circular sections on an ellipsoid (something for galapagos?)

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Permalink Reply by Yoann Mescam (Systemiq) on November 8, 2011 at 12:45pm

I think that Galapagos is better used in finding the optimal solution rather than a set of solutions.

By the way, what are the constraints you would like on your circular sections, because I would say there is an infinite number of them.

Permalink Reply by Jonathan Chertok on November 8, 2011 at 1:12pm

Hi Systemiq,

Thank you.

Can I ask you to elaborate a little on this? I mean, I understand the point about galapagos and the distinction here is that the number or sections are infinite but their cardinal orientation with respect to the object is fixed? I mean, I am just assuming here but I was hoping you can help me understand it a little better.

Thanks very much...

Permalink Reply by Chris Tietjen on November 8, 2011 at 2:01pm

"Any planar cross section passing through the center of an ellipsoid forms an ellipse on its surface, with the possible special case of a circle if the three radii are the same (i.e., the ellipsoid is a sphere) or if the plane is parallel to two radii that are equal."  Wikipedia.

 

Also...

Permalink Reply by Chris Tietjen on November 8, 2011 at 2:19pm

And this too...

Permalink Reply by Yoann Mescam (Systemiq) on November 8, 2011 at 2:48pm

Weird, Wikipedia says you can find circular sections (an infinity) only if 2 radii are equal , Indicatrix says there are exactly 2 if all radii are different, and the paper fold link shows something looking like an ellipsoid with 3 different radii but constructed with circles (only an ellipsoid-looking thing ?)

Permalink Reply by Yoann Mescam (Systemiq) on November 8, 2011 at 2:56pm

Playing a bit with Galapagos, you can find quasi circular ellipses as shown there :

showing the difference between min and max radii of the section, and variance of distance from the center for 1000 points.

 

Anyway, Jonathan you have to be a bit more specific in what you are looking for, perfect ellipsoids and circles ? With a tolerance ? Parallels sections ?

 

Permalink Reply by Jonathan Chertok on November 8, 2011 at 3:15pm

Hi Chris, Sistemiq.

Thank you.

I have to study these links a bit and apparently I have to take a look at some other material as well.

All I had to go upon originally would be /circular/ /sections/ through an ellipsoid. I did not have any information that indicated that it was necessary to have 2 radii the same or 3 radii the same but I would like to know if this is the case. Basically, I was hoping to find the sections /through/ (so the very nice screenshot would have to have the red part go through the ellipsoid if it is not already). But maybe this ellipse is passing through the ellipsoid which mean I have to specify somehow that it passes through more like a "section" (but I am not sure how to define this).

OK, so "quasi-circular" I think will not work. I would be looking for exactly circular and somehow they would have to pass far enough through the object but I am not sure how one would define this exactly.

Can you guys give me a couple of days to try and nail this down a little more? I can see if I can formulate this a little more specifically...

Goodwill,

Jonathan

Permalink Reply by Jonathan Chertok on November 8, 2011 at 3:25pm

Hi Sistemiq,

Interesting. Thank you.

I see, here you have ellipsoidal "sections" (in this case it is sort of a cut through the surface) and galapagos returns the "non-circularity" of the "section"? I am not so familiar with galapagos except in david's presentation but is it possible to get geometric results that would give only the sections on any ellipsoid that was provided which were exactly circular?

Would it be possible to restrict the cuts to those that were more architectural in a way - so that they went /through/ the surface according to some requirement and didn't simply /cut/ the surface?
I mean, ostensibly if the script existed could we submit examples of ellipsoids that were from variously restricted "classes" and then have the script provide us with the solutions? 

- Jonathan

Permalink Reply by Yoann Mescam (Systemiq) on November 8, 2011 at 4:26pm

More informations:

 

I took the centroid of a random ellipsoid with 3 different radii.

Galapagos works on 2 rotation axes of planes centered on this centroid. These rotations are the genes.

The fitness is the "circularity" of the sections of the ellipsoid by the planes. I measure that by sampling points on the sections, finding their centroid, getting max and min distance to centroid, and trying to minimize the difference between max and min. (As sections are ellipses, I think its accurate enough).

 

In this example, optimal section (one of the circles in the screenshot below) has a difference between min and max radii of about 9 e-5 , radii are about 10 units, so its not a perfect circle, but not so far.

 

Then I saw something on the net about families of circular cross sections, so I thought I could try to get some planes parallel to the optimal cut plane found by Galapagos, and cut the ellipsoid to see the results, screenshot shows that .

The radii delta is 1.47 e-4 on average, so it looks like its an infinite family of "circular" cross sections.

Of course there is (are?) another family, as the ellipsoid is symetric.

Important notice: I am not an expert at all in this stuff, just experimenting, so don't trust this at all.

Permalink Reply by Yoann Mescam (Systemiq) on November 8, 2011 at 4:51pm

To summarize:

• if ellipsoid radii are all the same, its a sphere, not interesting.
• 2 radii the same (spheroid) : you can cut it by planes orthogonal to the main radius axe and all cuts are circles.
• all radii different: there are 2 infinite sets (I think) of parallel circular cuts, finding the 2 planes of these families shouldnt require Galapagos. I may be totally wrong about this.

 

Permalink Reply by Chris Tietjen on November 8, 2011 at 6:33pm

all radii different:  intersect a sphere whose diameter is the length of the ellipsoids intermediate axis.  The infinite sets are then taken from parallel planes.

Permalink Reply by Yoann Mescam (Systemiq) on November 8, 2011 at 7:16pm

Nice Chris.

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