Hello ...

This is just a hypothetical question ...

How much HP is needed to consistantly run
12.00?
11.00?
10.00?
<10.00?

Assume an Automatic Transmission ... so driver skill is not important ...

 

The biggest variable will be launching the car. You can get to 12.0 with a little over 300 rwhp and you can get to 12.0 with 400 rwhp.. its all in the 60 ft. :seeya

 

True ... but if that is a constant ... there has to be some sort of formula? or general rule of thumb ...

 

You have to factor in gearing...
HP is not all..

A 400HP car with 2.73's is going to get stomped by a car with 325hp and 4.10's

 

Acceleration is an exponential function, measured in distance divided by the square of unit time. This means (in a laboratory sense) that it takes four times the power to cut an ET in half. In the real world, the behavior is even more exponential due to traction and wind resistance, and these two factors have an increasing effect as ET's get lower.

For a stock C5 (assuming 3200lbs, 300rwhp, and a 13.2ET - also assuming all other factors are constant except the engine output), ignoring traction and wind resistance, you would need roughly 360rwhp to pull a 12.0, 430rwhp to pull an 11.0, and 525rwhp to pull a 10.0. Factor in traction and wind losses, and these numbers need to be adjusted upwards - probably to something like 375, 455, and 570. You could confirm (or adjust more accurately) by checking the archives for rwhp/ET posts.

You can use the following formula to achieve numbers that don't take traction and wind resistance into account:

rwhp=52272/ET^2

If anyone's wondering how I came up with the formula, consider that ET's vary exponentially with respect to the weight to power ratio. Since we're not changing the weight of the car, however, weight can be ignored and we can simply use the inverse of the power:

(1/rwhp) x CONSTANT = ET^2
(1/300) x CONSTANT = 13.2^2
CONSTANT/300 = 174.24
CONSTANT = 52272

Now the constant can be used to solve for the rwhp needed for any given ET:

(1/rwhp) x CONSTANT = ET^2
52272/rwhp = ET^2
rwhp = 52272/ET^2

 

http://douglevinmotorsports.com/HP_calc.htm

Go there and plug in your data. It will solve it for you.

 

Another useful formula. Road & Track published in the last couple of months.

BHP=((mph/234)^3) X weight

Checked it out on several cars and it works well. Works better with other cars than C5's.

 

and weight . ..

 

Here are a couple more, Weight into HP, cube root, times 234=MPH. For ET HP into weight, cube root, times 5.825=ET. HP=(mph/234)3, X weight. Ric