Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Subtracting centrifugal acceleration from acceleration caused by movement

  1. Nov 8, 2011 #1
    I've been playing around with dead reckoning stuff on an rc plane by attaching a gps, 3 axis gyroscope, and 3 axis accelerometer to the plane.

    When the plane isn't turning, my algorithm works pretty well however when I enter a turn, the readings get way off. Specifically, the plane appears to be moving at a faster velocity than it actually is thus all turns are overshot.

    I believe I see this due to centrifugal acceleration and would like to cancel it out, but I'm getting quite lost, especially with 3d vectors. How might I go about subtracting this observed acceleration from acceleration caused by movement?

    Thank you
  2. jcsd
  3. Nov 8, 2011 #2
    Let me add what I have identified so far.. for at least identifying what I believe to be the acceleration component I need to subtract:

    F = ma[itex]_{c}[/itex] = (mv[itex]^{2}[/itex])/ r = mrw[itex]^{2}[/itex]

    The definitions are standard as of this wiki page:

    I know:
    (gps) speed = v
    (gyroscope) angular velocity = w

    I can calculate r via:
    (mv[itex]^{2}[/itex])/ r = mrw[itex]^{2}[/itex]
    v[itex]^{2}[/itex]/ r = rw[itex]^{2}[/itex]
    v[itex]^{2}[/itex] = r[itex]^{2}[/itex]w[itex]^{2}[/itex]
    r[itex]^{2}[/itex] = v[itex]^{2}[/itex] / w[itex]^{2}[/itex]
    r = [itex]\sqrt{v^{2} / w^{2}}[/itex]

    With r and w known, I can now calculate [itex]a_{c}[/itex] as follows:
    m[itex]a_{c}[/itex] = m[itex]rw^{2}[/itex]
    [itex]a_{c}[/itex] = [itex]rw^{2}[/itex]

    Am I anywhere close to being on the right track? :rofl:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook