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Correction

205.162.100.130

Posted by Steve Eddy on May 8, 2000 at 13:37:17

In Reply to: Your absolutely right......... posted by Sean on May 7, 2000 at 20:23:30:

Ok, the error in the previous post was that cone velocity doesn't change as excursion changes. It does.

mrbiggleswurth and I were discussing this subject some days back and during the discussion, I mentioned that velocity should increase if excursion increases. At the time he said no, that velocity remains the same, and acceleration increases. In my gut this didn't seem right but I deferred to mrbiggleswurth.

Anyway, we hooked up yesterday morning and rehashed the issue and got it straightened out. I think there was just some disconnect in our previous conversation. He was saying velocity remains constant for a given frequency and displacement (i.e. excursion). Apparently I hadn't got across to him that I was referring to increased (or decreased) excursion for a given frequency.

My apologies for any confusion the error in my post may have caused.

Maximum cone velocity and acceleration are:

v(max) = Aw
a(max) = Aw^2

Where A = maximum displacement and w (lower case Greek Omega) = sinusoidal angular velocity, or 2 * pi * frequency.

se

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