1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gyroscopic coupling

  1. Sep 28, 2009 #1
    I am not really solving an exercise for homework so hopefully this general forum is OK for my question.

    For my thesis, what is attitude control of a flying object with a reaction wheel, I am learning the basics from this NASA Technical Report:

    http://www.hot.ee/ronn/design_glob_anal_spacecraft_att_control.pdf [Broken] (I mirrored it because currently NASA website publishing this is down)

    Things are quite clear to me until eq (22) in PDF page 20, where the 3rd term is "gyroscopic coupling". I am unfamiliar with this phenomenon and can not grasp it from the equation. Could someone please explain me the meaning of this term? I do not see how can change of angular velocity depend only in the current velocity. Is it somehow the same thing as precession?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 29, 2009 #2


    User Avatar
    Gold Member

    I did some googling, and I did get the impression is that the expression 'gyroscopic coupling' is used for the case of experiencing consequences of gyroscopic precession.

    If so then the most extreme example of gyroscopic coupling was the flight behavior of aircrafts with a single rotatory engine. With a rotatory engine the cylinders are arranged radially, and the entire motorblock is rotating. That was good for cooling the cilinders, but it was a huge rotating mass in a relatively light aircraft. If the pilot tried to pitch up or down, the aircraft wouldn't pitch but yaw, and vice versa. To fly those aircrafts the pilot's had to anticipate the gyroscopic precession and compensate for it in advance.

  4. Sep 29, 2009 #3


    User Avatar
    Homework Helper

    For a helicopter, the pilots control inputs are moved about 90 degrees out of phase, to compensate for gyroscopic precession. A roll torque results in a pitch response and vice versa. Yaw isn't an an issue since it shares an axis with the main rotor.
  5. Sep 30, 2009 #4
    Hi, guys, sorry for the side orientated question,
    but i am also confused exploring the gyroscope theory.
    Could someone point out a web link with easy to understand explanations
    of gyro basics, in particular forced precession equations and gyroscopic moment.

  6. Oct 6, 2009 #5
    Ok. I seem to understand what the term is. So in this case the Wa is velocity of precession, h is the moment of impuls percessing and the product without inertia term is the torque of precession?

    Usually in textbooks they explain precession by applying torque to L and getting W from it, but it should also vice-versa, right?
  7. Oct 6, 2009 #6


    User Avatar
    Gold Member

    What is your question? What you wrote is not clear enough.

    Preferably, if your question is about a mathematical formula, enter the formula in LaTeX markup.
    Check out the https://www.physicsforums.com/misc/howtolatex.pdf" [Broken]

    Last edited by a moderator: May 4, 2017
  8. Oct 6, 2009 #7

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    This term arises because angular momentum is expressed in the body frame. This is of course a rotating reference frame, so the rotational equations of motion need to reflect this fact. Other names for this term include "inertial torque" (c.f. "inertial force", the fictitious forces due to frame acceleration/rotation) and "Euler torque" (c.f. the terms in Euler's equations).

    An easy way to see how this term arises is to use the identity that relates the time derivative of a vector quantity q as ascertained from the perspective of an inertial frame versus a rotating frame:

    [tex]\left(\frac {d\mathbf q}{dt}\right)_{\text{inertial}} =
    \left(\frac {d\mathbf q}{dt}\right)_{\text{rotating}} +\,\,
    \boldsymbol{\omega}\times \mathbf q[/tex]

    This is true for all vector quantities, not just position. So, apply it to angular momentum.

    [tex]\left(\frac {d\mathbf L}{dt}\right)_{\text{rotating}} =
    \left(\frac {d\mathbf L}{dt}\right)_{\text{inertial}} -\,\,
    \boldsymbol{\omega}\times \mathbf L[/tex]

    The first term on the left hand side is simply the external torque. The second term is gyroscopic / inertial / Euler torque.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Gyroscopic coupling
  1. Gimbaled Gyroscope (Replies: 4)

  2. Gyroscopic precession (Replies: 7)

  3. Principle of Gyroscope (Replies: 3)

  4. Gyroscope precession (Replies: 3)