# Frequency response of a capstan-like mechanical amplifier

Tension on a wire can be amplified by wrapping the wire around a powered rotating capstan, with output force governed by the linear capstan equation, Tout = Tin * emu*theta. Will this equation hold for rapidly varying input, if the wire has small elasticity?

I am interested in using this phenomenon to amplify power in a musical instrument, but I cannot find much information about how the capstan effect behaves when the input tension varies at audio frequencies. The wikipedia page for "Torque amplifier" mentions this effect being used for an audio output "electrostatic drum amplifier," but no citations are given and Googling "electrostatic drum amplifier" just returns the wikipedia page.

I am reasoning my way to an answer here - I have never seen this problem before so I could be completely wrong.

I think the issue is that you need to make sure the normal force is reduced as the tension on the "signal" side of the capstan is reduced. This is presumably why the slightly stiffer wire (as opposed to the rope) gave more predictable results at low frequencies. I would be surprised if you could do this without a great deal of distortion - the "normal force release" wave basically has to be able to travel around the capstan so the lower force is felt everywhere. If you have a capstan of diameter $d$ with $n$ turns on it, the length of the wrapped wire is $\pi n d$ . For a given density of wire (mass per unit length) $\rho$ you expect the speed of propagation of a wave to be

$$v = \sqrt{\frac{T}{\rho}}\tag1$$

The tension is a function of position along the wire - increasing exponentially from $T_1$ at the point where the signal wire meets the capstan, up to $T_2 = e^{\mu\theta}T_1$. The signal wave will travel faster where the tension is greater, and the total time taken will be

$$t = \int \frac{dx}{v}\tag2$$

Writing the tension as a function of position along the wire as

$$T(x) = T_1 e^{\alpha x}$$

we obtain

$$\log\left(\frac{T_2}{T_1}\right) = \alpha \pi n d$$ $$\alpha = \frac{\mu \theta}{\pi n d} = \frac{\mu}{r}\tag3$$

We now combine (1), (2) and (3) to get

\begin{align} t &= \sqrt{\rho} \int \frac{dx}{\sqrt{T}} \\ &= \sqrt{\rho} \int \frac{dx}{\sqrt{T_0 e^{\mu x /r}}}\\ &= \frac{1}{v_0} \int e^{-\mu x / d}dx\\ &= \frac{1}{v_0} \frac{d}{\mu} e^{-\mu n \pi d / d}\\ &= \frac{1}{v_0}\frac{d}{\mu}\sqrt{\frac{T_1}{T_2}} \end{align}

This answer can be simplified a little bit more - but the bottom line is that once we know the time, we have an estimate of the cutoff frequency of this mechanism (at higher frequencies, the capstan doesn't even "know" that a lower steering tension was sent to its wire). We can estimate the cutoff frequency as

\begin{align} f_{cutoff} &= \frac{1}{T} \\ &= \sqrt{\frac{T_1}{\rho}}\frac{\mu}{d}\sqrt\frac{T_2}{T_1}\\ &= \sqrt{\frac{T_2}{\rho}}\frac{\mu}{d} \end{align}

Finally, this means $$\bbox[5px,border:2px solid red]{f_{cutoff}= \frac{v_2 \mu}{d}}$$

A remarkably simple and pleasing result. If you make the capstan bigger, the cutoff frequency drops. A heavier wire (lower speed), frequency drops. Higher friction, higher frequency. It all makes sense.

I would love to see whether this is true...