Someone figured out that it takes 18-20% to run an A4,rear gears etc. I asume this ratio was done from a stock 345 HP engine ?
So If you lose 18% on a stock 345 hp A4, this means it takes 345 x .18 = 62 hp to turn the drivetrain.This formual would yeild 283 RWHP on a stock A4, which I think is about adverage? So THIS MEANS IT TAKES 62 HP TO RUN THE DRIVE TRAIN PERIOD !. SO why is the same formual used for higher HP engines?
For Example : A 500 RWHP car should only have 562 hp at the crane. not 500 x 18% = 590 at the crank, or a 1000RWHP car should have 1062 at the crank ,not 1180 at the crank. RIGHT ?

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This is a rather common debate here, but the answer is no.

Frictional losses of driven components increase as you increase applied force, since you are essentially increasing the applied load as well.

It doesn't take 62hp to move the drive train....heck, jack up your car and you can easily spin the drivetrain with your hands. Trust me, even if your first name is Hulk, your arm power doesn't have 62hp.

But yes, the drivetrain may take 62 hp to accelerate your car to it's maximum potential..why? because the load increased.

If that sounds confusing, remember, the whole reason you are upping power is to accelerate the car faster. To do that means putting more pressure and force placed against every moving part in the powertain, which means more friction, more heat produced, and more power absorbed (lost) in the process.

If you are driving a 300hp car down a street at a 30mph, and an otherwise identical, but 1000hp car down the street at the same speed, you would be correct, there would be no difference in the parasitic drivetrain loss between the two cars. This is the way most people tend to think of it. But in that condition, neither car is producing anywhere near their maximum hp either.

In the above scenario, if you both jump on it and go full throttle, although he will still definitely smoke you, his drivetrain is enduring considerably more load and producing more heat than yours is in the process, and that additional load is definitely consuming more power.

It is for this reason, hp losses are a percentage, and not a constant value.

 

Let's not forget wind resistance. As you go faster it increases so at 50 miles an hour you may lose 18% but at 150 its probably much higher. That will put additional drain on the HP to the rear wheels.

Things are much different on th etrack than they are on the dyno.

 

In this Mechanical Engineer's opinion, the 18-20 % rule for A4s and
15 % rule for MN6s have been empirically obtained and are good rules of thumb. If you believe the factory engine rating of 345 HP and run several MN6s on a chassis dyno getting 290 to 300 HP, then you settle on 15 % loss through the drivetrain that also includes losses due to rear CV Joints, rear wheel bearings and tires on the dyno drum. The dyno will also show more or less Torque and HP if lighter or heavier tires, wheels and brake rotors are substituted. What becomes an interesting debate is, "What happens to the loss percentage if the engine is modified so that both the engine torque and the RPM are increased by sy 10 %"? In the transmission and rear end, there are hysterisis losses in the gear teeth and bearings that increase with torque. There are windage losses from the lubricants that tend to increase with the square of the speed increase, and seal losses that are most likely constant with respect to torque and speed. In the long run, take the advice of Y2Kvert4me and just be concerned with how much RWHP is improved. Back calculating to engine HP is great for bench racing.

 

A 1000 rwhp car would make 1220 hp at the crank with an 18% loss.

 

The 18% is a little high, plus the IRS plays a part of robbing RWHP.

 

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