## Alloy wheels and fuel economy

Will switching to alloy wheels (typically 17% lighter than steel) noticibly improve fuel economy? It seems like we should be able to calculate a rough estimate if we had necessary data but i don't. Has anyone been there?
 Recognitions: Science Advisor What proportion of the mass of a typical car is the wheels? How much does a steel wheel (without the tire) weigh - say 20 pounds maximum? 20 x 4 x 0.17 = not much, compared with the total mass of the car. The main difference would be the performance of the suspension, because you are reducing the mass being bounced around by bumps in the road. But if you can go faster round corners, your fuel consumption is likely to increase rather than decrease.
 Wheels not only have to be accelerated linearly, but also angularly, which makes them harder to accelerate than regular car parts. Reducing rotating weight has a larger effect on fuel economy and performance than stationary weight. How much though...I never caluclated that. I'd be interested to know.

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## Alloy wheels and fuel economy

I doubt it is substantial but because the savings is all during acceleration, most of the savings is in "city" driving.
 But city driving is all stop and go, constantly accelerating and decelerating. I always thought that reducing the rotational mass would have a much greater effect than the cars overall mass. Due to the fact the wheels are directly propelled by the engine (well, not all depending on the drive).

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 Quote by Lsos Wheels not only have to be accelerated linearly, but also angularly, which makes them harder to accelerate than regular car parts. Reducing rotating weight has a larger effect on fuel economy and performance than stationary weight. How much though...I never caluclated that. I'd be interested to know.
Suppose the radius of gyration of the wheel (without the tire) is = k. and the radius of the tire is r.

At a road speed of v, the translational kinetic energy of wheel = $mv^2/2$

Angular velocity = v/r
Rotational kinetic energy of wheel = $m(v/r)^2 k^2/2$
= $(k/r)^2 mv^2/2$

k/r must always be less than 1. I would guess a typical value would be about 0.7 which makes the rotational KE about half the translational KE.

If you add the rotational KE, 1.5 times "nothing much compared with KE of the rest of the car" is still "nothing much".
 Recognitions: Gold Member Science Advisor depending upon the automobile..the increase in MPG is big... 1. anecdotal - friend of mine had a Chrysler 300 which had wheels and tires each weighed 50 lbs.,,cut the weight in half when we went to aluminum centerline wheels..the acceleration improved dramatically..opinion only 2. every pound of rotating weight you can remove, makes the car think you removed 10 pounds. so in the above case, removing 25 pounds per wheel made the driver ' feel" like it lost 1000 pounds, seat of the pants, of course.. empirically, the formula means your using less gasoline to accelerate the lower moment of inertia wheels

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 Quote by Ranger Mike 2. every pound of rotating weight you can remove, makes the car think you removed 10 pounds.
I think that number is based on reducing high speed rotating masses like the flywheel, crankshaft, conrods, pistons, clutch plates, camshaft, etc.

Unless you are measuring how much faster you can spin the tires, the math doesn't stack up for the road wheels.

Lower unsprung mass = better suspension performance = more power on the road without wheelspin, but that isn't going to improve your gas consumption.
 Recognitions: Gold Member Science Advisor Anytime you reduce weight of a vehicle , you get better MPG..it is a fact, even more savings is realized by reduction in ROTATION MASS... High speed rotation parts is relative and the shear mass of 50 pound tires and wheels are a huge factor. The calculations for the wheels are the same as regarding brake rotors, flywheels, and on a micro scale , crankshafts, cam shafts and axles..(gun drilled axles..yes, they do.) Pistons and connecting rods are reciprocating weight and the rule does not apply. ( ok- 50% of a con rod would apply for big end only, if you want to get specific)
 Recognitions: Science Advisor The "high speed effect" is proportional to RPM squared. If your engine is doing 6000 RPM and your road wheels 600, the rotationg engine parts are 100 times more significant than the wheels. Sure, people will go for ever last little saving. The old Mississippi steamboat racing captains used to make their crews shave their heads bald to reduce wind resistance, I believe. I would expect the psychological effect of gun drilled axles on the race team is much more significant than anything Newton's laws would predict.
 Recognitions: Gold Member Science Advisor AlephZero...good one!! ...i gotta remember that one! i was in germany and saw th oil pump impeller for a Formulal 1 engine..it was carbon fiber..just to reduce rotation weight..now that is extreme!! back to the math, typical race car in high gear will have tires rotating at 2016 ( 2.91 rear gear) to 3120 rpm ( 5.17 gear gear)...the real reason for going with light weight wheels is to minimize unsprung weight which can not be controlled when weight transfer occurs..the reduced moment of inertia is a good way to rationalize the purchase of a \$ 400 magnesium wheel t hat will get cracked and dinged by the driver..insane but true.. screw it ,,im gonna shave my head now!!
 Recognitions: Science Advisor I deliberately chose a lower wheel RPM because I was thinking about the maximum acceleration coming out of a sharp corner etc. This is all about how fast the engine can spin up. Rotating mass doesn't have any effect on max speed. Neither does non-rotatiing mass, unless you change the shape of the car to reduce the air resistance. Actually, air resistance from different shapes of wheel is another issue here - not my specialist subject, but I've been told that computational fluid dynamics models of cars include rotating wheels these days, to calculate the drag coefficient accuractely.
 Recognitions: Gold Member Science Advisor Excellent point AlephZero i was not aware of the newer programs capabilities...tires account for 40 percent of aero drag on open wheel cars...huge thanks
 Thanks AlephZero and everyone.

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 Quote by Ranger Mike depending upon the automobile..the increase in MPG is big... 1. anecdotal - friend of mine had a Chrysler 300 which had wheels and tires each weighed 50 lbs.,,cut the weight in half when we went to aluminum centerline wheels..the acceleration improved dramatically..opinion only
Considering the Chrysler 300 weighs about 4100-4300 lbs, I seriously doubt it made any "dramatic" difference in the car's acceleration or fuel efficiency. A person's "butt-o-meter" is notoriously inaacurate in guessing a car's performance characteristics... My guess is there was no significant difference in 0-60, 1/4-mile, or MPG numbers.

 Quote by Ranger Mike 2. every pound of rotating weight you can remove, makes the car think you removed 10 pounds. so in the above case, removing 25 pounds per wheel made the driver ' feel" like it lost 1000 pounds, seat of the pants, of course..
This might be a general rule of thumb for an engine component rotating at thousands of RPM, but definitely not true for wheels and tires. Losing 1000lbs on a car is VERY significant, that would be a 25% reduction in total weight for that Chrysler...

 Quote by Ranger Mike empirically, the formula means your using less gasoline to accelerate the lower moment of inertia wheels
I think you mean analytically or theroretically rather than empirically. Empirically would mean you tested it and measured a difference.

 Quote by AlephZero Actually, air resistance from different shapes of wheel is another issue here - not my specialist subject, but I've been told that computational fluid dynamics models of cars include rotating wheels these days, to calculate the drag coefficient accuractely.
It's true, race teams these days in F1 and NASCAR actually use "rolling road" wind tunnels, because both the wheels rolling at high speed and the road moving at the same speed as the wind makes a difference.
 Here's a test: http://www.carcraft.com/projectbuild...son/index.html I assumed the numbers were accurate enough to be representative and used a 1/4 mile calculator to assess the changes in the quarter mile times and speeds. After baselining the calculator, I subtracted the weight reduction of the wheels (82 lbs) from the entered weight of the car (3882 lbs for simplicity) and the calculator predicted a drop from 12.14 sec @ 111.8 mph to 12.06 @ 112.2 mph. Their result was 12.03 @ 113.11 mph; to get that same ET with the calculator required an extra 50 lb reduction. So the 82 lb reduction in wheel weight gave the same result as a 132 lb reduction in total vehicle weight, with 50 lbs worth due to reduction in rotating mass. This is pretty crude and certainly shouldn't be taken as gospel, but it indicates that the effect is there, just not the magnitude that is usually stated.

 Quote by AlephZero I would guess a typical value would be about 0.7 which makes the rotational KE about half the translational KE. If you add the rotational KE, 1.5 times "nothing much compared with KE of the rest of the car" is still "nothing much".
The empirical numbers suggest 1.6; close enough!