r34 racer

iTrader: (1)

I've been wondering for a while now why drivetrain losses are computed in terms of percentage of crank hp. The way i figure it, the drivetrain has a fixed, certain rotating mass that does not change with increased horsepower and therefore should take a set amount of the engine's power to rotate up to speed. Implying that a car loses a percentage of its power seems like assuming that the drivetrain picks up mass when HP is bumped up.

Consider this: a 300hp slightly modded Z and a 1000hp heavily modded Z, with dentical drivetrains. According to the percentage theory, the 300hp Z will lose approximately 45hp to the drivetrain (15 percent rear wheel drivetrain loss) while the 1000hp z will lose 150hp to the SAME EXACT DRIVETRAIN. In my book, that just doesnt make sense.

What im hoping someone will clear up is why drivetrain losses increase with increased HP instead of being a fixed HP lost. Or is this just a popular misconception?

Juztin

iTrader: (9)

Hmmm....

That is a very good point! Now that I think about it, since the drivetrain is a set mechanical system with no real variables (other than maybe chipped gear teeth, material strength due to wear & tear, and age), the drivetrain should really only take a certain STATIC HP to turn it that lessens as rotation increases, rather than a percentage of power.

My thoughts:

Drive Train RPM = DTR

From 0 DTR is where the most power would be required to begin to rotate the driveshaft and in turn the rear axles through the differential.

As the DTR increases, the amount of power to continue to spin and increase spin of the drivetrain should decrease. (Rotational weight effect)

So in turn:

As power output of the engine increases from say 287hp to a (hypotheical) 1000hp, the drivetrain will not be a percentage of drivetrain loss, but rather a static amount of hp which is lost to turn the drivetrain. Now as the drivetrain's DTR increases that static loss lessens because it will be easier for the engine to increase spinning the system. (Rotational weight effect)

So say with real world dyno runs reporting around 235ish hp. That's aroudn 50hp lost. (lets not take into account, elevation, humidty, etc crap). That 50hp lost will always be that static 50 hp. Not a percentage. So a 1000hp engine with the stock drivetrain should be putting out around 950hp at the wheels. Not 750 (percentage loss). As the rpms of the engine increase, the DTR will increase and that will free up power as it will be easier for the engine to keep increasing the DTR's cause of rotational mass and it's weight/momentum effects when spinning. So under acceleration you will start out at around 950hp, and then increase hp as the DTR's spin up and the load offset to spin the drivetrain becomes less. So 950.....958....965...970.....980....988...etc...
0 DTR...1000 DTR...2000 DTR.... 3500 DTR....etc..

Does this make sense to any1 or is it blasphemy?

wren57

iTrader: (34)

good point indeed, ive wondered this for a while as well...

iron1mike_alpha

im probly an idiot and completely wrong but i would say that your hp would start lower and increase up to that 50hp loss not starting at 50hp loss and having a decreasing loss with increasing rpm.

just my uninformed opinion

Silver Bullit II

With more torque, you need more heavy duty drivetrain (more friction, etc.) to handle the torque, Therefore the loss is calculated as a percentage of the available crank torque. You are correct, however, if the drivetrain mass is held constant. You might then, end up like my buddy who lost his big toe, when his clutch on his '69 heavily modified Z28 fragmented.

JMS in TX

bellox

I'm not a fluid dynamics expert but wouldnt the lube pose a variable resistance to the gears that increases with speed?

As I think about this, it will to a point then the gears will move so fast that they sling all the excess lube to the perimeter of the housing and rotate free.

Fascinating.

r34 racer

iTrader: (1)

Good point but think about this: say someone buys a lightweight flywheel for thier car. This mod doesnt free up much HP, maybe about 10ish hp or so. Most lightweight flywheels free up singnificant rotating mass, so wouldnt that counteract the increase in weight and friction caused by adding heavier clutch/driveshaft/diff components, at least to some extent? I find it difficult to believe that while a loss of a couple lbs will free up 10 hp, an increase of a couple lbs will sap 250hp from a 1000hp car.

r34 racer

iTrader: (1)

Never mind guys i figured it out with the help of a good friend.

The answer: Not all drivetrain loss is due to rotational mass sapping hp. It is about friction. Since mechanical friction is calculated as force of friction = mu times force applied (where mu is the coefficient of friction), it follows that as more and more force is applied, more and more power is lost as a certain percentage of power is taken to overcome friction. Thats why drivetrain loss is calculated as a percentage.