From MrEracer at corvetteforum.com

"We all need to look in another area for the answer to the drive train loss issue... The dirty little secret that no one seems to understand (and the Dyno Manufacturers will not tell you) is that the majority of the losses are not generated in the drive train of the car, the losses are generated in the dyno.

The typical chassis dyno is not an active load applying dyno, they are passive inertial drum dynos. That means that the system can only operate in a transient condition. As the drum is accellerated by the rear wheels of a car the drum is absorbing a portion of your horsepower in the form of inertial energy (called I-Omega losses). These losses are proprotional to the rate at which the drum is being accelerated and its mass (which is huge, approximately 3500#). The faster you accellerate the drum, the greater is the absorbed inertial energy... This is why your loss is greater with a lower rear end ratio. The lower ratio increases rear wheel torque causing the drum to be accellerated at a slightly faster rate and the result is more I_Omega loss absorbed by the dyno drum. With an assumed fixed percentage of loss (15% for MN6, 20% for A4), the result is falsely inturpreted as a slightly lower RWHP rather than a higher loss to the dyno. This is also true with higher HP engines and is the reason that the loss is a PERCENTAGE of total HP generated by the engine.

For those that do not believe this to be true, I suggest a simple test... Put your car on a dyno and do a normal 4th gear dyno run. Then repeat the run with the car in 3rd gear... The result will be a RWHP curve LOWER than the first one... Since the cars engine output is the same regardles of what gear it is in, the increased loss is due to greater inertial absorbtion by the dyno drum... Okay, okay, I can hear it now... "The losses are due to the extra gear set used in third gear and not in the dyno"... So that begs one more test... Conduct another dyno run with the car in 5th gear... Since this will accellerate the drum more slowly, the drum loss will be less and the RWHP will be HIGHER, even with the additional gear set in 5th gear... All this inertial energy is released in the form of heat after the dyno run is complete and the operator applys a huge brake to the drum bringing it to a stop...

As a side note, if it were possible to measure RWHP starting from high rpm and DECELERATING, the result is reversed since the drum would be releasing its energy on the decel giving your car a falsely HIGHER RWHP curve... This is not possible with a passive enertial drum dyno system, but any operator of an active load absorbing dyno knows these facts to be true...

So how much of the total loss is ligitimate drive train losses and how much is lost to the dyno? A typical gear set absorbes about 1.5% (just add up the stages). Bearing frictional losses and other losses will be about 3%... So your (MN6) car in 4th gear should not have more than about 5% actual drive train loss... That leaves about 10% that is generated in the dyno drum for the perceived total of 15%...

The proper way to conduct dyno testing is with an active load absorbing dyno that can measure engine output under constant speed conditions. This is normally done at specific engine speed points that can then be used to plot HP and torque curves for the engine... This method results in accuate hp readings regardless of where it is measured (flywheel or rear wheels)... It is also the reason that the Mustang dyno (an active load dyno) reads different from a passive drum style dyno... If the Mustang dyno were used properly by stopping at discrete speed points to measure steady state data, you would get actual RWHP data including about 5% of actual drive train loss...

How do I know all this? I am a degreed Mechanical Engineer and my job is to dyno test 2000 hp turboshaft helicopter engines every day... And I use an electric dyno to conduct steady state measurements and eliminate the I-Omega effects..."
Cheers
Shirl
SD Racing Enterprises

 

Yeah, I beleive it. I've always thought there was something missing in dyno measurements. I wonder if a GTEC would be more accurate in measureing HP.

DW

 

That's great and all, but the real use of a dyno is to determine gains and losses after modifications. It's a tuning tool, that's all.

Honestly, I don't believe that only 5% is lost and the rest is the dyno. It just doesn't make sense because this guy is not considering parasitic drag from the accessories, tires, of even the loss of power thru the fluids of a automatic transmission.


Dave

 

True those were left out. And all that matters is real world performance not dyno #s anyway. Has anyone used one of those hub mounted dynos yet? What if any is the difference in measurement?

 

Amen.

 

Y'know, it seems like it should be a pretty straightforward thing (for an ME, at least) to calculate and predict these 'I_Omega' losses, since the dyno drum itself is a more or less static quantity. Why wouldn't they just factor those losses into the final dyno numbers? Feh, it doesn't matter anyways since the phantom mental dyno I use gives me perfectly reliable results every time

the Loon

 

Sport Compact Car magazine did a comparison of dynapack vs. dynojet results a while back.

http://www.sportcompactcarweb.com/ed...ble/index.html

 

So this guy is saying that because he sees only a 5% loss on a 2000 shp turboshaft engine, that you should only see a 5% loss on an automobile also?

Sorry...apples to oranges. It's NOT the same thing.

Straight out of CarTest2000 for AUTOMOBILES SPECIFICALLY, not turboshaft engines for helicopters...

Mechanical Losses, Auxilaries: 2%
Mechanical Losses, Manual Transmission: 6%
Mechanical Losses, Automatic Transmission: 8%
Mechanical Losses, Differnential: 4%
Mechanical Losses, Axles and Shafts: 5%
Mechanical Losses, Torque Converter: 3%

MTX Total: 17%
ATX Total: 22%


I think this may be an example of the right knowledge in the wrong application

 

As I understand it, they do.

But, the calculation is hidden in the proprietary dyno software.

 

I don't really buy this. I have a degree in mechanical engineering, and this just doesn't seem to fly.

First off, I've never heard of anyone refer to anything as I-omega losses. The best thing I can think of is that he is trying to refer to rotational intertia losses. However, I-omega is angular momentum and isn't applicable in this situation. He seems confused.

Angular momentum would refer to how much energy is stored in the steel drum of the dyno. It would be easy to calculate, because the I (polar moment of inertia) is known, and the omega (angular velocity) is known. But hey, this doesn't cause any loss in the dyno, and is really only relevant at a constant angular speed. All you could really do with that is calculate how long it would take the drum to stop spinning. Not terribly exciting.

So, I think what's he's trying to say, with improper terminology, is that when the wheels are accelerating on the dyno, some energy from the wheels has to go into overcoming the inertial mass of the rotating drum. Yep, this is true. But his conclusion that this means the dyno numbers are inaccurate is wrong.

This type of thing is what is known as a modifying input, it changes the relationship between input (acceleration of the vehicle wheels) and output (torque reading). ANY engineer producing ANY measurement device knows about this sort of thing, and knows about ways to compensate for it. In the case of a dyno, you could statically calibrate the dyno (give it a known input and measure output, correct for the different), or, more likely, you could just calculate the known effect (since all properties of the drum are known, this is easily done).

I have a hard time believing that these dyno's do not compensate for this effect. I also really doubt anything this guy says because he claims that there is only a 5% drivetrain loss. A simple motor hooked up via a shaft to an interial mass will exhibit more than a 5% loss, and that doesn't account for a transmission, multiple gears, couplings, wheels and tires, etc. That's just silly.