My Math Forum 3-D projection problem (pesky angles)
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 June 20th, 2019, 07:42 AM #1 Senior Member   Joined: Jun 2019 From: USA Posts: 203 Thanks: 80 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.
June 24th, 2019, 03:24 PM   #2
Senior Member

Joined: Jun 2019
From: USA

Posts: 203
Thanks: 80

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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 Tags angles, pesky, problem, projection

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