How Much Is 250rwkw In Engine Kw?

[Adriano]

Cubes i agree with you 100%. I would like to see someone put a 600hp car on the dyno and hold it flat out at maximum torque for 1min, i bet the diff and the gearbox would be on the verge of being red hot internally.

[2rismo]

I still stand by a fixed drivetrain loss + a percentage.

Its obvious there's not a fixed loss.

Street cars can with no problems cruise around and not require diff and manual trans coolers.

Then look at the old Bathurst R32 GTR and you see diff and manual trans coolers hanging off them.

Increased power = increased friction within the driveline and motor its self; all of of which requires coolers to keep them reliable.

IF there is no added friction at WOT we would require trans and diff coolers when simply cruising around the streets. Simply not the case. WHICH suggests there is indeed a percentage value some where in the equation.

On the dyno; cruise with partial throttle. There's absolutely no way the driveline at this point is sucking 65kw.

There must be a fixed loss + a percentage.

Its IMPOSSIBLE for there to be a simple fixed value loss for all power levels. Its silly to think so.

I make reference to a car at cruise as it is comparable to one making 200rwkw and another 500rwkw. The 500rwkw may be at cruise when making 200rwkw.

Hope that makes sense.

It makes sense insofar as you've proved my point for me. Throttle position plays absolutely no part in this discussion so referring to WOT is meaningless. Partial throttle position or throttle position, cruise, full load etc has no bearing on the subject at hand. The questions then, are:

Q/ Does a car making 100hp at the crankshaft which produces 50hp at the wheels convert half of the original horsepower into other forms of energy?

A/ Yes.

Q/ Does increasing the crankshaft horsepower mean that there is extra load on the drive-line components?

A/ Yes. However marginal, it must.

Q/ Does this extra load increase the amount of kinetic energy converted into heat and noise and lost through the drive-train components?

A/ Possibly. Short of drive-line failure, it must be a ridiculously small amount - probably not measurable. Remember that we're working in theory as we don't have a test vehicle for this discussion so we have to assume an unbreakable drive-train for the purposes of the discussion.

Q/ Does this same car, now producing 1000hp but having the same drive-line convert the same, more or less horsepower into other forms of energy?

A/ The same.

Q/ What happens to the extra power (10x more in this example) produced by the engine and where does it go?

A/ All of it with the exception of the original 50hp lost as sound or heat plus a ridiculously small amount of extra heat and sound from extra bearing load etc is transferred to the driven wheels. Absolutely, positively nothing like the doubling and tripling mentioned earlier.

[2rismo]

Just to clarify, no-one is suggesting that there is a fixed loss for cruise mode - i.e. producing 35kW and losing 60kW would make for great fuel economy!

The example that started this thread clearly means maximum horsepower at its peak on a chassis dyno as discussed on this and every other car forum every day of the year.

[tridentt150v]

I believe drive train loss if it could be graphed would be a classic 'S' curve.

At very low output the loss is small because everything is moving so slowly that oil lubricates, heat doesn't build up, and things operate at an optimum efficiency.

As output builds - and this means revs - heat soak begins, oil and gears/cogs/bearings [inc. diff and axles] are placed under load and the friction component begins to use up engine energy.

At maximum output the loss plateau's, or even dips a little as gears/bearings/cogs etc can't radiate any more heat, they absorb the energy, but cannot dissipate it as friction [eventual meltdown].

So I think that Cubes is correct to a certain point, but somewhere on this 'S' curve there is a transition and then 2rismo theory takes over. That is, it doesn't matter after a bit if you have 300 or 1000hp the losses are increasing in value, but they are decreasing in overall percentage terms.

[2rismo]

Bingo. The reason I've stuck to my guns is because we're not talking about crusie, light load, regular driving. We're talking about max power measurements and their relationship to drivetrain loss when compared to the same car/drivetrain with more power.

[Cubes]

I believe drive train loss if it could be graphed would be a classic 'S' curve.

At very low output the loss is small because everything is moving so slowly that oil lubricates, heat doesn't build up, and things operate at an optimum efficiency.

As output builds - and this means revs - heat soak begins, oil and gears/cogs/bearings [inc. diff and axles] are placed under load and the friction component begins to use up engine energy.

At maximum output the loss plateau's, or even dips a little as gears/bearings/cogs etc can't radiate any more heat, they absorb the energy, but cannot dissipate it as friction [eventual meltdown].

So I think that Cubes is correct to a certain point, but somewhere on this 'S' curve there is a transition and then 2rismo theory takes over. That is, it doesn't matter after a bit if you have 300 or 1000hp the losses are increasing in value, but they are decreasing in overall percentage terms.

Sounds spot on to me.

Bingo. The reason I've stuck to my guns is because we're not talking about crusie, light load, regular driving. We're talking about max power measurements and their relationship to drivetrain loss when compared to the same car/drivetrain with more power.

I believe I wasn't clear enough in explaining.

The same car with a smaller turbo may very well make the same power as the car with the larger turbo at half throttle.

This is where I attempted to suggest a fixed loss theory is incorrect.

Part throttle or full throttle simply alters the amount of power; exactly the same as a larger or smaller turbo.

Obviously I wasn't clear enough.

[2rismo]

Wasn't that pleasant?

[Cubes]

Well we all came to an agreement. lol.

[tridentt150v]

My work is done here...................

[Adriano]

Just one more thought, driveline loss should be fixed=percentege of the torque, not the power, thats why it doesnt reflect obviously, as its rare to multiply the torque of a motor by say 4 times, but multiply the power by 4 times, happens all the time

[2rismo]

It is discussed as power even though a dyno measures torque and works out the power accordingly. The logic still applies. Plenty of people multiply the base torque level by many times with built and modified motors. It makes no difference to the discussion.

[Hardman69]

Some of it is near-constant drag (wheel bearings, brake drag etc), some is speed-related (oil drag in diff & box, inertia), some is friction, which varies with load (crownwheel & pinion is a big offender for this).. . . . .

The company I work for has done extensive testing, working out where "Our" car can make-back ground on fuel economy, compared to our traditional rival, and a lot of it is in drivetrain drag, because our drivetrain (especially diff) is VERY tight (to avoid backlash etc) and overly tough, which adds constant loss to the system.

Of course the TOTAL drivetrain loss increases with the KW transmitted through it, otherwise you'd need 50kw (for example) to get off the line!! And I don't think anyone would suggest they're making 50kw when they turn their wheels & tailshaft up on the hoist.. . . Also, the PERCENTAGE of power loss must reduce with increased engine power, the worst case being no movement, with enough power being put in at the clutch/stall converter to NEEEAAAAAARRRRRRLLLY move the car.. . . that's 100% of power lost to drivetrain & tyre drag.

As usual, the truth is in the middle, and silly boys on forums back themselves into a corner, getting more & more narrow in their arguments, trying to convince everyone (including themselves) that they're completely right.

sigghhhh.. . . .

Nice work on finding a middle-ground fellas ! ! ! !

[BaysideBlue]

Thanks for all your valuable inputs guys, especially 2rismo and Cubes.

[BaysideBlue]

It makes sense insofar as you've proved my point for me. Throttle position plays absolutely no part in this discussion so referring to WOT is meaningless. Partial throttle position or throttle position, cruise, full load etc has no bearing on the subject at hand. The questions then, are:

Q/ Does a car making 100hp at the crankshaft which produces 50hp at the wheels convert half of the original horsepower into other forms of energy?

A/ Yes.

Q/ Does increasing the crankshaft horsepower mean that there is extra load on the drive-line components?

A/ Yes. However marginal, it must.

Q/ Does this extra load increase the amount of kinetic energy converted into heat and noise and lost through the drive-train components?

A/ Possibly. Short of drive-line failure, it must be a ridiculously small amount - probably not measurable. Remember that we're working in theory as we don't have a test vehicle for this discussion so we have to assume an unbreakable drive-train for the purposes of the discussion.

Q/ Does this same car, now producing 1000hp but having the same drive-line convert the same, more or less horsepower into other forms of energy?

A/ The same.

Q/ What happens to the extra power (10x more in this example) produced by the engine and where does it go?

A/ All of it with the exception of the original 50hp lost as sound or heat plus a ridiculously small amount of extra heat and sound from extra bearing load etc is transferred to the driven wheels. Absolutely, positively nothing like the doubling and tripling mentioned earlier.

So is it safe to assume; a car makes 150rwkw and 206_engine_kw in stock form, which is 55kw drivetrain loss. if every thing else remain constant ie. drivetrain, tyres, dyno setting, temp and etc.

the same car now makes 250rwkw, its drivetrain loss or loss to other forms of energy will either remain at 55kw or increase marginally, but not exceed say 65kw?

same applys to 300rwkw and 330rwkw?

Edited May 25, 2007 by BaysideBlue

[sh@un]

If a car makes 250RWKW legitimately at the back wheels, you're looking at probably 251RWKW at the engine.

Drivetrains don't soak up much power at all.

i know i'm late

but is it too late to flame???

what a golden post- pls explain how a standard 32 GT-R makes around 170-5rwkW with the attesa fuse out then?

the bulk of the loss between the flywheel and the rear wheels on a chassis dyno occurs between the tyres and the rollers (slip/noise), with the gearbox, tailshaft, diff , driveshafts and wheels contributing the rest (mebe 5% between them)

good rule of thumb with 'Lines would be to multiply the r.w.k.W figure by 1.30-1.40 to get the rough equivalent at the fly.

[BaysideBlue]

i know i'm late

but is it too late to flame???

what a golden post- pls explain how a standard 32 GT-R makes around 170-5rwkW with the attesa fuse out then?

the bulk of the loss between the flywheel and the rear wheels on a chassis dyno occurs between the tyres and the rollers (slip/noise), with the gearbox, tailshaft, diff , driveshafts and wheels contributing the rest (mebe 5% between them)

good rule of thumb with 'Lines would be to multiply the r.w.k.W figure by 1.30-1.40 to get the rough equivalent at the fly.

that post surely deserves the flame.

but the

good rule of thumb with multiply the r.w.k.W figure by 1.30-1.40 to get the rough equivalent at the fly

has already proven to be misleading, in this thread. Edited May 25, 2007 by BaysideBlue

[sh@un]

i didn't even read the whole thread

Chassis dynos work on tractive effort and ramp-down inertial loads to come up with an a.t.w power figure- there are many things that the dyno operator can tweak to influence the final figure (ramp rate, correction variables, etc)

With sensible wheel sizes (16-19 inch, tops) with normal road tyres and normal pressures, and using a DynoDynamics rolling road dynamometer running in shootout mode only, I'd stand by a multiple of 1.30-1.40 giving a fairly accurate power range at the fly.

But what does it matter anyway? A Dyno is a tuning tool- may as well work out the power at the piston crown. It counts for s.f.a

[BaysideBlue]

i didn't even read the whole thread

Chassis dynos work on tractive effort and ramp-down inertial loads to come up with an a.t.w power figure- there are many things that the dyno operator can tweak to influence the final figure (ramp rate, correction variables, etc)

External variables like dyno tweaks have already been excluded from the equation. Assumed to be remain constant.

With sensible wheel sizes (16-19 inch, tops) with normal road tyres and normal pressures, and using a DynoDynamics rolling road dynamometer running in shootout mode only, I'd stand by a multiple of 1.30-1.40 giving a fairly accurate power range at the fly.

Multiples of 1.30-1.40 are reasonably accurate up to a certain power level for example; 150rwkw x 1.3 = 195kw, 200rwkw x 1.3 = 260 engine kw or even 250rwkw x 1.3 = 325 engine kw. these estimations are acceptable.

But what if you have 300rwkw x 1.3 = 390 engine kw ie 90kw loss through drivetrain? you would be question that. and what if you have 600rwkw x 1.3 = 780 engine kw, 180kw loss? what could have possibly robbed you of all that power? It couldnt have been the drivetrain otherwise it would be very inefficient wouldnt it?

But what does it matter anyway? A Dyno is a tuning tool- may as well work out the power at the piston crown. It counts for s.f.a

Its not a question of whether it matters or not. It about knowledge, finding the answer to a question, that many of us do actually wonder sometimes.

Edited May 25, 2007 by BaysideBlue

[sh@un]

and what if you have 600rwkw x 1.3 = 780 engine kw, 180kw loss? what could have possibly robbed you of all that power? It couldnt have been the drivetrain otherwise it would be very inefficient wouldnt it?

Already covered that- tyre slip against the rollers

I've seen cars with the boot loaded with friends or bags of sand/cement smoking the tyres against the rollers (we're talking mega power here)

Remember, the ramp rate is controlled by the load cells in the dyno (which can generate monstrous amounts of braking force)- the easiest outlet for the power is in the form of slip.

The only accurate way to compare figures is to measure on an engine dyno, if the air temps and relative humidity are consistent between engine tests- and even then it depends on weather both engine test rooms used SAE or DIN standards.

[BLK33T]

I think no one here really knows ... the truth seems to remain a mystery ... but they sound like good guesses tho ...

the only accurate way to get an idea of the amount of energy spent moving the mass of the drive train + the other factors such as heat, friction etc (as suggested) would be to do several tests comparing the engine and engine drive train combo <--- which no one is likely to try ... and that would only be reasonable if the next car in question had the same setup as the original test car/engine & drive train combo.

the hypothesis that the drive train under full load will rob a fixed amount of power + plus a proportional amount depending on load increase/ power + other factors such as fixed heat dissipation etc sounds the most accurate but it is probably much more complicated than that.

anyway the thread was a good read got me thinking ...