Final-drive ratio: gears 14, 3.94:1; gears 56, 3.35:1
None of these equate anywhere near 1:1. 6th gear(the closest to a 1:1 ratio) has a torque multiplier of 0.85*3.35 = 1:2.85. Meaning that the engine turns 2.85 rotations for every 1 rotation the drive wheels turn.
Since dynojet's measure drum acceleraton and engine rpm (from ignition firing), mathmatically it should compute the same horsepower regardless of gear. With one exception... That being friction. Since the higher the torque multiplier, higher the friction coefficient. The example below demonstrates the frictional losses on a turbcharged VW R32 running through its first 5 gears. 1st and 2nd gear are quite low due to turbo spool-up, but you can see my point in gears 3,4, and 5. Where each one shows progressively more power. This is due to less friction from less torque mulitiplication.
Dynojet's work based on the time it takes to move the drum (fixed known inertia), and ignition firing to determine engine RPM. From that, software determines the torque multiplication at the wheels, and converts it into nice readable graphs.
While these measurements are taken from the wheel, it is important to remember that the actual torque at the wheels is much higher than what the graph states. Since its job is to factor out the gear amplification.
Example, say a MS3 dynojet graph shows a peak torque value of 280ft/lbs while being dyno'd in 4th gear. At the wheel, the actual torque meeting the drum is roughly 1290ft/lbs. This is down converted for the graph because the computer knows how many engine rotations were made during the time it accelerated the drum to record that amount.
Then based on that corrected torque number, the simple (torque*RPM)/5252 = Horsepower formula can be applied and plotted.
