
efxr6wagon 



I've noticed that some dyno graphs plot tractive effort (in Newtons) rather than torque in Nm. You can see that some stats in the Fordmods Dyno Registry (under Resources) show the Newtons  typically in the 35004500 range for an NA I6  entered as Nm. Some just ignore the last digit and treat the first three digits as the Nm. I have looked into it and don't think that's correct. But I'm no physicist or engineer, so could someone please check my workings?
I found this website that gives the conversion: http://www.webtec.co.uk/techinfo/equati ... /eqn17.htm. I applied the formula to my own recent dyno results (pre J3 chip). I made 322 Nm of torque at the wheels. My tyres are 225/5516 for a radius (half the diameter) of 327mm (see http://www.tyreteam.co.nz/tyreinfo4.ht ... isoncars1). My diff ratio is 3.45. The auto transmission was in 3rd gear  a 1:1 ratio. And I worked out from the dyno data that it was slipping 8.8% at peak torque  the driveshaft was turning only 912 rpm for every 1000 engine rpm  effectively 1.0965 gearing. The formula gives me: 322 Nm x 3.45 x 1.0965 x 1000 / 327 = 3725 Newtons. Can anyone confirm this? If this is right, some businesses may be publishing incorrect performance stats  see http://www.bpracing.com.au/FordESeries.html. The EF Fairmont with 3.45 gears and histall would have about 310 Nm ATW, instead of 360 (from 3600 Newtons). And the EF 'Mont with 4.11s would be around 320 Nm ATW, rather than 440 (from 4400 Newtons). I'm not trying to take anything away from anyone (they're awesome cars), and I probably have it wrong, but am trying to understand how this all works. And I'm happy to admit that I'm way out of my depth . Any help would be appreciated.
_________________ 95 EF XR6 wagon, 17" FTRs, DBA, QFM, KYB/Koni, AU block, ported 4angled EF XR6 head, SS Inductions, Territory intake, 10:1 CR, Auckland 1258 cam, Coby headers, no cat, Tickford 2.5", 2850rpm stall, J3 chip 

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shyun 



I'm pretty sure the tractive effort is just the force the tyre is exerting on the dyno. That formula you used is right, but if you read closely it says that it's the engine torque. So you calculated the effective engine torque (If the driveline had a 1:1 ratio and taking into account drivetrain losses). Or this would also be the rw torque with a 1:1 driveline ratio. To work out the torque at the wheels you just multiply the Tractive effort * tyre radius (Torque=Force*Radius). Then divide by the drieline ratio to calculate the effective torque at the engine. Using your data (3.45/0.327)*1.0965=11.57, so for you effective Torque=T.E/11.57. So for most people i guess 10 is close enough to 11 or 12. That's probably why some end up quoting meaningless figures like 650rwnm n/a .
Another way to calculate it is to use the power at the wheels. Power (KW)(@RPM)=Torque(NM)(@RPM)* RPM / 9550. This also calculates the effective torque produced at the wheels if the driveline had a 1:1 ratio, or engine torque as they're the same with a 1:1 ratio. So if you assume say a 25% driveline loss your 322rwnm would be 4/3*322= 429Nm at the engine.
_________________ ED XR6, LeMans Red, 5Speed 

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