bitabur
Registered
Even if you don't want to dyno your bike, the liklihood is that your shift points will be almost the same as a stock bike. Get someone to post their baseline dyno and it should be pretty quick to analyze...
Anyway, I made some pictures, so I'll try to explain this as clearly as possible. Vertical axis is ft-lbs of torque, Horizontal axis is engine RPM.
Lets say you have this stock curve (and I know this doesn't look like the busa's curve, we'll get there):
If we want to calculate the shift point for this curve for the 1 to 2 shift for the busa (1st gear is 2.615:1, second is 1.937:1), we need only stretch (compress) the picture by (ratio 2/ratio 1) vertically, and by (ratio 2/ratio 1) horizontally, and then overlay the pictures. Stretching horizontally means you're matching the RPM scales of the pictures to the speed that the bike is moving, while stretching vertically means you're matching the scales of the pictures to the torque that is being put out at the crank. After stretching, we get something like this:
Then we can simply overlay them and see where they intersect:
So if this was your torque curve, your optimal shift point would be a little between 8000 and 9000 RPM. The true optimal would be a little higher because the vehicle speed drops between shifts.
Now, with a curve that looks more like the busa (at least at the end of the curve, which is the important part here):
The torque peak is later and doesn't fall off as much. In this case, the stretched image (still for the busa's 1-2 shift) looks like:
Now when I overlay the images, you can see that the curves never cross. In this case you would want to shift as late as possible (redline) to accelerate the fastest. I would guess that if you analyzed the busa's torque curves you would find this to be the case.
It is also important to note that to perfectly calculate this, a dyno would need to be run in each gear, with the dyno's resistance equal to the force required to accelerate the bike itself. Changes in the speed of accereration offset the dyno results, because the engine is also accelerating its own reciprocating parts.
Anyway, I made some pictures, so I'll try to explain this as clearly as possible. Vertical axis is ft-lbs of torque, Horizontal axis is engine RPM.
Lets say you have this stock curve (and I know this doesn't look like the busa's curve, we'll get there):
If we want to calculate the shift point for this curve for the 1 to 2 shift for the busa (1st gear is 2.615:1, second is 1.937:1), we need only stretch (compress) the picture by (ratio 2/ratio 1) vertically, and by (ratio 2/ratio 1) horizontally, and then overlay the pictures. Stretching horizontally means you're matching the RPM scales of the pictures to the speed that the bike is moving, while stretching vertically means you're matching the scales of the pictures to the torque that is being put out at the crank. After stretching, we get something like this:
Then we can simply overlay them and see where they intersect:
So if this was your torque curve, your optimal shift point would be a little between 8000 and 9000 RPM. The true optimal would be a little higher because the vehicle speed drops between shifts.
Now, with a curve that looks more like the busa (at least at the end of the curve, which is the important part here):
The torque peak is later and doesn't fall off as much. In this case, the stretched image (still for the busa's 1-2 shift) looks like:
Now when I overlay the images, you can see that the curves never cross. In this case you would want to shift as late as possible (redline) to accelerate the fastest. I would guess that if you analyzed the busa's torque curves you would find this to be the case.
It is also important to note that to perfectly calculate this, a dyno would need to be run in each gear, with the dyno's resistance equal to the force required to accelerate the bike itself. Changes in the speed of accereration offset the dyno results, because the engine is also accelerating its own reciprocating parts.
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