Balance point of a solid on its base triangle
AndreasA
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 stand trapezoids with heights:
h1 , h2, h3
Each of these heights may have different lengths. Typically h1 > h2 > h3, but in some case 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 think we 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?