Sin (x) = Sin (y) sqrt(2)

Subhotosh Khan said:
How did you get:

PC = \(\displaystyle \sqrt{2} \ \ \)*sin(y) ????
Click to expand...

It's one of the worst 3D pictures I've ever seen, but this comes from right triangle ACP. AO is on the z-axis, C is on the xy-plane.

The work is correct. I'm just curious why or how one would measure that angle y = PAC.
 
Dr.Peterson said:
It's one of the worst 3D pictures I've ever seen,...
Click to expand...
Thanks for the feedback.
Lol; I did the initial calculation in my head. And there is a reason for that ;-).
For why the accelerometer gives that angle. I don't know what the logic is here.
 
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Dr.Peterson said:
...I'm just curious why or how one would measure that angle y = PAC.
Click to expand...


I have to read once the full specs of the device. Maybe the issue comes from on how I - more correct a son - have chosen where the axes are.
 
When sketching you have to be careful not to draw things parallel that might not be parallel. This can easily lead to errors (and I think that you have made such an error in your equation). In particular, in your diagram line AC looks parallel to the "Y axis" when it isn't always the case (otherwise x would simply be 45°).

So the first step is to get the sketch fully correct. I've had a go at re-drawing (I swapped the axis labels to be more traditional!)...

sinx_siny.png
I am curious about what point "P" actually is, and why line CP is always purpendicular to the x-axis. Or maybe I should not have drawn the right-angle symbol at P? I guess this question is tied up with another question:- what is the device that actually measures angle y?
 
No matter what the angle is of the square, you can construct CP, with CP perpendicular to your X-axes. And CP is then part of the 2 right angles of the 2 right triangles So CP is perpendicular because I choose it like that.
More complex is why APC is right. CP is right on X axis, Z axis is right on XY plain. So CP is right on the ZX plane (this needs a prove?. ). And sinds AP lies in ZX plane and AP goes through CP that is also a right angle.
Clear?
Maybe to complete your picture (thanks for the picture btw) you could add the Z axes and indicate that the angle on APC Is also right.
The measurement device is a MPU6050.
Walter
Ps; your point D I labeled O for origin ;-).
 
Cubist said:
When sketching you have to be careful not to draw things parallel that might not be parallel. This can easily lead to errors (and I think that you have made such an error in your equation). In particular, in your diagram line AC looks parallel to the "Y axis" when it isn't always the case (otherwise x would simply be 45°).

So the first step is to get the sketch fully correct. I've had a go at re-drawing (I swapped the axis labels to be more traditional!)...

View attachment 21421
I am curious about what point "P" actually is, and why line CP is always purpendicular to the x-axis. Or maybe I should not have drawn the right-angle symbol at P? I guess this question is tied up with another question:- what is the device that actually measures angle y?
Click to expand...
As I understand it, A is on the z-axis, not in the xy-plane; and P is on the y axis, which was drawn horizontally for some reason.
 
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