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weight/power ratios....

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Old Jan 30, 2011 | 01:59 PM
  #1  
SoFlo Mustang's Avatar
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From: Naples, Fl
weight/power ratios....

So idk if this is common knowledge or if its an equation but.....

Lets say theres 2 cars. One weights more than the other by like 400 lbs or so, but its 100 horses(and x-amount ft-lbs more) more powerful. (Forget aerodynamics/conditions, same cars/places) Is there a certain ratio or amount of horses per pound that would make up the difference in a drag? Idk if im conveying my question well but.....

Is there a certain amount of horse power that would make up for each pound extra?
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Old Jan 30, 2011 | 09:25 PM
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conv_stang's Avatar
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From: Richmond VA
there are a few things that can make a difference...ie traction/transmission gearing/rear end gearing.but lets say all other things are equal.
a 4000 lb car with 500 hp = 8.0 lbs per HP

a 3500lb car with 500 hp = 7.0 lbs per HP

for thier performance to be equal with every thing else out of consideration. youwould have to add enough HP to the 4000 lb car to get its power to weight ratio down to 7.0 lbs per HP which works out to about 572 horsepower...

every car weighs different so there is no exact this many HP will make up for this many pounds. 50 pounds added to a 2000 lb car will need more HP to make up for the extra weight than a 4000 lb car will. just b/c that 50 lbs is a greater percentage of weight added to the 2000 lb car than the 4000 lb car.

hope that makes sense

Last edited by conv_stang; Jan 30, 2011 at 09:33 PM.
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Old Jan 30, 2011 | 09:52 PM
  #3  
Five Oh Brian's Avatar
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From: Pacific NW USA
Originally Posted by SoFlo Mustang
So idk if this is common knowledge or if its an equation but.....

Lets say theres 2 cars. One weights more than the other by like 400 lbs or so, but its 100 horses(and x-amount ft-lbs more) more powerful. (Forget aerodynamics/conditions, same cars/places) Is there a certain ratio or amount of horses per pound that would make up the difference in a drag? Idk if im conveying my question well but.....

Is there a certain amount of horse power that would make up for each pound extra?
What I've found with my Mustangs is that to get a one tenth of a second improvement in the 1/4 mile I can either remove 100 lbs or add 10 rwhp. To improve a quarter of a second takes either 250 lbs removed or 25 rwhp. Alternatively, you can mix and match of course.
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Old Jan 31, 2011 | 02:02 AM
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stupidgenius36's Avatar
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From: Denton, TX
I'm no expert but I think if the power/weight ratios match, then the cars should be equally quick. But that's assuming 2 *exact* cars. That may mean absolutely nothing after factoring in drag, traction, drivetrain efficiency/type, conditions, weight distribution, etc. P/W is a good indicator, but there's so much more that matters as well.
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Old Jan 31, 2011 | 06:25 AM
  #5  
Knight's Avatar
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From: Volo, IL
Originally Posted by Five Oh Brian
What I've found with my Mustangs is that to get a one tenth of a second improvement in the 1/4 mile I can either remove 100 lbs or add 10 rwhp. To improve a quarter of a second takes either 250 lbs removed or 25 rwhp. Alternatively, you can mix and match of course.
this works on a 13-14 sec car but the faster you are the amount needed grows exponentially.
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Old Jan 31, 2011 | 09:38 AM
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From: DMV
"More power makes you faster on the straights, less weight makes you faster everywhere." (including speeding past the gas pumps).

or the corollary:

"Straights are for fast cars, Curves are for fast drivers."

Last edited by rhumb; Feb 1, 2011 at 02:42 PM.
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