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Somewhat mathematically challenged person here.

I use Excel to generate points to feed a CNC Router to cut ellipses. (Strictly for personal use - nothing for sale - retirement learning process.) I calculate the Radian (Rad) by degree (Rad = Degrees * PI() / 180), stepping all the way around the ellipse, one Degree at a time.

I use the following two equations for calculate the X and Y points for each Degree step around.

X = (Width * COS( Rad ))

Y = (Height * SIN( Rad ))

(Width and Height are half of the diameter of their respective axis.)

I usually use 0,0 as my center point.

This works well for dishes and bowls where the major axis is perfectly horizontal or vertical. Now, I'm trying to branch out, and I'm hitting a snag.

Would there be a reasonable way to modify these equations to be able to skew the major/minor axis set, using, as an example, a 15 Degree skew? (If I can see how to do it with a specific skew angle, I think I can figure out how to adapt it to what I'm doing. That's why I suggested a 15 Degree skew angle.)

Any and all help (and constructive criticism for not learning enough to resolve this by myself) will be greatly appreciated!

David C

PS: For reference, I want to cut a "Star Burst" set if ellipses with constantly gently changing lights in each segment for a night light for each of my twin granddaughters. The electronics for the color changing lights are no problem for me. The equations for the ellipses, however.....

KReiser
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David C
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1 Answers1

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The equations for an ellipse in standard orientation is as you mentioned

$ x(t) = a \cos ( t) $

$y(t) = b \sin (t) $

where $a$ is the width and $b$ is the height, i.e. the ellipse semi-major and semi-minor axes lengths.

To rotate this ellipse by a skew angle $\theta$ apply the rotation matrix

$ R(\theta) = \begin{bmatrix} \cos(\theta) && - \sin(\theta) \\ \sin(\theta) && \cos(\theta) \end{bmatrix} $

To the point $( x(t), y(t) )$, to results in the rotated ellipse $( x'(t), y'(t))$. Explicitly, you have

$ x'(t) = a \cos(\theta) \cos(t) - b \sin(\theta) \sin(t) $

$ y'(t) = - a \sin(\theta) \cos(t) + b \cos(\theta) \sin(t) $

I hope I've answered your question.

Hosam Hajeer
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