1st-2nd: 7200rpm (JWT ECU, let's go to redline) 2nd-3rd: 6850rpm 3rd-4th: 6625rpm 4th-5th: 6625rpm

How did I get these numbers? Let's do the *feel* first, it's easy:

1. Get a dynochart w/ torque & hp vs. RPM

2. Know the gear ratios, these are the same for NA & TT 5spd:

1st: 3.21:1

2nd: 1.93:1

3rd: 1.30:1

4th: 1.00:1

5th: 0.75:1

3. Find the torque peak value, most dynocharts supply this number. Mine is 356.8 in this chart. BUT I want to maximize pull in each gear, so I'll want to start each gear around 3500rpm. That's the front of my torque 'peak' which stays pretty flat across to 4800rpm. Now we want to work out the gear ratios to put us at 3500rpm after each shift.

Got it? If I shift at these points, I will land back at 3500rpm & the front of a 1300rpm peak area to get the snappin' happenin' Peak torque will equal maximum acceleration for that gear, so I'm trying to maintain the highest amount of torque across a rev range.

Now, at the track, we're shooting for tangible numbers, measured by machine. Seat of the pants is out the window. Here's how to do the *real*:

1. Get a dynochart w/ torque & hp vs. RPM

2. Know the gear ratios, these are the same for NA & TT 5spd:

1st: 3.21:1

2nd: 1.93:1

3rd: 1.30:1

4th: 1.00:1

5th: 0.75:1

3. Gearing is a multiplier for torque, so even if the torque isn't as high chart wise towards the end, the gearing makes the mechanical difference & results in maximum acceleration. I'm just going to run down my chart so you can see how the process works, oh, here's the dyno chart again:

4. 3.21 0000(000)000.00 is nothing, literally, but the next chunk is the first shift point: 3.21 7200(220)706.20 -> 1.93 4327(348)671.64 at 7200rpm, the chart shows 220ft-lbs of torque. OK, I know, it really doesn't... think interpolation for a sec. I'm going to multiply the torque value by the gear ratio:

220 ft-lbs * 3.21 gear ratio = 706.2 ft-lbs of output shaft torque

5. Now we use the gear ratio finder from above to determine where this shift point will land us in the next gear:

6. At 4327 RPM we see the torque is 348ft-lbs. Multiply this by the gearing:

348 ft-lbs * 1.93 gear ratio = 671.64 ft-lbs of output shaft torque

This is significantly less (34.57ft-lbs) than the amount of power we shifted from. Why shift when the ground is still feeling more force then in the next gear? But, we ran out of RPM to work with, forcing a shift into the next gear.

For the next gears there's a lot of trial & error. I wrote a script to do most of the math for me, but it can be done on paper with a little more time. The easy way is to start graphing the torque from redline back in 50 RPM measurements. These numbers can be used through all gear readings, just use different multipliers for gearing. Keep doing the math until you find a shift point where the output shaft torque is the same or slightly less in the next gear or as close as possible. The first shift had a 34.57ft-lbs difference, the rest you can hit very close in the next gear. The shape of the torque curve will determine how close you can get.

If you want me to polish up the script & make it user friendly I can...

Uh, then again, I never found anything definitive on how to do this. I could be way out in left field. Cuss me if necessary.