ptuomov

I wouldn't use numbers from this study for the 928, but it demonstrates the various loss effects very well: http://citeseerx.ist.psu.edu/viewdoc...=rep1&type=pdf

At at 6000 rpm or so, my guesstimate is that the 928 s4 5 speed driveline loses about 10 hp to torque independent friction and 12% of the power to torque dependent friction. This is holding the load at a given rpm. One might be able to pick up a couple hp by pulling a modest vacuum in the crankcase, lowering the oil fill level, and running lower hot viscosity gearbox oil. One might also be able to pick up some torque-dependent efficiencies by polishing the gears.

At a later date, I'd like to estimate various inertia effects by altering the sweep rate.

Bulvot

The thing that you may be overlooking here is that force required to overcome driveline loss is not just a function of angular speed (RPM) and unchanging resistance. When you measure the ability of the motor to produce power, you are measuring it by applying a known opposing force and measuring the motor's ability to overcome that opposing force. When we're talking about a dyno, the opposing force is applied at the tires. This is a very important point because the driveline loss is not a static number, it is a function of the coefficient of friction times the force applied against the surfaces that are in friction with each other.

As the force of the motor increases, it applies greater force to overcome the opposing force faster than before. The two opposing forces (motor and dyno rollers) result in applying greater force against the surfaces in friction with each other in the driveline. Things like bearings, gear teeth, etc. Therefor, yes, it is impossible to avoid losing more power in the drive line if you are applying more power to it.

To oversimplify it, here's a basic description of how the force required to overcome friction increases as the force applied against the surfaces in friction with each other increases: https://www.engineeringtoolbox.com/f...nts-d_778.html

Drivelines have a much more complex set of calculations than the above link, but it provides the fundamental that the driveline loss calculations are built on.

Personally, I think dynos are best used when validating and tuning modifications to an individual vehicle. Not as a means to compare differently configured vehicles against each other. You could use it that way, but the variables are so numerous and flexible that you would never really know for sure how the results from two different cars truly compare. My opinion is that the only way to compare two cars is on the pavement. Even then, there is a large element external to the vehicle's capabilities that can drastically affect the results: the driver.

Anyway, unless you're going to identify and individually calculate power loss for each discrete component of the transmission, driveline and accessory system and for each level of wear on those pieces down to the individual ball/roller bearings and individual teeth....it's reasonable to use an estimate that the rest of the world has proven is in the ballpark and used almost universally.

If you want to be absolutely certain of the power at the crank, the only way to do that short of the above calculations is to remove the motor and test it on a stand.

ptuomov

+1

daveo90s4

"As the force of the motor increases, it applies greater force to overcome the opposing force faster than before."

Yes. Agreed But that was not quite my point. Nor I think the point Blau928 was making.

The contention is that a very very very big engine and a much smaller engine - both driving the same gearbox on the same dyno at the same speed will result in the same driveline power losses at that same rpm.

i.e. driveline power loss does not increase when you 'simply' make an OEM 928 engine more powerful - at any given rpm. At least, that is the hypothesis that is being explored here.

I'd love to see a proper physics - based explanation of how a drive line can 'know' or 'sense' how big or small an engine is that is driving it, and adjust its resistance accordingly (at any given nominated rpm). Maybe, but I can't see it being so...

Cheers, Sorry for the thread hijack.

Dave

ptuomov

What do you mean by "bigger" and "smaller" engine? If you mean displacement being different but power and torque per rpm being the same, it doesn't matter to the transmission losses. If you however mean more powerful with higher torque per rpm, then the more powerful engine will result in higher driveline losses when run at that higher power. This is because the load between the gears grows proportionally to the torque. If the friction coefficient is constant, the friction force and losses are proportional to torque.

The story is completely different from accessories. The accessories draw the same amount of power at a given rpm regardless of the engine torque. Therefore the friction losses in the gears, pulleys, and belts is independent of the engine torque. They are however dependent on the torque that the accessory device consumes.

daveo90s4

Hi Ptuomov. By 'bigger' and 'smaller' I am meaning 'more powerful and less powerful at a given nominated rpm'.

Maybe I'm not putting forward my my proposition as clearly as I should be.

If I have some sort of driven device that requires (say) 20 HP to spin it at 2,000 rpm with a 100 HP 4 cylinder engine spinning it, then what I'm suggesting is that that same device will still consume the same 20 HP when spun at the same 2,000 rpm with a 400HP V8 engine.. How can it be otherwise? The device that consumes the HP has not changed one iota.

I really am trying to see / understand how it could be otherwise (note in all case I am talking about losses at the same RPM between scenarios).

Happy to continue this interesting discussion off-line or on a new thread - am mindful this somewhat esoteric discussion may not sit well within Carl's original thread. Bed time here in Oz. :-)

Cheers

DaveO

ptuomov

On a load-holding dyno, the transmission + dyno package is an "accessory" that consumes about 300+ hp for a stock S4 6000rpm. Add two turbochargers and it consumes 2.5x that. There's no reason to think that friction losses to heat etc. inside the transmission would be the same in two scenarios.

Bulvot

Just to put it in perspective, my major was Applied Physics. And this is pretty elementary stuff in that discipline. I'm not just making it up or using google to take guesses at the dynamics involved.

You're correct that a static rotating assembly with a static force applied on both ends will have a static (assuming no build up of heat) loss of energy due to friction/heat/entropy. However, that's not how dynos work. They measure how quickly the motor can accelerate the drum, not whether or not it can maintain a speed. Which means that the more powerful a motor, the more power it applies to the drum (through the tires/wheels/transmission) and the faster it accelerates that drum. The drum's mass remains unchanged (although a load holding dyno can vary the resistance of the drum turning), but the power applied to it from the motor changes, which absolutely applies more force to the components in friction with each other, including the bearings supporting the dyno's drum. Which means more force is applied to the entire assembly than if it were just maintaining a static speed against an unchanging resisting force. In fact, as the power output of the motor changes across its RPM range, the driveline loss also changes. To make it even more complex, the coefficient of friction between components changes as their temperature changes. Make several runs on a dyno and the driveline loss on the last run will be different than on the first run (usually higher loss as it warms up).

You have to remember that it is extrapolating point in time/RPM power output by measuring acceleration of a known mass, but not actually measuring that power output while maintaining a steady RPM.

Carl Fausett

Please keep in mind that this was an eddy-current dynamometer. When we speak of accelerating a drum of known mass, that is commonly referred to as a inertial-drum dynamometer. Those two are the dominant chassis dyno types, and each has its own merits and demerits.

Among tuners, if you are shopping for the highest HP number available you usually hunt up a Dyno-Jet dyno because it is a inertial-drum type. The inertial-drum type dynos generates a little higher numbers than the eddy current type usually (although in the last 10 years I have noticed the calibration of the dynos by both manufacturers is better and the differences between them are less).

The eddy-current type dynos, like this one, are preferred when trying to get a tune just right, as the dynamometer can increase the load on the motor as it spins up to spend more time in each cell of the engine map. This is a bigger deal as the engine HP goes up. For example: the Meg, at 1113 HP would pass through to redline so fast on a inertia-drum model as to make seeing what is happening in each 200 rpm cell of the map just about impossible. But with an eddy-current dyno, load was increased during the pull to slow our sweep so we could see and record what we needed to. The demerit of an eddy-current type dyno is they typically yield lower numbers. This is why those that want a Hero Pull with highest possible HP numbers prefer a Dyno-Jet inertial-drum type.

Another benefit of an eddy-current dynamometer is the "simulation" setting, where the machine will make the pull to redline last a full 30 seconds instead of the normal 6 seconds. Really helpful if you want as close look at the map, or to see what the engine support systems (cooling, fuel, intercooler temps, etc) will do under more true-to-life conditions. I understand Dyno-Jet has now added that feature, which they do by adding auxiliary brake shoes to the drum, but I am told while it slows the wheels as planned, the accuracy of the load is not as accurate.

ptuomov

If one has a load-holding dyno, one should in principle be able to calibrate the inertial factor exactly by doing runs at different sweep rates on the same gear. I want to do that on John's dyno this summer.

Another interesting thing we've already found playing with the dyno is that the S4 engine responds extremely well to boost. The dyno power scales at least proportionally to manifold density, at least until one has to retard the ignition too much because of the knock limit.

Carl Fausett

A few more variables in drive train loss include the ratio of the gear-set in the diff (lower numbers, like 2.72:1 are more efficient and have less loss that the higher ratio gear-sets like 4.10:1); then there is which rear end gear-oil a person is using, and tire diameter. On a late-model, the PSD will shut off on the dyno, and I am not sure what that does to efficiency, if anything.

Frankly, surprised there is so much talk about this calculation. This was discussed on this very forum years ago during the supercharger wars and it was agreed (and we didn't agree on anything back then ) that 15% drive train loss for the Porsche 5-speed was the right number. Tim, Andy, DR, and I all used it.

Carl Fausett

That's what we have found as well. More specifically, we think its the lower valve overlap found in the S4 cams that makes the S4 engine respond so well to boost. Less valve overlap means higher cylinder pressures at lower rpms, improving bottom end torque. Normally, this would come at the cost of lowering the output at the top of the rpm band where we would like to see more scavenging of the cylinders. However, under rising boost pressure as rpm goes up, we still see plenty of scavenging and the HP and torque values do not fall off like a NA motor would.

ptuomov

It's true in our data also with bigger cams. The intake-exhaust absolute pressure ratio isn't too far from one for us because of turbos, so that's why the engine air flow and trapped mass respond to cams pretty much the same way as a normally aspirated engine.

The reason why I believe the power for S4/GT responds so well to boost is that the engine has a lot of friction losses. Some but not all of those friction losses scale with cylinder pressure at a given rpm. Those losses that don't scale with cylinder pressures become a smaller fraction of the gross output, making the net power scale somewhat faster than proportionally to the manifold pressure. Then this favorable scaling breaks down when the ignition needs to be retarded too much with higher boost.

andy-gts

Wow that puts it in perspective, Greg told me yrs ago that driveline losses were greater with the more powerful engine but 300hp at 6000 rpm and then your twin turbo consuming 2.5 x that ,,,,man that is a lot....you almost need to build the engines exponentially more powerful just to fight the drive train losses......geez...

ptuomov

Thats the dyno transmission combo that absorbs the power!