Suppose I have a plane (with a known weight, Mg). There are n points on this plane (all at known locations (xi,yi)), which each applies an upward force (that is, along Z) on the plane (where the force is of magnitude Ti).
The plane is in equilibrium.
I need to calculate the force which each point applies.
The equations I've come to so far are:
Sum of (xi * Ti) = 0
Sum of (yi * Ti) = 0
Sum of (Ti) = Mg
(I'll take the centre of mass of the plane/leg system to be the origin)
Recalling that I know Mg, and all of the (xi,yi) coordinates; I could easily solve the Ti values when there are just 3 points (as I have 3 equations). But suppose I add more points. How would I calculate all of these Ti values?
Thanks very much
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Find the centre of gravity of the table, and the centre of gravity of the legs assuming they all had equal weights.
If they are different points then equal weights would give a torque.
If the table rotates rigidly, and spring force is proportional to displacement,
then the equal weights are adjusted in proportion to the distance from the CG of the legs along the line to the CG of the table.
As a thought experiment, imagine all but three of the legs are exerting zero force. This would be a perfectly acceptable solution so long as it satisfies aburr's suggestion.
So there are at least n-2 solutions taking every triplet available. Clearly there is no single solution to the problem.
Also, just taking n=4 must have infinite solutions as any decrease in the force applied by one support can be balanced by increases elsewhere, again with aburr's restriction.
In short, you cant.
Paul
ozo has it right. Let k = nC3. Find the individual solutions for all k 3-point combinations. Come up with weights for these solutions. The end result will be all possible sums of solutions, constrained such that the total force on each point is >= 0.
In other words:
First calculate k size-n solution vectors s_1 through s_k, each containing (n-3) zeros and 3 non-zeros. Put them in a sparse n x k matrix called S.
Assign a weight variable to each solution, w_1 through w_k. As long as sum(w) = 1.0, S*w will be a solution at equilibrium, but it's an error for a negative weight to be applied to any point, since the plane is 'resting' on the points, so we have to restrain things so that all points are applying a non-negative force.
n equations like this (each S_x_y column is one of the k 3-point solution vectors):
S_1_1 * w_1 + S_1_2 * w_2 + .... S_1_k * w_k = f_1
S_2_1 * w_1 + S_2_2 * w_2 + .... S_2_k * w_k = f_2
....
S_n_1 * w_1 + S_n_2 * w_2 + .... S_n_k * w_k = f_n
1 equation for the w's.
w_1 + w_2 + w_3 + .... + w_k = 1.0
n constraints to keep the forces non-negative
f_1 >= 0
f_2 >= 0
f_3 >= 0
...
f_n >= 0
These essentially work out to a n convex hypertope in n dimensions in which the f vector could reside. The solution space is described by the combination of the k basis vectors in S, and the convex surface is derived from the inequalities on f_1 through f_n.
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by: ozoPosted on 2007-07-10 at 16:19:32ID: 19458660
If you've ever seen a table wobble, you'll now there is no unique solution.
You might find one set of solutions by picking any 3 legs that form a triangle around the CG and solving for 3 points.