Unlimited Top Speed Of F1LM

[Wilko21]

In the upcoming game Project Gotham Racing 3 the McLaren F1 LM is in the car. It is limited to 225mph in real life but in this game all of the cars have their speed limiters removed.

What i'm not sure of is how the LM is limited. Is it limited electronically? Or is it limited by it's rear wing. And also if it's electronic how fast do you think it could go?

[Peloton25]

It's my understanding that the LM is drag and/or gearing limited to 225 mph - whichever way you want to phrase it. Therefore, you would have to change the gearing of the car, increase the power, or reduce the amount of drag to theoretically give it a higher top speed.

>8^)
ER

[Wilko21]

Thanks. And another thing. Since the LM has 680hp as opposed to the F1's 627, does it accelerate faster?

[Peloton25]

Yes, the LM does accelerate faster. Here's a breakdown of in-gear acceleration times compared between the LM and F1 taken from CAR Magazine.



Besides the additional power, the LM also weighs less than a standard F1 by approximately 60kg.

>8^)
ER

[theflinger]

Wow!
100-120 takes 3 vs. 4.5 seconds. That's a huge difference!

My guess is that the LM spec engine in the road car would go 250 flat.

F= CdAV^3/150,000
x rwhp = 5.75(241^3)/150,000
x = 536, so road car has 20 or more to play with for tire losses. 627 horsepower * .88 to get rwhp, because .15 is a value for low horsepower cars, and apparently this figure becomes less accurate with the more horsepower you have. I wouldn't be shocked if it dyno'd on SAE corrected dynojet at 560 or more.
680 * .88 -20 = 578.
578 = 5.75(v^3)/150,000
15088695 = v^3
v = 247.106343

Good enough guess.

[amanichen]

Originally Posted by theflinger

Wow!
100-120 takes 3 vs. 4.5 seconds. That's a huge difference!

My guess is that the LM spec engine in the road car would go 250 flat.

F= CdAV^3/150,000
x rwhp = 5.75(241^3)/150,000
x = 536, so road car has 20 or more to play with for tire losses. 627 horsepower * .88 to get rwhp, because .15 is a value for low horsepower cars, and apparently this figure becomes less accurate with the more horsepower you have. I wouldn't be shocked if it dyno'd on SAE corrected dynojet at 560 or more.
680 * .88 -20 = 578.
578 = 5.75(v^3)/150,000
15088695 = v^3
v = 247.106343

Good enough guess.

Where are you getting those numbers from? Where do you include gearing? Please say real world test data or I'm going to shoot myself.

[theflinger]

Which numbers? You doubting my horsepower or CdA figures? Look em up yourself if so.
The formula is derived from a base drag equation including the density of air.
Gearing has nothing to do with it -- I said the road car, so you might have to swap in for a slightly higher sixth to bring the power peak in at 250, but gearing has nothing to do with it in this case.

[F1 monster]

Can you tell us the equations you use, what each of the variables is, and the values you use for each of the variables?

Thanks!

[amanichen]

Here's a post on another forum where I discuss drag and a top speed calculation:

http://www.mclarenfreak.com/forum/in...p?showtopic=12

I document all my variables, explain my equations, and explain where all of my numbers come from.

If you could do the same, it would make things easier to understand =)

[theflinger]

As far as I can tell, your equations are correct. I don't think that including the torque/wheel radius is necessary. My base equation was: F = 0.5CdAρV^2 , p = rho (density of air). Converting to get horsepower required took me to my final equation. Just a bunch of unit manipulation and standard density of air.

"this means that the engine does NOT produce its maximum force when the wheels are spinning the fastest."
Absolutely true in a lot of cars. From what I read in Driving Ambition, however, the McLaren did hit its top speed while in the "Sweet spot" because although its sixth gear is overdrive, the car is so damn fast it is not really overdrive the way my car, which tops out around 4000 rpm in 4th gear is.

It was explained to me by a few texts that way. Since my original equation gives you the force necessary, my final equation also gives you that, where the force is how much force you can put out at the rear wheels, measured in horsepower. That way eliminates the tire and such, because you are using a measured value (dyno), although for cars that have not been dynoed like the mclaren, you need to estimate driveline losses.

I don't know how else to explain this. I'm simply calculating the drag that the road car experiences, then applying that in a ratio type comparison with the added horsepower of the LM spec engine.

F1 Monster - CdA of the road car is 5.75, although I forget (Driving Ambition is downstairs!) what the CD and A are. CD is somehwere between .31 and .33, but you can look that up if you like. All I have with me at the moment is the combination of the two . I explained my RWHP equation previously.

I dont know what else to tell you.

[McLaren F1 Guy]

Wow...those are pretty impressive math terms And the theoretical top speed is very cool. I'll have to remember that the LM's engine could pretty much hit 250

[Thorst13]

What good is a theoretical topspeed?
The LM could not do that speed anyway because it's aerodynamics arent up to it.
The LM is more of a circuithandler, but still the sickest car in the world.

[F1 monster]

F1 Guy, thanks for the explanation.
Amanichen, your comments?

[Thorst13]

Originally Posted by F1 monster

F1 Guy, thanks for the explanation.
Amanichen, your comments?

Are you reffering to me?
In that case lets cool down the sarcasm as I'm simply pointing out the fact that it's not possible to run 250mph with a LM as it is and I'm sure you all already knew that, but nevertheless still a fact.


It's really an endless discussion if were about to discuss theoretical topspeeds on McLarens as their engines are what you make of them.
But, trying to "modify" this work of art is almost close to idiotic as I trust McLaren has already done extensive testing on that matter.

[Peloton25]

It sure doesn't look like he was referring to you Thorst.

>8^)
ER