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 ?
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 ?
On
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 $x=\sqrt{3}y$.
Update, per request:
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.
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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