Thaddeus

I heard somebody say once that reducing the weight of your automobile by 100 lbs is the equivalent of adding 10 horsepower. Is that correct? Is there a more reliable benchmark? I suppose there is a methematical relationship between horsepower, weight, wind resistenace, etc, that would predict how much faster a given car will accelerate given a reduction in weight. Is there one?

If there is no formula, is the rule of thumb given above more or less correct?

TIA

Thaddeus

Dave

There IS a formula...sorry ...I don't know it. However, the ratio of HP to weight means more than HP alone. If a car weighs 3000 lbs and has 150 HP then, of course, it has a HP to weight ratio of 20:1 (or is it 1:20). In this case a weight reduction of 100 lbs would equal a 5 (or so) HP increase.As I stated above, there is a formula that can estimate how fast a given weight can accelerate (to 60 for instance), or the HP required to propel a given mass at a given speed. These formula do not take into account aerodynamic drag or parasitic losses (ie alt, waterpump) or drivetrain inefficiencies. Hp to weight is a good rough estimate. HTH

[ 06-24-2001: Message edited by: Dave ]

Jon Eismond

For motorcycles, losing 8 lbs of weight gains you one hp. I have seen a formula for computing time 0-60 give hp and weight. However, this is non-linear function, since drag or air resistance is as square of variables. Will try to look it up and post.

Bradl3y

(3^(whp/w))*234=1/4mile mph

whp=wheel hp, not flywheel or crank hp

w=weight, wet with driver, tested weight.

So, 250 wheel hp on a 3100lb car would bring about 101mph trap speed at the end of the quarter mile.

Cars with a broader torque curve will usually have a higher mph and cars with a narrower or peakier torque curve will have a lower mph than provided by the formula.

I use this alot when fantasy racing at work. It saves alot of money and I don't have to justify it with the wife.

More information can be found at www.turbofast.com.au

Cheers,


Ethan

Tabor

you must have imputed your formula wrong

(3^(whp/w))*234=1/4mile mph

so, a car with 200 whp and 3000lb weight:

(3^(200/3000))*234=
3^(0.06667)*234=
1.08*234= 252.7MPH

Thaddeus

I am not familiar with this form of mathematical notation. What does the ^ signify?

TIA

Thaddeus

cale

^ means to the power of, or raised to.
2^2 = 4
3^2 = 9

Ahh the math minor comes in useful yet again
I would look at the HP/weight question, atleast related to trap speed this way, there is going to be a change in kinetic energy of the car, starting from 0 and going up to .5m(v^2) where m is the mass (wet, driver, whatever goes across the line) and v is the final velocity. Because there is a change in energy work is done, and by getting the differential of work over time, we get power, which is HP. Thus, discounting air resistance, rolling resistance, and basically everything thats in the real world

Power (in HP) = (.5 m * (v^2))/(time*746)
1HP = 746Watts
mass is in KG = 2.2 lbs.

since we are interested in distance this becomes:
Power (in HP) = (.5 m * (v^3))/(d*746)

d = distance in meters.
1/4 mile = 1320 ft. = 402 meters

Then again, that is in an ideal world with no air resistance, and a whole lot of constants because I figured no one here really wanted a post filled with differentials and the like, as it is its probably wrong as my HS physics is a bit hazy.

cale

Sorry, forgot this little bit, that final velocity will be in meters per second, NOT mph, a conversion is:
V (m/s) = 2.2369 V (MPH)

cale

OK, I'm sure you all will kill me for the 2nd reply to my inital reply, with all that physics I forgot what I was trying to achieve:

Weight(in lbs) = ((H/P) * 1492 * distance) /
( (Speed(in mph)/2.2369) ^ 3 )

so assuming a 1/4 here, and your speed through the trap is 48 m/s (105mph)

1 lb = .2452 hp (which is going to be high because there are no resistances)

Hope that atleast provides some insight somewhere

Bradl3y

Tabor,

Yeah, somewhere I got the cubed root thingy messed up. 200/3000=.06667 repeating or whatever.

Take the cubed root of that (that's where I came up with 3^, which was what I understood would represent the cubed root in text, but I guess I'm wrong. I have yet to go to college :-) )

We are now left with .4054801. Multiply this number by 234 and you are given your rough estimate in mph.

Hope this helps and sorry for the confusion. If anybody can let me know what the correct text was to post on this, please let me know.

Ethan