}+Vt$

where $\scriptstyle \ n_{1},\ n_{2},\ n_{3}$ are the roots of the cubic equation

(12)	$MBn^{3}+\left(MY+FB\right)n^{2}+FYn+FG=0$

If $\scriptstyle \ n$ be a pair of roots of this equation of the form $\scriptstyle \ a\pm {\sqrt {-1}}b$, then the part of $\scriptstyle \ x$ corresponding to these roots will be of the form

(13)	$e^{at}cos\left(bt+\beta \right)$