Those numbers don't make sense at all, sorry.
If you apply the simple math and count whatever percentage on top of your numbers, say 18% for drivetrain loss, you'll arrive at a 410 making roughly 420 horses and 625 ft lbs of torque. No way.
This need to be computed with the rear gearing and trans gearing in the equation, and know if it is torque per axle, to get back to the flywheel.
If the dyno gives torque per axle (i.e., per wheel..the answer to which I do not know) then:
Total torque to the rear axle is 2x that, or 1060 ft lbs.
Divide that by the rear gear ratio (3.91 in this case) to get to the driveshaft torque (with no losses).
Then if the trans is in top gear, then you ignore the trans gear ratio and the driveshaft torque is the flywheel torque (with no losses).
Divide by [1-(loss%/100%)], which is .82 for an 18% total driveline loss.
Comes out to 330 ft-lbs net torque at the flywheel for 18% driveline loss, with all the losses in the engine like water pumps, fans, air cleaner, hot air under hood, etc.
