# How to find the minute points of an ellipse clock, knowing the minor axis and the major axis?

I want to make an analogic clock, not circle, but ellipse. So the distance between minute points is not constant. I guess it grows proportionally with the division of major axis with minor axis.

How can i find these points on the ellipse ?

• Do you want to find these points geometrically or algebraically? Geometrically, if you already have the ellipse, you can just draw a line from the center at the correct angle and see where it intersects. – user856 Jul 2 '13 at 16:46
• Why was the answer removed? – nbubis Jul 2 '13 at 16:52
• @nbubis, because there were errors. Perhaps the solver will fix them and undelete, or perhaps not. – vadim123 Jul 2 '13 at 16:53
• @RahulNarain I want to find the x and y coordinates. Easy to say what you suggested, but i would have to do this 13 times (15 minus the obvious 2 values, the rest of the points are symetrical). So i have to find a formula. Thank you:) – darkdante Jul 2 '13 at 17:35
• According to me, if you think that you want a circle like clock on the ellipse, then the answer will simply be the actual clock (if the clock represents 15 minutes in 90 degrees). But if you want the pointers equally spaced then it might change it's orientation. And if you are thinking about the first one then the layout of a simple circular clock is what you will get, with pointers hugely spaced on major and crammed as we go towards minor. – Rohinb97 Jul 2 '13 at 17:44

To get the points, parametrized by time, first note that $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and $$\frac{x}{y}=\tan\left(t\frac\pi{6\text{ hrs}}\right)$$ which solved simultaneously yield the parametric curve $$(x,y)=\frac{ab\left(\sin\left(t\frac\pi{6\text{ hrs}}\right),\cos\left(t\frac\pi{6\text{ hrs}}\right)\right)}{\sqrt{b^2\sin^2\left(t\frac\pi{6\text{ hrs}}\right)+a^2\cos^2\left(t\frac\pi{6\text{ hrs}}\right)}}$$ Example:
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Suppose the ellipse is not rotated, i.e. $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
"3" is easy. "2" should satisfy $$\frac{y}{x}=\tan\frac{2\pi}{12}=\frac{1}{\sqrt{3}}$$ or $x=\sqrt{3}y$. Substituting we get $$\left(\frac{3}{a^2}+\frac{1}{b^2}\right)y^2=1$$ Solve for $y$, then $x=\sqrt{3}y$.
To calculate the point at 14 minutes, that is angle $\theta=\frac{2\pi}{60}$. To calculate the point at 13 minutes, that is angle $\theta=2\frac{2\pi}{60}$. To calculate the point at 12 minutes, that is angle $\theta=3\frac{2\pi}{60}$. Once $\theta$ is known, start with $\frac{y}{x}=\tan\theta$, solve for $x$ (or $y$), then substitute into the ellipse equation.