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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?
Attached Images CoordinateSystemLabelled.jpg (13.5 KB, 1 views) Tags angles, pesky, problem, projection Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Arie2002 Math 8 April 6th, 2015 01:16 AM Drasik Geometry 4 October 11th, 2014 07:08 AM zak100 Geometry 1 April 6th, 2014 08:55 AM iclestu Applied Math 1 September 1st, 2010 01:41 PM velcrome Linear Algebra 0 April 28th, 2008 03:43 AM

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