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SubZer0 · Jan 2, 2012, 10:38 PM

Hi all,

Can anybody explain the Transport Theorem (http://en.wikipedia.org/wiki/Rotatin... tionary_frames) in more non-physicist terms? I simply can't wrap my head around the visual of this theorem, which has the gist of:

d/dt f = df/dt (angular velocity) x f(t)

Where f(t) represents the position in time.

I simply cannot visualise the result. If I use an angular velocity vector of [0, 1, 0] (rotation around y axis), a position of [0.5, 0, 0], and position change of [0, 0, 0], the resulting derivative is [0, 0, -0.5]. I would at least expect a time derivative with changes of x, and z, instead of just z. Am I not understanding the actual *meaning* of the result? If df/dt = [0, 0, 0], I'm assuming that the rotation should result in a derivative vector of something like [x, 0, z], where x and z are some non-zero values?

Thanks!

ebaines · Jan 3, 2012, 07:08 AM

Is this what you are referring to? From Rotating reference frame - Wikipedia, the free encyclopedia :

$<br /> \frac d {dt} f = [(\frac d {dt})_r + \Omega \ \times ] f<br />$

At the instant a point is at (0.5,0,0) and rotating about the y axis its instantaneous velocity is in the z direction only.