lt4 fd

Why the heck is it a percentage... I mean if you have a 200hp motor with a certain drivetrain that has 170rwhp assuming 15% loss, that means it takes 30 hp to drive the drivetrain right? So if I had a 500hp engine with the exact same drivetrain it would suddenly take 75 hp to drive it???

96z

Injuneer Fred posted once that driveline losses decreased as the HP increased in his car.
It cant be an actual number for all cars so a percentage is used to apply to all cars.

Z28tt

You could mathematically model it, if you knew all the variables. A drivetrain will take x amount of hp to spin at a specific rpm, which will be a fixed number. When you add load to it (i.e. accelerating a dyno, or the car in the real world), parts start slightly deforming. When parts deform, the gears don't mesh (slide locally) as efficiently, bearings have more resistance since they aren't aligned as well, etc. The more load you put into it, the more parts will deform, and the more loss you'll experience.

Drivetrain losses are never a straight percentage, but people like to say they are, since it makes the dyno numbers look larger A stock 2002 Z28 w/ a T56 does not have a 15% driveline loss. If it did, you'd have to convert all that lost energy into heat. 40 kW of heat is a ton. Nope!

A.

Injuneer

As noted above..... drivetrain losses are NOT a straight percentage. That is just a number quoted for convenience of quick conversion of rear wheel HP to flywheel HP. There is both a "fixed" and a "variable" component of drivetrain loss.

To keep it simple, think of driveline losses as consisting of two different components:

Inertial:

This is the energy required to bring the mass of the drivetrain to peak rotational speed. It is a "fixed" value. As long as the mass of the rotating parts doesn't change, and you bring them up to the same RPM, the energy absorbed is "fixed", and is not a function of the HP being transmitted.

Frictional:

Energy is lost to friction in bearings and gear surfaces. The friction loss is roughly proportional to the loads being transmitted. Double the torque and you double the forces acting on gear faces and bearings. So frictional losses are directly proportional to torque/HP.

My data points for a 381 LT1 driving through a Street Twin, T56, 3" chrome moly driveshaft, 12-bolt 3.73's, and 17" wheels with 315 drag radials were:

Motor: 425rwHP/486fwHP = 12.6% loss
1-stage N2O: 555rwHP/633fwHP = 12.3% loss
2-stage N2O: 670rwHP/762fwHP = 12.1% loss.

When you consider that the 12-bolt is rated at about 7% loss, it accounts for a good portion of the losses. Bending the torque around a 90-degree corner exacts a price.

When I converted to the TH400, the convertor slip (at least I think that's what it is - I know squat about auto trannies) became significant, increasing with the torque being transmitted:

Motor: 390rwHP/486fwHP = 19.8% loss
2-stage N2O: 600rwHP/762fwHP = 21.3% loss

Z28tt

Fred - Thanks for posting some real numbers on crank vs rear wheel torque. The fixed number on the drivetrain loss would depend on the acceleration rate on the rear wheel dyno, vs the engine dyno. Do you recall what the rpm/second sweep rate was set for the engine dyno? I think we were using 200 rpms / second sweep. Was the chassis dyno a Dynojet inertial, a Mustang eddy current, or something else? Good data points ya got there

Andris

lt4 fd

Wow, big difference in the manual and auto tranny's. How much of that difference would you think the gear vendors unit has? That IS what you have though right? Do you ever split it up into the 6 gears you could or is that needless? And thanks for clearing that whole drivetrain loss thing up for me.

Injuneer

Those numbers were taken with only the TH400 in the car, before the GV was added. It has a non-locking convertor that flashes to 5,000rpm under 800ft-lb. The DS was still a 3" chrome moly, but a bit longer, the rear was still 12-bolt 3.73..... can't remember what wheels/tires, but most likely still the 17x11 AFS replicas with 315 DR's. The slip appeared to be around 10-11%.

I have some chassis dyno numbers that were done after the GV install, but by then we had added about 30HP to the nitrous, and swapped the tuning from alpha-N to speed-density, so the rwHP no longer corresponds to any pulls on the engine dyno. Those pulls were also done with 4.10's and 15x10 Welds with 28" QTP's.

I'd have to go back to the printouts from the dyno pulls to find the sweep rate.... just don't remember. Damn engine/car has more dyno pulls than 1/4-mile passes...... but that will change in 2003 . Just need to find a way to drop a few 100# from the 3,760# race weight....... the car is a pig for a coupe.

Z28tt

Ya know what would be interesting? If we all figured out the moment of inertia for various drivetrain parts - such as factory axles, heavy 35 spline axles, various driveshafts, clutches, transmissions, rims, tires, etc. Once we have that data, it'd be much easier to get an accurate drivetrain loss figure. What would be the best way - connect a small fixed HP electric motor, and measure how long it takes to accelerate to X rpms, and back calculate inertia from the acceleration rate? I think for the complex parts (wheels & tires) it'd be hard to figure it out from sections, but for shafts & flywheels, once you know the material, it shouldn't be too hard.

Mindgame

Just my thought here but I think it'd be way too difficult to calculate. You have losses in the engine to contend with too. Friction, different types of lubricant viscosity, bearing clearances, geez!!
I'd say that this is a good candidate for some simple empirical formula. Just need a good long list of test subjects. I can pull up some of my old roller dyno sheets if you guys are interested? Anyone else, post em here.

-Mindgame

Stephen 87 IROC

HP is a calculated number from torque. Torque is the actual work being done and HP is the rate at which it is done.

Transmissions use a torque converter not a HP converter. Since it's torque that's being transmitted through the driveline, decreases because of friction and tranny design will automatically decrease HP. The amount lost will depend on the amount inputed.

You don't see many small 4 cylinder cars with automatic transmissions. There are a few but not many. The main reason is that the engine doesn't produce a lot of torque to effectively use an automatic transmission. The driveline losses would be too high to be practical.

A high HP car is producing enough torque that although there is still losses through the driveline, they're not enough to be a problem.

BBB

This is a real pet peeve to me also. I think that drivetrain % loss is solely a magazine tool used to make themselves not look stupid. I think that GM Hightech is one of the worst offenders. Their latest issue figues that every car loses 20% so that a 400+ hp car now looks like a 500+hp car "at the crank" .The only way to really tell is to run on both an engine dyno and dyno for RWHP with each mod.

When I had the TH 400 rebuilt for my 67 Chevelle the builder told me that the easiest way to go faster would be to change to a Powerglide set-up. He said that although bullet proof the 400 gobbled about 4 dozen horse. The two speed ate up only about 25 hp and the 350 was in between.

BBB

MadMax350

I can see where the percentage issue could easily be confused and not make any sense when it comes to the trans at least. Rear ends, on the other hand, seem to make a little more sense to me and I will try to offer my knowledge of the field to explain why the percentage theory fits here.

The typical 10-bolt rear end unit accounts for about 4% of drivetrain losses. This doesn't sound like much, but you might ask how would a bigger engine result in more power being sapped out by the rear end when compared to a smaller engine. I think the answer lies in friction. The pinion wipes across the surface of the ring gear on more than one axis during operation. There is a very thin film of oil present between the meshing gears, but, like the roller bearings, friction is still present in the unit. The more power that you put to the rear end unit, the more load that will be exerted on internal components resulting in increased friction. This exponential rise in power-in vs. power-out loss is why the percentage value works.

I guess the same thing can happen inside transmissions. The more power you put into these things results in more loading on bearings and gears and such which will result in higher power losses because of friction.

Eric Bryant

Eh, not quite. You've got coulombic frictional losses that are proportional to force, but you've also got viscous frictional losses that are going to be proportional to rotating or linear speed. Within a manual trans or rear axle I'd expect the coulombic friction to be dominant, while a automatic tranmission probably exhibits a complex blend of the two (due to the pump).

Lowend

Now to really confuse this - different chassi dynos tend to produce different numbers...
A Dyna-Pack (hub-bolt design) tends to show lower HP numbers than a Dyna-Jet (roller design).
I would imagine that different wheelbearings and types of tires (on a roller dyno), tranny & rearend lubes would also effect the HP loss %.
The 15% number exists as a mere estimate, I have seen some cars loose as much as 25% and as little as 10% on chassis vs engine dynos (same temp and elevation).

Injuneer

Eric:

I prefaced my remarks with "To keep it simple.......". If this is going to come down to some sort of esoteric pissing contest I won't even bother attemting to explain it so most people can understand it.

Fred