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?
>>
>>
>> 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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