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Old 03-24-11 | 05:59 PM
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I_like_cereal
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From: PDX

Bikes: Reynolds 853 Jamis Quest 1990s

Originally Posted by furballi
In SI unit, you need to use kg for m (mass of hub, rim, spokes, freehub, cassette, tire, and inner tube), and meter for r (radius from center of hub to the outer most edge of the rim). Why at the edge of the rim? Because the tire also has mass, so we can cheat by approximating the center of mass located at the outer edge of the rim. The resulting moment of inertia I has a unit of radian per sec. Note that I=mrr is valid for a thin cylinder with mass concentrated at the outermost edge.

The actual rear wheel also has a sizable amount of mass from the spokes, cassette, and freehub...somewhere between a disc with uniform thickness (I for disc=0.5mrr) and a thin cylinder with mass at the outer edge (I for cylinder=mrr). I would calculate I of the wheel using 0.90 or 0.95mrr to compensate for the spokes and the cassette/freehub.


i=.95*m*rr??

333.6mm is the radius of a 700c wheel with a 23mm tire. Which is what I have.

I convert 1645.88 g to kg, then convert 333.6mm to m

I=.95*0.167775739 kg s / m*.3336m
I=0.0531714872 kg s???
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