Equation of the Surface of Fusée Barrel The fusée is a device used in certain mainspring-driven clocks, to level-out the torque that the mainspring applies to the movement. A mainspring without correction produces a torque that declines roughly linearly with the angle through which it has unwound. Although the very purpose of a clock's movement is to produce rotation at a rate independent of the magnitude of the applied torque, there is still a second-order error: the greater swing of the balance-wheel, or pendulum, or whatever the regulating mechanism is, is not perfectly offset by the increased speed of the motion; and a clock can be made more accurate by installing some kind of mechanism that in some way levels out the driving torque.
Anyway the fusée is a mechanism for doing precisely that; and consists of a ... not cone ... I don't know what the name for it is ... a concave quasi -cone with a quasi-helickal track of gear teeth on it. A chain - somewhat like a bicycle chain (but usually much finer) - is wound onto it and the barrel of the mainspring in such a way that when the mainspring is fully-wound it is pulling at the fusée barrel at the narrowest part; and as the mainspring unwinds, the chain also winds off the fusée barrel in such a way that the radius of the point on that barrel at which it is pulling is always such as to compensate for the diminution of the torque of the mainspring. A good book on horology (such as Grimshaw's classic treatise - probably the definitive horological text) will probably have a thorough description of this device, with diagrams & all.
The question is - assuming the torque of the mainspring does decrease linearly with angle-through-which-it-has-turned, what is the equation giving the radius of the track on the barrel as a function of angle about it's axis? 
Or might it be easier to formulate it as a function of unwinding angle of the mainspring? I think probably not ... but if someone can show that the other formulation has a great advantage, I would be extremely interested. But I doubt there is ... infact it's essentially just introducing distance along the track as a variable and getting the equation in terms of that instead. Actually, it's undoubtedly easier - but to construct one we need $r$ in terms of $\theta$ (or vice versa) explicitly, really.
They have other applications also - I have read that some high-grade crossbows have them, for easing the arming of them.


So, referring now to these pictures: what would be the equation of $r$ in terms of $\theta$ polar co-ordinates along the track on the surface of these barrels?
As can see also, it doesn't absolutely have to be a chain & betoothen track - a cable & smooth track works also.
 A: As seems to be happening regularly, the thought that goes into framing the question for presentation to an audience goes also towards clarifying it! So the solution has come to me more quickly than I at first thought it would. It also brought the pleasant surprise that a certain integral that looked intractable turned out to be actually quite tractable.
Set $l_0$ = the length displaced at the point at which the force of the mainspring falls to 0, & $r_0$ = the minimum radius of the track on the fusée barrel, ie its radius at commencement of unwinding of the mainspring, & $F_0$ = initial tension in the chain or cable.
$$k(l_0 - l)=\frac{F_0 r_0}{r}  .$$
Because $k=F_0/l_0$
$$l_0 - l=\frac{l_0 r_0}{r}  .$$
Set $a=\sqrt{l_0 r_0}$
$$\frac{dl}{dr}=\frac{a^2}{r^2}$$
$$\frac{dl}{d\theta}=\frac{a^2}{r^2}\frac{dr}{d\theta}$$
$$\sqrt{r^2+(\frac{dr}{d\theta})^2}=\frac{a^2}{r^2}\frac{dr}{d\theta}$$
$$r^2+(\frac{dr}{d\theta})^2=\frac{a^4}{r^4}(\frac{dr}{d\theta})^2$$
$$\frac{dr}{d\theta}=\frac{r}{\sqrt{\frac{a^4}{r^4}-1}}$$
$$\int_{r_0}^r\sqrt{\frac{a^4}{r^4}-1}\frac{dr}{r}=\int_{\theta_0}^\theta d\theta$$
The integral in $r$ is, perhaps surprisingly (it was to me anyway), actually quite tractable - it's actually just a scaled & transformed case of
$$\int\sqrt{e^x-1}dx =2(\sqrt{e^x-1} -\arctan\sqrt{e^x-1}). $$
Set $\theta_0=\frac{l_0}{r_0}$.
$$\theta=\frac{1}{2}(\sqrt{\theta_0^2-1}-\arctan\sqrt{\theta_0^2-1}-(\sqrt{\frac{a^4}{r^4}-1}-\arctan\sqrt{\frac{a^4}{r^4}-1})$$
So the problem can be dedimensionalised by setting a characteristic angle 
$$\theta_0=\frac{l_0}{r_0}$$
& characteristic length
$$\sqrt{l_0 r_0} .$$
Set $\rho=r/a$, whence the initial value of $\rho$ is $1/\sqrt \theta_0$.
If $r$ is only varying slowly wrt $\theta$, ie at the narrow end of the barrel, we have simply 
$$\theta = \int_{1/\sqrt{\theta_0}}^\rho\frac{d\rho}{\rho^3}=\frac{1}{2}(\theta_0-\frac{1}{\rho^2}) .$$
Not making that simplifying assumption, we have
$$\theta=\frac{1}{2}(\sqrt{\theta_0^2-1}-\arctan\sqrt{\theta_0^2-1}-(\sqrt{\frac{1}{\rho^4}-1}-\arctan\sqrt{\frac{1}{\rho^4}-1}) .$$
It's notable that in the approximate solution $\rho$ can increase without limit, whereas in the accurate one it cannot exceed unity. If we set $\rho=1-\epsilon$ ($\epsilon$ <<1), and also $\sqrt{\theta_0^2-1}-\arctan\sqrt{\theta_0^2-1}=\Theta_0$, then when $\rho$ is nearly 1
$$\theta≈\frac{\Theta_0}{2}-\frac{1}{2}(2\sqrt\epsilon-(2\sqrt\epsilon-\frac{8\epsilon^{3/2}}{3}))$$
$$\theta≈\frac{\Theta_0}{2}-\frac{4\epsilon^{3/2}}{3} ,$$
so we have an abrupt termination of order 3/2 in the growth of $\theta$ at $\rho=1$ ... ie $r=a$.
The total amount of chain or cable that becomes unwound is 
$$a\int_{1/\sqrt\theta_0}^1\frac{d\rho}{\rho^2}=a(\sqrt\theta_0-1) = l_0-a .$$
The amount of the total length $l_0$ left unutilised, then, is $a$, whence the torque is a fraction 
$$\frac{a}{l_0}=\frac{1}{\sqrt\theta_0}$$
of its initial value; however the radius on the fusée barrel at which this force is now applied is multiplied by the factor 
$$\frac{a}{r_0}=\sqrt\theta_0 ... $$
so the torque is the same at the end as at the beginning.
It's notable that the function (in the exact solution) expressing  $\theta$ in terms of $r$ cannot be inverted in closed-form to give $r$ in terms of $\theta$. But this doesn't really matter, as the function the way-round that it is is just as suitable for actually  designing one of these things on a drawing-board.
Looking at this solution qualitatively it can be seen how when $r$ is only slightly greater than $r_0$, the exact solution cleaves closely to the approximate one ... although does not quite tend asymptotically to it, as $\frac{dr}{d\theta}$ does not tend to zero at $r_0$. As $r\rightarrow a$, however, the particular shape of the exact solution kicks-in, with the $\sqrt(a^4/r^4-1)$ 'rescuing' it from the clearly absurd behaviour of $\theta$ tending to a finite value as $r$ increases abundantly, and the $\arctan$ providing for an intuitively realistic $\epsilon^{3/2}$ termination, rather than the absurd $\epsilon^{1/2}$ one that it would have without the $\arctan$ or something similar to cancel the first order term in $\epsilon$.
When I say 'exact' solution, there are still actually one or two troublesome matters that have been neglected in my analysis. One is the shift in the chain in the axial direction as it winds/unwinds. I decided that as the shift can easily be set up to be the same on both barrels (indeed, it would be only natural to do so), it would probably only affect the total length of chain used - it would need to be longer by ~$z_0^2/2l_0$, where $z_0$ is the total axial displacement, ie the length of each barrel (presumed equal). The other is that the chain makes a changing angle to the to the line joining the centres of the barrels ... but I do think that effect is probably negligible. If anyone wishes to present a solution that incorporates these effects ... be my guest! But I would be truly amazed if you could still get a closed-form elementary-function solution. I was surprised enough to find that mine was such, as it stands!
