T Rex Bimota YB5

[ribbert]

  Any idea of the loss of hp at the crank through the drive? I could guess maybe 35hp

According to the dyno technician that did mine, a loss of around 18% is not uncommon at the back wheel.

Pete, wouldn't the loss at the back wheel be a constant, not a percentage?

Noel

[giantkiller]

I was thinking the same thing. If you have a 10hp motor you wouldn't lose much. But those turbo guys must have a brake locked up.

[Bill_Rockoff]

wouldn't the loss at the back wheel be a constant, not a percentage?

No, the more you put into a mechanical system that has friction losses, the more you lose. I can't think of a way to build something that always needs a fixed amount of power to move. A fixed amount of FORCE, maybe, but the faster you apply that force, the more power it is.

If you had a system with a constant loss of, say, 10 hp, you wouldn't be able to move it at all with a 10 hp motor. I'm having trouble even picturing a thing you couldn't budge with a 10 hp motor.  That's a motorcycle that you and half a dozen friends couldn't push, because each of you makes maybe 1 hp max.

But on a normal actual system of parts, if you put in a little power, you lose a little and get the rest out; if you put in more power, you lose more.

18% is pretty typical for a rear-wheel-drive car with a manual transmission. You lose a bit in the transmission gears, a bit more than that in the rear diff, and a little in the drive-shaft and half-shaft's CV joints. A solid-axle car does a little better because there are no half-shafts. A transverse-engine car, like most front-wheel-drive setups or like a mid-engine transverse car, does a little better than that because there's no 90-degree change of rotation in the diff, all the shafts rotate in parallel axes. A typical torque-converter automatic does a little worse. A motorcycle with a chain or belt drive does better (I've read "15%" but a shaft drive is back to 18% or more, depending on whether its driveshaft has a 90-degree gearset at each end or just at one (longitudinal crankshaft like a Honda ST or a Moto Guzzi or a boxer BMW.)

[ribbert]

No, the more you put into a mechanical system that has friction losses, the more you lose....

True, but is the increased loss proportional to the increased power?

Noel

[Pat Conlon]

So Pat has the 1380 cured the oil consumption problem?  And I don't remember what happened to cause the rebuild?

The top end coked up from excessive oil. I had the 82mm Hank Scott Prolite short skirt pistons.
I decided to change course and use the conventional Ross 83mm long skirt pistons.
All is well since.

[oldktmdude]

wouldn't the loss at the back wheel be a constant, not a percentage?

No, the more you put into a mechanical system that has friction losses, the more you lose. I can't think of a way to build something that always needs a fixed amount of power to move. A fixed amount of FORCE, maybe, but the faster you apply that force, the more power it is.

If you had a system with a constant loss of, say, 10 hp, you wouldn't be able to move it at all with a 10 hp motor. I'm having trouble even picturing a thing you couldn't budge with a 10 hp motor.  That's a motorcycle that you and half a dozen friends couldn't push, because each of you makes maybe 1 hp max.

But on a normal actual system of parts, if you put in a little power, you lose a little and get the rest out; if you put in more power, you lose more.

18% is pretty typical for a rear-wheel-drive car with a manual transmission. You lose a bit in the transmission gears, a bit more than that in the rear diff, and a little in the drive-shaft and half-shaft's CV joints. A solid-axle car does a little better because there are no half-shafts. A transverse-engine car, like most front-wheel-drive setups or like a mid-engine transverse car, does a little better than that because there's no 90-degree change of rotation in the diff, all the shafts rotate in parallel axes. A typical torque-converter automatic does a little worse. A motorcycle with a chain or belt drive does better (I've read "15%" but a shaft drive is back to 18% or more, depending on whether its driveshaft has a 90-degree gearset at each end or just at one (longitudinal crankshaft like a Honda ST or a Moto Guzzi or a boxer BMW.)

Thanks for your answer Bill. That's what my reply would have been but you have explained it far better than I could have. Every dyno figure that I have seen, always quotes the rear wheel hp loss as a percentage. During the first dyno run that I did, my drive chain had a tight spot in it and this showed up as a intermittent loss of around 5 to 8 hp. Just shows how  relatively small changes in friction values can effect the output figures.   
   Regards, Pete.

[giantkiller]

So Pat has the 1380 cured the oil consumption problem?  And I don't remember what happened to cause the rebuild?

The top end coked up from excessive oil. I had the 82mm Hank Scott Prolite short skirt pistons.
I decided to change course and use the conventional Ross 83mm long skirt pistons.
All is well since.

I have an APE block and a Wisco block (cylinders). I can't remember I think one was 1380 and one was 1450? I've had them for while. Was thinking of using the Ross pistons. And seeing if Randy might want to build me a couple of monster motors. In a couple of years If I can gather enough money.

[Flynt]

wouldn't the loss at the back wheel be a constant, not a percentage?

doesn't make sense...  let's say it is a constant 15 HP.  Would the bike make negative Hp below the "drivetrain loss" Hp of say 15Hp?  It is, of course, a percentage.  It is not truly linear however... but that's beyond the scope of this discussion.

Frank