Acceleration at the center of the triangle

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I am trying to calculate the acceleration at the node between the three points, photo attached. The points (1,2,3) measured accelerance via tri-axial acceleromter and now with the help of that info I want to calculate the acceleration at the center point. How can I proceed with this problem? Is there any function available to calculate the acceleration in matlab?
  5 Comments
Abhikaran Amitkumar Bhatia
Yes, the goal here is to calculate the acceleration at the screw with the help of the accelerance data collected by those three sensors. There is no rotating motion. Not the position and the sensors are not collecting data considering the gravity so they are only capturing the motion.
James Tursa
James Tursa on 16 May 2022
Edited: James Tursa on 17 May 2022
"... the sensors are not collecting data considering the gravity ..."
Not sure what this means. If your device is in contact with the ground then the sensors are certainly sensing the reaction force to gravity, even if the device is perfectly still on the ground. What do you mean "they are only capturing the motion"? Are you only considering the horizontal component of the acceleration? Or ...?

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Answers (1)

John D'Errico
John D'Errico on 16 May 2022
Edited: John D'Errico on 16 May 2022
If you have some known parameter at the three corners of a triangle, then the best estimator of that parameter at the center of the triangle is the average of the three values. This assumes the relation varies linearly over the triangle, and since you provide no reason to assume anything else, that is the best possible estimator.
In fact, you you have some uncertainties in the three parameter estimates at the corners, the mean should also be the minimum variance estimator.
And that means your answer is simple, just take the mean. And that is trivial.
acc_center = (a1 + a2 + a3)/3;
There is no explicit "function" necessary. Just one line of code.
  4 Comments
Torsten
Torsten on 16 May 2022
I think the center of gravity of the triangle is the center of the circumcircle - thus the distance to the three points is the same and we again are at John's formula.
Abhikaran Amitkumar Bhatia
I attached one more photo of the actual case.Maybe that might make my explaination more clear.

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