Posted by Indiana Ken on November 06, 2012 at 18:38:32 from (66.249.234.243):
In Reply to: True Horse Power posted by bill mart on November 05, 2012 at 21:40:15:
Quoting Removed, click Modern View to see
With respect I would like to comment on comparing the 6.0L to the 6.6L engine.
I am not familar with the 6.0L or the 6.6L engine or who manufactures them, so I am not taking sides here. I do maintain that the measure of an engine's capability is indeed HP. However, for those that insist on using torque as a measure the process is as follows:
1) Obtain the torque and RPM values for each engine at the engine HP peak.
2) Choose an RPM value that you wish to compare the torque output at.
3) The engine with the greatest torque output wins.
For example:
Lets assume we want to compare torque at 1700 RPM. You stated that the 6.6L will hold 500 ft-lbs at 3000 RPM, that would be 286 HP. However, the HP peak typically occurs after the torque begins to fall off. So lets assume the 6.6L can put out 450 ft-lbs at 3400 RPM, that would be 291 HP. For our example we will call it the HP peak. Since we want to compare torque at 1700 RPM we need a 2:1 gear reduction. At the output of the gear reduction the engine will produce 900 ft-lbs of torque at 3400 engine RPM. Note, (900 x 1700) 5252 = 291 HP, we are using torque for comparison but it is based on peak HP.
Repeat the process for the 6.0L engine. Take the torque value at the HP peak and multipy by the gear reduction required to reduce the RPM to 1700. If the 6.0L has greater HP than the 6.6L you should find it will produce greater torque when compare at a common RPM.
As other posters have mentioned there are loses in transmissions. However, the loses are small and in the above example both vehicles have transmissions, so the loses tend to cancel out.
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