Question about Weight to power ratio

[DinanM3_S2]

WPR is a good way of estimating performance, but definitely not absolutely conclusive. Its probably even more useful in non-drag racing. In drag racing, lighter cars will generally have a slightly easier time launching, but will generally lose out in the long run to a stronger car. On a track however, a heavier car generally won't be able to corner as quickly as a lighter car, so the weight element of WPR is more important.

A Bugatti Veyron is a monster in a straight line because of its monstrous amount of power, but on a road course the Pagani Zonda F, Koenigsegg CCX, or possibly even an Enzo will probably win.

[AMGalltheway]

just remember horsepower sells cars torque wins races

could you try doing a WPR with torque instead of HP?

[CassiesMan]

Originally Posted by psych0hans

You also have to take into account the aerodynamics of the car... You could put a 1000+hp in almost any car, but if its not aerodynamic it wont do much good. Thats the difference between a 600+hp Skyline and a mclaren f1, the Mclaren will win mainly on accord of its shape etc...

Aerodynamics are for people who can't build engines
-Enzo Ferrari

[KillaKaze024]

Originally Posted by AMGalltheway

just remember horsepower sells cars torque wins races

could you try doing a WPR with torque instead of HP?

Ya sure, ill post it when ever I finish it.

[CassiesMan]

Originally Posted by AMGalltheway

just remember horsepower sells cars torque wins races

could you try doing a WPR with torque instead of HP?

...

Kawasaki ZX-14: 102 lbft of torque
Dodge Viper SRT-10: 535lbft

Who do you think is going to win?

Torque may win races, but it is a function of Horsepower. With no horsepower, you have no torque. So again, it has nothing to do with just one factor.

[AMGalltheway]

okay but massive weight difference there
and no horsepower is torque at speed

[CassiesMan]

Torque=(5252xHP)/5252

So if HP is 0, then Torque = 0/5252, and that is 0. Torque is a function of horsepower.

So again, it takes far more than just any one given number to determine the outcome of any contest of performance.

[AMGalltheway]

http://member.rivernet.com.au/btaylo...ginePower.html
heres a dandy article about it( although long probably best to just skim through it)

and your right in that it takes more than one factor to determine performance

[BeZerK2112]

I tried to play with the equation and put torque into it. I got some interesting numbers and posted them here. Heres a link. Take a look at it.

http://www.automotiveforums.com/vbulletin/t700356.html

[BP2K2Max]

Originally Posted by CassiesMan

Torque=(5252xHP)/5252

So if HP is 0, then Torque = 0/5252, and that is 0. Torque is a function of horsepower.

So again, it takes far more than just any one given number to determine the outcome of any contest of performance.

that can't be right. multiplying something then dividing it by the same number doesn't do anything.

to find the power at a given rpm you take the tq times the rpm you're looking to find power at, then divide that by 5252. for example my car makes 220 lbs of wtq at 2,000 rpm so 220x2000/5252=83.777 whp at 2,000 rpm. to find torque you multiply hp at a specific rpm x 5252 and divide that by the rpm you found hp at to see what the torque is.

you're right though the two are directly related. look at any dyno the hp and tq always cross at exactly 5252 rpm.

[-The Stig-]

Originally Posted by CassiesMan

Torque=(5252xHP)/5252

So if HP is 0, then Torque = 0/5252, and that is 0. Torque is a function of horsepower.

So again, it takes far more than just any one given number to determine the outcome of any contest of performance.

Um... you're almost right.. but slightly off bro!

This is how you get Torque:

T = 5252 x HP

rpm

And Horsepower:

HP = rpm x T(torque)
5252(constant)

[CassiesMan]

Meh, close enough. I was never good at memorize equations.

[-The Stig-]

Ha, I had to google it to make sure I was right.