I've been building hot rods for going on 20 years, and in conversation with someone I had mentioned that what I had been taught through reading and speaking with other people was that for every pound of rotational weight that you could save on brakes & wheels it was equal to 10#'s of sprung weight on the chassis. But he says that the conversion is more like 4:1, does anyone here know the correct answer?

I've searched all over the data base, google & other sites & can't find a concrete answer on what the conversion factor is with unsprung / rotational weight vs sprung weight?


Help!!!!!!!!!!

I am confused what the 10:1 or 4:1 benefit ratio refers to. As far as acceleration is concerned, it doesn't matter where the mass is, a # is a #, sprung or un-sprung. Weigh the entire car and apply the equation a=F/m to the whole mess. Rotating weight reduction is good since it reduces the inertia that must be overcome to get it started rotating but dumping weight in general is much more important.

Unsprung weight reduction becomes important in obtaining a responsive suspension that can be made to stick to the road. Big traction and handling implications since a lighter suspended weight responds much faster than a heavy one thus the suspension can conform to changing terrain quickly and maintain contact with the road. Don't know how a simple ratio can be associated with that affect though. Here is a web site that explains the concepts better than I did.

This is rotational weight as applied to acceleration, like going to light weight wheels & brakes. Everything that I've found on the net was about un-sprung weight as it pertains to handling and not acceleration.


All the old hot rod magazines that I use to read according to my memory always stated that for every pound that you could save on un-sprung rotational weight such as wheels was the equivalent of saving 10#'s of sprung weight on the chassis.

The only thing that I could find that showed any acceleration gains due to light wheels was

this

Like I said above, the inertial effect of rotating mass is not that much of a problem. 2:1 is more like it if you want to put it in ratio terms. Here is a site that explains the physics.

The easiest way to handle a rotating inertia, when making performance calculations, is to convert it to an "equivalent" weight. In other words, you can think of this as an additional weight, tossed into the trunk, which would have the equivalent effect of increasing the rotational speed of an item as the car accelerates. This would be in addition to the static weight of the car.

The equivalent weight of a wheel and tire assembly would be equal to the rotating inertia of that assembly divided by the square of the effective radius of the tire. (The effective radius is the distance up from the track surface to the axle centerline as the car launches.)

The equivalent weight of a driveshaft would be the driveshaft inertia times the square of the axle ratio and divided by the square of the effective tire radius.

The equivalent weight of a flywheel would be the flywheel rotating inertia times the square of the transmission ratio and times the square of the axle ratio and divided by the square of the effective tire radius. Note that a reduction in flywheel inertia has more of an effect in first gear than in second.

It is important that the inertia be expressed in units compatible with the calculation. For instance, if the inertia is expressed in pounds mass inches squared, the calculation will yield an answer in pounds mass, which is what you want.

This doesn't have much to do with "sprung mass." The sprung mass is normally considered to be that portion of the car supported by the suspension. The unsprung mass would then be the rest, including suspension linkage, wheels, tires, axle, etc.

Just looked at the site you referenced, Willys, and there's nothing wrong with it, of course, but I just wanted to point out that the "equivalent mass" to which Mr. Martinez makes reference is not the same as the "equivalent weight" which I used in my post.

First, though your high school physics teacher would object, let's use "mass" and "weight" interchangeably. A one pound mass will have a weight of one pound while on the surface of the Earth, which is where most of us plan on driving our cars. So, as far as the numbers are concerned for us, it doesn't make any difference.

In the automotive industry, the "equivalent mass" refers to that additional car mass which would have the same effect on acceleration performance as the rotating inertia of a component that is accelerated, rotationally, as the car accelerates. This is what I defined with the equations in my post.

So, when Mr. Martinez describes an additional weight added to the tire periphery, the incremental increase in inertia would be equal to the product of that weight and the square of the tire radius. To get the equivalent weight I defined, you divide by the square of the tire radius and you're right back to the value of the incremental weight.

Mr. Martinez, however, lumps this back with the incremental change in static weight (what you see on the wheel scales) and calls this the "equivalent weight." This, of course, is double the static weight increment, but it is also double that value which the automotive inductry commonly calls the "equivalent mass."

Not faulting Mr. Martinez' work, understand. He's free to define terms as he wishes. But, if you do further reading, I believe you're far more likely to see "equivalent mass" considered apart from any changes in static weight.

Again, note the effects of gearing on the equivalent weight. While changes in wheel inertia have very little effect, the same cannot be said for changes in flywheel inertia. Usually, inertia reduction achieved by the machining of a steel flywheel can be easily calculated, using equations found in handbooks or physics texts. Manufacturers of aluminum flywheels will usually supply you with inertia values for their products. It is then possible to calculate the full benefits of the change.

Well from talking to some of my engineer friends, we've come to the decision that you can't put a good static number on it.

As far as rotational mass goes we've figured that it really depends on where on the wheel the weight is placed. If the largest portion of wheel weight is located at the center of the wheel it will accelerate faster than a wheel that has most of it's weight located on the outer edge of the wheel.

I appreciate all the help, but there is just no static number that you can apply to this, because each different type of wheel has it's mass concentrated in different places. Each different type of rotational component has it's own mass located in different places. So basically we can all spit out all kinds of numbers all day and each can be right and wrong at the same time.

All I know is that my car is a couple of tenth's faster on the drag strip when I have the Pro-Stars with skinnies on the front and slicks on the back than when I'm running forged 17" x 11" rear wheels with 315 Drag radials on the back & 17" x 9" wheels in the front with the 275's.

this is why i wish i didnt have the tendancy to fall asleep in math class in high school... however. i love these threads... i wish my teacher could have showed us how stuff related to REAL life problems....

I am not surprised that Jamnut gains a couple of tenths with drag slicks and skinny tires up front over a street tire setup..has to do with difference in weight of a rotating mass...that and the slicks give a wee bit better bite off of the line..

Think of accelerating a heavy flywheel versus a lite flywheel..the lite flywheel takes less horsepower to accelerate versus a heavy flywheel so a lite tire and wheel takes less effort to accelerate than a heavy tire and wheel..

Now the issues of handling sprung and unsprung weight goes to suspension compliance and keeping the wheels on the ground so you can get the power down..and steer that thing in the direction you wish to go.

Lighter is better provided of course we can come up with a tire that has sufficient section width to handle the horsepower we are attempting to get to the ground..

There are formulas for calculating the power required to move a certain weight but what we need to know is just how these changes will affect what we are trying to accomplish..

Bottom line is that if we save weight in the rotating mass of the vehicle we will see performance gains..

__________________
I would rather make it work than make it popular..And if it does not work it will not be popular..

It is all relative depending on the entire combination.Lets say that car number one is a 6 second NMCA prostreeter,such a car has to accerate it's wheels and tires from a standing start to about 1800rpm in about 6.8 seconds.Now lets look at car number 2,which is a 14 second cruiser.It has to accelerate it's wheels and tires from a standing start to about 1100 rpm and has 14 seconds to do it.As you can see,taking 1 pound of weight off the wheels of a prosteeter would have much more of a dramatic effect.Both cars could weigh 3000#,but the effects would be much different.
Now lets look at driveline and transmission components located after the clutch and torque convertor.Lets use 2 new cars for this one.Car number 1 is a 10 second bigblock camaro,it weighs 2900#,and has a big old 468 under the hood.That car has to accerate it's driveshaft from a standing start to about 6500rpm in about 10 seconds.Car number 2 is a Toyota Supra with a tricked out turbo 6 banger.I also runs 10's.I has to accerate it's driveshaft from a standing start to about 8500rpm in 10 seconds.As you can see the toyota would gain more from removing driveshaft weight then the Camaro would despite the fact that they are both 2900# cars running 10's with about 600horsepower.
Now lets look at 2 more cars,and evaluate the effects of changes in engine rotation and flywheel weights.The first one is a 10 second 2900# camaro with a bigblock chevy and a powerglide.The second car is identical in every way,but it has a 5 speed stick shift trans.The powerglide car has to accelerate the engine from about 5000rpm's to 6500rpm's twice.The first time it does it it has about 3.5 seconds to get the engine up to 6500rpm,the second time it does it,it has about 7 seconds to go from 5500 to 6500 rpm.Car number 2 has to accelerate the engine from 5000 rpm's to 6500 rpm's 6 times.The first time it is almost instantaious since the clutch is disengaged.The second time it takes about 1.5 seconds,the third time it takes about 2.5 seconds,and the 4th time about 3.5 seconds,and so on.In this cars,the stick shift car would benifit much more from reducing rotating weight then the 2 speed car would.Both cars are identical in every way except the number of gear ratios the transmissions have,yet they would respond totally different to a change in crankshaft and flywheel weight.
So now that I have you all confused,I hope I cleared it up for you

One has to compare apples to apples when evaluating a change to a car..

Example our old t-bucket..I changed from a steel wheel and a 9" wide tire to an aluminum wheel and a tire of a different brand still 9" wide and 30" tall..nothng else changed (LIGHTER TIRES AND WHEELS ONLY) and picked up about 0.50 seconds in the quarter with that car..

the car revved a lot quicker and I needed to change gears earlier..All that due to a tire and wheel change..

REDUCTION IN ROTATING WEIGHTS WILL BENEFIT PERFORMANCE OF A VEHICLE.. GOT IT..!!!

For linear, or translational, motion an object's resistance to a change in its state of motion is called its inertia and is measured in terms of its mass, in kg. When a rigid body is rotated, its resistance to a change in its state of rotation is called its rotational inertia, or moment of inertia. This resistance has a two-fold property. (1) The amount of mass present in the object and (2) the distribution of that mass about the chosen axis of rotation. In general, the formula for an object's moment of inertia is CM = kmr2 where k is a constant whose value varies from 0 to 1. Different positions of the axis result in different moments of inertia for the same object; the further the mass is distributed from the axis of rotation, the greater the value of its moment of inertia.

Go here for a discussion of the physics involved:

http://online.cctt.org/physicslab/c...larmomentum.asp

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I would rather make it work than make it popular..And if it does not work it will not be popular..

Good examples, Super. And, they all become obvious when you convert the rotating inertias to equivalent weights, as I described in the first of my two posts above. This is the technique employed by those who sit at their desks for 40 hours each week, making performance calculations for the industry.

Think of it this way would a concrete wheel that weighs 200 pounds takes more hp to turn
than a wheel weighing 22 lbs. I think you know the answer. Sean Hyland put big wheels nd brakes on an Acura and lost 12 hp and couldn't figure out why