# Can you solve unknown triangle from shared hypotenuse?

• I
• FGD
FGD
My question comes from an accelerometer attached to a rotating plate. C and T are the centrifugal and tangentinal accelerations. X and Y are the accelerometer x, y axis readings. The x and y axis are not perfectly aligned with the C,T accelerations and are rotated by some arbitrairy unknown angle. I can get the magnitude of x, y which should be the same as C,T. (I think). But then I would like to know the amounts of T and C in each of the x, y axis. With all that said, is there a way to get the values of C, and T? Thanks in advance..

You cannot solve for C and T. If you draw a semicircle with H as the diameter, then any pair of C and T with the right angle vertex on the semicircle will work. So there are an infinite number of C,T pairs that satisfy your condition.

FGD, Nik_2213, phinds and 1 other person
FGD said:
But then I would like to know the amounts of T and C in each of the x, y axis.
With all that said, is there a way to get the values of C, and T?

It seems that the posted diagram and your question are not very well related.

Could you add the x and y readings of the accelerometer to determine one unique acceleration vector, which could then be decomposed into radial and tangential acceleration vectors?

pbuk
phyzguy said:
You cannot solve for C and T. If you draw a semicircle with H as the diameter, then any pair of C and T with the right angle vertex on the semicircle will work. So there are an infinite number of C,T pairs that satisfy your condition.
Thank you for confirming there is no solution. I was hopeing there would be some way I did not know about to figure this out. :)

Lnewqban said:
It seems that the posted diagram and your question are not very well related.

Could you add the x and y readings of the accelerometer to determine one unique acceleration vector, which could then be decomposed into radial and tangential acceleration vectors?
I guess I could have drawn the diagram with the right angle corners together and the hypotenus as the magnitude. I thought drawing it the way I did would be less confusing. Thanks for the advice.

Getting (C^2+T^2) is trivial, but then you have an arc of possible solutions...

FGD
I think what you are saying is that
neither the x-axis nor the y-axis of the accelerometer is directed radially
(and the other is not directed tangentially).
Is it that the z-axis of the accelerometer is translated from the z-axis of the disk?

Are you allowed to re-orient the plate so that it is vertical but not rotating?
You could manually change the angular displacement of the plate
You could then mark up the plate so that you know where the x-axis points.
You could repeat for the y-axis.

Lnewqban and FGD
robphy said:
I think what you are saying is that
neither the x-axis nor the y-axis of the accelerometer is directed radially
(and the other is not directed tangentially).
Is it that the z-axis of the accelerometer is translated from the z-axis of the disk?

Are you allowed to re-orient the plate so that it is vertical but not rotating?
You could manually change the angular displacement of the plate
You could then mark up the plate so that you know where the x-axis points.
You could repeat for the y-axis.
Yes you understood the issue I was having with alignment. Your idea is really good! With that I should be able to tell more or less the exact position and orientation of the x,y axis. Smart idea! Thanks.

Lnewqban

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