Lower Wheel Weight = FE Increase? - Page 2 - Fuelly Forums

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Old 03-18-2008, 09:35 PM   #11
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you have to remember as well that the spinning weight is only going to do good as long as it's perfectly balanced, otherwise it's turning in to heat in your bearings, and vibration.
Apparently back in the '80's there was an article about a guy who went thru a balanced every rotating part of his car, from the engine to the wheels making it as close to perfect as possible.
you also have to remember that unsprung weight is going to take more energy to go over bumps and move out of potholes.
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Old 03-19-2008, 06:26 AM   #12
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Folks throw around the rule-of-thumb that 1Lb of rotating mass = 10Lb of regular mass. If that's true, then the lighter-weight rims would be equivalent to removing 280Lbs from the car. But, is it true?...
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Old 03-19-2008, 06:34 AM   #13
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I believe that is said for racing purposes, though I'm not entirely sure whether it means for acceleration or for handling (gyroscopic effect)

Trying to justify weight as inertia storage is like trying to justify playing a lottery as a savings account. No matter how sophisticated your technique is, statistically you lose.
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Old 03-19-2008, 08:20 AM   #14
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Folks throw around the rule-of-thumb that 1Lb of rotating mass = 10Lb of regular mass. If that's true, then the lighter-weight rims would be equivalent to removing 280Lbs from the car. But, is it true?...
Let's calculate the kinetic energy of a wheel.

Assuming a wheel+tire is a solid disk with uniform mass distribution, then its moment of intertia is I=m*r^2/2, where 'm' is its mass and 'r 'the radius. Assuming the car is moving at speed 'v', then the angular speed of the wheel is w=v/r. So the kinetic energy of the wheel due to the rotational motion is

K1 = 1/2*I*w^2=1/4*m*v^2.

The wheel also has kinetic energy due to its translational motion, which is

K2 = 1/2*m*v^2

So the total kinetic energy of the wheel is

K = K1+K2 = 3/4*m*v^2

Comparing this to a same mass m inside the car with only translational motion, whose kinetic energy is 1/2*m*v^2, the rotating wheel with uniform mass distribution is only equivalent to '1.5*m' of non-rotating mass. If we consider that the mass distribution of the wheel is not uniform, the wheel is at the most equivalent to '2*m' of non-rotating mass as far as kinetic energy goes. So saving 10 lbs on the wheels is probably equivalent to saving 15-20 lbs in the car, more or less.

I think the "1Lb of rotating mass = 10Lb of regular mass" probably refers to the handling benefit of reducing unsprung weight. But according to my calculation, changing to light-weight wheels probably won't provide as much FE benefit as one may hope for.
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Old 03-19-2008, 08:46 AM   #15
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sweet. thanks spacepilot. now i don't have to buy those titanium lug nuts
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Old 03-19-2008, 08:54 AM   #16
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sweet. thanks spacepilot. now i don't have to buy those titanium lug nuts
Lol. Titanium lug nuts, I didn't know they existed. I'd rather spend my money on some cheetah blood. Smear it on the valve cover and the engine gains 20hp.
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Old 03-19-2008, 02:49 PM   #17
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Trying to justify weight as inertia storage is like trying to justify playing a lottery as a savings account. No matter how sophisticated your technique is, statistically you lose.
This is actually untrue when you throw in a *minimum* speed into the equation. For any given slope, the maximum speed coasting down it is when the gravity force propelling it is equal to the drag and rolling resistance forces. The more shallow the slope, the greater the mass-to-drag must be in order to maintain a given speed. Granted, this would be valid only at speeds high enough for aerodynamic forces to be the primary loss.

For the sake of argument, imagine a car that can maintain 58 mph down a given slope of the freeway. The speed limit is 65 mph. Obviously you'd need the engine on and burning fuel (very inefficiently I might add) to maintain 65 mph. Now add a couple 250 lb passengers. . .and say, for the sake of argument, you can maintain 65 mph down the same slope. I think it's fair to say that the extra fuel you burn getting up to speed with the additional weight will be less than the fuel you save by EOC'ing. The extra fuel you burn will be at a higher load, and thus a lower BSFC than the fuel you'd be burning in the downhill portion.
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Old 03-19-2008, 03:56 PM   #18
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Yeah, and sometimes your numbers come up on the lottery too, but there's all those times where they don't.
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Old 03-19-2008, 06:01 PM   #19
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I believe that is said for racing purposes, though I'm not entirely sure whether it means for acceleration or for handling (gyroscopic effect)

Trying to justify weight as inertia storage is like trying to justify playing a lottery as a savings account. No matter how sophisticated your technique is, statistically you lose.
think of a wheel as a flywheel to an engine, heavier flywheel more inertia it stores (thus lower idle rpm) BUT has longer throttle response time. lighter flywheel and you have to have a higher idle rpm to keep it running smooth BUT the throttle response is alot quicker so it revs faster, and when you let off the gas the rpm will imediately drop 9or very close) VS a heavy flywheel that you could let off and it will coast before it reaches idle again.


so how does this apply to wheels? well lighter wheels less built up inertia so it will be easier for the engine to accelerate said car but said car wont coast as much (good for city driving in stop and go) heavy steel wheels, will coast longer once thier going but will also take more fuel to get spinnign faster.


how much, well i asked my physics teacher this one day and he said that there wouldnt be much of an improvement over replaceing steel wheels with aluminum.
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Old 03-19-2008, 06:39 PM   #20
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Yeah I would say just guessing that the weight if all at the surface of the tire tread would have a 2x weight of the non-rotatiing weight of the vehicle thus the 1.5 makes sense other than the fact that you are lightening the rim so it would have even less effect than if you were lightening the tire since the rim is in more on the tire it has less rotating effect. But if you loose 5lbs per rim that is like 20*1.5 or 30lbs lighter vehicle so there will be a little effect if the car is light.

The flywheel in the engine being lighter will affect acceleration more because it spins a LOT faster than the wheels espically in first gear and can really help the car accelerate faster with less wasted fuel if you shift a lot. It also will impart more power pulses to the gears if lightened possibly breaking down the oil film in the tranny and causing gear wear.
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