15,391 reputation
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bio website quantdec.com
location Northeastern US
age 14
visits member for 2 years, 11 months
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Consultant (environmental and spatial stats a specialty), expert witness, and teacher. I can be reached through (outdated but still valid) links posted on my web site.

Twitter: @WilliamAHuber // ASA-P website: http://amstatphilly.org/


Why waste time learning, when ignorance is instantaneous?

--T(iger) Hobbes.

For any complex problem there is a simple solution. And it's always wrong.

--[Mis?]attributed to H.L. Mencken by Dava Sobel, Longitude.


Jun
2
awarded  Nice Answer
Apr
30
awarded  Necromancer
Apr
30
awarded  Nice Answer
Apr
30
revised Generating convex polyhedron from face planes?
added 50 characters in body
Apr
30
comment Generating convex polyhedron from face planes?
@level1807 Good catch! Although I haven't tested it, it looks like a bug to me--evidently from a typographical error. The intention was to to compare the chopped values to zero, not to chop the results of a floating point comparison (which makes little sense anyway). I will go ahead and modify the code in this answer. Thanks again.
Apr
30
comment Generating convex polyhedron from face planes?
@level1807 I cannot say: as you can see from my example, I had no trouble with $50$ faces. The possible explanations for what you are encountering could range from differences among MMA versions through floating-point errors through some kind of bug in my code. It could be related to the limitations described in the introduction to this answer. If you want to resolve this you will need to find as simple as possible an example that can be reproduced and then offer it in a new question. Good luck!
Feb
6
comment Plot a 2D vector path onto a surface
@rm-rf Thank you for the tip and the kind words. I am over-extended moderating two other SE sites. Although both have elected two more moderators in the interim, the work is still too much. I hope to return here at some point.
Jan
17
awarded  Yearling
Jan
16
comment How to check if a 3D point is in a planar polygon?
@Rainer Assuming the points are very close to coplanar, the smallest eigenvalue will be close to zero and much smaller than the other two. The eigenvectors associated with the other eigenvalues--that is, the first two principal components--thereby span a plane passing close to most of the points. This gives two solutions: (1) The cross product of the first two principal components will be a normal vector for that plane. (2) The third principal component will be orthogonal to the other two, whence it too is a normal vector for the plane.
Nov
7
comment NMinimize usage
@Frederik You are right: I made that typo early on and because everything worked, I did not catch it. Your suggestion $x = 1+y^2$ accomplishes what was intended and everything goes through as planned. Thank you for reading this post so carefully and well!
Nov
7
awarded  Nice Answer
Oct
24
comment Insert $+$, $-$, $\times$, $/$, $($, $)$ into $123456789$ to make it equal to $100$
@xzczd Thank you! That is a useful insight and looks to be a likely explanation for the failure.
Oct
5
awarded  Nice Answer
Aug
24
awarded  Good Answer
Aug
3
awarded  Nice Answer
Jul
6
awarded  Nice Answer
Jun
12
awarded  Good Answer
Jun
11
awarded  Enlightened
Jun
11
awarded  Nice Answer
May
24
comment AstronomicalData and Planetary Heliocentric (x,y,z) Velocity Components
+1 Consider a central difference instead: it should be more accurate. It makes a substantial difference, by the way: it affects the third sig. fig. in your example, because it reflects the net planetary acceleration during the course of 12 hours.