Converting Dyno nm to Engine nm?

[Dan355]

Just wondering if anyone knows how to convert a torque figure from a dyno run to a estimated figure for engine or at wheels? I was told by someone you need to devide it by your diff ratio but they werent 100% sure. Is this right?

[JD]

i dont know but try ringing turbo tune or another place that has a dyno and ask them they should know.

[shaggerz]

You do indeed need to divide it by diff ratio, but also the ratio of the gear you are in when the dyno run is done.

Example: 3.08:1 diff and dyno run done in second gear, which might be 1.63:1. Lets say the dyno reads 420NM torque at the wheels after those reductions, so divide that by the diff ratio first, which gives you 136.36NM at the tailshaft. Divide that by the 2nd gear ratio and you get 83.66NM at the flywheel.

Note that this torque reading is still lower than the real torque figure because the engine still has to drive the power steering pump, water pump, alternator, etc.

Also note that I am 100% NOT CONFIDENT in my analysis there, but that is how I understand it would work.

I would be thrilled if someone could either confirm or deny that what I have said is accurate, made up dyno'd torque figure aside.

[snake_VL]

i'll do a bit of maths here:

using some conversions i found on the net (don't know if the 3820 is accurate)

Power (hp) = (torque (Nm) x Rpm)/3820

so if you get a dyno torque of 400Nm at say 190km/h (i'm plucking these figures out of my arse here lol) in a 1:1 ratio of the box (usually 4th for manual) you can work out the axle rpm from the tyre circumference

190km/h is 52.778 m/s for a 2021mm circumference tyre 235/45/17
thats 1566.89 axle rpm

and power at the wheels = 164.07hp

assuming the power loss in the diff is zero (there would be some but it would be small compared to the gearbox loss)

1566.89*3.08 (the diff ratio) = 4826 gearbox and engine revs (because of the 1:1 ratio

so 164.07hp (assumed to be equal remember) = 4826rpm x torque/3820

so torque at the engine would be 129.87Nm

which is equal to 400Nm/diff ratio 3.08

SO....................

after all that yes, all you do is divide by diff ratio, as long as the car is in a 1:1 gear on the dyno

[snake_VL]

so some dyno printouts have calculated engine torque on them? because some dyno printouts i have found seem to have accurate torque readings on them eg. 240Nm for a RB30E

[Dan355]

Thanks snakevl. Will print that off and try to understand it a little more lol. Yea my nm is in the thousands i was like "wtf" when i first read the sheet. So i now have to find out ratios for my th700. Not sure if it was done in second i would say it was. Thanks again.

[snake_VL]

no worries, so all you do is divide by diff ratio then divide again by the gear ratio from the dyno run. this will give you an approximation of how much engine torque is making it to the rear wheels. there will be more in the engine than you calculate because of the power losses (and therefore torque losses) through the drivetrain. so it will be equivalent to a rwhp figure for power if you know what i mean

[Dan355]

Yea cheers mate