Chassis dyno losses - please critique my thinking.

[Chargermal]

So you mean..by higher TORQUE at same RPM...the losses are greater? Remembering HP is a function of Tq and RPM?

If so...what is it about Tq that impacts on driveline losses?

[RDY4WAR]

More power, with all else equal, will accelerate the drivetrain quicker which means more parasitic losses. It's neither a static number nor static percentage due to the laws of physics at play.

[digger]

So you mean..by higher TORQUE at same RPM...the losses are greater? Remembering HP is a function of Tq and RPM?

If so...what is it about Tq that impacts on driveline losses?

The forces cause more friction especially if it’s a roller dyno that drives the tyres into the roller harder. All the meshing forces within gears and radial loads increase with higher torque

[Circlotron]

So you mean..by higher TORQUE at same RPM...the losses are greater?

Hypoid type diffs that I mentioned earlier, the ones that have the pinion shaft below the level of the axle, have quite a degree of sliding contact between the mating gear teeth. The lower the pinion shaft the more pronounced this is. This sliding makes for a certain amount of friction, and the greater the torque applied to it the more difficult it is for those gear teeth to slide past each other. That's why gearset losses are stated as a percentage, not a fixed hp amount. The more you put through it the more you lose. Evidence of this is the harder you work it the hotter it will get.

[F-BIRD'88]

You also get more deflection between any mating mechanical elements or load bearing elements the more power applied to it.

[agertz1]

If you really want a heartbreaker, try drag racing ? Friend has a 68' Dart nice survivor. 273" , 2bbl and torkflite. He beat a $50k GTO.
Mental HP. counts.

[Chargermal]

So, what is being asserted, is -

For a given RPM, and using identical drivelines.......the driveline transferring more torque to the wheels will show a greater loss.

I’d be interested in which has the greater parasitic effect in this scenario - RPM or Torque?

[Rick!]

So, what is being asserted, is -

For a given RPM, and using identical drivelines.......the driveline transferring more torque to the wheels will show a greater loss.

I’d be interested in which has the greater parasitic effect in this scenario - RPM or Torque?

As you increase RPM beyond the torque peak, does torque go up or down?
Chassis dynos measure torque and do math to display hp.
Chassis dynos are really good for tuning to the torque peak.
Once past that, torque goes down and assuming the gear mesh in question has adequate lubrication, heat generation rate will decrease but the torque transmission loss through that element will remain at a percentage of input torque. Circlotron has stated this twice already. If you go back to your thread from 8 years ago, a gentleman clearly stated what he measured as losses from an engine dyno to a twin eddy current dyno. Once you see the heuristic data folks have been contributing and remove the anecdotal stuff, it's pretty clear what the range of torque transmission loss is running a car on a chassis dyno.

I suggest thinking about it for another 8 years and asking the same question again.

[Fastvette94]

I see a mix of answers and I wanted bring up an example case that might clarify where the ‘losses’ reside. This is all for fun bench racing with friends and drinking a few cold ones.

At the track your vehicle will accelerate based upon a ratio of forces and masses.

a = Force / Mass

Assuming torque converter is 95% slip ratio and the rear tires are slipping 2-3% on the curve but on the left side of peak and steady, let’s define some forces and mass.

Net Vehicle Force = Force applied to rear tires - rolling resistance- aerodynamic drag.

NVF = [(FWT - F1) * F2 x Nt x Nd x 1/LWR] - (.015 x vehicle weight) - (.025 x vehicle mph x vehicle mph)

All of this is approximations of course, how else are we to calculate stuff without a super computer, right? For discussion only. This is what I prefer to use for back of napkin calcs.

Mass = Vehicle Mass + Rotational Inertia. The rotational inertia changes depending on the current gear. It’s to the square power so a lot more is lost in 1st than final drive.

FWT = Flywheel Torque ( w/ accessories on) ft lbf
F1 = Static losses like seals, contacting steels and clutches, trans oil pump. (Typically use 10 here)
F2 = losses from shaft/gear misalignments (gets worse with more torque due to deformation) (typically use .83-.85)
Nt = transmission gearing (2.52, 2.48, 1.8, 1.54, 1, etc)
Nd = differential gear ratio (4.32, 4.11, 3.23, 2.73 etc)
LWR = loaded rear wheel radius (in feet)

Take away here is the force at the ground doesn’t include the rotational inertia which is contained in the mass that the forces work to overcome. Your poor tires have to handle this amount. If you have a 100 pound socket extension you will still get the same torque applied to the bolt vs a 1 lb extension. The rate that you can spin it is obviously changed as you divide by mass/inertia.

The ‘drive train’ losses is mostly a percentage of applied torque with a much smaller static element assuming reasonable power output. (F1 and F2 above)

[Chargermal]

Rick - Can I respectfully submit that the answer to my question is not present in the thread from 8 years ago..as it was a different question. .nor is it in your last response. I’d prefer it if you excused yourself from my posts in future...clearly you lack the maturity and tolerance to contribute without judgement. Nothing I have asked or written deserves such a patronising and snot-nosed response.

—————————

Fastvette - Thank you for the insight however I’m afraid some if what you wrote is beyond my understanding.

What I’m trying to do is seperate the losses created by increased rotational speed through transmission fluid, universals and diff oil (RPM) - from the losses created by increased force transmitted through these components (Tq).

I’ll ask another way -

If a driveline consisting of a 4000 stallconverter (non lock up), a 727 TF transmission, and a Dana 60 diff is said to lose 110HP - is it possible to say ‘x’ % is due to rotational friction, and ‘y’ % is due to Tq induced friction?

And if it is....which is likely to be the bigger contributor at ‘realistic’ peak power limits?

Or is it a piece of string?