SPLN SUX Posted December 3, 2009 Report Share Posted December 3, 2009 I'm trying to get some clarification on how to calculate the rotational inertia of a wheel mounted to a tire. Do you: A - figure them as a hoop (tire) and disk (rim) and add your results together B - figure them as one large disk (since it is filled in the center) c - figure them as one large hoop (since most of the mass is in the outer portion) Here's the dilemma... Right now I run a 31" tire weighing 40lbs, on steel wheels weighing 28lbs, giving me a total rotating mass of 68lb on each corner. I want to drop the wheels to 16lbs, and am trying to figure out how big of a tire (height and weight) I can run w/o upsetting the rotational inertia force that is currently on my drive train. Lets see how many of you smart asses are really smart. I bet the guy that knows this drives a Fox body. :bangbang: Quote Link to comment Share on other sites More sharing options...
Guest Removed Posted December 3, 2009 Report Share Posted December 3, 2009 i'll wait....... i've done this before just for a 2 tenths at the strip..lol Quote Link to comment Share on other sites More sharing options...
AJ Posted December 3, 2009 Report Share Posted December 3, 2009 (edited) I bet the guy that knows this drives a Fox body. :bangbang: or a vette http://library.thinkquest.org/16600/advanced/rotationalinertia.shtml Rotational energy is linearly related to Inertia. It depends on the distribution of the mass, but rotational inertia (which you'll have to fight to speed up or slow down) increases linearly with mass and squared with radius (distance from centerline). example Simplifying the wheel to a hoop with no spokes (as in only the rim portion) increasing from 17" to 18" assuming no change in weight (mass) and assuming rolling diameter remains constant (so angular velocity, rpm, remains the same) there is a 12% increase in rotational intertia. Athough in reality the tire may be less massive since it would have to be lower profile to maintain OD so this would counteract this a bit. Going from 40 to 45 lbs, keeping the same radius and ignoring the mass distribution for simplicity would increase a maximum of 12.5%. Edited December 3, 2009 by AJ added info Quote Link to comment Share on other sites More sharing options...
SPLN SUX Posted December 3, 2009 Author Report Share Posted December 3, 2009 i'll wait....... i've done this before just for a 2 tenths at the strip..lol LOL how did I know :bangbang: Seriously though, which way do you do it? Quote Link to comment Share on other sites More sharing options...
Guest Removed Posted December 4, 2009 Report Share Posted December 4, 2009 or a vette http://library.thinkquest.org/16600/advanced/rotationalinertia.shtml Rotational energy is linearly related to Inertia. It depends on the distribution of the mass, but rotational inertia (which you'll have to fight to speed up or slow down) increases linearly with mass and squared with radius (distance from centerline). example Simplifying the wheel to a hoop with no spokes (as in only the rim portion) increasing from 17" to 18" assuming no change in weight (mass) and assuming rolling diameter remains constant (so angular velocity, rpm, remains the same) there is a 12% increase in rotational intertia. Athough in reality the tire may be less massive since it would have to be lower profile to maintain OD so this would counteract this a bit. Going from 40 to 45 lbs, keeping the same radius and ignoring the mass distribution for simplicity would increase a maximum of 12.5%. correct sir! Quote Link to comment Share on other sites More sharing options...
SPLN SUX Posted December 4, 2009 Author Report Share Posted December 4, 2009 Yes but whats the proper way to calculate it? Separately, or as a whole, and which equation(s) do you use? I'm starting to think the ring equation is the way to go, measuring from the outside of the rim to outside of the tire as the portion of mass. Quote Link to comment Share on other sites More sharing options...
Guest Removed Posted December 4, 2009 Report Share Posted December 4, 2009 its figured as a whole for rotational mass.. theres lots of math that goes into it.. just have to sit down and do it and you have to figure it more then one way too! as height increase's,width, rotational mass does as well. but i know if i remember correctly, under 30" tire height,under 10" width , it no more then 2"(height wise) and 18-22 pounds as a hole..but to make it the same or less, i had to drop over 22 pounds of mass going 2" taller, and 3 " wider, and still gained inertia mass over all. have have the formula'a saved in the old computer, its out in the shop.. i hope i still have it. but i found a calculator for this if i remember right. all you needed was a scale and a tape measure ...lol have you seen this one? http://www.analyticcycling.com/WheelsInertia_Page.html Quote Link to comment Share on other sites More sharing options...
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