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Thread: Calculate Inclination from Pitch and Roll

  1. #1

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    Question Calculate Inclination from Pitch and Roll

    Hello!

    My english is not so good but I hope you can understand me.

    My problem is that I have values for pitch and roll angle (range: +90° ... -90°). pitch is inclination from vertical at x-direction, roll is inclination from vertical at y-direction. Now I need to calculate the resultant inclination from vertical, but I don't know how.

    Can somebody help me?!
    Last edited by thorondor; Oct 15th, 2012 at 06:18 AM.

  2. #2
    Only Slightly Obsessive jemidiah's Avatar
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    Re: Calculate Inclination from Pitch and Roll

    Unfortunately I cannot understand you. Specifically...
    (1) What does "inclination from vertical at x-direction" mean?
    (2) What does "inclination from vertical at y-direction" mean?
    (3) What does "calculate the resultant inclination from vertical" mean?

    "Pitch" and "roll" are technical terms in aviation, but I don't think you're using them with that meaning. (Specifically, see the relevant Wikipedia page.) My best guess is that you're actually using something similar to spherical coordinates and want to do something similar to the following: given the angles theta and phi in this image compute the angle between the ray drawn and the y-axis. In this specific case, the three coordinates can be computed with a little trigonometry as

    x = r sin(theta) cos(phi)
    y = r sin(theta) sin(phi)
    z = r cos(theta)

    The angle with the y-axis can then be computed using the dot product. Set r=1 and dot the above (x, y, z) with the positive y-axis vector (0, 1, 0) to get sin(theta) sin(phi). In general,
    A dot B = |A||B| cos(T)
    => T = arccos(A dot B / (|A||B|))
    [arccos = inverse cosine]

    where T is the angle between A and B. Here, we've used A = (x, y, z) with |A| = 1 and B = (0, 1, 0) with |B| = 1, so this equation gives
    T = arccos(sin(theta) sin(phi)) = angle between +y axis and the vector

    Again, I cannot understand you so I don't know how relevant the above is. Most likely the same or similar ideas can solve your actual problem.
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    Bertrand Russell

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