# Thread: Circle Involute - Solving "y" for any given "x"

1. ## Circle Involute - Solving "y" for any given "x"

Having familiarized myself with the parametric formula of a circular involute (i.e. x = a(cos t + t sin t), y = a(sin t - t cos t)), I was wondering if it is possible to derive an equation which will solve y for any given value of x.

I currently visualise the parametric formula as two adjoining right-angle triangles, the second being derived from the x/y values of the first. However, due to the implicit relationship exhibited between these two structures, I fear that this problem far exceeds the elementary trigonometry I currently understand.

As I'm reasonably sure that this problem cannot be solved using Trigonometry alone, any help or advice as to how I should approach a solution would be very much appreciated.

2. Originally Posted by Gemma1971
Having familiarized myself with the parametric formula of a circular involute (i.e. x = a(cos t + t sin t), y = a(sin t - t cos t)), I was wondering if it is possible to derive an equation which will solve y for any given value of x.

I currently visualise the parametric formula as two adjoining right-angle triangles, the second being derived from the x/y values of the first. However, due to the implicit relationship exhibited between these two structures, I fear that this problem far exceeds the elementary trigonometry I currently understand.

As I'm reasonably sure that this problem cannot be solved using Trigonometry alone, any help or advice as to how I should approach a solution would be very much appreciated.
The short answer is that it can't be done. The involute of a circle is a spiral, in which there will be infinitely many values of y corresponding to any given value of x. So it is unreasonable to expect a formula for y in terms of x. That is why the equation has to be given in parametric form.

3. Thank you for the replay Opalg.

This makes a lot of sense; I now see that the involute spiral regularly intersects both x and y axes as it unwinds ad-infinitum.

My next question, which may well be as dumb as the first, is whether or not it is possible to determine the exact point at which the unwinding spiral intersects a circle of given diameter? I'm guessing that this problem has a finite solution on the basis that there can be at most a single intersection, but whether this is a trivial matter is another thing altogether

As the centers of both involute and arbitary circles are coincident, I guess what I'm actually asking is whether or not it is possible to determine which angle, as used within the parametric formula, will yield a specific distance/radius from the center of the involute circle?

4. Originally Posted by Gemma1971
My next question, which may well be as dumb as the first, is whether or not it is possible to determine the exact point at which the unwinding spiral intersects a circle of given diameter?
It turns out that this has a nice solution (assuming that the centre of this "circle of given diameter" is the same as the centre of the spiral). Rather than the diameter, I'll use r to denote the radius of this circle.

It follows from the given equations for x and y that $\displaystyle x^2+y^2 = a^2(1+t^2)$. The distance of the point (x,y) from the origin is $\displaystyle \sqrt{x^2+y^2}$. Hence the point at which this distance reaches the value r occurs when $\displaystyle r = a\sqrt{1+t^2}$. Express that formula in terms of t to see that this happens when the parameter t has the value $\displaystyle t = \sqrt{\Bigl(\frac ra\Bigr)^2 - 1}$.

5. Thanks ever so much Opalg; you have managed to pitch this at a level that even the mathematically-challenged like I can understand it

Devoid of the assumptions that often accompany an established discipline, you certainly have a rare gift for enlightening the ignorant; it's funny how I have learned more about involution from your two posts, than I have within the past three weeks of schooling.

Just goes to show that with all due respect, the student isn't always at fault

Once again, thanks very much for your time and effort: you are officially my hero-of-the-week.

Best regards, Gemma