CdA of a GT40 - Calculate your optimal gearing...

[MikeP]

Thanks Arnie. Now I know why I'm so slow!
Cheers
Mike

[Julian]

OK, well here are the aero figures from John Horsmans book.

He gives frontal area as 17.58 sq feet. But there are two Cd figures quoted, .312 and .35, and it's not clear what car configuration they relate to.

So, kW needed to overcome aero drag at different speeds are, based on the above two Cd's:

100 km/h 6.7 kW, 7.5 kW
200 km/h 53.3 kW, 59.8 kW
250 km/h 104 kW, 117 kW
320 km/h (200 mph) 218 kW, 244kW

Now these power figures should roughly correspond to rear wheel kW on a chassis dyno to take into account tyre and frictional losses. Plus a bit more power is needed for front wheel tyre and friction losses, but I don't know how to estimate that. Well, you could do a coast down test I guess, which would give a fair idea of losses, both aero and frictional. Just GPS log the deceleration from your highest speed of interest and that can be used to determine reasonably accurately the combined total of areo, frictional and tyre rolling drags on the car. Any spare power your engine produces at a given speed can then be put to use accelerating the car. Which is the fun part!

Pretty easy to do 250 km/h (156 mph), but you can over double the required power to do 320 km/h (200 mph)!

Start your bench racing...

[Julian West]

Interesting.... so if we need 244kw to do 200mph, thats 327 RWHP, which is fairly easily obtainable.....??? what's the problem...

[Russ Noble]

Originally Posted by Julian West

Interesting.... so if we need 244kw to do 200mph, thats 327 RWHP, which is fairly easily obtainable.....??? what's the problem...

I guess the problem is you have to accelerate it to that speed, which will take quite a few more horses because you'll be starting to run out of available road real fast! It's not the power to maintain a fixed speed, it's the power that's needed to get there in a limited length of road (or track) in the first place.