Wheel Horsepower to Crankshaft Horsepower
Wheel Horsepower to Crankshaft Horsepower
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Discussion

Pasteurised

Original Poster:

324 posts

182 months

Monday 12th January 2015
quotequote all
I apologise if this has been done to death but I wondered if anyone could shed some light on this for me.

Let's say my car, in stock form makes 130 at the wheels and 150 at the crank.
If the car is then tuned and now makes 185 at the wheels is it now 205 at the crank?

I've seen people use percentages but that doesn't make sense to me, since it implies that the drive line saps more energy as the power is increased.

SuperchargedVR6

3,138 posts

250 months

Monday 12th January 2015
quotequote all
The dyno software calculates the drivetrain losses during coast down, so it's not just a simple percentage. I've seen lighter wheels and brakes affect the readings. WHP or KW should be the figure quoted, but people would be disappointed.

Edited by SuperchargedVR6 on Monday 12th January 13:49

gavsdavs

1,259 posts

156 months

Monday 12th January 2015
quotequote all
Pasteurised said:
I apologise if this has been done to death but I wondered if anyone could shed some light on this for me.

Let's say my car, in stock form makes 130 at the wheels and 150 at the crank.
If the car is then tuned and now makes 185 at the wheels is it now 205 at the crank?

I've seen people use percentages but that doesn't make sense to me, since it implies that the drive line saps more energy as the power is increased.
If the car makes 100 at the fly and it translates to 80 at the wheels, you've lost 20 percent. For those figures it's also 20 hp, but it's incidental. It sensible to think of it in terms of 'percentage loss' as opposed to 'absolute amount loss'.

underphil

1,357 posts

240 months

Monday 12th January 2015
quotequote all
who cares about how much power there is at the flywheel? - at the wheels is all that matters

xRIEx

8,180 posts

178 months

Monday 12th January 2015
quotequote all
Pasteurised said:
I've seen people use percentages but that doesn't make sense to me, since it implies that the drive line saps more energy as the power is increased.
In a way it does - friction is a resistive force and increases as relative velocity increases. I doubt it's as linear as e.g. "20% is always lost" but it's much closer than e.g. "18bhp is always lost".

SuperchargedVR6 said:
WHP ... should be the figure quoted, but people would be disappointed.
This - it's the net result at the wheels that is going to influence performance. My Impreza apparently has 275bhp, but drivetrain losses are likely to be greater than, say, a RWD BMW that also has 275bhp.


And while we're at it, can we convince people to stop say "BHP at the wheels" etc.? A brake was a device that measured horsepower at the crank, so wheel- and brake- are mutually exclusive.

GroundEffect

13,864 posts

186 months

Monday 12th January 2015
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It is definitely a percentage.

AngryPartsBloke

1,439 posts

181 months

Monday 12th January 2015
quotequote all
underphil said:
who cares about how much power there is at the flywheel? - at the wheels is all that matters
How are you supposed to reduce drivetrain loss if you have no idea what the loss is?

Pasteurised

Original Poster:

324 posts

182 months

Monday 12th January 2015
quotequote all
underphil said:
who cares about how much power there is at the flywheel? - at the wheels is all that matters
Someone who wants to learn smile

xRIEx

8,180 posts

178 months

Monday 12th January 2015
quotequote all
AngryPartsBloke said:
underphil said:
who cares about how much power there is at the flywheel? - at the wheels is all that matters
How are you supposed to reduce drivetrain loss if you have no idea what the loss is?
So the correct answer to the question is, "powertrain engineers care."

kambites

71,594 posts

251 months

Monday 12th January 2015
quotequote all
I think the unfortunately answer is that it's neither a fixed amount of power nor a fixed percentage of the power being developed, but rather a hybrid of the two. Losses definitely do increase with both load and speed, but not linearly. It also depends on things like what gear you're in and how much slip you have across the differential.

Luigi123

29,052 posts

213 months

Monday 12th January 2015
quotequote all
kambites said:
I think the unfortunately answer is that it's neither a fixed amount of power nor a fixed percentage of the power being developed, but rather a hybrid of the two. Losses definitely do increase with both load and speed, but not linearly. It also depends on things like what gear you're in and how much slip you have across the differential.
This make sense. If you were driving gently the engine may only be making 20bhp so you couldn't have a straight 20bhp transmission loss.

zeppelin101

724 posts

222 months

Monday 12th January 2015
quotequote all
Your drivetrain friction can change incrementally between two runs so back-calculating flywheel horsepower from measured wheel horsepower is a non-starter.

Losses are largely a function of friction and mass. Engine / gearbox friction, gearbox / prop mass etc etc The only things that stay fixed there are the masses. The friction is a function of fluid temperatures and can change by a rather large amount for a small change in temperature depending on the design of the components / wear and so on.

For me, calculating back engine horsepower from measured wheel horsepower doesn't make sense. The losses can be changed upstream through changes in wheel mass etc or lower viscosity oils or changing the mean temperature but working back to engine horsepower when everything in between is changing dynamically? No thanks.

Pasteurised

Original Poster:

324 posts

182 months

Monday 12th January 2015
quotequote all
Frictional losses increase with rpm, so in my first example I am making peak power at 6000rpm where the 20hp losses are being measured, so at 3000rpm there would be only a 10hp loss etc

zeppelin101

724 posts

222 months

Monday 12th January 2015
quotequote all
Pasteurised said:
Frictional losses increase with rpm, so in my first example I am making peak power at 6000rpm where the 20hp losses are being measured, so at 3000rpm there would be only a 10hp loss etc
Not necessarily no, friction follows a polynomial model not linear. It still takes some amount of torque to get an engine to start turning from 0 rpm, for instance (which is also temperature dependent!)

underphil

1,357 posts

240 months

Monday 12th January 2015
quotequote all
AngryPartsBloke said:
underphil said:
who cares about how much power there is at the flywheel? - at the wheels is all that matters
How are you supposed to reduce drivetrain loss if you have no idea what the loss is?
Whether you measured 30bhp or 40bhp losses - it's a total drivetrain loss, rather than an itemised breakdown so wouldn't really help
- you still have the same list of items that you could improve, and the change in measured wheel horsepower would tell you of any improvement

john banks

276 posts

220 months

Monday 12th January 2015
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Quote from an old contact who used to design dynos:

"Firstly I don't subcribe to the fixed overall % increase with increased wheel power. The resaon I say this is that the parts that do change with a greater power input such as the gearbox, diff, tyre (deformation change only) will only contribute around 3 - 5% of the total 'losses or drag' as measured/displayed on a roller style chassis dyno. Much of the losses aren't in fact losses, but unmeasured power requirements. At constant speed, power is calculated by measuring the torque reaction of the retarder which is trunnion mounted & the rpm @ which the torque is produced (P = Torque * RPM (roller rpm!) / 5252 or 9549 depending on imperial or metric measurments) However there is power required just to be able to turn over the rollers & retarder, windage of the retarder (which has rotors very similar to a centrifugal pump) boundary layer windage from the rollers, parasitic drag from bearings, seals & grease (same goes for the gearbox) etc. So just to be able to turn over the rollers requires power in itself that goes unmeasured by the retarder. The power unmeasured increases with speed in an exponential fashion. The other power losses (or drag as it is displayed on SUN/MAHA dyno's) is a combination of heat & noise from the tyres, gearbox, bearings etc, windage from rotating components such as wheels/tyres, brake disks, & shafts. Other factors which cause different drag/losses is the seperation of the rollers, those dyno's which have the rollers further apart cause greater drag, as the tyre deformation is greater, as it causes the tyre to be 'pinched' the same goes for smaller diameter rollers. The faster the 'road speed' the greater the losses, as the heat/noise losses goes up, as does windage of the afore mentioned parasitic absorbtions."

AWHP can overestimate losses compared to RWHP. My Subaru on some dynos "lost" much less than some RWD cars on others. WHP means little between dynos of different types and is on no way a true representation of acceleration even on cars of similar weight as I proved ridiculously on an airfield day that followed a dyno day.

underphil

1,357 posts

240 months

Monday 12th January 2015
quotequote all
john banks said:
Quote from an old contact who used to design dynos:

"Firstly I don't subcribe to the fixed overall % increase with increased wheel power. The resaon I say this is that the parts that do change with a greater power input such as the gearbox, diff, tyre (deformation change only) will only contribute around 3 - 5% of the total 'losses or drag' as measured/displayed on a roller style chassis dyno. Much of the losses aren't in fact losses, but unmeasured power requirements. At constant speed, power is calculated by measuring the torque reaction of the retarder which is trunnion mounted & the rpm @ which the torque is produced (P = Torque * RPM (roller rpm!) / 5252 or 9549 depending on imperial or metric measurments) However there is power required just to be able to turn over the rollers & retarder, windage of the retarder (which has rotors very similar to a centrifugal pump) boundary layer windage from the rollers, parasitic drag from bearings, seals & grease (same goes for the gearbox) etc. So just to be able to turn over the rollers requires power in itself that goes unmeasured by the retarder. The power unmeasured increases with speed in an exponential fashion. The other power losses (or drag as it is displayed on SUN/MAHA dyno's) is a combination of heat & noise from the tyres, gearbox, bearings etc, windage from rotating components such as wheels/tyres, brake disks, & shafts. Other factors which cause different drag/losses is the seperation of the rollers, those dyno's which have the rollers further apart cause greater drag, as the tyre deformation is greater, as it causes the tyre to be 'pinched' the same goes for smaller diameter rollers. The faster the 'road speed' the greater the losses, as the heat/noise losses goes up, as does windage of the afore mentioned parasitic absorbtions."

AWHP can overestimate losses compared to RWHP. My Subaru on some dynos "lost" much less than some RWD cars on others. WHP means little between dynos of different types and is on no way a true representation of acceleration even on cars of similar weight as I proved ridiculously on an airfield day that followed a dyno day.
so best use a hub dyno ? (and take into consideration wheel and tyre size/weight)

john banks

276 posts

220 months

Monday 12th January 2015
quotequote all
Personally I view a dyno as a tuning tool, and relative changes in power second to second or minute to minute matter rather than absolute figures, but I'd rather tune on a large expanse of tarmac. I'm only interested in turbo engines though, so that alters the relative merits.

GroundEffect

13,864 posts

186 months

Monday 12th January 2015
quotequote all
underphil said:
so best use a hub dyno ? (and take into consideration wheel and tyre size/weight)
Of course.