Fatman0203

For every lets say 100 lbs off your car how much horsepower could you say you added? I was thinking every 100 lbs = 5 or 10 hp? Maybe if that, is there a formula for this?

RotaryAXer

iTrader: (4)

Well Force = Mass * acceleration.

Lets say you can accelerate at 0.7g. If you drop 100 lbs and keep the same power...(relates to force) you can accelerate at a higher rate.

So:

Force/mass = 0.7g

Force/(mass-100lb) = X g.

This is not giving you an increase in HP. It increases your acceleration. You could relate this so that to have the same acceleration with an increase in power instead of a decrease in weight.

Basically:

X g * mass = Force+Y (higher force)

To relate force to power isn't direct. It depends on gears, tire diameter and rpms you are at. RPMs change so it is only possible to get a single number at a single rpm.

Fatman0203

Ok im lost, what can I measure with I lose weight? That I can measure on a consistant basis? My acceleration g's?
Also if I plug everything in, Im still lost help me out please.

DamonB

Yes, you can measure acceleration g's. It's like adding a lightweight flywheel to an engine: It can accelerate faster because there is less mass but the actual power of the engine is still the same. By making the chassis lighter the engine can accelerate the car more quickly since it doesn't have to use as much of its power fighting inertia.

Fatman0203

I understand that, now how can I measure this in a semi simple way? I want to see what weight reduction does, say 100 lbs? Say 200 lbs?

RotaryAXer

iTrader: (4)

Yes you measure acceleration. That is really all you will be effecting by recucing weight.

Lets give it some numbers.

Force~1000lbf
Mass=2500lbm

a1 = Force/mass => a = 0.4 g

Now

a2 = Force/(mass-100 lbm) => a = 0.4167 g

Now

Force + Y = a1 * (mass) => Y = 41.75 lbf

To relate that force to engine power is less exact than relating to torque. We will relate it to torque.

Tire Dia = 22.5 in
RearGear = 4.1:1
2nd Gear = 2.1:1

IncreasedTorqueAtEngine = Y*(TireDia/2)/4.1/2.1 = 5.17 lb*ft

So effectively you have increased the torque by 5 lb*ft.

Fatman0203

How do you know what your force is? Also would this line be consistant or as I get lighter it would start being 6 or maybe 7 pounds.

DamonB

You're just not going to get a nice neat number like you are trying to.

Force=Mass times Acceleration.

Mass is constant. To compute acceleration all you need to know is the rate in change of speed over time. This is not going to spit out a nice number you can equate to horsepower as your rate of acceleration at any given time is related to how slow/fast you are already going. Therefore that number is constantly changing.

The only blanket statement that can be made is that a reduction in weight will always bring about an increase in acceleration. You can't directly equate that to a power number though. You'd have to compute it at every speed and every interval and so the power number would be all over the place.

Here's the formulas you need to know. If you don't know algebra you're screwed:

s= displacement (distance)
vf= final velocity (final speed)
vi= initial velocity (speed you started at. From rest= zero)
a= rate of acceleration
t= time

s= vi(t) + .5a(t)^2

vf= vi^2 + 2a(s)

vf= vi + a(t)

You can compute anything you want to know from those formulas. I used to spend a lot of time doing so...

Umrswimr

"Generally" speaking, drag racers equate 100 lbs to 0.1s or 10HP. Of course, this is EXTREMELY general as Damon has explained above, but it seems to bear out for most reasonable circumstances. Obviously adding 3000 HP will not make your car 30 seconds faster.

Take 100 lbs out of the car, and it will run the 1/4 mile 0.1s faster.

Fatman0203

So I guess its just easier to do power to weight ratio huh?

Umrswimr

Basically, yes. Use the forumulas Damon listed above and you can solve just about any Newtonian physics problem...

What ARE you trying to do, anyway?

edmcguirk

Right, what they are saying is that 100LBS = 10HP in a 1/4 mile drag race.

If you change the distance or type of racing you get completely different numbers.

ed

adam c

iTrader: (2)

If you are looking for something really simple, that won't be exact, you can use a simple general equation.

The FD weighs around 2800 lbs. If you remove 100 lbs, that is roughly 3.5% of the weight, so your car may accelerate roughly 3.5% faster. If you want to equate that to a HP gain, you could say that it would take 3.5% less HP to accelerate the car (as quickly as before) at the new weight.

I know that isn't perfect, but I think it answers his question.

If you truly are, as your name implies a "fatman", losing some of your weight will have the same effect as removing it from the car.

DamonB

Not any more accurate than anything else. Doesn't take into account initial speed, momentum or drag.

All cars no matter what the power have the highest rate of acceleration at slower speeds. Pick any gear you want (keeping the car within its power band obviously) and accelerate from 30-50 mph and then do the same from 60-80 mph. You'll always do the 30-50 quicker even though the change in speed is 20 mph in both cases.

adam c

iTrader: (2)

Certainly not more accurate than anything else. Just simpler, which I think is what he wanted