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The True Benefit of Skinnies

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Old Jul 14, 2006 | 06:06 PM
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Default The True Benefit of Skinnies

When normal front wheels are replaced by skinnies, the weight (mass) reduction will be equivalent to the removal of some greater static mass. This becomes obvious when you consider that the total energy in a rotating wheel consists of two parts, one of which is the kinetic energy of an equivalent static mass. The other part is the rotational energy of the wheel. I have seen claims that when skinnies replace heavier wheels, each pound saved, is equivalent to the removal of two, or four or eight pounds, of static weight. I have looked unsuccessfully for verification of these claims so I am asking for help with my calculations. The largest estimate of benefit results from assuming that the wheel mass is concentrated at the rim rather than distributed throughout a solid disk. Using the rim-loaded model of a wheel, my calculations indicate that the total energy in a wheel (translational plus rotational) is exactly double the translational (kinetic) energy of an equal non-rotating mass. If the solid-disk model is used, then the rotational energy of the wheel will equal one-half the wheel’s translational energy. This suggests to me that each pound saved when skinnies replace fatties is equivalent to, at most, two pounds of static weight. My excel spreadsheet is attached. Comments and corrections would be very much appreciated.
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Old Jul 14, 2006 | 06:46 PM
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Are you saying that to rotate skinnies (compared to other tires) takes less energy away from accelerating the vehicle, so vehicle can accelerate quicker...?
makes sense;

note also that skinnies have less drag due to less rolling resistance.
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Old Jul 14, 2006 | 07:07 PM
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I think your 2 lb estimate is probably pretty close. Many people like to quote the 8lb figure when getting lighter wheels, but that seems really high. When I switched to 4lb heavier wheel (same diameter, same tire), I noticed no significant difference in 1/4 times. If the wheels added a corrected 128lbs (using the 8 lb figure) the car should have been a tenth or so slower on average.

I have heard of people slowing way down with heavier wheels, but I think there are other factors involved in those cases. (way heavier wheel tire combo, thin sidewalls, more weight farther from the center of the wheel)

This is my non-scientific 2 cents.
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Old Jul 14, 2006 | 07:28 PM
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Are we talking each pound saved off of the front of the car, or each pound saved from the rotational mass that propels the car (e.g. the rear wheels in our case). I would think the weight saving ratio would be different for each. Thus, putting skinnies on the front would equate to a greater weight saving ratio than would putting skinnies on the rear (assuming traction stayed constant)... or I could have it backwards. Either way, I don't think putting heavier wheels on all the way around would be a constant 8:1 (or whichever ratio you choose) at the front and at the back.
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Old Jul 14, 2006 | 07:54 PM
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Originally Posted by ArcticZ28
Are we talking each pound saved off of the front of the car, or each pound saved from the rotational mass...
Lighter wheels, front or rear, will benefit acceleration by some amount greater than removing an equal non-rotating mass. I simply focused on front wheels (called "skinnies" by drag racers) because there is usually less freedom to reduce the weight of rear wheels. The exact benefit factor has undoubtedly been calculated by real physicists but I haven't been able to find their work.

Last edited by Gary Z; Jul 14, 2006 at 08:10 PM.
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Old Jul 14, 2006 | 11:46 PM
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Originally Posted by Gary Z
The exact benefit factor has undoubtedly been calculated by real physicists but I haven't been able to find their work.

Thats something you might find in some obscure SAE article?
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Old Jul 15, 2006 | 01:42 AM
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Originally Posted by Gary Z
When normal front wheels are replaced by skinnies, the weight (mass) reduction will be equivalent to the removal of some greater static mass. This becomes obvious when you consider that the total energy in a rotating wheel consists of two parts, one of which is the kinetic energy of an equivalent static mass. The other part is the rotational energy of the wheel. I have seen claims that when skinnies replace heavier wheels, each pound saved, is equivalent to the removal of two, or four or eight pounds, of static weight. I have looked unsuccessfully for verification of these claims so I am asking for help with my calculations. The largest estimate of benefit results from assuming that the wheel mass is concentrated at the rim rather than distributed throughout a solid disk. Using the rim-loaded model of a wheel, my calculations indicate that the total energy in a wheel (translational plus rotational) is exactly double the translational (kinetic) energy of an equal non-rotating mass. If the solid-disk model is used, then the rotational energy of the wheel will equal one-half the wheel’s translational energy. This suggests to me that each pound saved when skinnies replace fatties is equivalent to, at most, two pounds of static weight. My excel spreadsheet is attached. Comments and corrections would be very much appreciated.

that is probably right but when you put wider and bigger in diameter tires in the rear for drag racing, it probably ends up being a wash when it comes to total rotating mass (less up front, more in rear)
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Old Jul 15, 2006 | 02:28 AM
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More linear vehicle acceleration due to less tangential friction from surface contact of road, less resistance to acceleration due to lower rotational interia, and faster acceleration due to decreased vehicle weight...

Cons? Can't corner as well. Racing wheels don't hold up to daily driving as well.
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Old Jul 16, 2006 | 11:35 AM
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Further analysis of this question has convinced me that the theoretical maximum benefit to straight-line acceleration of reduced wheel weight is 2 for 1. Furthermore, the theoretical maximum cannot be achieved in practice. The maximum equivalent static weight reduction will be approximately 1.5 pounds for each pound of reduced wheel weight. The often repeated claims of much larger mass-equivalence factors are misinformed. For anyone who might be interested, I have attached my analysis. It is an algebraic argument at the level of high-school physics.
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Last edited by Gary Z; Jul 17, 2006 at 03:56 PM.
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Old Jul 18, 2006 | 01:12 AM
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Also note that race wheel are small in diameter compared to normal wheels. For example, 14 inch rims instead of 18 inches for lower interial resistance.

But then again... 18's would carry you down hills faster and provide smoother coasting.
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Old Jul 18, 2006 | 08:25 AM
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Also remember that their is a fair amount less wind resistance due to a smaller frontal area and (typically) lower cd then a wide wheel/tire. Not anything awestriking, but significant nonetheless.
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Old Jul 21, 2006 | 03:15 PM
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Mickey Thompson quotes 8# static for each 1# of rolling mass. Sounds high to me, but in the case of rear wheels you also have less drivetrain loss with lighter wheel/tire which would equate to more HP to the ground.

I know in my race car, when I switched to 31x10.5s w/o tubes from 32x16's w/tubes I lost like 12# per wheel. I trapped about 200-300 rpm less due to less load on the converter.
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Old Jul 21, 2006 | 10:58 PM
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Originally Posted by 1fastWS6
Mickey Thompson quotes 8# static for each 1# of rolling mass.
This claim on the MT Website was the source of my interest.

Last edited by Gary Z; Jul 21, 2006 at 11:51 PM.
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Old Jul 26, 2006 | 11:46 AM
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Originally Posted by 2002_Z28_Six_Speed
Also note that race wheel are small in diameter compared to normal wheels. For example, 14 inch rims instead of 18 inches for lower interial resistance.

But then again... 18's would carry you down hills faster and provide smoother coasting.
Ok, im no physicist however, you common logic in diagnosing the discussion at hand....you start off with a 275/40/17r rim and tire on the front weighing in at say 30 lbs (whatever) not only is the 30 lbs static weight but the car must also obtain the tq necessary to rotate the 30 lbs at a given velocity. so, the amount of tq necessary to rotate the tires will depend on the level of the surface (dragway being flat), therefore no resistance there, second the surface area of the tire is in direct proportion to the amount of friction between tire and ground in this case being a standard 9 or 9.5 in rim you have that much friction for rotation, third you also have that same amount of surface area which will be causing although mynute, still some amount of wind resistance. so when you compare changing to skinnies think of it as deducting a a number whatever it is use X from each hinderence in this case an example would be...... surface friction-x, wind resistance-x and slope-x....in doing this you see that when you go to a skinny you lose surface area there for creating less friction between tire and the road therefore increasing rotation, you also shave weight therefore decreasing parasitic drag and also decreasing the amount of tq necessary to propel the tire of a larger size to the same velocity. lastly you would also lose wind drag given the loss of surface area on the rim and tire....now you can look at this in 2 ways, in switching from skinnies if it takes x amount of force to obtain a velocity of A with a larger tire it would only take you x-(your loss of resistance) to obtain that same velocity of A or if you keep the amount of force necessary to obtain velocity A with a larger tire and switch to a skinny you would then be able to exceed velocity A due to less tq necessary to rotate and propel the car.......this also differs from comparing the drive tires of a rear or front wheel drive car becuz the discussion of skinnies eliminates the necesisty of friction for traction and also eliminates the tq being delivered directly to the skinnies and there for having to calculate propultion of the object and discussing at what point the coefficient of friction would become 0.

lastly in reguards to what 2002_z28_6speed said about down hill although the parasitic drag would be the same the weight in the tire and the rotational mass would actually benefit car being that we are now calculating in the affects of gravity on a negative slope where as in the preceeding we have been basing our expirement and anaylsis after a level plane. so of course the heavier the object the larger the force acting on the object and being on a negative slope a larger tire generating a higher amount of rotation inertia would increase in velocity quicker then that of less static wiehgt. so to acturatly discuss this we must take into consideration the amount of variables that we are adding into the equation i.e. slope, forces of gravity, size, wieght, car, dragway or auto X etc.


...anyway thats jus my anaylsis while i am at work so take it for what its worth

Last edited by YellowMonster87; Jul 26, 2006 at 11:52 AM.
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Old Jul 26, 2006 | 01:25 PM
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Don't forget it also allows you to go straight through the waterbox without having to worry as much about tracking water up to the staging lights.
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Old Jul 28, 2006 | 12:57 AM
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Originally Posted by slow67
Don't forget it also allows you to go straight through the waterbox without having to worry as much about tracking water up to the staging lights.

Or you could go around, and not be an ******* to other racers also.
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Old Jul 28, 2006 | 01:44 AM
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FWIW i raced a guy a few months ago that had the same car as me, same mods, only he ran skinnies and i ran stockers, and i had a guy in the car with me who weighed ~160lbs and he pulled on me pretty good.. so good i thought he was running an m6 untill i talked to him and saw it was a 2.73 geared auto.
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Old Jul 31, 2006 | 01:34 AM
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I rememember reading somewhere if you have two hoops of equal mass and diff diameters, assuming a wheel+tire as one ring of mass contrentrated at a radius, that the larger hoop will go down a hill faster than the smaller hoop. Due to torques I believe.

Anyways all you want for racing is a wheel that can have enough meat with the smallest weight and diameter possible. Not good for street use but hauling *** down the track.
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Old Jul 31, 2006 | 09:50 PM
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Rotational energy is proportional to the square of the rotational velocity. Furthermore, rolling resistance is inversely proportional to diameter. Therefore, a larger diameter wheel will rotate slower, and have less rolling resistance.

Now, in terms of skinnies... we have a narrower wheel which has less rolling resistance not only due to dimensions, but also because it is likely a harder compund that is not made for high traction.

Typically, a skinny is also riding on a 14-15" rim, as opposed to the stock 17" rim. The weight of the wheel is most likely concentrated closer to the rotational axis. Therefore, if you were to calculate the rotational energy saved, the effective disk/hoop diameter could be reduced.

For street use, it is typical to go to larger brakes for better stopping power. For track use, one can go to smaller diameter brakes to reduce the rotational mass (swap LS1 down to LT1).

As an added bonus, front wheel weight loss also contributes to improving the balance of the car.

One could carry on for hours with suggestions as to what factors should be calculated for. To truly get the most accurate number, you would have to run physical tests. Before I leave this alone, tire pressure plays a HUGE role in calculating this mess too.

Cheers,
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Old Jul 31, 2006 | 11:52 PM
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Originally Posted by 2002_Z28_Six_Speed
I rememember reading somewhere if you have two hoops of equal mass and diff diameters, assuming a wheel+tire as one ring of mass contrentrated at a radius, that the larger hoop will go down a hill faster than the smaller hoop. Due to torques I believe.

Anyways all you want for racing is a wheel that can have enough meat with the smallest weight and diameter possible. Not good for street use but hauling *** down the track.
Pretty much the end results but with the above mentioned hoops the smaller diameter hoop will achieve its max rpm quicker. the larger will EVENTUALLY pick up to a faster velocity but we are concerned about acceleration. with equall mass but different dia. the two rolled down a slope as in an equal race the smaller one will jump ahead a little at first. its been too long since I sat in a university physics class so I can't give an inertial equation off hand. Your right just thought a little quirk needed to be added to help out.
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