Re: Balance point of a solid

• To: mathgroup at smc.vnet.net
• Subject: [mg113571] Re: Balance point of a solid
• From: Ray Koopman <koopman at sfu.ca>
• Date: Wed, 3 Nov 2010 02:54:14 -0500 (EST)
• References: <iam332\$jvn\$1@smc.vnet.net>

```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?
>
>
> 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. 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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