Acceleration at the center of the triangle

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Abhikaran Amitkumar Bhatia
Abhikaran Amitkumar Bhatia el 16 de Mayo de 2022
Editada: James Tursa el 17 de Mayo de 2022
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?
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Abhikaran Amitkumar Bhatia
Abhikaran Amitkumar Bhatia el 16 de Mayo de 2022
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 el 16 de Mayo de 2022
Editada: James Tursa el 17 de Mayo de 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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John D'Errico
John D'Errico el 16 de Mayo de 2022
Editada: John D'Errico el 16 de Mayo de 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.
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Torsten
Torsten el 16 de Mayo de 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
Abhikaran Amitkumar Bhatia el 16 de Mayo de 2022
I attached one more photo of the actual case.Maybe that might make my explaination more clear.

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