Calculated drivetrain loss

There seems to be normal that drivetrain loss is expressed as a percentage, but I cannot understand why this can be correct.

Let`s say it takes 20HP to overcome the resistance of the drivetrain, before you get anything out to the hubs that can be measured.

If your engine produces 100 hp, you should see 80hp at the hubs. 10% loss.

But if the engine produces 200 hp, you should still se similar losses I would believe.

More load on the drivetrain will probably increase the friction somewhat, but not so much that you loose double the amout with double the power.

Makes no sense, please give me some input on this subject.

I have been looking for answers on this without success.

I dont`t care about the calculated engine power, but would like to understand the subject.

Good explanation here:

https://www.youtube.com/watch?v=E3F7oUQwwFs&t=1484s

Thank you for the video, however I feel it does not satisfy my curiosity on this subject.

I do not believe that double the power doubles the loss, think about the insane amounts of heat that would cause in the transmission.

Does the resistance/loss really have a linear relationship to power?

I have a hard time believing so.

Ready to be convinced otherwise.

It's 99% BS and guesswork, used by '"tuners", engine builders, dyno' operators and cry-babies to "explain"why their xxxhp monster is a yyy kitten.

It's true, there are drivetrain losses, but as you pointed out, if they were anywhere near the claimed figures there would be a car-shaped molten mass in the middle of the shop.

Probably the biggest factor is how engine builders use their dyno's, or more correctly the configuration compared to in a vehicle. With few exceptions builders will stack the deck in their favour, by using electric water pumps, open intakes with no filter or other erestictions, open exhausts and, in some instances, carefully sized exhaust manifolds when it will use cast in the vehicle, lower engine tempratures, and other tricks - basically SAE, if you wish to look that up.

In the vehicle the engine has to drive the water pump and sometimes a cooling fan, draw air through the filter assembly, pass the exhaust out through less efficient exhuast manifolds and the exhaust stem, etc. ALL these are parasitic losses, and when an engine builder, or OEM, quotes DIN, it's replicating the installed power that might be expected. Depending on the engine, this is usually around a 15% difference, on it's own, but can be much higher in some instances. You may have noticed that some vehicles get very close to, even exceed, their rated power when dyno'd, this is partly because the factory figures are a bit conservative, and a "minimum" figure.

The transmission can also affect the values.

With a manual in a direct 1:1 gear, the frictional losses in it are minimal - mostly from the lightly loaded, free-running gears and the viscous friction of the oil. If a non-direct gear is used, there will be a small increase in sliding friction and tooth deflection, but it's still minimal. However, in some instances, such as heavy vehicles, the energy/power loss in the gearbox can be enough to require additional cooling - still a lot less than some realise. Same principle applies with the differential gears.

With older automatics, there is a degree of "slippage" in the torque converter, this 'steals' power in the form of heat, as does the internal oil pump - with "lock-up" converters that loss is removed. The most extreme example is when an engine is "stalled" against the converter, ALL the power the engine is delivering at the flywheel is going into heating the transmission fluid! It's also why "street" cars with "loose" converters have trouble cooling the fluid - there is a lot more slippage. It's why one should run a "lock-up" when possible, even - especially - on high power, high stall "pro-street" cars!

Another factor is the wheel/tyre (tire) package, beside the inertial loading (see below), all tyres absorbe energy by deflection of their structure - the obvious one is the sidewal flexing, but less obvious is the flexing of the tread, itself. This flexing, and heating, is yet another parasitic loss.

The last (that comes to mind) is that inertial dyno's are just that - they use a roller of known totational inertia, and by measuring the change in roller rpm Vs time, the torque, and hence power, required to do that can be calculated. The problem is it's a strictly inertial calculation, and while it has the benefit of, for the good ones, measuring the total inertia on over-run when slowing down, to take account of the wheels' mass, drive train, etc, it can't account so well for the tyres' etc.

With eddy, or water, brake types, an acc'n rate is usually used - depending on the dyno', this will usually be a set rpm per second acc'n rate or a "road" speed acc'n rate. Changing the acc'n will change the net torque/power, because the faster the rate, the more torque is required to acc'ate the engine, etc. The highest values will be achieved in steady state.

Even with otherwise identical acc'n rates, tunes, weather, etc, there is usually be a "higher" power figure for hub than for rollers, because of the wheel assembly's inertia. Even as simple a change as tyre pressures, tyre type/compound, and/or light weight wheels can make a difference with the rollers.

Then there is the "correction" applied for air pressure, air temperature, atmospheric humidity, etc - different people may use different corrections, even when it may be built into the machine.

So, some losses may be proportional, some may worsen with rpm, and some may reduce.

My personal opinion is that ANY dyno' is just a tool, and should be used on a comparative basis when making changes. You should also not make direct comparisons between different dyno's.