Figuring FWHP from RWHP ?

[427dart]

I pulled up my Desk Top Dyno files from when I built this engine and pulling up the power Graph chart they show Flywheel and Wheel HP.
The WHP numbers on there were almost exactly like the chassis dyno test yesterday and on the FWHP 550 was the top number!

Who knows...it runs great though!

[user-23911]

F1-engines transaxles used to have lots of cooling problems as they need about 50Kw of cooling capacity

Have you any idea what 50Kw represents as heat?

50 1 bar heaters.

It's just not happening.

Drivetrain losses go up with road speed, not engine output.

[Steve.k]

We see at least 100hp from Dyno to car in terms of hp difference. Pretty much all the time.

[gruntguru]

First, the drive train losses aren't simple frictional losses that produce heat. As a sanity check, 50 hp is 37 kW - picture 37 1000 watt radiant heaters stuffed into the trunk. You might get to the end of a 1/4 mile pass without a fire but a stock car would melt down into a puddle of slag in a couple of laps so it can't be that simple. There are frictional losses in all of the bearings, in the rolling resistance of the tires, and power used to whip up the transmission and rear end lubes, but this all totals about 5-10 hp. It does change with vehicle speed, but not by much. A drag car can pretty much ignore it, and a stock car just adds a differential and manual trans cooler. For now ignore the torque converter if there is one, I'll come back to it at the end. So where does that 50 hp of drive train loss go? Think of all of the spinning parts of a transmission, flywheel, clutch, driveshaft, ring and pinion, axles, and the driven wheels and tires as one giant flywheel. This starts at rest and has to be spun up to some final speed as the car goes down the drag strip, and that takes power that now isn't available to accelerate the car so this power is the drive train loss. It isn't really lost, it is just stored in the rotational inertia of that giant composite flywheel, at least until the car comes back to rest when it gets converted into heat in the brakes. Now think of your 500 hp car and assume it runs a 10.0 sec quarter mile at 136 mph, and say the power lost spinning up that flywheel is 50 hp. Now make it 800 hp at the same weight and gearing so it runs 8.55 at 159 mph. Now it has to spin that flywheel up in 8.55 / 10.0 = 0.855 or only 85.5% of the time, and it has to spin it up to 159 / 136 = 1.17 or 17% faster rpm. Same car but now the drive train loss has grown because it has more hp. I can't say if it stayed at 10% and grew from 50 to 80 hp exactly, but assuming a constant percentage is a simple and not terribly inaccurate estimate.

Now, torque converters filled with trans fluid. They are large and heavy and make a huge flywheel all by themselves and they never reach 100% coupling without a clutch and often don't get below 10% slip until at least 1000 rpm over the stall speed (every brand and size is different, this is the general idea), so they do convert a lot of engine hp into heat in the fluid so the actual frictional losses can be 10's of hp as well - that's why drag cars need such big trans coolers. Plus, the spinning internals of an auto trans make a much bigger flywheel than a manual trans, which with the extra losses involved in driving the pump in the transmission account for the larger drive train losses of an automatic compared to a stick car.

Some chassis dynos can load the car at constant speed. None of the inertial losses apply under those conditions. Sure the Dynojet graphs are done in "flywheel" mode but AFAIK Dynojet apply a "correction" which assumes a typical moment of inertia for engine and driveline.

Dyno Dynamics do this even though they use an absorber/retarder. They also have a feature that will ramp-up, ramp-down and then software-average the two plots.

[ijames]

First, the drive train losses aren't simple frictional losses that produce heat. As a sanity check, 50 hp is 37 kW - picture 37 1000 watt radiant heaters stuffed into the trunk. You might get to the end of a 1/4 mile pass without a fire but a stock car would melt down into a puddle of slag in a couple of laps so it can't be that simple. There are frictional losses in all of the bearings, in the rolling resistance of the tires, and power used to whip up the transmission and rear end lubes, but this all totals about 5-10 hp. It does change with vehicle speed, but not by much. A drag car can pretty much ignore it, and a stock car just adds a differential and manual trans cooler. For now ignore the torque converter if there is one, I'll come back to it at the end. So where does that 50 hp of drive train loss go? Think of all of the spinning parts of a transmission, flywheel, clutch, driveshaft, ring and pinion, axles, and the driven wheels and tires as one giant flywheel. This starts at rest and has to be spun up to some final speed as the car goes down the drag strip, and that takes power that now isn't available to accelerate the car so this power is the drive train loss. It isn't really lost, it is just stored in the rotational inertia of that giant composite flywheel, at least until the car comes back to rest when it gets converted into heat in the brakes. Now think of your 500 hp car and assume it runs a 10.0 sec quarter mile at 136 mph, and say the power lost spinning up that flywheel is 50 hp. Now make it 800 hp at the same weight and gearing so it runs 8.55 at 159 mph. Now it has to spin that flywheel up in 8.55 / 10.0 = 0.855 or only 85.5% of the time, and it has to spin it up to 159 / 136 = 1.17 or 17% faster rpm. Same car but now the drive train loss has grown because it has more hp. I can't say if it stayed at 10% and grew from 50 to 80 hp exactly, but assuming a constant percentage is a simple and not terribly inaccurate estimate.

Now, torque converters filled with trans fluid. They are large and heavy and make a huge flywheel all by themselves and they never reach 100% coupling without a clutch and often don't get below 10% slip until at least 1000 rpm over the stall speed (every brand and size is different, this is the general idea), so they do convert a lot of engine hp into heat in the fluid so the actual frictional losses can be 10's of hp as well - that's why drag cars need such big trans coolers. Plus, the spinning internals of an auto trans make a much bigger flywheel than a manual trans, which with the extra losses involved in driving the pump in the transmission account for the larger drive train losses of an automatic compared to a stick car.

Some chassis dynos can load the car at constant speed. None of the inertial losses apply under those conditions. Sure the Dynojet graphs are done in "flywheel" mode but AFAIK Dynojet apply a "correction" which assumes a typical moment of inertia for engine and driveline.

Dyno Dynamics do this even though they use an absorber/retarder. They also have a feature that will ramp-up, ramp-down and then software-average the two plots.

Yes, that should get you comparable numbers to an engine dyno minus the drive train friction losses, without any of the flywheel effect. However, when you compare an engine dyno to a drag strip you have the full flywheel effect at the track. I guess that's another unstated assumption I made, that what you really care about is the et and what whp you need to run that time, and then what engine hp you need to get that whp.

[MadBill]

First, the drive train losses aren't simple frictional losses that produce heat. As a sanity check, 50 hp is 37 kW - picture 37 1000 watt radiant heaters stuffed into the trunk. You might get to the end of a 1/4 mile pass without a fire but a stock car would melt down into a puddle of slag in a couple of laps so it can't be that simple. There are frictional losses in all of the bearings, in the rolling resistance of the tires, and power used to whip up the transmission and rear end lubes, but this all totals about 5-10 hp. It does change with vehicle speed, but not by much. A drag car can pretty much ignore it, and a stock car just adds a differential and manual trans cooler. For now ignore the torque converter if there is one, I'll come back to it at the end. So where does that 50 hp of drive train loss go? Think of all of the spinning parts of a transmission, flywheel, clutch, driveshaft, ring and pinion, axles, and the driven wheels and tires as one giant flywheel. This starts at rest and has to be spun up to some final speed as the car goes down the drag strip, and that takes power that now isn't available to accelerate the car so this power is the drive train loss. It isn't really lost, it is just stored in the rotational inertia of that giant composite flywheel, at least until the car comes back to rest when it gets converted into heat in the brakes. Now think of your 500 hp car and assume it runs a 10.0 sec quarter mile at 136 mph, and say the power lost spinning up that flywheel is 50 hp. Now make it 800 hp at the same weight and gearing so it runs 8.55 at 159 mph. Now it has to spin that flywheel up in 8.55 / 10.0 = 0.855 or only 85.5% of the time, and it has to spin it up to 159 / 136 = 1.17 or 17% faster rpm. Same car but now the drive train loss has grown because it has more hp. I can't say if it stayed at 10% and grew from 50 to 80 hp exactly, but assuming a constant percentage is a simple and not terribly inaccurate estimate.

Now, torque converters filled with trans fluid. They are large and heavy and make a huge flywheel all by themselves and they never reach 100% coupling without a clutch and often don't get below 10% slip until at least 1000 rpm over the stall speed (every brand and size is different, this is the general idea), so they do convert a lot of engine hp into heat in the fluid so the actual frictional losses can be 10's of hp as well - that's why drag cars need such big trans coolers. Plus, the spinning internals of an auto trans make a much bigger flywheel than a manual trans, which with the extra losses involved in driving the pump in the transmission account for the larger drive train losses of an automatic compared to a stick car.

Some chassis dynos can load the car at constant speed. None of the inertial losses apply under those conditions. Sure the Dynojet graphs are done in "flywheel" mode but AFAIK Dynojet apply a "correction" which assumes a typical moment of inertia for engine and driveline.

Dyno Dynamics do this even though they use an absorber/retarder. They also have a feature that will ramp-up, ramp-down and then software-average the two plots.

Gear drive efficiency is measured as output power (which of course also varies with RPM) divided by input, and per the table in this document: http://www.roymech.co.uk/Useful_Tables/ ... iency.html , ranges from as lowv as 70% in some hypoid gears to 99% for the best spur gears.

Those ijames numbers are way low. Even with coolers the size of our 700 HP* engine's oil cooler, the diff and gearbox of our road race car run at 250-280°F. Per attached graph, the power loss difference alone between various diff gear designs can be over 10 HP at under 6,000 RPM. (*Obviously, it doesn't spend 100% of a lap at peak power, so the average load is probably no more than half that.)
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9 inch Vs. 12 bolt.jpg

[ijames]

I agree, Bill, my example friction numbers are too low and Naukkis' numbers are probably closer to today's reality. I was going from memory from some old OEM reports from many years ago so the total hp was much less, sigh. The principles haven't changed but I'm still more numerically calibrated for the 300-400 hp street cars my friends had back in the 70's, and not today's 700+ hp showroom stock monsters and the race engines I come on here to read about. Back to lurking for me.

[Erland Cox]

Ijames, your explanation was the best I ever read.
I have seen that lots of the losses are in wheel speed on rolling roads with dual drums.
The tire wedges itself more and more between the drums.
On the same car with different axle ratios the losses increase with wheel speed.
The test should be done in direct drive in a manual gear box.
There are no loaded gears and less losses then even if the difference is not that big.

Erland

[arlancam509]

i should continue to lurk, you should not, ijames.

[Big Al]

Te = Tw + Fr + Ta / nt
Te = Enginetorque (Nm)

Tw = Wheeltorque (on dynopaper) (Nm)


Fr = (Nf * Nk * b / r) / ut
Fr = Roll resistance (Nm)

Nf = Normalforce (gravity) (m/s^2)
Nk = Gravity on wheels (kg)
b = Cr * r
Cr = Roll coefficient
r = Wheelradius inc slip (m)


nt = nv * nd
nt = Total Efficiently powertrain (%/100)

nv = efficiency gearbox (%/100)
nd = efficiency diff/diffs (%/100)


Ta = (Id * w') + (Ih * (w' / ut))
Ta = Power torque (inertia) (Nm)

Id = moment of inertia powertrain (+clutch) without flywheel (kg/m^2)
Ih = moment of inertia wheels (kg/m^2)
w' = angular acceleration Engine (rad/s^2)
ut = Total total gear ratio (transmission * diff)

My own calculations on how to calculate the engine torque from wheel torque in a dyno.
Translated from Swedish so ...

[peejay]

F1-engines transaxles used to have lots of cooling problems as they need about 50Kw of cooling capacity

Have you any idea what 50Kw represents as heat?

50 1 bar heaters.

It's just not happening.

Drivetrain losses go up with road speed, not engine output.

A rule of thumb is that an engine needs as much cooling capacity as the amount of horsepower it is making, power from fuel burnt being roughly evenly split between usable power at the flywheel and waste heat, with the rest being spent out the exhaust system as heat and exhaust energy.

50kw engines do not need very large radiators.