gmdh555 Posted October 14, 2005 Share Posted October 14, 2005 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 Link to comment Share on other sites More sharing options...
Myles Posted October 14, 2005 Share Posted October 14, 2005 I think it's more like a 2-1 ratio than 8-1. I can't remember the relevant physics off the top of my head though... Project Scope-Creep is live... Alcester Racing 7's Equipe - 🙆🏻™ Alcester-Racing-Sevens.com Link to comment Share on other sites More sharing options...
oldbutnotslow Posted October 14, 2005 Share Posted October 14, 2005 Oily knows as he worked out my saving on the lightened flywheel. Grant 😬 183 BHP of Black and 'Stone Chip' excitement. 😬 here Link to comment Share on other sites More sharing options...
jono Posted October 14, 2005 Share Posted October 14, 2005 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. Link to comment Share on other sites More sharing options...
db Posted October 14, 2005 Share Posted October 14, 2005 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** Link to comment Share on other sites More sharing options...
Normans_Ghost Posted October 14, 2005 Share Posted October 14, 2005 Right db so whats the answer? Norman Verona, 1989 BDR 220bhp, Reg: B16BDR, Mem No 2166, the full story here Link to comment Share on other sites More sharing options...
susser Posted October 14, 2005 Share Posted October 14, 2005 Eureka Thass good stuff !! These go faster wheels what weigh a ton make the chavs even slower that I'd realised. Smashin. Link to comment Share on other sites More sharing options...
gmdh555 Posted October 14, 2005 Author Share Posted October 14, 2005 Right db so whats the answer? What Myles said Link to comment Share on other sites More sharing options...
Myles Posted October 14, 2005 Share Posted October 14, 2005 Gad! I'm surprised there is any left of the 1st class physics degree - I've apparently only drunk enough in the last ten years to neutralise a 2nd... Project Scope-Creep is live... Alcester Racing 7's Equipe - 🙆🏻™ Alcester-Racing-Sevens.com Link to comment Share on other sites More sharing options...
db Posted October 15, 2005 Share Posted October 15, 2005 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. 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** Link to comment Share on other sites More sharing options...
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