Rotational Mass

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.

http://audiforums.com/forum/showthread.php?t=130606

I've heard that 10:1 bit of wisdom before, but I don't think it stands up to scrutiny. It is absolutely true that a pound saved in the wheel or tire is worth more than a pound saved somewhere else, but not a factor of 10 more. I did a "back of the envelope" calculation using basic physics principles, and what I find is that a 1 pound savings in wheel and/or tire weight is equaivalent to about a 1.5 pound savings in the non-rotating weight of the car. What I did was compare the power needed to accelerate a 1 pound mass in a straight line versus the power needed to increase the rotational veocity of a wheel with a 1 pound mass concentrated at the rim an equivalent amount.

Here's the anaysis. To keep the math simple, let's assume the following:

1. The diameter of the wheel is 18", or in other words the radius of the wheel � is 9 inches, or 3/4 ft.
2. The radius of the tire � is one foot.
3. All the weight savings for a 1 pound reduction in wheel weight is concentrated at the outer edge of the rim - that is, 3/4 ft from the center. This is a bit unrealistic, but is the most favorable assumption with respect to trying to show the advantage of reducing rotating mass.

The instantaneous power needed to linearly accelerate a given mass (m) is given by P(linear) = m*a*v, where a = the acceleration, v is the instantaneous velocity.

The power needed to increase the rotational velocity of a wheel is P(rotation) = I*alpha*omega, where I = rotational moment of inertia of the wheel & tire, alpha is the rotational acceleration (in radians per sec squared), and omega is the instantaneous rotational velocity (in radians/sec). Alpha is related to the linear acceleration of the car by a = alpha*R, and omega is related to the car's velocity by v = omega*R. So the power required for rotational acceleration can be expressed in terms of a and v by: P=Iav/R^3. As for the value of I: for a 1 pound mass concentrated at a distance r from the center of rotation, I = mr^2. Putting it together: P(rotation) = I*alpha*omega = mr^2av/R^3. Since r = 3/4 ft and R = 1 ft, that means you have P(rotation) = 9/16 * mav.

Note that P(rotation) is 9/16 of the amount of P(linear). Now if you can save that 1 pound of mass in the wheel you save on the power needed to both spin the wheel up to speed and to acelerate on down the road, so the total savings in power because of the lighter wheel is (1+9/16)*mav, or just more than one and a half times the power savings if the weight was some place other than the wheel. So while saving a pound in the wheel is definitely more advantageous than saving a pound of weight in the rest of the car, it's no more than about 50% better (not 900% better as claimed).

Of course this says nothing about the other improvements that come about from a ligher unsprung weight - namely faster damping of vibrations and less transfer of energy from the wheel into the car when you hit a bump.


I found some more information on wheel weights, rotational mass and weight reduction. This website has a spreadsheet with calculations that take EBAINES' formula (thanks again) for rotational mass and acceleration with tires using 3 calculations: one for uniform density (wheel and tire weight is uniform from center to edge of tire,) one for the wheel and tire weights separately, and one assuming more weight at the edge of the tire.

http://www.the-welters.com/racing/rotational.html

Long story short: Using the least advantageous calculation (Wheel and Tire calculated separately,) going from a 23lb. wheel to an 18lb. tire saves you an equivalent 50lbs. of chassis weight.

1 lb. (Wheel) = 2.5 lb. (Chassis)

So at the very least, going from a stock B5 Avus wheel (27lbs.) to a 17lb. wheel would be equivalent to shaving 100 lbs. from the chassis, or 0.1 sec from your 1/4 ET.

Finally, go here....
http://www.the-welters.com/racing/rotational.xls
http://www.csgnetwork.com/speedocalibcalc.html

http://www.w8ji.com/rotating_mass_acceleration.htm

Rotating Mass, Available Horsepower, and Acceleration


There are several common questions about rotating mass. Let's look at a few of them.

If I use a lighter crankshaft, how much power is gained?

If cluster gear weight of a transmission is reduced, how much will acceleration improve?

Does replacing a heavy stock steel flywheel with a light weight aluminum flywheel improve acceleration?

How much ET can be gained from using lighter wheels?

These questions can be answered if we know the weight change, the distance out from the center the weight change occurs at, the speed (RPM), and the time period over which the RPM change occurs.

What does a rotating mass actually do?
A rotating mass does not really consume energy. The mass just stores energy and eventually returns energy to the system or converts it to some other form of energy. The energy storage can be helpful, not do anything at all, or be harmful. With a little time and thought, we can understand how changes in rotating mass will affect available horsepower in a vehicle. Available horsepower in turn affects acceleration in a very predictable manner.

Four things determine the effect of rotating mass on our vehicles:

How quickly and often a rotating mass speeds up or slow down
How heavy the rotating mass is
The rotating weight's distance outwards from the centerline
How fast the weight spins
Here are the important things to worry about:

If we push energy into the rotating mass and pull energy out several times, obviously we move more power around than if we made a slow smooth change in speed.
The amount of weight is the least important thing! If we double the weight we only double the stored energy.
Distance weight is from the center line is very important, because it determines the weight's circular velocity (speed)! Stored energy goes up by the SQUARE of the radius change. If we replace a 4 inch diameter hollow driveshaft with an eight inch diameter tube of exactly the same weight, it has four times the stored energy!
The faster we spin the weight, the more energy it takes to move it and the more energy we must remove from that weight to slow it down. If we double the RPM, we multiply stored energy four times. Again it is a square of the change, just like number 1 was a square.
If we reduce mass from twenty pounds to ten pounds keeping the same distance out and same peak RPM, we reduce stored energy to half the original amount. Reducing weight is a one-for-one change.

If we reduce diameter by half while keeping the same weight and RPM, stored energy will be 1/4 the original stored energy. This change is a square. Twice is a "four times" effect. 2*2=4. Four times is a sixteen time effect on stored energy. 4*4=16

If we cut RPM in half, we would reduce stored energy to 1/4 the original amount. Once again this is a squared change. Change RPM three times, and the stored energy changes nine times. 3*3=9

We should carefully think about what this means when we change things. Some changes are worthwhile, some are not.

Wheel Changes

Let's assume, just for an example, all of the weight in a wheel is at the outer edge and remains at the outer edge. If we reduce a wheel's diameter but keep the overall weight the same, the wheel is a spinning ring with smaller diameter. The smaller diameter increases the wheel's RPM at the same vehicle speed. The smaller diameter also moves the spinning weight closer to the center.

Let's say we cut diameter in half. Now think about how fast the wheel spins. RPM will be twice what it was at the same speed. The half size diameter reduction spins the wheel twice as fast, and that would increase stored energy to four times the original amount if the weight was the same distance out.

The weight isn't the same distance out. The spinning weight is now half size. This 1/2 size reduction decreases stored energy by four times!

Because the same weight got closer to the center, but the increase in RPM increased stored energy, and payback is the same for both nothing changed.

In this example, we gained nothing at all with this change. We also lost nothing by the size change.

Lightening the tire or wheel would reduce stored energy, especially if the weight reduction was at the maximum distance out from the center. Here is an example where we want to make something as light as possible on the OUTER edge, not near the (wheel) center. Spending money on smaller or lighter rotors to save rotating weight is not a good use of money, because the rotating weight is close to the hub of the wheel. Unless the rotors are huge and we take weight out of the outermost edges of the rotors, things will not change much. (A light rotor is good for reducing un-sprung weight, and that helps keep our tire's in contact with the road. It also reduces vehicle weight. But this is a different problem. Here we are talking about rotation, not the bounce inertia or "dead weight".)

If we spent money on the same weight reduction in the wheel, reducing weight out a little further away from the center, we would do much better. We would be removing weight further out from the center, where it does the most good.

Which brings up an important point we almost never hear mentioned, a lower weight part might not be lighter at the outside edge. It might be lighter in the center, where the weight reduction doesn't mean much.

If we spent our money on a lighter tire we would be getting the very most return for the weight change. The tire's weight change is mostly outside between the rim edge and the tread area. We get maximum effect from the change!

Think about this carefully. If we buy a lighter tire, we know for sure the weight comes off the most critical area. If we buy a lighter rotor, it is close to the center and for the same weight change the return is much less.

The wheels also speed up and slow down gradually. With an 11-second car, we have 11 seconds to speed the wheel up. Most of the horsepower pushed into the wheel is pushed in near the end, when acceleration is least. Since we have more time to push the bigger amount of energy into the wheel, it takes less horsepower than we might expect. A little ways down, I'll show you how to determine the power if you know the speed, weight, and time.

Drive Shaft Example

Now let's think about a drive shaft. The driveshaft is a fairly thin hollow tube. Nearly all drive shaft weight is at the outside, since it is of course hollow. The shaft also turns at the same RPM no matter what the driveshaft diameter, because the RPM is set by the rear end ratio, tire diameter, and vehicle speed. If we make a driveshaft lighter and keep everything else the same, the vehicle acceleration change is most often insignificant.

Why insignificant in most cases?

In the first place, the drive shaft is small in diameter. With a small diameter, less energy is stored for a given weight. In the second place, a driveshaft is really not that heavy. A steel Mustang driveshaft weighs somewhere around 30 pounds, so we just can't take that much weight out.

Also, the driveshaft spins up gradually and smoothly over a long period of time. It accelerates fastest at slowest speeds, and that is when it needs the least energy to spin up. Because it has a long time to spin up, is a small diameter, and doesn't weigh much the driveshaft does not remove very much horsepower at any instant of time. Despite what we are told, a change in driveshaft weight has at best a very small effect on acceleration. Likely any change is immeasurable in a street/strip car.

Now a lighter shaft certainly can help in a very light vehicle or in a road race car where instant change in applied power is required, but it really won't change much in a 3000-pound 11-second car, except how fast dollars leave your wallet!

Crankshaft

A crankshaft is a bit worse than a drive shaft. A crankshaft accelerates and changes speeds in every gear, so it is constantly storing and returning energy to the system. In low gears it spins up pretty fast, spinning up from "launch" RPM to shift RPM. This spin up repeats at every shift. The crank also has to be heavy to support the pounding and tugging of the pistons and rods as they accelerate and decelerate, so we are dealing with some weight.

Fortunately the crank diameter is small. A 3 inch stroke requires only a 1.5 inch throw radius. Unless we make a huge change in OUTSIDE weight in the counterweights, in most engines making the crank lighter makes very little sense. The dumbest thing to do is hollow out the crankshaft center because it is the smallest rotating diameter area. Don't believe this? Download the following technical paper from the Scat Crankshaft website. Lightweight Crankshafts- Performance or Deception

Scat has it 100% correct. Many bench racers, and even some crankshaft manufacturers, exaggerate a good bit! They remove weight where it makes little difference in stored energy, but might make a difference in strength. Some transmission experts worry about the wrong thing also. If we worry about the outside edge weight of the largest-diameter fastest-spinning parts that speed up and slow down at every shift, we are worrying about the correct parts. If we worry about parts that speed up at the rate of the driveshaft, we would be wasting our efforts.

The purpose of the examples was to give you a feel for what to look at first. Any weight reduction is good for horsepower to weight ratio, but some weight reduction has a bigger payback. Things that change speed often, change speed rapidly, and/or are heavy at a large distance out from the center...make the most difference. Look there first.

The last "things" to worry about are small diameter "things" that change speed a smaller amount, change speed over a longer time, and change speed less often. They will have much less stored energy. If we want to reduce rotating mass we should look at the heaviest things that speed up and slow down most often, spin the fastest, and are large in diameter with most of the weight at the outside edge.

Flywheel

A flywheel can be fairly heavy, and the weight is a good distance out from the center. It spins at crankshaft speed, and it has to slow back down at every up-shift. The flywheel can affect acceleration, but it can affect it two different ways! In a light car with very fast 60-foot times, a lighter wheel can slightly improve 60-foot times. This is because the launch is often at full throttle, the car generally has a steep gear, and we want to plant the tires hard into the track without encouraging spin. The tires hook hard, and usually have a very soft sidewall that absorbs shock. We want the engine to quickly match the RPM needed to move the rear wheels, and not overpower the available traction. It is a wide open throttle high-RPM launch.

A typical street-strip car is different. Generally we can't launch at wide open throttle, the tires are stiffer walled, the suspension is heavier, and things just don't hit as hard. We actually want a heavy wheel (and a heavy crank) to smooth out the power. This lets us have a much more controlled launch, and smoothes out any sudden application of throttle. An aluminum wheel, especially when the car is severely traction limited and heavy, can really hurt 60-foot times. A light aluminum wheel not only makes a street car hard to drive, it hurts at the track. It is especially bad with a heavy street machine.

Why do things work this way?
First we have to understand what power and energy are, and what rotating mass does with that power or energy. Contrary to popular belief, rotating mass does not consume energy. A rotating (or moving) mass stores energy. This effect is very much the same as pouring energy in a bucket, much like charging a capacitor in an electronics circuit. Virtually all of the stored energy, except for that lost by conversion to heat, is still there and available to do work at some time in the future. That future where energy is returned might be milliseconds later and help us out, or it could be some considerable time later and waste energy. This is why time is very important.

One example of useful energy storage is the flywheel and crankshaft of a car. The force on the crankshaft is in pulses. A common four cycle V8 has four power cycles per crankshaft revolution, and there are 100 turns of the crank per second. At 6000 RPM an 8-cylinder 4-cycle has 400 power pulses per second. The flywheel (along with the harmonic dampener and weight of the rotating assembly) smoothes these pulses out by storing and releasing the pulsed energy from the explosions in the cylinders. The result is a smooth rotation that will not tear gears up, vibrate the car, or beat on bearings.

We should always remember rotation, or movement of a mass, does not actually destroy energy. If it did, the earth would have stopped spinning millions of years ago! The key to understanding how weight changes affect performance is to understand some very simple basic energy flow in the system.

Definitions:
Energy

Energy is the capacity of a physical system to perform work. Energy exists in many forms like heat, mechanical, electrical, and others. According to the law of conservation of energy, the total energy of a system remains constant. Energy may be transformed into another form, but it is constant within a system.
For example, we all know two pool balls eventually come to rest after colliding. They stop moving only because the applied energy (from moving the cue stick) is eventually converted to heat (from friction with air and the table) and sound (which is not very much of the energy loss). The ball movement along the table's felt surface and through the air transfers energy outside the two moving balls to the air and environment around the table and into the table itself. The temperature of the table and air rises ever so slightly, because the applied energy moves outside the system we "see"! Since the heat energy is spread all around in a very large area, we don't notice the temperature rise. We just notice the balls quickly quit moving.

Another example is our car's brakes. The energy stored in the moving weight of the car is converted to heat by friction of brake pads rubbing against metal rotors attached to the rotating wheels. This converts stored energy (the engine put into the weight of the vehicle) into heat, and the heat (containing all of that energy) radiates out into the air. Most of what we actually do in a car is move heat around.

Newton's first law

A mass continues in its state of rest, or continues uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it.

Old guys like Newton sure had a lot of time on their hands to think about simple things, but they got it right. A rocket coasting through outer space is a good example. It will go on forever in a straight line unless it hits something, or unless gravity or some other force pulls it in a new direction. The earth wants to move in a straight line, except gravitational attraction to the sun bends its path constantly. A bullet reacts the same way, except friction with air and gravity changes the direction and speed gradually over distance.


Newton's second law

The acceleration produced by a particular force acting on a body is directly proportional to the magnitude of the force and inversely proportional to the mass of the body.

We push harder and/or longer, and something moves faster. If it is heavier, we need to push longer or harder (or both) to obtain the same speed. It takes more energy to accelerate a heavier object to the same speed as we might move a lighter object to that same speed. We can either apply more force or apply the same force over a longer time to make something move faster. It is all about TIME times the FORCE, or the amount of TIME that FORCE is applied. This is why those big showoffs can eventually move a large boat, a railroad car, or an airplane. All it takes is low friction and enough time and someone who can't move a Volkswagen with two flat tires can roll a 10-ton railroad car.



Acceleration, Energy, and Power

Acceleration, by definition, is a change in direction or speed. If we slow something down it is acceleration, just in a negative direction. If we turn a vehicle or any other mass in a new direction, it is really acceleration at a new angle or in a new direction. This is why we can compare or define braking and cornering in G-force (g's), just as we do with "taking off" acceleration.
We apply force (and this means we apply energy) over time to accelerate an object. If we want to spin a top, we apply force off-center from the axis and at right angles to the axis. The top stores the energy we apply, and continues to rotate. Over time the stored force is converted to heat from friction and the top gradually slows until it finally stops.

Force is pressure or power. The product of the time we apply the force and the amount of force is the energy. Power over time is a very useful measure of energy. Power alone is not. Let me give some examples:

"Watts" are a measure of power, much like horsepower. "Watts" alone are not energy, because a watt does not include time. A watt is only power level.

If we include one hour's time we would have a watt-hour. Kilowatt-hours, watt-seconds, watt-hours, and other combinations of power level and time define electrical energy or work. This is why we billed for kilowatt-hours at our homes! If we were billed for plain old "watts", it would not tell anyone how much "work" we bought. Watts is a true scalar (single dimension) measure, just as horsepower is. Both indicate a force or the ability to do work, but both lack any inclusion of time so we have no idea how much work was done, or could be done.

Horsepower is a function of RPM and Torque, just like watts are volts times amperes. Horsepower is an ability to do work, but doing actual work requires time. Torque is pressure, and since it does not include speed it is not a very useful measure of system power or the ability to accelerate or move weight. Despite what we hear, crankshaft torque is not directly related to moving something off the line or pulling a heavy load. Up at the engine, it is really all about horsepower. The horsepower (torque at a certain RPM) is eventually converted through gears and other mechanical devices to a new torque value at a different RPM. Eventually all we care about is the rotational pressure on the contact patch of our tires that thrusts our car forward. A 800 lb/ft torque at 2000 RPM engine does not accelerate a vehicle as well as a 400 lb/ft engine at 5000 RPM, because horsepower is a product of torque and RPM. The higher RPM engine can be geared to provide more forward pressure at the wheels, in short the higher RPM engine in this example has more horsepower.

If you notice, ET calculators don't ask for torque. This is because torque does not quantify the ability to do work. ET calculators ask for horsepower, because horsepower clearly defines an ability to do work.

Joules are another common measure of work. A joule includes both time and force (pressure). A single joule is one watt-second, or the equivalent of one watt applied for one second. A single joule could be 10 watts applied for 1/10th of a second (10*1/10 = 1), the product of time and force only has to be ONE watt-second to make one joule. If we applied TWO watts for 1/2 second, we have the same work. Two watts for 1/2 second is one joule (2*1/2=1).

Horsepower can also be stated in kilowatts. One horsepower is approximately 0.7457 kilowatts, or 745.7 watts (the exact value is 0.745699872 kilowatts). This means 746 watts for one second is 746 joules and that is one horsepower-second! One kilowatt is 1.341 horsepower.

Many European engines are rated in kilowatts instead of horsepower, you've probably seen that. A 300-horsepower engine would be about 223.7 kilowatts. Your house probably consumes between 2 to 5 kilowatts of average power, depending on how large it is and how you heat or cool. This is somewhere between 2-1/2 to 7 horsepower of average power. Think of what would happen to the power grid if we converted all our cars and trucks, like the Greenies want, to run on electricity! We would run out of electricity very quickly.

How many joules are in 1492 watts when applied for 1/2 second? 1/2 times 1492 or 746 joules! 746 joules is one horsepower-second. We could rate our engines in joules if we needed to include both power and time.

Horsepower and Acceleration
We know horsepower alone is not a measure of work, so we must know the time a certain horsepower is applied (or removed) to know how it affects acceleration. Fortunately there are horsepower calculators that predict ET for a given power. These calculators work because they know the distance, they know the applied horsepower (they assume it is constant), and from that they can calculate speed and elapsed time. They do this because they assume the power is applied constantly and they calculate the speed change over time. From the speed and time, they get the distance. When they see 1/4 mile (or 1/8th mile) they stop calculating and display the speed and the time taken to reach that speed and distance.

Now here is an interesting thing. It takes a certain number of horsepower-seconds (certain energy applied) to reach a certain speed for a given weight. If we make the vehicle twice as heavy, it takes twice as many horsepower-seconds (twice as much energy) to go the very same speed.

For example, go to this link:

http://www.race-cars.net/calculators/et_calculator.html

Now let's apply 100 HP to go 1/4 mile in a 1000 pound vehicle. We went 108.6 MPH in 12.55 seconds. Now let's say we have a 2000 pound car. To have the same speed and time, we have to also double the applied force. If we apply 200 HP in our 2000 pound car we have exactly the same ET and MPH! Now we know why insurance companies, in the late 60's, often limited insurance to a car with 10:1 weight to horsepower ratio or more. They didn't care if it was a 4,400 pound Super Bee Dodge with a 425 HP hemi or a 315 HP 3200 pound Hurst Rambler Scrambler, the insurance companies wanted weight to power over 10:1 ratio or you could not buy insurance. 10:1 weight-horsepower is at very best a 108.6 MPH at 12.55 seconds car! My American Motors 10:1 Weight-HP Hurst S/C Rambler, as a documented fact, set a new national ET record of 12.54 seconds in the 1/4 mile back around 1970.

Rotating Mass

Let's say we want to change the drive shaft rotating mass to improve power available to the rear wheels. We all know most of the weight in a driveshaft is at the outer edge. It is a hollow tube. Let's say the original shaft weighed 30 pounds, and we want to change it to a 15 pound aluminum shaft. The drive shaft is 3.5 inches in diameter.

We can go to another calculator to find the joules stored in the driveshaft! When we know the joules, we know the horsepower-seconds sapped from moving the car. Let's say the engine peaks at 6000 RPM at the end of the 1/4 mile, and that took 13 seconds.

Go to this calculator:

http://www.botlanta.org/converters/dale-calc/flywheel.html

The original driveshaft weighed 30 pounds and we had to spin it to 6000 RPM. If we input that, we see it consumed (and stored) 5310 joules. 480 ounces in a 3.5 inch diameter RING (hollow center) and 6000 RPM.

That is 5310/746 = 7.12 horsepower-seconds to spin the shaft to 6000. Since the time was 13 seconds, the shaft soaked up 0.548 horsepower distributed over that 13 seconds.

Now we change to the aluminum shaft. Everything is the same except the weight, it is now 15 pounds or 240 ounces. Using that flywheel calculator we find we used 2655 joules. This is 2655/746 = 3.56 horsepower-seconds. Over 13 seconds, we "stored" .274 horsepower. The net gain in available energy over 13 seconds was about 1/4 horsepower.

Here is the real rule of how this works....

If we are spinning up a very large diameter mass, or a very heavy mass, and we do it rapidly, we sacrifice a lot of available power. If we are spinning up a very small diameter mass, especially over a longer period of time, we give up less power at any instant.

The change from an aluminum flywheel to a steel flywheel is much more pronounced than the change of the same weight in a driveshaft because the aluminum wheel is much larger in diameter. We also speed and slow the flywheel as we accelerate and shift, instead of smoothly spinning the thing up like a driveshaft.

The truth is for drag racing, unless we have a God-awful fast car or a road race car where we have to instantly change power, an aluminum wheel barely makes a perceptible change over a steel flywheel. The aluminum wheel can actually be slower in a drag car, because the applied power is not as smooth. It is harder to get a light aluminum flywheel out of the hole, and that can easily offset any small "available power" change.

Summary
This is an approximation designed to give you a reasonable feel for how a change in rotating mass affects acceleration. We can see the power extracted to spin a weight up is not very much if we do not spin it up too quickly, or if what we spin is not very heavy and/or very large in diameter. The "feeling" most people cling to (and parrot) is that "heavier rotating mass kills acceleration". This is generally not true at all. Most things we fret about make no appreciable difference in the grand scheme of things. I would never bother changing from steel to an aluminum driveshaft in my car, because my car takes 11 seconds to go 1/4 mile. The car weighs 3000 pounds, and this means I might save 20 pounds of weight and 1/2 horsepower lost to spinning that weight over the length of the track. $400 is not a good investment at all for 1/2 horsepower over the length of the track, or the extra 1/2 horsepower applied for 11 seconds I have to extract at the end and convert back to heat with my brakes.

I don't really have to worry about how fast things spin up at this point. I don't care if the crank is 12 pounds lighter out of 50 pounds. I don't care if the driveshaft is 15 pounds lighter out of 30 pounds! Right now that $400 to $1000 would go a lot further if it made 20 more engine horsepower, or removed 60 pounds of static weight. When I start running out of easy power, then I will spend money making expensive things lighter. The big problem right now is traction, so right now I want to smooth the power out. The last thing I need is to make the car more critical for launch RPM by using a lighter flywheel or shock the tires more by using a lighter driveshaft. The first major weight reduction will be the front K members, because that would remove weight from the front and effectively add a larger percentage of weight to the rear wheels! The last weight reduction for my car will be an aluminum flywheel or driveshaft.

The best way to reduce rotational mass is by using 17" tyres. The lightest is OZ Allegerita 17x7x48. It weighs 6.25kg per piece. Coupled with CS3 205/50/17 93W XL (8kg per piece) and lightweight lugs, total weight per wheel will be no more than 33 pounds.

Thats a reduction of 16-17 pounds per wheel from stock 18x7x46 GT wheels and A10 215/45/18.

The total rotational mass reduced from this is 167kgs while not compromising handling capabilities.

Wow....that is a lot of difference. Around 30kg plus.

Yes, alot of difference considering 8kgs less per wheel (32kgs in total) and rotational mass of 167kgs. Totality = 199kgs. That is a 1 second off acceleration times 0-100kmh, 0-400m etc etc.

lightest 17 inc rim is SSR type C bro.. 5.8kg if nt mistaken..

bro, jason king oso got ssr type c.. tiawan saibon made

I thought RE30 is lightest...?

Wedsports carbon fibre is lightest... Ard 2kg... Price ard 20k

Elan, yes for 17", maybe that could be lighter.

Rev, yes for 18" it is the lightest, but when it comes to 17", it is not.