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I have constructed a "force plate" to reconstruct the location (x,y) of a steady state contact point force f. The system consists of four fixed compression load cells L1, L2, L3, and L4 and a thick aluminum plate. The aluminum plate is free to move along the z-axis only (see figure below). I have tested the system using a universal precision tester and it performs very well. Using the following equations I am able to reconstruct the location of the force based on reading from the load cells:

$$x = \frac{\sum(x_if_i)}{\sum(f_i)}$$ $$y = \frac{\sum(y_if_i)}{\sum(f_i)}$$

However I would like to create a computer model and simulation that works in the other direction. For example, solve for the force at each load cell when the magnitude of the contact force and it's location are known. It would be nice to have ideal simulated data to compare with the actual measurements taken from the precision tester.

I have spent a couple of days trying to find a solution (using sum of moments at each cell), but I have been unsuccessful. I am an electrical engineer and forces are a little beyond my comfort zone. A fellow mechanical engineer said the problem may be statically indeterminate.

Force Plate Setup

Is this problem unsolvable using traditional methods? Any ideas? Thanks for reading.

EDIT

Based on Wolphram jonny solution I have attempted to solve the system of equations using the matrix method below. This method works for all points except when y = 0 ... So close! enter image description here

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    $\begingroup$ You exhibit a system of equations. You solve the reverse problem by inverting the system of equations. It is more compact written in matrix form, which is also the way you will find inverting linear systems of equations addressed in every introductory text on computational methods. $\endgroup$ – dmckee Nov 23 '14 at 3:10
  • $\begingroup$ @dmckee Thanks for the input. I have used the matrix form to solve electrical circuits but I am unsure how I can extend this to forces. I can find the inverse of the above equations but can't bridge the gap. I will look into introductory text. $\endgroup$ – atomSmasher Nov 23 '14 at 5:17
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to solve the problem you need at 4 linear equations (or the two linear ones you have plus another nonlinear one). Otherwise, the solution is undetermined, you have more variables that linear equations and the number of solutions is infinite. Here are the two extra equations that you need.

1) the total force equald the applied force (otherwise it will move vertically):

$f ={\sum(f_i)}$

2) the second condition is that the plate does not tilt (zero torque), and you get (assuming it does not tilt horizontally):

$0 =\sqrt{(x-x_1)^2+(y-y_1)^2}f_1+\sqrt{(x-x_2)^2+(y-y_2)^2}f_2-\sqrt{(x-x_3)^2+(y-y_3)^2}f_3-\sqrt{(x-x_4)^2+(y-y_4)^2}f_4$

Let me know if you do not know how to solve a system of linear equations. But brute force always works (this is, isolating one variable in one equation, put it on the next, isolate the second variable, etc.

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  • $\begingroup$ Wolphramjonny and @dmckee I added a matrix as an edit. Does it look like I am on the right track? I ask because MATLAB is returning unrealistic values that sum to larger than total force. $\endgroup$ – atomSmasher Nov 23 '14 at 15:36
  • $\begingroup$ the matrix seems fine, let me check if the is some mistake in one of my equations, or I'll try matlab to check what is going. Actally, if you already implemente din in matlab, could you send the the code? It will save me time and also might find some error on the implementation (asuming it was not mine) $\endgroup$ – Wolphram jonny Nov 23 '14 at 16:39
  • $\begingroup$ I found it! You are correct, just needed a plus sign before the third term in the second equation. Thank you SO MUCH for your help! $\endgroup$ – atomSmasher Nov 23 '14 at 16:43
  • $\begingroup$ Great! I am happy to hear that $\endgroup$ – Wolphram jonny Nov 23 '14 at 16:45

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