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Problem

A electrical engineer needs a new coil and decides to make one from scratch. He hasn't decided the radius or length of the cylinder on which the coil will be wound. Define a function $f(r,l)$ where $r$=(radius), $l$=(height of the cylinder). There have to be exactly 10 full cycles of winding in total independent of its height.

Attempt to solve

We can define our curve in the form of a parametric equation:

$$ r(t)=\begin{bmatrix} r\cos(\frac{20\pi t}{l}) \\ r\sin(\frac{20}{l}) \\ t \end{bmatrix} $$

where $l$ and $r$ are constants, and $x,y,z$ are unit vectors.

$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} $$

Plot of $r(t)$ when $l=10,r=3$ & plot of reference cylinder

3d graph of functions as described

Now we want to define a function for the length when the parametric equation is known. Length can be defined with the integral

$$ f(r,l)=\int_{a}^{b} || r'(t) || dt $$

$$r'(t)=\begin{bmatrix}x'(t)= \frac{d}{dt}r\cos(\frac{20\pi t}{l})=-\frac{20\pi\sin(\frac{20\pi t}{l})}{l}\\ y'(t)=\frac{d}{dt}r\sin(\frac{20 \pi t}{l})=\frac{20\pi\cos(\frac{20\pi t}{l})}{l} \\ z'(t)=\frac{d}{dt} t=1 \end{bmatrix}$$

$$ f(r,l)=\int_{a}^{b} \sqrt{(x(t))^2+(y(t))^2+(z(t))^2}dt $$

$$ f(r,l)=\int_{a}^{b} \sqrt{(-\frac{20\pi\sin(\frac{20\pi t}{l})}{l})^2+(\frac{20\pi\cos(\frac{20\pi t}{l})}{l})^2+(1)^2}dt $$

Now not too sure about integration limits since my understanding of what's going on here is not that good but I guess that we should be integrating from $0$ to $l$

$$ f(r,l)=\int_{0}^{l} \sqrt{(-\frac{20\pi\sin(\frac{20\pi t}{l})}{l})^2+(\frac{20\pi\cos(\frac{20\pi t}{l})}{l})^2+(1)^2}dt $$


Since we only had to define the function and not calculate anything with it I haven't tried to calculate any length since the integral looks like something I do not want to integrate by hand.

Anyway it would be highly appreciated if someone could point out the flaws (I am pretty sure there are flaws in this) or if this is correct I would like to still know the reason why integrating the tangent length will result in the length of the curve?

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  • $\begingroup$ The last integral is a piece of cake: all that mess in the square root is just a constant number (constant wrt $\;t\;$ , of course), equal to $\;\cfrac{400\pi^2+l^2}{l^2}\;$ ... $\endgroup$ – DonAntonio Jan 9 '18 at 22:05
  • $\begingroup$ The parametrization I would choose would be slightly different, but afaict the results are the same as what you get (including the integration limits - but check out the answer below re. integration). $\endgroup$ – NickD Jan 9 '18 at 22:25
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    $\begingroup$ You forgot to mention what the result of the function $f$ should be... $\endgroup$ – immibis Jan 10 '18 at 1:12
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    $\begingroup$ @ChaseRyanTaylor it's fixed now. It seems this types of errors seem to get more frequent when your tired. $\endgroup$ – Tuki Jan 10 '18 at 10:09
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    $\begingroup$ If you're doing this kind of thing a lot as more than just an academic exercise, I'd recommend getting a en.wikipedia.org/wiki/Pocket_Ref or similar. It has a whole section on winding things around cylinders in various ways. $\endgroup$ – Perkins Jan 10 '18 at 19:07
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You are working a little harder than you need to. But, you are so close to the finish.

$f(l,r)=$$\int_{0}^{l} \sqrt{(-\frac{20\pi r\sin(\frac{20\pi t}{l})}{l})^2+(\frac{20\pi r\cos(\frac{20\pi t}{l})}{l})^2+(1)^2}dt$

Use the trig identity $\cos^2 t + \sin^2 t = 1$

$\int_{0}^{l} \sqrt{(\frac{20\pi r}{l})^2+(1)^2}dt$

Notice that the integrand doesn't rely on $t.$

$\frac {\sqrt{(20\pi r)^2+l^2}}{l}\int_{0}^{l} dt\\ \sqrt{(20\pi r)^2+l^2}$

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If you imagine unrolling the cylinder the wire is the hypotenuse of a right triangle. The base is ten times around the cylinder and the height is the height of the coil. You can find the length of the wire by Pythagoras. There is no need for integration.

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    $\begingroup$ That's a really nice way to think about it +1 $\endgroup$ – Lamar Latrell Jan 10 '18 at 8:21
  • $\begingroup$ ÷1 Exactly what I first thought of when I saw this question. $\endgroup$ – Adrian Jan 11 '18 at 14:48
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From $$r(t)=\begin{bmatrix} r\cos(\frac{20\pi t}{l}) \\ r\sin(\frac{20\pi t}{l}) \\ t \end{bmatrix}$$We get $$ || r'(t) ||= \sqrt {1+r^2(\frac {20\pi }{l}})^2 $$ Therefore the length of the coil is $$ \int_{0}^{l} || r'(t) || dt = \sqrt {\;400r^2\pi^2+l^2}\;$$

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