weight vs. hp

Is hp to weight a linear relationship or is it something else, the reason I ask is because on this "junkyard wars" this guys says that hp is the square of weight. So he says that if your car weighs twice as much as my car, it will need four times the horepower to keep up with it. Is this true?

dont know if that relation is correct.

but HP and weight are related in some ways. if your car has 100hp and weighs 1000lbs, and my car has 50hp and weighs 500lbs, we should have a dead even race, but if i had 100hp and weighed 500lbs, id beat you by a mile.

So then your saying that horepower to weight is a linear relationship?

If so then does anyone approx. know the ratio of horepower to weight? (how many lbs. = 1hp)

So then your saying that horepower to weight is a linear relationship?

If so then does anyone approx. know the ratio of horepower to weight? (how many lbs. = 1hp)

Damn good question! pretty interested myself

I am curious too. I remember hearing this discussed over at Honda Tech before...but I completely forget what the forumla was. I'd really like to know as well.

I know that the (horsepower/weight) gives you the power to weight ratio..but I'm not to sure what the squaring does for the forumla.

As for lbs = 1hp...I KNOW I've seen that in a post somewhere at Honda Tech...was a long time ago. I didn't answer the question, but I am searching now..

Edit: Just for anyone interested...a little page that discusses power-to-weight ratio's. It's off of "HowStuffWorks"..

http://auto.howstuffworks.com/horsepower3.htm

Just make a scatter plot using ms excel with the Weight as a function of Horsepower, F(x) => W = f(hp).

Put Horsepower in 1 column, weight in another column, the more cars you include in the listing, the better closer your equation will be.

make a scatter plot. see if it appears to be linearly or exponentially related. You can then plot a regression line, and determine the slope for the line. You'd get some type of equation, y=mx + b, or something else if it is exponentially or power related. I'm sure it's gotta be power related.

if you're too lazy to plot, look,

http://forrest.psych.unc.edu/research/vista-frames/help/lecturenotes/lecture12/regression.html

This

You see they plotted the weight of many cars with their respective horsepower. Their equation was Wt = .37 + .02*Hpwr
Their correlation was R = +.92, which is fairly good, i fyou're taken statistics.

But I think it would be biased because taking that this is all based on statistics, all cars are aerodynamically shaped differently, but this would be a rough estimate of what you are looking for.

I'll make a graph later today and see what equation I get.

To do it this way you would have to have all cars that were evenly matched right? Because you would have to somehow include how fast the car is. If you just plot horsepower vs. weight for different cars, they won't make a good line or exponential.

What would work would be to take a car with a certain horepower. Say 100.

Then on the graph, change the weight of the car from increasing to decreasing. And on the y axis, quarter mile time.
If the line is straight then it is a linear relationship, if it curves it is exponential.

I found a quarter mile time calculator so I am going to make a graph and figure it out.

To do it this way you would have to have all cars that were evenly matched right? Because you would have to somehow include how fast the car is. If you just plot horsepower vs. weight for different cars, they won't make a good line or exponential.


It would be impossible to have something be equally matched because even in the same exact make, model, trim, one 95 Civic EX will still be different than another 95 Civic EX due to minor differences such as weight, even off by a few ounces, or maybe the room temperature was hotter in the room when one car was built compared to another, maybe they had to scrape off more metal on one than the other to make things fit. The only thing we can do is make a rough estimate of the relationship we see between all these cars.

Testing every single car and logging all their information would be the most accurate, but for simplicity sake, we get the information from a few random sample of cars and take that as how the majority of the cars would behave. The sample that is taken has to be very random in order to account for all cars, like not neglecting, VWs, Kias, Hyundais, pickups, and suv's.

If you take the quarter-miles of all the cars, not all cars have the same gear ratios, so with gear ratios changing, that wouldn't be the best estimate either unless you disregard that difference. BTW, the equation they use for that quarter mile time calculator was derived by statistics too. :smile:

Drag would be another issue. for example racing a scion xA with an xB, they're both the same horsepower, same engine, probably a little different weight by about +100lbs, but one is box shaped and another is hatchback.

I think if you do the quarter-mile vs weight graph, that would be the best estimate. You can then calculate acceleration from the quarter-mile time and the 1/4 mile distance taking the (final velocity - initial velocity)/time.

Get actual quarter-mile times instead of the calculated ones because things perform less than how they should theoretically.

So how much hp per lb do I have in a 2300lb Civic with 200hp?

So how much hp per lb do I have in a 2300lb Civic with 200hp?

well thats just simple math 2300/200 = 11.5LB:1HP

With that math a 6000lb car with my same 200hp would have 30hp per pound and be faster...

you would have .086 hp per pound I think
so the 6000lb car would have .033 hp per pound if it had the same 200 hp
correct me if I am wrong but it seems to work

you would have .086 hp per pound I think
so the 6000lb car would have .033 hp per pound if it had the same 200 hp
correct me if I am wrong but it seems to work

yeah that works. being inversely proportional. now we gotta get something going with weight vs quarter-mile, or acceleration. I think it can be safe to assume the car with the highest hp/weight ratio would win the race assuming both drivers are equally good.

I plotted the 50 values taken from modernracer.com of weight vs the quarter mile times and got almost no correlation. It makes sense because some people like to put small engines into big cars.

I wouldn't mind the math; I never liked it. But then common sense is really all there is in determining power to weight ratio. An object which is heavier would need more power to move it compared to something much lighter. So, if 2 cars with the same hp, but one is heavier than the other, were to go together, the lighter one would go faster because less weight is being moved as compared to the heavier one. That is why Honda itself confirms that as far as power to weight ratio is concerned, the 1999 Honda City (a car which is only available here in Asia) can match the 1999 Honda Civic SiR, whose engine is a B16 as opposed to th D15 of the City. At least in the initial stage both cars would be evenly matched; the SiR would eventually push ahead in the upper gears (3rd to 5th) eventually though.

hahah just a few days ago in math, i was debating with the teacher when the hell id ever need to know in real life functions and y=mx+b....

Fate has a way of biting you in the ass i guess... :evillol:

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

my friggin brain hurts now

Just :banghead: It'll stop soon... :biggrin:

Buzz1167
Jon N

my friggin brain hurts now

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

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

Probably a ton of other determining factors though.

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.

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.

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

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

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.

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?>

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. :nono:

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

Mathily yours,
Buzz1167
Jon N

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

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.

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

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. :cwm27:

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

I think I can sum all this math up which is scaring everyone:

THE RELATIONSHIP IS ALMOST LINEAR, IE, A .08 DIFFERENCE, WHICH ISNT VERY LARGE.

That means (and you have to work in %, not exact weight): If you increase the weight by 10%, you have to increase engine power by 10%, or vice versa. Simply saying 100kg to 10hp is naive. What if the car had 3000kg weight to begin with and 100kilowatts? Would taking 100kg off increase the performance time by '10 kilowatts'? No.

In a 1,000kg car, take just 50kg weight off and you get a 5% performance increase.

horsepower numbers arent really numbers alone, they are factors of torque, your torque makes the hp so any formula should use torque as a factor not hp the formula has something to do with torgue and weight = HP but you must remember any loss in drivetrain (estimated 20% loss in HP) aerodynamics are a factor and many more

The formula I have posted isnt just some blind stab in the dark. Many people have come up woth similar formulas (using 197 instead of 199) and written in different forms. It does work but it is approximate. And the relationship between time and hp is not linear becuase of the fact you just stated, aerodynamics. And the numbers are measured at the wheels. The problem is getting from crank to wheel hp without a dyno.

The numbers I gave at the end of my post said that 100lbs was approximatly 2/10 sec in the quarter mile. And that doesnt give the correct relationship, it just happened to be .2 in that instance. If you have a really heavy car, and remove 100lbs you dont loose very much, however, if you have a really light car and remove 100lbs you get alot (relatively).

When I put my formula into MS excell and got a trendline this is what it spat out (I held hp constant).

y = 34.863x^(-0.6526). X = weight and Y = time. So:
T = 34.863 * (weight)^(-.6526). It had a correlation of R^2 = 1 (perfect).
So the relationship depends on the initial weight of the car, it cannot be stated that 100lbs decrease = any set amount of performace. Thus 100lbs doesnt equal any amount of horsepower, or any amount of time decrease, it is different for each weight. The problem is that the change in times is from .1sec to .3 seconds and are dependant on horsepower. So if you wanted a even Greater approximation, you could say that it was close to .2 seconds in the 1/4 per 100lbs, but that would be an even worse approximation.

Thus the % form of HyperS's idea would probably be alright, except that is kind of a gross generalzation, even more so than mine. And defining "preformance" is even harder. I know its not 5% ET becuase that would just be wrong. if a 2800lb car looses 5% (140lbs), its definatly not going to gain 9/10's in the 1/4. (using the numbers from crx's accord). And an effective 8hp (5%) increase is still 3/10's of a second, which I doubt. Maybe if he had measured the WHP on the accord, the numbers would come out more correctly. Take 5% of 120hp (160*.75) and you get "effective" 6hp increase, whitch still seems high, but more plausable.

HTH
Buzz1167
Jon N

buzz - yea my accord is a 4spd automatic.