How to parametrize eliptic curve?

Question

def curve(t, radiusX, radiusY):
    x = sin(t*pi/2) * radiusX
    y = cos(t*pi/2) * radiusY
    return (x,y)

Following Python pseudo-code will draw a part 1/4 of an ellipse.
However, I am looking to stop drawing at given Y, as I don't need full 1/4 of this ellipse.

Can this function be parametrized such that it will stop drawing at given Y, without affecting the curve < Y?

An alternative approach is to draw using the current function and then cut the resulting object to given max height from the bottom, which I am trying to simplify.

I think the direction of the solution is to keep y at constant step using y = t * radiusY while x should be a transformed formula that will tell us what's the proper x given our y?

The term "elliptic curve" has a different meaning in Mathematics. What you want is an "arc of an ellipse". — jjagmath, Dec 15 '23 at 19:46
From the given Y you can compute the corresponding value of t and stop there. — Intelligenti pauca, Dec 15 '23 at 19:48
it would be against the whole idea of parametrization, to be honest — underflow, Dec 15 '23 at 19:51
Just solve for $t$ in $Y = \cos(t\cdot \pi/2)\cdot \text{radiusY}$ — jjagmath, Dec 15 '23 at 19:52
is y = t * height x = radiusX/radiusY*sqrt(radiusY*radiusY-y*y) a proper formula? it seems to bug out at lower height (at least) — underflow, Dec 15 '23 at 20:09
ok, I believe the above formula is correct, but I had unrealistic combination of radiusX/radiusY/height. I am working on a CAD model right now, and they will get auto-calculated as I progress the model... — underflow, Dec 15 '23 at 20:48
I think height can also be represented in angle form, which will be easier — underflow, Dec 15 '23 at 20:49

John Alexiou · Answer 1 · 2023-12-16T21:22:42.400

0

Use

def curve(phi, radiusX, radiusY):
    r = 1/sqrt( sin(phi)**2/radiusX**2 + cos(phi)**2/radiusY**2)
    x = sin(phi) * r
    y = cos(phi) * r
    return (x,y)

and now phi represents the polar angle of the ellipse, with 0 at the 12 o'clock positon and moving clockwise. So if you only want 1/4 of an ellipse use phi = 0..pi/2

This follows the polar coordinate definition for a centered, axis-aligned ellipse

$$r(\theta)=\frac{1}{\sqrt{\frac{\sin^{2}\theta}{a^{2}}+\frac{\cos^{2}\theta}{b^{2}}}}$$

and then $x = r \sin \theta$ and $y = r \cos \theta$.

edited Dec 16 '23 at 21:22

answered Dec 15 '23 at 21:53

John Alexiou

13,816

thank you, I went with y = t * height x = radiusX/radiusY*sqrt(radiusY*radiusY-y*y) which turned out very well – underflow Dec 15 '23 at 22:00
This won't work if the ellipse extends to the negative x areas. – John Alexiou Dec 15 '23 at 22:05

How to parametrize eliptic curve?

1 Answers1