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I am interested to find a particular connection between a right triangle and an arithmetic spiral.

In a right triangle, when one side is $0.8$ and the other side is $0.6$, the hypotenuse is $1$. Trivially, when one side is $1$ and the other side is $0$, the "hypotenuse" is $1$.

Suppose we use the equivalent values for a rotating circle whose origin is $(0,0)$ on the $(x,y)$ system. Let's say a line is drawn from the origin along the $y$-axis (the radius) so it's the "radial velocity".

When the angular velocity of the circle is $1$ and the radial velocity is $1$, what is the length of the spiral?

Similarly, when angular velocity is $1$ what are the spiral lengths for radial velocity $0.8$ and radial velocity $0.6$ respectively?

I am wondering if anyone has already solved for this, and whether the result is interesting in advance.

argonaut
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  • The spiral would have a simple equation in polar coordinates of form $r=k\theta.$ So maybe look up "arclength in polar coordinates" by google or on Wikipedia. – coffeemath Aug 08 '14 at 11:07

2 Answers2

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I'm not sure what exactly you are asking here. In particular, you never specified how many turns of the spiral you wanted. Let me just provide enough information for you to do some further exploration of your own.

Suppose we have an Archimedian spiral which we can parametrize in polar coordinates as $$(r(t),\ \theta(t)) = (at, bt)$$ Here $a$ is the radial velocity and $b$ is the angular velocity. The arclength of this parametrized curve from the origin, as a function of time, is given by $$s(\tau) = \int_0^\tau \sqrt{r^2\left(\frac{d\theta}{dt}\right)^2 + \left(\frac{dr}{dt}\right)^2}\ dt = a\int_0^\tau \sqrt{(bt)^2+1}\ dt$$ Solving the above integral gives us $$s(t) = \frac{at}{2}\sqrt{1+(bt)^2} + \frac{a}{2b}\sinh^{-1}(bt)$$ Alternatively, as a function of $\theta$, the arclength is given by $$s(\theta)=\frac{a}{2b}\left(\theta\sqrt{1+\theta^2} + \sinh^{-1}(\theta)\right)$$ Note that from the latter formula, we see that for a given angle $\theta$, the arclength depends only on the ratio $\frac{a}{b}$ which is physically reasonable. That means we can keep either angular velocity or radial velocity fixed and only consider the other, which is what you appear to be doing.

In any case, you can figure out all the arclengths you want by substituting the appropriate values for $a$, $b$, $t$, $\theta$

EuYu
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  • Yes, I omitted an important point: The circle stops rotating after 1 radian. Is the spiral length well known when the pencil mark's radial velocity equals 1? (And 0.8, and 0.6). – argonaut Aug 08 '14 at 12:45
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When the pencil mark's radial velocity equals $1$, the spiral length equals $1.147793553$
When the pencil mark's radial velocity equals $0.8$, the spiral length equals $1.219427735$
When the pencil mark's radial velocity equals $0.6$, the spiral length equals $1.356964015$

This is not very remarkable!

argonaut
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