My 16 inch wheels were 6 pounds heavier than my 15 inch wheels...

But my 15 inch tires were 2 pounds heavier than the 16 inch tires (same OD-24.2").

But my car is getting better gas mileage and accelerates easier.

I think this is because overall I dropped 4 pounds per wheel/tire combination.

 

I think the reality is a combination of what you and mahout said.

Mahout seems to go out of his way to make wheels seem insignificant. While, it is true a drop of a single lb on the wheels has less effect than a lb on the tire, its effect is still proportional to its distance from the center.

For the sake of this post, I'll use 18lbs as the weight of the wheel and also the weight of the tire as stock. And also, for the sake of simplicity, we'll put the weight of each object on its edge.

the amount of instaneous force needed to rotate stock tires would be (18*8)+(18*12). Or 360 in*lbs.

Your wheel dropped 6 lbs, but moved in 1 in. Your tire increased 2 lbs in weight. So, (12*7.5)+(20*12). Or 330 in*lbs.

A change of 30 in*lbs or 2.5 ft*lbs. That's what the drive axle will see. Turning's "gyroscopic effect" may see the same thing (though, I'm not sure what the calculation would be).

Again, weight change in the wheel has less effect, but its still an effect.

 

Since the wheel/tire and axle are considered a rigid fixture such that the model is merely a large diameter axle the force is applied at the edge where the tire meets the opposition both front and rear.. So the force is applied postively on the drive (front)axle and negatively of the driven (rear) axle. The only difference is the force applied to the front is greater by the amount of acceleration. The resistance to rotation of all wheels is the same.

 

I'd agree that the resistance to rotation is the same but the force needed to rotate a wheel from the axle vs from the outside edge is different. so while the wheel may resist rotation no matter whether it is on the front or rear, it will take more effort to rotate the front wheels. So in this case, having lighter front wheels would be more beneficial (effective/important) than the rear wheels.

~SB

 

It is going to take the same amount of energy to rotate the wheels of the same weight regardless of whether they are drive wheels or just bearing weight... The same amount of braking force is needed to reduce the speed of the wheels also if they are of the same weight, as well as lateral force needed to change direction... I can assure you that if you change to lighter front wheels without doing the same on the rear you are going to be spinning your front wheels more than you would if you had also put lighter wheels on the rear, negating any gain out of the hole... At higher speeds in taller gears the heavier wheels are going to have the same effect in slowing your rate of acceleration regardless of which end of the car they are on based on the additional power needed overcome the mass even though it is in motion.. I learned this a long time ago messing with old British motorcycles.

 

This has turned into a fantastically technical discussion. I want to see more of these!

 

x infinity...

 

I agree with this on slowing down but, on acceleration, the engine will have to work harder to move the car forward as it has to do more work to rotate those front wheels due to the fact that it is applying the force to the axle and not the outside edge of the wheel. On the rear wheels, there is no axle so by pulling the car forward, the engine is using the outer edge of the tire to rotate the wheels (b/c the car is in a fw motion). as far as linear forward motion, there is no difference between front or rear but in terms of rotational motion on each wheel, it (still in my mind) HAS to be different because of where the force is exerted in order to make the wheel turn. (front - center of the wheel, back - outer edge of the wheel)

~SB

 

If you were to take a bicycle that was motorized using a roller on the end of the engine crankshaft making contact with the tire tread and another motorized bicycle using a belt and pulley arrangement to drive the wheel at the hub with both geared to the same amount of revs of the engines to rotate the wheels at matching revs they would both use the same amount of power to attain and maintain the same speed... I do understand what you mean about the effect of power applied to the wheel at the axle requiring more torque to rotate the wheel but that is due to the gearing more than the part of the wheel power is applied to.

 

Good comparison.
As long as the wheeland tire is ridgid to the axle and the resistance to the rotation is at the OD there cannot be a difference. The drag that induces rotation at the rear is precisely the same as at the front to rotate the tire. Course, if you don't allow the rear to rotate freely, as by applying the emergency brake, that indeed would be different. Ditto front.

 

Speaking as someone with a degree in physics...

Assuming that all tires are in contact with the pavement during acceleration, all wheels will need to reach the same angular velocity within the same amount of time. Since the wheels are of the same type, there is no difference in their moment of inertia (aka their resistance to a change in angular velocity).

The work required to get a wheel turning to a certain angular velocity w from rest is:

1/2 I w^2

where I is the moment of inertia of the wheel.

Since both the moment of inertia and final angular velocity are the same, there is no difference in the amount of work (which is directly analogous to the energy required).

 

It should also be noted that lighter wheels will have little to no effect on your mileage unless you accelerate at the same rate as you did with the heavier wheels.

 

By conservation of energy, the amount of work done in order to move a mass up to a certain veolcity on a level road is equivalent to the kinetic energy of the mass.

- For a rotating mass, such as a wheel with tire on,

Kinetic energy = 1/2 x m x v^2 + 1/2 x I x w^2

since w = v / r

therefore

Kinetic energy = 1/2 x (m + I / r^2) x v^2

for the extreme case, with all mass at the edge of the rim

I = m x r^2

which will further simplify :

kinetic energy = 1/2 x (2m) x v^2

- For a non rotating mass

Kinetic enegy = 1/2 x m x v^2

where m : mass of wheel (kg)
v : velocity (meter/sec)
I : mass moment of inertia (kg M^2)
r : radius of the wheel
w : angular velocity (radian/sec)

- Therefore it takes more work (the worst scenerio is 2 times) for a rotating mass comparing with a non rotating mass to get up to the same speed.

Damn, I still remember my Physics........

 

ok... I'm somewhat but not totally convicned. I'm going out on another wild tangent here but say the front wheel was 500lbs. wouldn't it put more strain on the engine and transmission to rotate that wheel than it would if the 500lb wheels were on the rear?

~SB

 

But don't forget the rear wheels are attached to the chassis, which is in turn attached to the transmission and engine. Therefore the engine will see the same strain as if the weight is at the front wheels.

 

Once I started, I just can't stop........

So for Fit Sport M/T with no navigation, curb weight: 2520 lbs

average driver weight: 160 lbs

Fuel weight (half filled) : 50 lbs

Say if we can reduce 10 lbs per wheel by getting lighter tires and wheels. With the assumption of the best scenario of rotating mass kinetic energy is twice of non-rotating mass (with all mass at edge see previous post). Our estimated performance increase (again best scenario):

(4 x (2 x 10)) / (2520 + 160 + 50) = 3%

My guess may be 2% performance gain in real life.

 

Well, in short, no. But I see what's puzzling you, and thinking through puzzles like this is a great way to learn physics. Steve244 gave you the right answer above. When you bring a rotating mass up to a certain speed by applying force to the outer edge you do indeed require less force than you would require if you applied the force closer to the center of the mass, but you have to apply that force over a correspondingly greater distance when you apply it to the outer edge. Think of it in linear terms: you can apply a large force to a mass over a short distance and bring that mass to a certain speed by doing so, or, you can bring that same mass to the same speed by applying a much lower force over a correspondingly greater distance. Same thing applies when you're dealing with a rotating object.

 

Wait, I'm no physicist, and don't pretend to understand everything here, but a calculated 2-3% increase in performance does not seem significant in a 117hp vehicle in terms of acceleration.

So, I'm guessing the perceived gains in performance center mostly around handling and braking?

 

Nope. Unless the wheel bearings were unequal it takes just much effort to turn the wheel in front or rear. Anything more results in acceleration or deceleration. As long as the wheels were same weight too.
You might have a case considering the front hubs are heavier than the rear hubs in effort to turn the wheel but the difference is so slight as to be microscopically different. Course that has no real bearing on whether the lighter wheels goes front or rear; there's no real change there unless you change hubs.

 

I seem to remember 1 lb or rotating mass translated to 4 lbs of static. (back when I had the ITR)

couple other things to think about... less unsprung weight is easier to control for your suspension. Takes less spring to keep them on the ground, and takes less shock to dampen a lighter spring.

weight of your flywheel/crankshaft/axles... and whatever else needs to be spun to accelerate also factors in.

Id imagine a set of spoon 15s and a lightwheight flywheel would spunk up the fit quite a bit.