Veloce Raptor

Actually it does, Greg. Can your sanctimony and try driving some different cars on track besides your own sometime.

And you're right: comparing an E36 to an S2000 is apples and oranges....sort of like comparing a 911 and a Viper.

mark kibort

Greg makes a great point.

faster, better, better means faster, STOP!!!!!

VR, re-read my last post with the math calculations.

Out of a turn, if two cars have the same hp they will have the same torque at the axles and will produce the same accelerative forces. (this means both will get off the turns equaly as well, as fast as quick, etc etc)

Does this make sense, as you are starting to argue like my 4 year old.

mk

Veloce Raptor

I would also need to know where the max torque is made, and for how long (the curve).

Greg Smith

This has nothing to do with what cars I've driven on track. The question doesn't even involve another car.
"if you had a car with a flat torque curve from 2000-5000 RPM, would you chose a gear closer to 2000 RPM or 5000 RPM for maximum accelaration?"

You don't need to answer the question with a question, it's all there!

mark kibort

you are incredible. you talk in riddles.

Go to the GTRS vs viper graph. put the HP curves on each other and lower the vipers torque curve downward to match. (ill have to find that 438rwhp Viper curve, as that would be an easy graph to speak from)

anyway, we are saying that say the curves are ALL the same shape. same same same. only the values are different.

does a torque value even matter??? high low, the same ,etc??? if the HP curve is the same shape in the area of operation, then the total accelerative forces will be the same

mk

Greg Smith

Neither matters, but that information is already in my question since I specifically said HP was linear, and torque can be derived from HP...

Veloce Raptor

Sorry, bub, no it's not. Viper? 2000 rpm. Z06? Depends. E36 M3? 5000 rpm.

Bob Rouleau

All else being equal, and both cars being geared optimally there should be no difference between the 500 Lb/ft 4500 RPM max engine and the 9000 max RPM, 250 lb/ft engine since in the latter case proper gearing would mean that both had the same torque at the rear wheels. The high RPM engine would be turning twice asa fast but the final drive ratio would be twice as high numerically to multiply torque.

OTOH what happens if both cars are on a treadmill, will they take off?

Veloce Raptor

Interesting. And ironic.

mark kibort

ding ding, 2 stars!

now, why is VR having such a hard time understanding?

VR, yes it is Ironic, as i have already outlined what you might be talking about.
(higher torque engines sometimes have flatter HP engines).

I think you like to fight just to fight!

Ive given you the tools, what you do with them is your decision.

But, be rest assured that Bobs example is dead nuts on what we are talking about. out of a turn, its equal HP that determines exit acceleration forces.
same HP, same rear wheel torque through the gears , regardless of engine torque values.

Mk

Veloce Raptor

And once again, Mark, you missed the point of the OP by a MILE.

Greg Smith

Exactly. Do you understand that Dave?
No, it's 5000 rpm regardless of which car. They all make the same peak torque from 2k-5k rpm. If you were at 2000rpm in a viper say you'd be in 4th gear at 60mph. For max acceleration, would you rather be at 2000rpm in 4th gear(60mph), 3000rpm in 3rd(60mph), or 4500rpm in 2nd(60mph)?

The answer is 2nd[obviously], it has to do with the gear multiplication.

How about this:
Say you're in a Viper driving in a vacuum with the same flat torque from 2000-5000rpm. If you were to floor it in 4th gear at 2000rpm's at 60mph vs 4th gear at 120mph at 4000rpm, which would give the car greater acceleration? Don't over think this, ignore things like tire and gear friction, etc...

EDIT-Good lord this site is running slow.

mark kibort

Ok, be patient. Explain what your point is. and a "mile" to boot???

I'm waiting to hear this one. I cant even imagine what the heck you are talking about. However, if i was to guess, maybe it has to do with variations on the exit of turns due to the different shapes of torque curves and you dont want to be surprised with some burst of torque. ???? Or you just are so lost, you cant even stand it and you are trying to use a good defence as a good offence.
With you back peddaling with " better exit ", "faster exit", "quicker exit" speeds, who knows what the F you are talking about. But, lets hear it.

But trust me, what i have explained is right on the money and in terms that even you could understand.

As an observer, do you know how many heated arguments you have been in on? All that I have seen have one thing in common. You stay vague to start the fight and get vague'er as you go forward, propetuating the arguement.
We clash, because I provide details, graphical proof and logic. Yet you find fault in all of it because no one can understand what the heck you are talking about! Just an observation.

when I provided the cliff notes to the "war and peace" novel you said i wrote. (by the way, it was 4 short paragraphs). you couldnt even respond with your point in an intellegible way. slow down, clue all of us mortals in on what the heck you DO mean then? what is your point? I hope you know mine by now!


mk

Veloce Raptor

Greg, you cannot ignore all those other things. That was the point of the OP. Street car, street tires, road course, same HP, one car has markedly different TQ (and curves) than the other. Also, have you ever driven a Viper on track?

Mark, as I said, it is obvious you are taking it personally, and are starting to subtly delve into the multiple personal insults. Relax. Get over yourself.

mark kibort

Greg thanks for the validaiting post, but your last paragraph has some issues.

remember, (hint to VR) because acceleration = power/mass x velocity, acceleration goes down inversely with velocity. so, at the higher speeds, even at the same hp, the acceleration rates will go down proportional with speed.

Greg, your first paragraph was correct and you could have left it at that.

Mk