SleeperX

I found this write-up on another forum, but I think you guys might find it very informative. Of note, it claims that one pound of unsprung weight is equivalent to a ten pound reduction in vehicle weight..

http://hondaswap.com/general-tech-ar...art-2-a-29058/

Unsprung Weight - Part 2

By: Eric Albert

Introduction

In the first part of this series, we took a look at the effects of high unsprung weight on suspension and handeling. In this part, we will look at rotating mass. Be careful not to confuse unsprung mass with rotating mass. Reducing both is good, but they are not the same. Let's take a look.

Rotational Inertia (or Momentum)

Rotational inertia is a concept a bit more difficult to deal with than unsprung weight. Inertia can be thought of as why a car wants to keep rolling once moving, or remain in place once stopped (unless you forget to set the parking brake on that hill). I believe the terms momentum and inertia are interchangeable. The term “flywheel effect” also refers to these concepts. In a car, there are a number of rotating masses which require energy to accelerate. Up front, ignoring the internal engine components like the crankshaft, we have to worry about the flywheel, clutch assembly, gears, axles, brake rotors and wheel/tire. Out back its a little simpler (for FWD) with just the brakes and wheel/tire contributing most of the mass.

The more mass an object has, the more energy it takes to accelerate it. To accelerate a rolling object such as a wheel, you must both accelerate its mass plus overcome its rotational inertia. As for braking, you must overcome its rotational inertia plus decelerate its mass. By reducing the weight of the vehicle's rotational mass, lightweight wheels provide more responsive acceleration and braking.

Before continuing with our informal analysis here, I want to point out something very important about rotational inertia. We’ve all seen the ice skating move where the skater starts spinning. She pulls her arms in and speeds up, then extends them again and slows down. Why is this? Well, the further a mass is from the center of rotation, the faster it must travel for a given angular speed (how many degrees of an arc it traverses per time unit). The faster anything moves, the more energy it has, so when the arms are pulled in, conservation of energy says that the rotation rate must increase due to equal energy being applied to the same mass over a smaller diameter. Applying this to wheels and tires, which have most of their mass spread as far as possible from the rotation center, I think you’ll agree that it naturally takes more energy to accelerate them. Example: Take a two identical masses, but one is a solid disk of diameter D, the other is a ring of diameter 2D. The ring will require more force to accelerate it (in a rotational manner). Therefore a heavier rim with a smaller diameter could have less rotational mass than a lighter rim of a larger size, and accelerate faster with the same force applied.

The effect of rotating mass can be calculated using Moment of Inertia (MOI). MoI is related to not only the mass of the rotating object, but the distribution of that mass around the rotational center. The further from the center, the higher the MoI. The higher the MoI, the more torque required to accelerate the object. The higher the acceleration, the higher the torque required.

Because of this, the weight of rotating mass such as wheels and tires on a car have a bigger effect on acceleration than static weight such as on the chassis on a car. When purchasing new wheels and tires for a performance car, it can be useful to compare the effects of different wheel and tire combinations. This is especially true when considering upgrading to larger wheels or tires on a car.

The use of light-weight alloys in wheels reduces rotational mass. This means that less energy will be required to accelerate the wheel. Given that each pound of rotational mass lost provides an equivalent performance gain as a 10 pound reduction in vehicle weight, the benefits of light alloy wheels on vehicle performance cannot be overlooked.
For example:
A reduction in the weight of the rim/tire assembly of 5lbs x 4 (all around the car) is equivalent to a 200lb weight reduction in vehicle weight (thats worth 0.200 in the 1/4 mile)

So What's the Point?

The point of this discussion is as follows: There is a great deal of rotational mass to deal with in a car and tires and wheels may only make up half of it. Estimates for weight (o.k. for comparison since they’re all in the same gravity field, therefore the mass would be a similar ratio)
Front: Rear:
Wheel/tire: 30-35 lbs each 30-35 lbs each
Flywheel: 15-20 lbs
Clutch: 15 lbs
Halfshafts: 7-10 lbs each
Gears: 5-7 lbs
Rotors: 3-5 lbs 3-5 lbs
Misc: 3-5 lbs 3-5 lbs
------------------------------------------------------------------
Total: 115-148 lbs 76-90 lbs

So a couple pounds here and there on wheels and tires will make a difference, but that difference is magnified because that weight is placed further from the axis of rotation than any other mentioned (remember the ice skater). All these masses must be accelerated, so any reduction is a good thing. Now you know why we always say don't get those 18" rims for your civic. Not only are the heavier, they have a larger overall diameter. Even with lower profile tires, most plus sizing leaves us with a slightly larger wheel.
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2K6E55

Nice. Thanks for sharing.

phatmitzu

Nice write up! Thanks for posting.

SleeperX

Interestingly enough, he states that larger lighter wheels could cause your car to accelerate slower than heavier smaller wheels. Can this really be true?

chiromikey

it can absolutely be true considering that most of the mass on the larger wheel is further away from the hub.

unsprung and rotational mass plays a big role. i've played with lightweight things such as wheels, rotors, and even pulleys and flywheels on motorcycles and have been able to document dyno gains from each. obviously, the motor isn't making more hp, it's just wasting less on rotational mass between itself and the ground. numbers higher than a 10:1 ratio are often discussed regarding the removal of total weight to unsprung weight. when it comes to motorcycles, there's no better improvement than lightweight wheels. no other single mod can improve acceleration, deceleration, and handling all at the same time and although the implementation of this theory may not be as beneficial for cars, it definitely applies.

BenzoBoi

Very informative post...thanks!!!

zoink

I've forgotten all the physics I learned

See calculation below, if the weight of tires and weight of rims are the same, does it mean that 265/30/19 has ever so slightly (negligible) less rotational mass than 265/35/18?

Center of the hub to the ground = Rim radius + (Tire Width X Sidewall AR / 100).

265/35/18 = 321.35mm from center of the hub to the ground
265/30/19 = 320.35mm from center of the hub to the ground

What is the formula again to calculate the power to rotate a tire+wheels in relation to its radius?

And what negligible is negligible?

Thanks for any info from physics expert

SleeperX

Effect of Wheel Mass on Acceleration Eduardo Martinez 2001
What is the effect of a change in wheel/tire mass? Several people have quoted factors of x6 or x8 to determine the equivalent non-rotating mass variation. That is, an increase of 10lbs/wheel could be as detrimental to performance as carrying Arnold Schwarzenegger and Scooby Doo as passengers.

I decided to refresh my Physics and came up with the following conclusion: from the point of view of acceleration, an increase of X in wheel or tire mass is no worse than an increase of 2X in passenger mass. Not 6x, not 8x, just 2x worst case. This is why.

At any given speed/gear combination there is maximum torque T available at the wheels. The torque does two things: (1) opposes the rolling resistance and the aerodynamic drag and (2) accelerates the car. The equation that describes the equilibrium is:

T = I*u + M*r*a + D*r

*: denotes multiplication
T: torque at the wheels
I: total moment of inertia of the rotating parts
u: angular acceleration of the rotating parts
M: total mass of the vehicle
r: external radius of the tires
a: linear acceleration
D: total drag and rolling resistance

The first term, I*u, is the torque used in making the wheels rotate faster. The second, M*r*a, is the torque used to accelerate everything (wheels included) in the direction of movement. The third is the torque used to cancel the drag and rolling resistance.

Another way of writing the equation above is:

T/r = I*u/r + M*a + D

Now the terms are forces. The left side is the force available at the contact patch.

If the tires are not slipping, the angular acceleration and the linear acceleration are related in the following way:

a = u*r

Replacing in the equation and moving things around, we get

T/r - D = (I/r^2 + M)*a

^: denotes exponentiation (r^2 means "r squared")

We can say that the left side is the force available for acceleration. Such force accelerates an "non-rotating equivalent mass" E,

E = I/r^2 + M

Now suppose that we increase the mass of the wheels and tires by an amount X. Both the total mass and the moment of inertia will increase; let's call the new mass M' and the new moment of inertia I'. Obviously,

M' = M + X

What about I'? Well, the moment of inertia of a "punctual mass" [m] (a mass concentrated in a point) at a distance [r] from the axis of rotation is [m*r^2]. That is,
the moment of inertia depends critically on the distance between the mass and the axis of rotation.

In a real wheel+tire combination the mass is distributed in different amounts at different
distances from the center. In order to compute the total moment of inertia we would need to know the mass distribution and use integral calculus. We can do a simpler thing though. We can make the pessimistic assumption that all of the mass increment is
located on the periphery of the tire, that is, at a distance [r] from the center. This assumption is pessimistic because in a real wheel some of the mass will be located closer that [r] and will contribute less to the total momentum (it is not too pessimistic though: most of the mass is located pretty far from the center, if not at the periphery). So now we can compute I',

I' = I + X*r^2

The new "equivalent mass" is,

E' = I'/r^2 + M' = I/r^2 + M + 2*X = E + 2*X

In other words, from the acceleration point of view, the equivalent non-rotating mass increment corresponding to an increment X in rotating mass is - at worst - 2X.

NOTE: After doing some approximations and assumptions about mass distribution in a typical wheel+tire combo, I believe that 1.7X is a better approximation. A 10lb/wheel mass increase would not hurt acceleration worse than carrying RinTinTin.

http://www.audiworld.com/tech/wheel13.shtml