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Rotational mass


gmdh555

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I have heard it said that a 1kg reduction in rotational mass (i.e. wheels/tyres) is equivalent to an 8kg reduction in static mass. Can I give any credence to this, including the specific ratio quoted? For example, my new wheels/tyres weigh 12kg less than the predecessors - is this directly translatable into 96kg reduction when considering power-to-weight?

 

Cheers,

Gareth

 

 

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I have heard very similar figures sugegsted with regard to push bikes but the physics may a little different. Certainly when I upgraded my push bike to lighter wheels, wow - my legs could certainly feel the difference.
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The moment of inertia of a disc* (which a wheel isn't, but hey), is 1/2 mr^2. Importantly it is proportional to m.

 

Applying a net torque, n, to rotationally accelerate at w, Newton tells us:-

n = Iw

Required torque, therefore would appear to be proportional to m, to spin up the wheel.

 

Torque is also used to longitudinally accelerate the car - where

Nr = Ma where M is mass of car and a is longitudinal acceleration.

 

However, the wheel also forms part of the mass of the car which must be accelerated linearly. So the mass of the wheel counts twice in the effort put into accelerating the car (N+n).

 

So the rotating masses count twice.

 

 

* uniform disc, radius r, mass m.

 

My ... Preciousss!

Member #109**

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I think I must have drunk the same amount, just started off with a proper analytical degree, not one of the soft sciences like physics.

 

😬

 

I wonder if you can get it up to 8-times by including rotational effort around another pair of axes? It might mean that you can out-perform the chav whilst rolling into the ditch, too. *confused*

 

The other effect of lighter weight is in increasing tyre efficiency (force per unit load) so if you are lighter, a smaller (and so lighter) tyre suffices, and you go round corners much better. But then you all knew that.

 

My ... Preciousss!

Member #109**

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