The Stovebolt Site Forums The Shop Area Engine Shop Factory vs. Dyno torque

What is the difference between present day factory rated torque figures in the fancy dealer brochures and the torque at the back wheels on a dynometer?

This past weekend a friend had a Predator upgrade chip installed in his 2003 Dodge diesel. It is factory rated at 550 ft lbs and it was dyno'ed before and after the chip was installed. Before it was installed it had around 750 ft lbs and after it was installed it had 937 ft lbs of torque. The dyno result was taken in 5th gear which is direct in the Dodge 6 speed transmission.

Why would the factory rate it at 550 if it is indeed 750 at the rear wheels?

I thought it would be less at the rear wheels.

It sure did pull with the chip installed. We were towing a tandem axle car trailer with a 94 Pontiac Sunbird on the trailer. We weighed 14,000 lbs at a scale and this thing drove down the highway at 70 mph and passed like there was no tomorrow.

Well, you do have the gears in the rear axle to multiply the torque from the engine. Assuming drivetrain loss is 27%, the 550 foot-lbs at the flywheel with 3.73 gears come out to 550*3.73*.73=1500 foot-lbs at the axle, divide by two or so for the wheel:dyno ratio
torque=750 ft-lbs
Back calculating gives you about 688 ft-lbs at the flywheel with the new chip.

Why divide by two? Aren't the drums in a chassis dyno connected together?

I puzzled over these numbers as well. The chassis dyne is going to be a measurement of dyno drum torque and 750 lb-ft is meaningless without also knowing the drive wheel diameter and the dyno traction drum diameter and having verification of the rearend ratio.

I have done work on chassis dynes for transmission test stands for Borg Warner and if there is anything the big auto makers are (in)famous for it is specsmanship.

Lets say the dyno rollers are 16 inches in diameter and truck tires are 32 inches (easier math :p )

That means that 750 lb-ft at the drum is 1500 lb-ft at the rear wheel / 3.73 = 402 lb-ft at the flywheel. My experience tells me that most drivetrains of this type are actually much higher than 95 - 97 percent efficient which in the worst case would be 402/.95 ~= 425 lb-ft of real torque at the flywheel.

That sounds about right . . . if the dyno drums were smaller, that would mean higher engine torque. I bet the drums are smaller than 16 inches, I have seen them between 12 inches and 14 inches most commonly which would equate to something a bit closer to 500 lb-ft at the flywheel.

zatmakesense?

I fudged the efficiency to fit the numbers. The factor of two is the dyno:wheel ratio, like Ken said. That should have been measured and corrected for by the dyno operator, but probably wasn't. Sorry about the mud, I was in a hurry.

I looked up the factory gear ratio on Dodge's web site and the 27% loss is a very typical number I hear from high performance guys who care about their rear wheel horsepower.

The drivetrain losses due to gears and bearings are fairly insignificant, but the loss from the rolling resistance of the tires is big. It is the dominant loss usually, followed by accesories, which are usually unconnected for the factory advertised numbers. The alternator, fan, water pump, and power steering all draw quite a bit of power. If you disconnect the accesories, overinflate the tires, and run an electric water pump on the dyno you will see a much better driveline efficiency.

Thanks guys, but doesn't the dyno take this into account? I thought the operator had to input the tire size and rear axle ratio? I will ask my friend if they did.