? weight reduction = ? horsepower gain

S

S2_ITBVW

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I have been digging around the web looking for "rule of thumb" estimates about the relationship between weight reduction and horsepower gain. I know that reducing dead weight doesn't increase horsepower, but there is less that the horsepower has to do. I have seen in a few places where folks have suggested that a reduction in "dead" weight of 10 lbs. creates a performance increase that is equal to 1 hp. I have also seen folks suggest that a reduction in rotating mass is equal to 4 times that of dead weight. So, for example, removing 2.5 lbs. of rotating mass would be equal to removing 10 lbs. of dead weight and would again equal about a 1 hp increase.

I know it has to be waaay more complicated than this. But, I'm wondering if these "rule of thumb" statements are in fact somewhat accurate.

Before any of you digital weenies pounce on me, I'm the furthest thing from an engineer (i.e., Psychologist, yikes!) so be gentle. I know not what I say . . .

Thanks!

Dave
 
Attorney here. I'll let the numbers guys pummel me later.

But, it's actually very easy to figure the net "gain" in IT from weight loss.

The target power to weight multiplier in ITB, your class it looks like, is 17 lbs per 1 hp.

So, to net an effective 1 hp vis a vis other cars in your class, you need to lose 17 lbs (but not lower than the minimum of couse!).
 
For all intents and purposes Jeff's correct on the weight with respect to your class. As to the rotating weight, well, someone has made an assumption that has averaged, or maybe overestimated, the radius on the rotating weight. But besides that, in the driveline of an IT car there are some things you can legally do to reduce rotating weight. You can surface the flywheel, get a light clutch and pressure plate, use a lighter ring / pinon (any ring gear and pinon is allowed), but that is about it. You can't go crazy and lighten the crank, rods, pistons, use an ally flywheel, and so on. Those little gains add up, but I'm not sure how you'd come up with a general rule:

1 lb of rotating weight = 4 lbs of static weight

Can't be correct. Think about it this way - with weight and balance concerns you want to reduce weight far from the center of mass, therefore, weight at the furthermost edge of the car would be worth more than weight around the C.G. How would that fact jive with the rotating mass statement?

Some rules of thumb are worth more than others.
 
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"where" the rotating mass is makes a big difference. physics class was a long time ago but it is a relatively simple equation to calculate the "work" or power being used to accelerate a rotating mass.

on a "typical" straight, what type of shifting are you doing vs. what type of acceleration?

for my ITB crx, i might get onto a long straight (mid-ohio or road atlanta) and accelerate from say 55 to 110 mph. and this would be while shifting from 3rd to 4th to 5th.

now the rotating mass on the wheels are only "accelerating" on a rotational basis once from 55 to 110 or doubling. but the rotating mass on the motor might be going from 4500 to 6500 rpm's three (3) different times.

i sort of think of this as a really good steak in a nice restaurant. i have paid nearly $50 for a 16 oz steak before, would i spend $50 to remove a pound of weight from the car? yes.

and although you can balance the flywheel, i would confirm the rules about surfacing it. look for lower weight pressure plates and clutches.
 
It's simple, each pound you loose off of one rim, you loose off the other three. So each pound you loose on a rim is equal to taking four pounds off the car. That said, it depends on where that wheel weight is taken off as to how it reduces the polar moment of inertia.

As an aside Ron, lots of modern cars come with non-surfacable dual mass flywheels, so that method of reducing rotating weight isn't always an option. When these wear out, you have to buy a new one at about $800 a pop or more.
 
Don;t forget the effect of weight on braking and cornering changes as a function of the square of the speed.
 
JS154 said:
Don;t forget the effect of weight on braking and cornering changes as a function of the square of the speed.
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If the weight lost is in the rim barrel then, yeah it really helps that. But if the weight lost is in the rim spokes, as is often the case, or even worse in the hub section, then there's not much gained by a lighter rim. In those cases it's the simple fact that 3lbs lost on one wheel is multiplied by four wheels for a 12lb less weight on the car.
 
shwah said:
The light and under driven accessory pullies also reduce rotational inertia.
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That reminds me, we've got to make a water pump pulley for the stang to replace the 2.9 lb steel one. I'm sure we can do one up in ally at around 1.2 lbs or so.
 
Numbers junkies:

EK=1/2 I w^2 (Kinetic Energy = one half of the Moment of Inertia times the square of the rotational rate)

Rotational rate must be in Rad/sec, so take RPM times .1047

I= 1/2 m r^2 (Moment = one half the mass times the square of the radius)



While the calculations themselves are cumbersome, the point is simple. As stated above, the location of mass reduction (in relation to the center of rotation) is MUCH more important than the amount of mass removed. Also though, the speed at which the body rotates is extremely important as well.

That all being said, I like Jeff's original answer, because mass reduction is much more critical in a low-mass/low-power car. The only thing I'd add is that HP gain % could be calculated as the reciprocal of the reduced weight %.

A 100hp car at 1700lbs that drops 100lb- 1600 lb/1700 lb= 94%
Reciprocal of 94% is 106.25%
106.25% x 100hp = 106.25 hp
So you effectively gained an equivalent 6.25hp.

This of course is ONLY relative to acceleration in a straight line. The lighter car will be able to turn better and stop faster also, all other things being equal.

Yeah, I know. I type too much. Bad thing is that I talk even more...
 
plus one for a square being in there... something going 100mph has 4 times the kinetic energy of when it was going 50mph.

and the faster you go, the less you are fighting mass(yes, mass NOT weight) to accel and the more you are fighting air. (air resistance is exponential btw, worsening this effect) therefore the effect of lightning any one car will change track to track.

and as someone touched on, the effect of lightening is a percentage thing on the hp and weight sides of the equation. Real world example. When I took the clutch fan off my 100hp light rotating mass 12a powered Rx7 I felt the difference. When I took the flex fan off my 450hp heavier rotating mass SBC I couldn't tell the difference at all!
 
Okay, I'd say it depends on where that weight is reduced. It's not a simple answer IMO as a formula states it is. ESPECIALLY in Club Racing. Many "tests" are not from a driver consistent enough to really show the true data.
 
Thanks for all the good information. This is great stuff.

To summarize . . .

1.) The weight of rotating mass is not the only factor that is important, but where the mass is located is also important. For example, reducing the weight in the center of a wheel will not create as much gain as reducing the weight on the outside of the wheel. The “radius” of the mass is a big deal. So, the new 9.5lb Kosei wheels were a good idea . . .

2.) The elements related to the engine that accelerate and decelerate every time you clutch and shift make a big difference because the engine has to work to get them up to speed over and over and over again, even as you move through the gears when the car is accelerating in a straight line.

3.) Little gains are more important for a low powered ITB car. All other things being equal, if you add 1hp to a 100hp car that is, relatively speaking, a much bigger gain than adding 1hp to a 200hp car (1% vs. .5% respectively).

4.) The “rules of thumb” I listed aren’t that accurate. It helps to know the kinds of changes to make and their relative impact. But, I guess this reinforces the fact that the only way to really know what you have achieved is to head to the dyno.

Thanks again!
 
S2_ITBVW said:
But, I guess this reinforces the fact that the only way to really know what you have achieved is to head to the dyno.
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But the dyno only tells part of the story. It would show no change at all if the mass you removed is not rotating. And does nothing to measure braking and turning gains.
 
I also do not think the changes in weight and it's true impact for us is linear. I've often thought it would be interesting to see a real life graph of how the weight actually impacts various components of racing - acceleration, braking, cornering, tire wear over multiple length sessions, and so forth. Sounds like a good project for you Dave. LOL
 
gran racing said:
I also do not think the changes in weight and it's true impact for us is linear. I've often thought it would be interesting to see a real life graph of how the weight actually impacts various components of racing - acceleration, braking, cornering, tire wear over multiple length sessions, and so forth. Sounds like a good project for you Dave. LOL
Click to expand...

Well, F=ma, so any reduction in mass will require less force for the same acceleration, or will allow more acceleration for the same force.

If you make large changes in mass though, you'd certainly have to re-optimize your setup to take full advantage. Mass changes would affect weight transfer and roll couple pretty significantly, so things like ride height, alignment, and tire pressures would *possibly* need to be changed to extract the full benefit. While there may be a diminishing rate of return, I think that most of our minimum weights are such that we should be in the (mostly) linear region of the curve.

Bottom line- I can't thing of any real IT-compliant mass reduction that is a bad thing, so shed those unsightly pounds wherever and whenever possible.

I know that I've found a no-cost method to lose about 35 pounds off the car/driver combo in a way that makes my wife very happy- it didn't cost anything AND I'm in better shape than I have been in 15 years... But now I do have to re-set corner weights. :)
 
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