Re: Balance point of a solid

*To*: mathgroup at smc.vnet.net*Subject*: [mg113595] Re: Balance point of a solid*From*: Ray Koopman <koopman at sfu.ca>*Date*: Thu, 4 Nov 2010 03:58:31 -0500 (EST)*References*: <iam332$jvn$1@smc.vnet.net> <iar4jk$gmi$1@smc.vnet.net>

On Nov 3, 12:56 am, Ray Koopman <koop... at sfu.ca> wrote: > On Nov 1, 3:00 am, Andreas <aa... at ix.netcom.com> wrote: >> >> Hi everyone, >> >> This isn't strictly a Mathematica question, but once I formulate a way >> to think about the problem, I'll use Mathematica to generate solutions >> to specific situations. Besides, this seems like the smartest forum >> going. So... >> >> Background >> >> Start with a equilateral triangle as the base of a solid. The triangle >> has sides: >> >> a1 = a2 = a3 = 1 >> >> At right angles to each side of this base triangle stands a trapezoid >> with heights: >> >> h1 , h2, h3 >> >> These heights may have different lengths. Typically h1 > h2 > h3, >> but in some cases any or all of them could have equal values. >> >> Of course, all the h's stand at right angles to the base triangle. >> >> Now connect the points of h1, h2, and h3 and we have a solid composed >> of: >> >> 1 equilateral triangle (the base) >> 3 different trapezoids all with their base lengths =1 >> 1 triangle on top. >> >> I can find the area of each trapezoid like this: >> >> A = area >> A1 = 1/2 x a1 x (h1 + h3) >> A2 = 1/2 x a2 x (h1 + h2) >> A3 = 1/2 x a3 x (h2 + h3) >> >> I can find the volume like this: >> >> Volume = V >> >> V = A1 x A2 x A3 >> or >> V = (a1 x a2 x a3 x (h1 + h3) x (h1 + h2) x (h2 + h3)) / 2^3 >> >> Problem: >> >> I need to find the point on the base triangle upon which I can balance >> the solid. >> >> This is not the centroid of the base triangle, because the solid would >> have more weight or volume towards it's highest side, typically h1. >> >> Simply described, I want to find the balance point on the base >> triangle such that if the solid rested on a pin at that point, the >> base triangle would remain parallel to the floor. >> >> My guess is that I need to calculate the center of gravity of the >> volume, then project a line to the base triangle where it would >> intersect the base triangle at a right angle. Basically just dropping >> the center of gravity to the bottom plane. >> >> Maybe a bit tedious, but I should have all the information to >> calculate this. Given the right angles and known lengths I think I >> have enough information to calculate all angles and lengths of the >> solid. >> >> Will this work? >> >> How would I calculate the center gravity for the solid described? >> >> Follow up question... >> >> If this solves the problem for a 3 dimensional solid of this nature, >> how can I extend the solution to a 4th, 5th, or nth dimensional >> space? >> >> Thanks to all. >> >> A > > This is a top-of-the head response, so it may be completely off, but > why not think of the solid as two pieces: a flat-topped triangular > column, height = min(h1,h2,h3), and a tetrahedral cap. Sorry, but the cap isn't necessarily a tetrahedron. > The centroid of each piece is the simple average of its vertices. > The centroid of the whole thing is the weighted average of the two > centroids, with the weights being the volumes of the pieces. > > I'm not sure how you want to extend the solid to n dimensions. > Can you be a little more specific?

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