I seem to have a problem visualizing the addition of angular velocity components for rotation in a tilted plane and hoping someone will explain any basic errors I am making.

Please see diagrams below where I've drawn a circular tilted plane described by a vector R in a $3D$ $XYZ$ reference frame. Then I've drawn 3 circular projected shadows of that plane in the $XY$,$XZ$ and $YZ$ planes.

Let's assume that the vector R moves in a complete circle in 1 second, therefore the angular velocity $\omega$ (ie. 360 degrees/sec or 2π radians/secs).

I'm assuming that vector $R$ would also draw those circular shadow planes at the same time so therefore the angular velocity of the components of vector $R$ in $XY$,$XZ$,$YZ$ planes would also have an angular velocity of $\omega$.

But that doesn't make sense because if I added those component angular velocity vectors using pythagoras theorem we would have 'Angular Velocity Vector $ R' = \sqrt{\omega_x^2 + \omega_y^2+ \omega_z^2} $ and that does not equal to $\omega$.

I can't figure out what I'm doing wrong.

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  • $\begingroup$ I'm having trouble with the paragraph starting "But that doesn't make sense ...". Can you elaborate? Add details? $\endgroup$
    – garyp
    Nov 14, 2022 at 15:14
  • $\begingroup$ I think I've made a mistake with the right-hand rule for direction of the angular velocity normal to the XZ plane. It should be pointing in the -y axis direction. I'm going to correct that image. Basically, I am saying that the magnitude of the angular velocity in each of those shadowed planes seem to be 'w' (same as the angular velocity in the tilted plane). But that shouldn't happen because if angular velocity is a vector, and if I vector added all those red w vectors, the resultant vector magnitude in the tilted plane would not be w. $\endgroup$
    – Dubious
    Nov 14, 2022 at 17:46
  • $\begingroup$ I've just edited the image but I think it doesn't make any difference to the question I've posed. $\endgroup$
    – Dubious
    Nov 14, 2022 at 17:52
  • $\begingroup$ FYI - x^2 → $x^2$ and \sqrt{x} → $\sqrt{x}$. Also \omega → $\omega$. Enclose all math in dollar signs $ ... $ for inline expressions and double dollar signs $$ ... $$ for paragraph equations. $\endgroup$ Nov 14, 2022 at 17:56
  • $\begingroup$ The projection of a tilted circle on a plane is an ellipse, and I am not sure using this process is helpful here. $\endgroup$ Nov 14, 2022 at 17:58

2 Answers 2

  1. The projections are not circles but ellipses.
  2. The projection of a vector does not rotate with uniform velocity even if the original vector does.
  3. The magnitude (length) of the rotation vector designates its speed and therefore if you add up the three projected lengths using a Euclidean length then you would get

$$ \omega = \sqrt{ \omega_x^2 + \omega_y^2 + \omega_z^2 } $$

this is because the length of a vector in Euclidean space is invariant to rotations.

You can add up the areas of the projected ellipses in a vectorial way to get

$$ \text{Area of circle} = \sqrt{ {\rm A}_x^2 + {\rm A}_y^2 + {\rm A}_z^2 } $$ where ${\rm A}_i$ is the corresponding area of each ellipse.

  • $\begingroup$ I've just added Diagrams A & B and a projection onto the XY plane (ie. an ellipse as you've mentioned). But do you see my confusion regarding the rotation of the red dot versus the projected yellow dot? For Diagram A, the red dot will rotate at some constant 2π rad/sec, while the yellow dot should also rotate at 2π rad/sec even if its path is an ellipse. Now look at Diagram B which is a side view showing the red dot circular plane tilted at 45°. When I decompose the 2π rad/sec vector into components, the angular velocity normal to the XY plane is a constant 2πSin(45) rad/sec not 2π rad/sec. $\endgroup$
    – Dubious
    Nov 15, 2022 at 0:07
  • $\begingroup$ Basically, the physics (as I've interpreted it) doesn't match the mathematics. My logic is going wrong somewhere as I've either got the physics or maths wrong (or both). $\endgroup$
    – Dubious
    Nov 15, 2022 at 0:17
  • $\begingroup$ Putting arrows along the path does not make the path a vector. If you are looking for a geometric interpretation of the angular velocity vector, then read a bit on geometric algebra and pseudo vectors. $\endgroup$ Nov 15, 2022 at 1:10
  • $\begingroup$ Not sure what you mean? I've put arrows along the path to show the direction of rotation of red and yellow dots (consider them as being point masses), therefore by the right-hand-rule you can then see the angular velocity vectors (ie. the blue and red arrows pointing about positions S and S' in diagrams A and B). $\endgroup$
    – Dubious
    Nov 15, 2022 at 16:48
  • $\begingroup$ I made a typo error in my previous comment as I meant the black and red vector arrows, not the blue and red arrows. $\endgroup$
    – Dubious
    Nov 15, 2022 at 17:00

I think it's my confusion regarding the definition of angular velocity. I've assumed that rate of change of angular displacement of that yellow dot around the geometric centre of an ellipse can be used as some average angular velocity = 2π rad/sec. But that is plainly wrong because angular velocity is instantaneous about a circular path.

Therefore, I cannot use the ellipse because a position vector of that yellow dot from the geometric centre of the ellipse is not necessarily a 'radius' and not always at right angles to the orbital velocity.

I am therefore misinterpreting the definition of angular velocity and applying it incorrectly to the angle traversed around a non-circular orbit in a certain period of time.

In fact, I can decompose the angular velocity (black vector) into components as shown in Diagram B.


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