Speed Addicts Rejoice!

xctasy

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The Tech section has Bowlings formulae for drag and horsepower calculations. I've used the same sort of formulae for my calcs. Its an eye-opener for those interested in these things.

I have been using a formula for the last 5 years for calculating the maximum speed of any car.
The formula for power required is:-

1. Top Speed required in mph.Multiply it by itself 3 times.

Then multiply by the cd (or drag factor). A Pinto is about 0.48, an intermediate Falcon (66-70) about 0.48, a Thunderbird 0.35, a Fox Mustang 0.44 or 0.36 if its got an SVO body kit. Early 60's XK Falcons are quite slippery before the T-bird roof get things messed up...more like 0.40 than the 0.48 or so of a 65 Falcon. Early Sprints were likely to have the least drag.

2. Then multiply by frontal area Note that a Pinto is about 20.2 ft2, an early (late 60's) intermediate Torino/Fairlane is 24.2 ft2, and a fat bodied Mustang (71-73) about the same. A Fox Mustang is around 20.8 ft2. A Maverick could be as low as 21 ft2 for an early two-door, or over 21.5 ft2 for a post 74 dodgem bumper number.

3. Then divide all that by 147733, the sea level density of air on a 70 deg F day.

That gives you the net rear axle power required.

4. Then you add the "coast-down" rolling HP loss ( As a guide, a Honda CRX with 2000 pounds and 195 wide tires, consumes 8 hp at 150 mph. A 3500 pound Falcon with 245 section tires consumes 17.4 hp at 150 km/h.)

5. Net flywheel power, from years of data from dyno reports all over the world, needs to be between 1.27 to 1.33 times greater to cope with the drive train losses in a manual to automatic rear drive car. Multiply the result by 1.27 - 1.33. A Falcon with a T5 is 1.27, an auto without a lock-up clutch is 1.33.

6. That lot gives you the minimum flywheel net HP.

For example, a 1981 Mustang 200 may have a 0.44 drag factor, 20.8 ft 2 frontal area and 245 tires. It may weigh in at 3000 pounds, and have a T5 gearbox.

The owner violently reworks his six with a 2V head and a turbo charger. He wants a 150 mph top speed for an open class race he's planning on.

How much power does he need?

(150*150*150*0.44*20.8 ) all divided by 147733. That needs 209 hp at the road wheels.

Then do an estimate on a 195 mm wide tire on a 3000 lb car at 150 mph. Take that Honda CRX's 8 hp, and multiply it by 3000/2000...this gives 12 hp rolling resistance. Then multiply the 12 hp by the % increase in section width. That's about 15 hp. You can make it easier by just using mm*lb*mph*8
.............58 036 680

209 HP+15kW= 224 HP needed on an engine dyno

Then the drive train loss is around 1.27. That's a need for over 285 hp net at the flywheel.

This means the hot little six needs a 10 pound boost turbo unit and a bit of work on the cam and strength of the internals. Or a nearly stock Cobra SVT 5.0!

Formulae for power due to drag and drive-train loss is

cd*FA *mph*mph*mph * 1.27
. 147733

Then add tire loss

mm*lb*mph*8
. 58 036 680


Now go to it. You can see how much hp you need at any speed from 0 to whatever you hope you can manage!
 
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