QUERY FOR PROPELLER HEADS - Driveline Loss - What If?

[HOdbleFman]

Treynor, you are assuming the coefficient of friction of the surface the boys are using is constant. As Torquemonster said, it is a different story when the coefficient is linear or even a parabolic function. Differential fluid, due to increasing internal shear stresses, does not have a constant or even linear coefficient of friction. Therefore, horsepower loses are not directly proportional to speed, but due increase with speed.

 

[Larry Macedo]

I'm not an engineer nor do I want come off as one, but I do have a few questions. If the 500hp engine and the 1000hp engine are turning the same mass and are terminal at the same speed, why does the 1000hp motor exert more energy other than just reaching the terminal speed in a shorter period of time? Sorry, I asked a lot of questions in class too...

 

[Torquemonster]

Hi Larry - I was too busy drawing pictures to ask many questions in class - but I guess that if the 1000hp engine spins to terminal speed much quicker - then it will produce more heat doing so - therefore there will be more loss of power between crank and wheels.... ???

I think that extra loss could be small tho - maybe the 3% figure above is close to the mark - but I feel 40hp is a bit light for the basic loss - but then it may well be correct.... what do you think?

anyone know someone that does engine dyno's - they'd have some case examples.

 

[Torquemonster]

I found this interesting article http://www.superstang.com/horsepower.htm

This is interesting about dyno readings....

Short version: The higher the RPM of the vehicle the more parasitic drag that exists inside the transmission and rearend, regardless of acceleration rate. Therefore more power is spent to spin the drivetrain at high RPM.

Another thing to note: The reported HP (flywheel -or- rear wheel), is highly dependent on acceleration rate. The higher the acceleration rate, the lower the reported HP. So on a dynojet-type dynamometer (fixed inertia), higher horsepower will accelerate the drum quicker, which will increase drivetrain inertial losses, thereby reporting even lower RWHP numbers. This means that the percentage of horsepower drivetrain loss through a Dyno-Jet type dynamometer will actually grow as horsepower increases at a rate larger than a "loaded inertia" dynamometer.

This deserves more explanation. Typically you will find a car that produces 200 horsepower on a Dyno-jet (fixed inertia) will only produce 180 horsepower on a Mustang (loaded inertia) dynamometer - 10% less. However as horsepower increases the numbers (as a percentage) will become closer. So a vehicle making 800 horsepower on a Dynojet may only make 740 horsepower (7.5% less) on a Mustang dynamometer. The percentage has become closer. This is the result of time being reduced (with the dynojet dynamometer) in the equation of work over time.

Real World Example:
(from Muscle Mustang & Fast Ford Magazine, Febuary 2003, Article titled "Mass VS. Myth" by David Vizard)" the example is very interesting - but I don't know how to copy the graphs....

"In the article they dyno'd a car in both 1st gear and 4th gear (this was with a manual transmission). Normally all dyno runs are done in whatever gear provides a 1:1 ratio through the transmission (3rd in Automatics, 4th in Manuals typically). What was amazing was how much less rear wheel horsepower was generated when the car was in first gear. Take a look at the chart and the authors notes (quote: "Because of the rapid rate the engine internals and components back to the wheels are accelerated the power absorbed is greater. As can be seen the difference in rear wheel output between first and fourth gear is an amazing 85 hp and a staggering 140 lbs.-ft of torque.")

Real World Example #2: (added 9/29/2003)

The November 2003 issue of Car Craft has a great article titled "The Brutal Truth" by Jeff Smith (page 40). In the article they place two engines on an engine dyno and then dyno the engine again once it is installed in the vehicle. One engine is a 357 cubic inch Ford Windsor engine and the other is a 455 Buick. The Ford 357 was installed in a 63 Comet using an AOD transmission and Ford 9" rearend with 3:50 gears (exact combination of drivetrain in my Mustang except my AOD is a non-lockup which means it is even less efficient or should lose even more horsepower). The 455 was installed in a 70 Buick GS with a Muncie 4-speed and a 12 bolt rearend with 2.73 gears. The point of the article was to show how things like a belt driven cooling fan or poor vehicle exhaust could affect the engine output in the vehicle but was equally as valid for showing drivetrain induced power loss.

Without reading any further it would be my assumption that the Ford combination would lose a larger percentage of power through the drivetrain. Not only is the AOD an Automatic it is also a 4 speed Automatic that has substantial weight and rotating mass. The 9" rearend is also larger and heavier than the 12 bolt Chevrolet.

The Ford 357 produced 371 horsepower on the engine dyno at 5,000 RPM. On a 1990 Mustang that came stock with an AOD and a 3.27 8" rearend (more efficient than the 9") the rear wheel horsepower is typically 180 hp. That represents a loss of 45 horsepower given the rated 225 flywheel horsepower on that vehicle. Using this 45 horsepower and even giving it another 5 for the 9" rearend the 357 would have produced 321 peak horsepower on the chassis dyno. Well, it didn't! Even after removing all the factors that could have contributed to extra power loss in the vehicle (removing the belt powered cooling fan) the chassis dyno only showed 283 hp. In fact over the entire power curve the difference between the engine dyno and the chassis dyno was 24%. This provides more evidence that the power loss through common drivetrain remains a percentage even as power is increased rather than remaining a static loss value.

The Buick 455 produced 329hp and made 280 through the drivetrain at 4,500 RPM. The average drivetrain horsepower loss in this vehicle was 18.3%. This can be accounted for by the fact that the 4 speed Muncie is more efficient (require less power to accelerate) as well as the 12 bolt rearend being more efficient than the 9".

now to contradict that.... the same article closes with...

"What's the Loss?

When we began to contemplate flywheel horsepower figures for our different combinations, we bantered about some concepts that can be deceiving. Most enthusiasts have been exposed to the idea that flywheel and rear-wheel horsepower can be equated by factoring in a given percentage for drivetrain loss - the drag that occurs from all the items between the flywheel and the rear tires. You may have seen factors such as 15 percent for stick-shift cars and 25 percent for automatics, applied by dividing rear-wheel horsepower by either 0.85 for stick-shifts or 0.75 for automatics.

Now take a time-out and consider the following. Our original baseline indicated 195 hp at the rear wheels, which when divided by 0.85 equates to 229 flywheel horsepower, and implies that the drivetrain is absorbing some 34 horsepower. On the other hand, our combination of blower and traditional bolt-ons netted nearly 340 hp on the Blood Enterprises dyno, which when divided by 0.85, equates to 400 flywheel horses, and implication that the drivetrain is now absorbing 60 hp.

Nothing has changed between the flywheel and the rear wheels on our '93 LX, so does it make sense to figure the drivetrain is now absorbing nearly twice as much power? Such a concept just doesn't jibe in our little brains, so we asked a couple of people in the biz what they thought. Lee Bender of C&L Performance and Paul Svinicki of Paul's High Performance are both well versed in evaluating Mustangs on the dyno, and they both agreed that extrapolating drivertrain horsepower loss via percentages is flawed. Lee believes that the stick Mustangs experience roughly a 35hp loss through the drivetrain, whether they make 200 hp or 400 hp. He did explain that ultra-high-powered vehicles - typically race cars - can be and exception to this rule, but that's a topic for another time. Interestingly, a 35hp loss for stick-shifted drivetrains is strikingly similar to the difference between Ford's horsepower ratings and the rear-wheel numbers we've observed on dynos across the nation. Hmmm..."

So we're back to square one - confused?

 

[Torquemonster]

Here is a link that shows a graph overlaying engine power with rwhp - but note the comment under the graph that it showed 30kw (40hp) less than on ANOTHER chassis dyno - showing the above article is right - diff dyno types get different results...

anyway - interesting to note the relationship of engine power to rwhp as rpm's increase...

http://homepage.swissonline.net/3000gt/engine_problems.html

 

[Torquemonster]

I think this is the best link yet on the subject - now we are getting somewhere!

http://www.sdsefi.com/techdyno.htm

here's a brief quote from it...

"The worst scenario case for a rear drive setup is on the order of 12.5% in 4th gear, not the 20 -25% often published. If 25% was being lost in the drivetrain, the oil would boil in the differential housing in short order and aluminum transmission cases would fatigue and break from the temperatures generated. On a 200 hp engine, something on the order of 37,000 watts would have to be dissipated out of the transmission and differential housings. Obviously, this is not the case."

 

[Torquemonster]

The above link links at bottom of page to this one - and this finally gives us something we can apply:

"Converting wheel bhp to flywheel bhp and vice versa
To reflect the fact that % losses are high for low powered cars and vice versa I use the following equations which have been found to correlate well with real world transmission losses.

FWD cars - add 10 bhp to the wheel figure and divide the result by 0.9

RWD cars - add 10 bhp to the wheel figure and divide the result by 0.88

4WD cars - add 10bhp to the wheel figure and divide the result by 0.84

To estimate the expected wheel bhp from a known flywheel bhp just reverse the equations

FWD - multiply flywheel power by 0.9 and then deduct a further 10 bhp

RWD - multiply flywheel power by 0.88 and then deduct a further 10 bhp

4WD - multiply flywheel power by 0.84 and then deduct a further 10 bhp

Remember, these percentages are not "gospel" - they are good realistic averages."

The rest of article is at http://www.pumaracing.co.uk/power3.htm

BOTTOM LINE - Take RWHP add 10hp and divide by 0.88 to gross up. However the rwhp is basically a guide only as no rwhp dyno can be accurate. The spikes we see in some graphs are also covered in that article and prove that the software is inadequate - an engine dyno does not look like jagged saw teeth. That explains what Ron was experiencing in his SRT10 - it probably wasn't his car - it was the dyno software!

 

[phiebert]

Typically you will find a car that produces 200 horsepower on a Dyno-jet (fixed inertia) will only produce 180 horsepower on a Mustang (loaded inertia) dynamometer - 10% less. However as horsepower increases the numbers (as a percentage) will become closer. So a vehicle making 800 horsepower on a Dynojet may only make 740 horsepower (7.5% less) on a Mustang dynamometer. The percentage has become closer...

I have had much more radical differences on Dynojet vs. Mustang Dyno. My car dyno'd at 342/400 (RWHP/TORQUE) on a Mustang but at 552/678 (RWHP/TORQUE) on a Dynojet. Same car, on Nitrous with no mods inbetween. I'm just making this point that comparing a Mustang dyno to anything is a waste of time...which is pretty much what I think of a dyno that measures a NOS sprayed V10 with exhaust and intake mods at 342 RWHP! That same Mustang Dyno showed numbers of 370 RWHP and less torque than my car for a drag car running in the 8's.

 

[Miles B]

Remember that on a dyno that holds the engine speed constant (by applying a braking force), the driveline is not *accelerating* and is therefore NOT absorbing power in the form of rotational energy. Frictional losses, though, still exist.

It is *possible* to *approximate* the losses through your drivetrain, if you are prepared to build an inertia dyno with a brake (most probably a current brake), and do a lot of testing and work. If you get the car up to a constant speed, then stand on the clutch, and apply the brake on the dyno drum, accurately measuring the energy it absorbs, you can carefully calculate an "approximate" moment of inertia for your drivetrain.

Most of the articles fail to realise the drivetrain absorbs energy, not as heat, but as rotational kinetic energy, just as a flywheel does. People have been looking at huge flywheels as a means to store energy for a long time now.

This quote: "A change in wheels/tires also does not affect true, wheel hp" from the sdsefi page is actually completely WRONG, assuming by "wheel hp" they mean road wheel and not flywheel. When they say "results varied by almost 4 % by doing the runs with a different wheel/tire combination", I am NOT surprised!! Big heavy wheels absorb energy and store it!

Take the light wheels off your 10 speed racer. Put on some truck wheels of EXACTLY the same diameter. Now you can't argue it is just a gearing change. Now hit the road and try to pedal. The wheels will take a large portion of your "flywheel/pedal hp", leaving little of it to hit the road. Electrical guys, think of it as your flywheel hp being a voltage source, and the wheels and road being a load in series. The bigger the load the wheels are, the less power dissipated across the road.

I favor the approximations of "10 or 20hp plus about 10%". I haven't done the maths, but from experience working these things out, that sounds about right. I have done the maths for motorcycles. Judging from the power lost to rotating mass, and the relative sizes of each of the components, I would say a 10% rotational loss is probably very close to right. Like the article says "If 25% was being lost in the drivetrain, the oil would boil in the differential housing in short order and aluminum transmission cases would fatigue and break from the temperatures generated"... they realise frictional losses over a handful of hp are not possible, they just fail to realise where it really goes.

I have a hunch another reason why automatic gearboxes lose more (apart from the TC, which you would be surprised how efficient it really is), is because of the nature of their gears.. planetary arrangment could be a much higher moment of inertia.

I will ALWAYS support an inertial dyno more than a brake dyno. This comes from my background in motorcycles. Take 2 identical engines.. one guy has put on super light magnesium/carbon wheels, a lighter chain and sprocket etc.. the other guy has all the heavy cast stock items. The one with the fancy driveline WILL put more hp to the road. An inertial dyno will show this, as it shows the bike's power under load while accelerating. The brake dyno will show identical numbers, as it tests the bike under load but not accelerating. Which of these is closer to the real world? I don't know about you, but I don't run many races under load but not accelerating... only 2 situations I can think of - up a very very very steep incline, or at maximum speed - are examples of the bike being under load but not accelerating. The numbers end up being larger than you think, and VERY significant. So I bring this knowledge over to cars, and to me, it applies as much or even more so. The wheels are much bigger, and in the case of the Viper, a LOT LOT heavier. Where the bike only has a chain, the car has all of the axles, shafts, and a differential.