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June 20th, 2019, 07:42 AM   #1
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3-D projection problem (pesky angles)

(Note: changed Greek variables to Latin for simplicity typing; "." represents the dot product)

I ran a quick static calibration experiment on some accelerometers. Now I'm trying to apply the results correctly.

Problem statement:
I have a fixed RH Cartesian unit vector coordinate system x,y,z (though I only care about z), and another RH Cartesian unit vector coordinate system X,Y,Z attached to a rigid object.

I have measured two angles.
a is the angle between the z-X plane and the Z-X plane.
b is the angle between the Y-z plane and the Y-Z plane.
I don't care about the third angle (basically "yaw"), because I am only interested in projections on the z axis.
a and b are defined such that:
If b = 0, then
{z.X = 0
z.Y = sin(a)
z.Z = cos(a)}
If a = 0, then
{z.X = -sin(b)
z.Y = 0
z.Z = cos(b)}
Problem is, I have some cases where both a and b are unavoidably non-zero.

How do I find the projections (z.X, z.Y, z.Z) as a function of both a and b?

If anyone can help lead me to the answer (or give it with an explanation), I would be extremely grateful. I have tried looking at z cross X and z cross Y unit vectors, even pulled out some of my index notation vector identity tricks, but so far I haven't found anything helpful.

BTW, I am aware that when a = +/- pi/2, b may be undefined, and vice versa, but it seems like this shouldn't matter in terms of the projections on z.

Also, if it turns out I could have solved this as the product of two trivial rotation matrices, I'm going to be mad, lol.
DarnItJimImAnEngineer is offline  
June 24th, 2019, 03:24 PM   #2
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From: USA

Posts: 203
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Alright, getting warmer and managed to draw the picture correctly.

$\displaystyle \alpha$ is the angle between the X-Z plane (green) and the X-z plane (yellow) as shown.
$\displaystyle \beta$ is the angle between the Y-Z plane (red) and the Y-z plane (magenta) as shown.

Looking for the projections of z onto X, Y, and Z.

It looks like it's just:
$\displaystyle -sin(\beta)$
$\displaystyle sin(\alpha)$
$\displaystyle cos(\alpha)cos(\beta)$

Was it really that simple? Can someone verify, please?
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DarnItJimImAnEngineer is offline  

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