weight vs. hp

[Buzz1167]

Ok, I dont know about all the middle discussion, so this could have been pointed out. But I have the official "physics way" to describe what your asking. F = M * A or Force = mass * acceleration or F/M = acceleration. Acceleration is what you want to find out in order to find out how racable your car is. This equation results in N/Kg, or M/(S^2). Or in SAE units, HP/Lbs. So if you divide the HP by the weight, (assuming the same units) you get its power to weight ratio, witch is the most important relationship for racing on the street.
However, if you plan to get up to 100+ mph and run someone down, there is much more inportant equation, which is based from the first one but differently applied. Since the velocity has a limit of some number x, the acceleration has a limit of 0, thus as A -> 0 the equation becomes F - F(f) = 0. Where the F is the power you have, and the F(f) is the summation of forces acting against you, such as drag and friction. So given a praticular horsepower and drivetrain 2 cars will top out at the same speed only if their drag/friction coeficients are the same, thus even if someone took a ford windstar and got it down to 2000lbs, it would still loose !!!in the long run!!! vs a full interior windstar with some bolt ons. Mass is not a function of top speed.

HTH
Buzz1167
Jon N

[Ricochet]

my friggin brain hurts now

[Buzz1167]

Just It'll stop soon...

Buzz1167
Jon N

[Milliardo]

Originally Posted by Ricochet

my friggin brain hurts now

That's why I gave up on math a long time ago.

[Prelewd]

This comparison has been thrown around quite a bit, but i've heard 100lbs ~ 10HP.

Probably a ton of other determining factors though.

[crxlvr]

Originally Posted by Prelewd

This comparison has been thrown around quite a bit, but i've heard 100lbs ~ 10HP.

Probably a ton of other determining factors though.

100lbs is the same as 10hp, is completly wrong.

look simply go by this method. Get your car weighed(lbs), dont go by the weight in the door jam, it is not correct. then get your car dynoed(hp)

example my crx is roughly 2200lbs with me and all my stuff, and with my engine should produce roughly 175-180hp.

that means my car moves 11.43-11.11LBS for each hp it produces. the lower i can make the weight, even with the same amount of power the quicker i become. so lets say a corvette moves 20lbs per HP it makes(in theory) if i can get my crx to move 20lbs for its 180hp compared to the near 350-405hp the vette makes, i will be going much faster than the vette.

[Prelewd]

Originally Posted by crxlvr

100lbs is the same as 10hp, is completly wrong.

look simply go by this method. Get your car weighed(lbs), dont go by the weight in the door jam, it is not correct. then get your car dynoed(hp)

example my crx is roughly 2200lbs with me and all my stuff, and with my engine should produce roughly 175-180hp.

that means my car moves 11.43-11.11LBS for each hp it produces. the lower i can make the weight, even with the same amount of power the quicker i become. so lets say a corvette moves 20lbs per HP it makes(in theory) if i can get my crx to move 20lbs for its 180hp compared to the near 350-405hp the vette makes, i will be going much faster than the vette.

like i said man, hearsay.. i think you misunderstood, or i didn't explain clearly enough. you remove 100lbs from your car and it's the same as adding 10HP. this has it's boundaries, and will differ on every car, but it's just a basic comparison that's i've heard.

[Ricochet]

I heard for every 100lbs you remove it's .1 seconds faster in the 1/4 mile..

[Buzz1167]

Ok I'll fix you all up. First Ill start off with some conversions. 1mi = 1.609km: 1mph = .447m/s: 1kg = 2.2lbs: 1lb = 4.448N: 1m = 3.281ft: 1Hp = .746kw.

Were going to use some arbitrary variables here, such as weight = M, Accelleration = A, Distance = S, Speed(Velocity) = V, Time = T.

1/4Mile = .40225km or 402.25m
2200lbs = 1000kg

S = .5*A*T^2
402.25m = .5*A*T^2
We want T, but we need A to do it.

So F = M * A.
145ft-lbs = 196.47 J / T (Watts)

Were also going to make a bad assumption here that the force is constant (becuase it isn't) but if we dont assume that it is, the math gets even worse.

196.47 W * T = 196.47 J
So put that back into the equation
196N-m(J) * T / (1000kg) = .196 * T = A
So now we sub back into the original equation
402.25m = .5 * (.196 * T) * T^2
402.25m / (.5 * .196) = T^3
CubRt(402.25m / (.5 * .196)) = T = 15.99 Sec

A More useful formula would be to combine all the steps and say this:

CubRt((LBS/2.2)*2*402.25)/(Hp * (1.355) *1.355)) = T
Or much more simply:
CubRt((LBS*199) / HP) = T

I find that the multiplier for Horsepower (with relation to torque) is about 1.355, So in order to use HP instead of torque you just tack on the extra 1.355, So the torque term becomes 1.355 * HP and I find that the HP calculations are much more accurate becuase of the RPM ranges you are in when you drag. The torque formula is off becuase HP is what really drives most of the drag race and becuase the number is almost always lower.
But anyway, if you didnt understand my last post, just take this one for granted... and use the formula.

Based on my formula Crxlvr should run about 13.6-14 on the track Assuming you can get fairly good traction.

Does anyone have a car track time they could test that on? (use the HP version)

Im Sure I wrote something down wrong, so If I did, and anyone gets through it, please correct me.


Edit: Times with different Weights
160hp constant
CubRt(199*2200lbs / 160hp) = 13.98sec
CubRt(199*2300lbs / 160hp) = 14.19sec
CubRt(199*2100lbs / 160hp) = 13.77sec

250hp constant
CubRt(199*2200lbs / 250hp) = 12.05sec
CubRt(199*2300lbs / 250hp) = 12.23sec
CubRt(199*2100lbs / 250hp) = 11.86sec

Average change? About .2sec per 100lbs, of course times like .2sec are also dependant on how consistant the driver is... Less weight also = less traction, so a little less than .2 probably in reality.

CubRt(199*3300bs / 133hp) = 17sec (My crv in its stock form, ouch)


HTH
Buzz1167
Jon N

[454Casull]

Speed vs HP and weight is not a very accurate relationship, because two weights can be the same numerically, but one can have slower acceleration than other. Imagine two exact cars with everything the same. Now, drop 100lbs in the backseat of one car. Now replace the rims of the second car with, say, depleted uranium so that each rim now weighs an additional 25lbs. This would mean an additional total weight of 100lbs. But the second car will be slower.

I'll leave it up to you to figure out why.

[crxlvr]

Buzz, i ran my best of 3 times in my accord at a mind blowing 16.8 seconds, my car has 160hp and 160lb/ft, not sure on the exact weight but i believe it to be in the range of 2800lbs with me and everything else in it.

i couldnt really follow all yuor math, but if you plug it in, does it make sense?>

[Buzz1167]

Geez, each wheel is another 25lbs? Your like adding 100lbs to the total rotational mass, of corse its gonna be slower;
Lets look at that relationship.

***MATH WARNING***
We can assume for simplicity that the wheel is a solid rotiational cylinder, even though it isn't, we'll approximate. A solid disc's moment of intertia is .5Mr^2, and a hollow disk is just Mr^2. Well say that since its neither (it has some pices in the middle, but most of its on the outer rim. Well approximate 4/5Mr^2. So if you double the mass, (25lbs is probably double for most wheels, maybe even more for some light ones.) the I is now equal to 8/5Mr^2, or effectivly doulbing its resistance to motion, keep in mind that there are 4 wheels and your doubling them all, and youve got a pretty substantial loss in time. However, lets say that you get smaller wheels (including smaller tires - same aspect ratio), and you decrease the R from 17in/2 to say 15in/2 thats a 1" decrease in R and a small decrease in weight. So lets compare that where M is about 18lbs and R goes from 8.5 to 7.5.
4/5*18*8.5^2 to 4/5*17*7.5^2. 1040.4 to 765. Which is a 26% decrease in I, remembering that I is enhanced by 4 b/c of the whole 4 wheel situation.
Thats doesn't translate into having a 26% decrease in time, by any means, but it does mean that it takes less horsepower to accelerate the wheels, and gets you Some time at the track.
The same with the weight situation, if you add 100lbs to the back, you arent gaining much, becuase now you have increased the weight to accelerate, but (in a FWD car) you haven't really added much traction, however, if you add the same 100lbs in the engine compartment (and not to a rotating mass) you will get a higher Frictional force between the tires and the road, resulting in better launching; but the results in the long run could be either good or bad, depending on how much that 100lbs is in relation to the car.
***MATH WARNING OVER***

You have to assume some things constant or the equation will be hella big, and that is one of them, the fact that most wheels (on the cars were talking about) are approximatly the same diameter and weight. I know that if you decrease unsprung weight that its better for your time, than decreasing the same amount of weight, say in the trunk, but this is for "normal cars" and is only an appoximation, If you have a better formula, please inform me.
Crxlvr, What car are you running that has 160ft-lbs and 160hp? Thats like really odd, usually most import cars have less torque than they do hp. So if you take out the hp converter (that I threw in at the end to approximate torque) and put in your real torque (becuase its the same at hp), I get 16.78sec. However, I think your engine has some potential that your not using.

In your 175hp crx, thats about 2200lbs, what are you running? like low 14's?

Mathily yours,
Buzz1167
Jon N

[Prelewd]

The H23 has similar torque/hp, but you are right Buzz, it's uncommon of 4 cyl. engines

[crxlvr]

i havent run the car yet, as i havent finished rebuilding the motor, so when i get it down and at a track ill let you know. but i ran my accord at 16.8sec and approx weighs about 2800lbs.

[Buzz1167]

Actually its .02 seconds off. 16.78 Me thinks (using the torque numbers)


I think its really cool that with like 2 equations you can guess (pretty well) what a car is going to run, unless of course you go crazy on the mods, like 4' wide slicks and 1000hp.

It also depends on where you are racing, a hot day and a cold day can change your times, also getting lucky with the launch and screwing one up is porbably the biggest issue. But Im suprised that the HP calcuation (by itself) is far off, Do you have an automatic accord?

Buzz1167
Jon N