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A series of ellipses which are scaled and rotated at each step will form the illusion of a logarithmic spiral. Years ago, I saw this on some recreational math site and lately I have been seeing this in discussions of spiral arm formation in galaxies. Like this illustration

enter image description here

If the scale and angle are selected correctly, two successive ellipses will be tangent and that the tangency points will form an exact log spiral.

I'd like to draw this in a CAD program with tangent ellipses but I am having trouble with the math. For a given rotation angle, it should be possible to find the correct scaling factor to have tangency. Do you have to use the tangent angle relative to the vector to the origin at the point of tangency to solve this?

ETA: I googled to learn more about the properties of ellipses to try to find some aspect to exploit to solve this problem and found two properties that seemed relevant. One was a ratio between distances involving the tangent and the major axis and the other used the angles between lines from the tangency point and the focus points and the tangent line which are equal. The latter one helped me draw an approximate diagram of the problem. To solve the problem would involve having those pairs of angles equal for both ellipses while preserving the eccentricity.

Also, any two ellipses with a common center can have either 0, 2 or 4 intersections and what I'm trying to find is a second ellipse at a particular angle to the first that has only two points of intersection while preserving the shape. So there must be something about the two ellipses being tangent that I need to understand to solve this.

bevbh
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    See whether this helps. – Ng Chung Tak Mar 14 '21 at 00:06
  • Yes, that is exactly what I was looking for. Thank you. I love the links to the animations too. – bevbh Mar 14 '21 at 02:02
  • I successfully calculated and drew the inner ellipse in my test case but when I calculated the tangency point using the formula in your comment in the linked question, I got the wrong answer. Swapping the sine and cosine gave a correct looking answer. Did you measure your tilt angle the opposite way? In the inserted screen capture from Mathematica it says theta=60 but the tilt looks like 30 to me. I'm having some confusion with the notation because t is usually used for the parameter in radians but the questioner used it for the tilt angle. – bevbh Mar 14 '21 at 07:12
  • The 1st picture is $\theta$ while the 2nd picture is $t$. – Ng Chung Tak Mar 14 '21 at 07:55
  • $\tan \theta=\dfrac{C(t)}{S(t)}$ – Ng Chung Tak Mar 14 '21 at 08:23

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