Lower Wheel Weight = FE Increase?

[Mightymouse13]

So here is my first post. I'm just getting into the Hypermiling gig and have a quick question. It should be an easy one for the FE guru's here. I currently own a 01 Mitsubishi Mirage with an automatic transmission. I currently get 35mpg.

The tires I am running on are already descent as far as rolling resistance goes (Sumitomo HTR 200). I was thinking about getting lighter wheels to increase my FE . There is obviosly fuel to be saved by lowering the total wheel weight....but how much?

My current steel wheels weigh in at a porky 20lbs each! I was thinking of getting some aftermarket aluminum rims that weigh 12.8lbs each.

With a savings of about 7lbs per wheel...would I notice much of a FE improvement? Are we talking like a 2mpg increase? Or would it be more like 0.2? I was wondering if anyone has first had experience with this.

If the fuel savings are that minimal, it might not be worth it for me to justify the $500 purchase.

Thanks in advance!

[DarbyWalters]

If you could find some "good condition" used replacement wheels it might be worth it. I would think that $500.00 could be spent better for a better return...or just spent on fuel.

Since you have an automatic...you could invest in a beefier Torque Converter that locks up a bit earlier and holds rpm slippage to a minimum...that might be better than just lighter wheels...Good Luck

[bowtieguy]

Go to www.tirerack.com. i saw some close-out wheels for $65 each in my size(185 65 14).

BTW, do you like your tires? i was thinking of buying those exact tires for myself.

[Mightymouse13]

Thanks for the replies! The tires are pretty good. They hold the road pretty good in wet/dry conditions above 45 degrees or so. As for FE, I can't compare them to anything with this car...I put the tires on right after I bought it.

Thanks for mentioning tirerack! I didn't even think of them. They have the wheels I was interested in for $69 each. However with shipping, mounting, and balancing it still comes to $400.

Just for the sake of argument, what kind of FE increase would 7lbs a wheel get you?

[bowtieguy]

Originally Posted by Mightymouse13

Thanks for the replies! The tires are pretty good. They hold the road pretty good in wet/dry conditions above 45 degrees or so. As for FE, I can't compare them to anything with this car...I put the tires on right after I bought it.

Thanks for mentioning tirerack! I didn't even think of them. They have the wheels I was interested in for $69 each. However with shipping, mounting, and balancing it still comes to $400.

Just for the sake of argument, what kind of FE increase would 7lbs a wheel get you?

okay, just a suggestion...

wait until the next time you need tires, then order the wheels and tires from tirerack and mount them yourself! hopefully they'll still have a low priced wheel for you.

[JanGeo]

7lbs per wheel - you may not even notice any fuel savings - how much does the vehicle weigh? The rotational mass will not amount to much either - again 20lbs per wheel compared to 2000+ lbs for the vehicle. You will have better tire traction on bumpy roads since you have less weight bouncing around but that is about it.
The Boron treatment can't last forever if the metal wears with time but it does sound interesting.
Synlube on the otherhand uses 2-3 micron size particles of graphite, teflon and moly that I have in my motor and tranny for 16337 miles and have not changed it yet and will not have to in a long time - now that translates to big savings for all teh oil and filters I will not have to by and as for mileage try 20 miles from the gas station tonight at 51.8mpg.

[RoadWarrior]

IMO, they would probably help mpg in heavy stop-start city traffic, though it would probably be lost in the noise in the general unpredictability of the traffic, unless you just had a long stretch of stop signs every block or two, with minimal other traffic to have to wait for, then it might show up as 0.5 to 1mpg different in a car that gets good city mileage.

On the highway, the difference will be in the aerodynamics not the weight. It seems like hubcovers off vs on is worth 1-2mpg on Marvin for example.

IMO if the wheels are upsized without widening the tire, as well as losing weight, you might see a gain from less rolling resistance. A lower profile has a bit less sidewall flex to waste energy in deforming. However, you'll see people running 13 inch 75 profile tires, that's just to get the skinnier tires I think. If the tire takes high pressure well then it might work out. If it doesn't it will roll like a tank track. Got a Tempo in the family with some 75 profile tires on that are like that, 40 PSI in the damn things and they still look a bit flat, and the damn thing is harder to push than Marvin, which has nearly 1000lb and another inch of tire width on it. They are great for the winter though, and apart from the low speed steering effort don't feel too stodgy for handling, odd that.

I've also got a suspicion that tires with a solid center rib will do slightly better for mpg.

[VetteOwner]

heavier the wheel, it will cosat better as it has "built up" torque. lighter wheels will allow it to accelerate easier thus using less gas but less coasting torque.

soo id say if you do alot of city driving then lighter wheels

alot of highway then maybe just leave it as is.

[samandw]

This is a question I've been thinking about myself. A heavier wheel will require more energy to accelerate. . . but also extend coast distance. You're essentially "storing" energy in the wheel when you're moving. If you brake hard to stop, you lose it all. . .but if like most of us you gently coast to a stop (EOC) that extra energy would allow more distance traveled with no fuel burned. Maybe I need to run some calculations. . .

[RoadWarrior]

Weight only counts for you in two motorsports...
Boyscout soapbox derbies and demolition derbies.

[Ryland]

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.

[s2man]

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?...

[RoadWarrior]

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.

[spacepilot]

Originally Posted by s2man

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.

[s2man]

sweet. thanks spacepilot. now i don't have to buy those titanium lug nuts

[spacepilot]

Originally Posted by s2man

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.

[samandw]

Originally Posted by RoadWarrior

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.

[RoadWarrior]

Yeah, and sometimes your numbers come up on the lottery too, but there's all those times where they don't.

[VetteOwner]

Originally Posted by RoadWarrior

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.

[JanGeo]

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.

[1993CivicVX]

Originally Posted by VetteOwner

heavier the wheel, it will cosat better as it has "built up" torque. lighter wheels will allow it to accelerate easier thus using less gas but less coasting torque.

soo id say if you do alot of city driving then lighter wheels

alot of highway then maybe just leave it as is.

Wait a minute, let me get this straight. Aside from the effect an overall reduction in weight has by having lighter wheels, there is an effect on acceleration and coasting by having lighter wheels? Is that why my Honda coasts more poorly than my friend's Honda? (And all this time I was afraid it was from wheels not turning freely) I guess it makes sense, since the axles are turning the wheels. Interesting to know at any rate. So just to clarify, if you are doing a lot of city driving, your fuel savings with lighter wheels would be more noticeable because it will accelerate easier (using less energy) beyond the fact that accelerating is easier when the overall mass is lighter?

[spacepilot]

Originally Posted by 1993CivicVX

So just to clarify, if you are doing a lot of city driving, your fuel savings with lighter wheels would be more noticeable because it will accelerate easier (using less energy) beyond the fact that accelerating is easier when the overall mass is lighter?

Yes, please see my calculations above. If you reduce 10lbs of mass on the wheels, it's roughly equivalent to removing 15-20lbs from the non-rotating part of the vehicle while keeping the heavier wheels, as far as the kinetic energy goes. In another word, removing 1 lb from the wheels goes further than removing 1lb from elsewhere. But removing mass from the non-rotating part of the car first probably is more cost-effective than switching to light-weight wheels.

[1993CivicVX]

Originally Posted by spacepilot

Yes, please see my calculations above. If you reduce 10lbs of mass on the wheels, it's roughly equivalent to removing 15-20lbs from the non-rotating part of the vehicle while keeping the heavier wheels, as far as the kinetic energy goes. In another word, removing 1 lb from the wheels goes further than removing 1lb from elsewhere. But removing mass from the non-rotating part of the car first probably is more cost-effective than switching to light-weight wheels.

Thanks spacepilot. My brain does not convert symbols into meaning very well. As a result I fail at math.

[VetteOwner]

yes city driving you might see a small increae (i wouldnt expect a miracle but) highway you would hardly notice a thing...