Ripp did a dyno pass with a 392 in awd and make 325 wheel
Ripp did a dyno pass with a 392 in awd and make 325 wheel
That was, unfortunately, in 4HI because of some technical issue about the traction/stability control systems not allowing a rear-wheel-only dyno run.Originally Posted by 2k13jk
My 27% drivetrain loss assumption gives about 345 hp and 350 ft-lbs of torque at the rear wheels for the 6.4 Hemi. This seems like a reasonable difference between 4WD and RWD numbers. Maybe? No idea, actually
Last edited by UselessPickles; 04-05-2015 at 12:21 AM.
Pickles. Is power loss in the drivetrain always a percentage of engine hp/torque that scales with the engine or does it take a fixed amount of effort to turn the drivetrain at a given rpm/gear? In other words would a 1000 hp engine give up 250 hp to turn the driveline based on a 25% loss factor; a 500/125; 100/25 and so on or do all engines start with a -X loss due to the hp/torque required to turn the drivetrain?
All depends on the efficiency of the transmission, differentials, number of axles being driven. 27% seems a little high to me. I've had my truck dynoed when I first bought it-- it has about 140hp at the flywheel at 2600 rpm and it showed about 120-125hp at 2600rpm redlined at the rear wheels or about 14.29% loss. Which is pretty good.
Both
My brother (grad student in physics) and I actually explored this question (and many others) by working through the relationships between things from the engine, through the drivetrain, all gear reductions, through the tires, to the dyno drum, as a complete system being accelerated in a power pull sweep on a dynojet style dyno. I'll spare you the math and try to just describe the findings relevant to your question.
The context of what I say here is limited to a vehicle on a dynojet style dyno, going through an rpm sweep power pull with a given transmission gear, axle ratio and tire size. Things change as you change transmission gear, axle ratios and tire size. I do not attempt to cover this here.
There's two main components to drivetrain loss:
1) Loss due to friction.
2) loss due to moment of inertia of all the drivetrain components.
For a given drivetrain, transmission gear, axle ratio, etc., the amount of the engine's torque that is "consumed" by friction is practically constant. Doesn't matter how fast the drivetrain is spinning or how quickly it is accelerating. So this component takes a "fixed amount of effort" as you worded it.
The loss due to moment of inertia is directly proportional to how quickly the drivetrain is being accelerated. On a dyno (at least, on a dynojet dyno in a power pull sweep), the drivetrain is accelerated at a rate directly proportional to the amount of torque the engine produces (minus the the amount of torque required to overcome drivetrain friction). This component is essentially a "percentage of the engine hp/torque".
Due to the way dyno power pulls are done (on Dynojet and similar dynos, at least) with a fast sweep through the RPM range (quick acceleration) with a fixed dyno drum mass, the loss due to moment of inertia is much more significant than loss due to friction. So for a given drivetrain, the amount of drivetrain loss will be close to the same percentage, but that percentage of loss will go down some as you increase the engine's power because there is the friction component of the loss that is constant. That frictional loss becomes a smaller percentage of the engine's total output as you increase the engine's output. The specifics of how much the overall percentage will change depends on the specifics of the particular vehicle's drivetrain friction and moment of inertia.
There's lots more details I could try to get into about what/how a dyno technically measures, differences between different types of dynos, etc., but I think what I've said so far is good enough for the context of this discussion (measured drivetrain loss on a dynojet dyno).
One interesting note worth mentioning, though, is that a steady dyno pull (where the dyno applies torque to hold the RPMs steady, records the amount of torque, moves up to another RPM, rinse and repeat) completely removes the drivetrain's moment of inertia loss from teh equation because the drivetrain is not accelerating while taking the measurement. This would result in higher dyno number, and a completely constant amount (not percentage) of drivetrain loss, because friction is now the only source of loss. This does not produce realistic results, though, because most people want dyno results to gain an understanding of how the vehicle will accelerate... not how much work it can do at a steady speed.
Your truck is not a Wrangler. Stock Wranglers typically show about 30-33% drivetrain loss. JeepLab's dyno results for the stock Wrangler show a 33% loss. So 27% loss with the Hemi is not unreasonable. Since the 27% loss came from a 5.7 Hemi dyno chart, I actually expect that the 6.4's drivetrain loss will be a bit less than 27% (see above explanations), but nowhere near something like your truck's 14%.
Also keep in mind that there's a difference in transmission with the Hemi Wranglers, so that could be a significant factor in the difference in drivetrain loss between stock and 5.7 Hemi.
Basically, like I said at the beginning, we don't have any exact data fro the 6.4 Hemi Wrangler, so I'm just doing the best I can with the data I can find to create a close approximation. Don't take the exact simulated numbers/times as absolute truth. Just look at overall general trends as an approximated relative comparison.
If you haven't actually removed your engine to dyno to figure out bhp then you really cannot tell how much you've lost due to the drivetrain. It is all guess work.
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