Theoretical question.
Discussion
Alfanatic said:
I don't mind, as long as they provide some constructive feedback on the table I posted and explain where it goes wrong, if it is wrong.
I know there's a constant relationship between power, torque and speed, but since torque at the wheels is dependant on gearing, power must be too, since there's a constant relationship between them.
I'm more than happy to accept that Parkers is right, but you'll need to be a bit more convincing, like showing me how the relationship works with an equation or two, and how the gearing is already included, if that's not too much trouble?
So, instead of saying "do the maths", you could show us the maths that proves you're right. Or you could just bang your head against a wall again like you did last time someone disagreed with you. Your choice, I'd preferred it if you chose to enlighten me though. I don't mind being proven wrong.
So how about it? What's the maths that supports Parkers?
EDIT: If it helps any, my maths is telling me that it would be power at the wheels that matters, which depends on torque and wheel rpm - or road speed, but if you change gear, the wheel rpm and road speed don't change at that instant, but if the torque goes down, you now must have less power at the wheels, so acceleration will reduce.
This may be flawed reasoning, but it is why I'm struggling to accept what you are saying at face value. I'm not an expert, and I'd like an expert to explain this to me. As would, I am sure, the OP.
I understand your confusion, you can't get your head around the fact that if the wheel speed is constant and you up-shift to a longer gear, the torque must drop and thus the power must drop as well.I know there's a constant relationship between power, torque and speed, but since torque at the wheels is dependant on gearing, power must be too, since there's a constant relationship between them.
I'm more than happy to accept that Parkers is right, but you'll need to be a bit more convincing, like showing me how the relationship works with an equation or two, and how the gearing is already included, if that's not too much trouble?
So, instead of saying "do the maths", you could show us the maths that proves you're right. Or you could just bang your head against a wall again like you did last time someone disagreed with you. Your choice, I'd preferred it if you chose to enlighten me though. I don't mind being proven wrong.
So how about it? What's the maths that supports Parkers?
EDIT: If it helps any, my maths is telling me that it would be power at the wheels that matters, which depends on torque and wheel rpm - or road speed, but if you change gear, the wheel rpm and road speed don't change at that instant, but if the torque goes down, you now must have less power at the wheels, so acceleration will reduce.
This may be flawed reasoning, but it is why I'm struggling to accept what you are saying at face value. I'm not an expert, and I'd like an expert to explain this to me. As would, I am sure, the OP.
Edited by Alfanatic on Thursday 5th May 09:46
However, if you look at the Parkers curves, the torque roll-off at high rpms is so severe, that up-shifting and brining the revs down, actually brings the engine torque value up so much, that it counteracts the drop in torque multiplication by the gearbox. Example (numbers are not based on anything real):
200Nm(engine)@6000rpm x 1.50 gearing (3rd) = 300Nm(wheel)
300Nm(engine)@4500rpm x 1.00 gearing (4th) = 300Nm(wheel)
If you make that shift at a constant road speed, then the power figure is constant at the wheels because torque and wheel rpm are fixed.
doogz said:
200Nm is 147.5lbft.
Which at 6000rpm is 169bhp.
300Nm is 221lbft.
Which at 4500rpm is 189bhp.
The torque at the wheels might be the same, but power is related to the speed at which the engine is turning, not the wheels.
Yes, but I just pulled those figures out my ass! I made the assumption that this was roughly representative of a constant engine power as depicted in the Parkers example Alfanatic was disputing. Was really just trying to highlight trying that an increase in engine torque can offset a loss in torque multiplication by up-shifting.Which at 6000rpm is 169bhp.
300Nm is 221lbft.
Which at 4500rpm is 189bhp.
The torque at the wheels might be the same, but power is related to the speed at which the engine is turning, not the wheels.
Change the second example rpm from 4500 to 4000 and both examples are 169hp with 300Nm wheel torque.
blank said:
Power at wheels is equal to power at engine (ignoring transmission loss which will be near enough a set percentage anyway) due to conservation of energy.
There is a link to the maths earlier in this thread or a link at the start of the Parkers thread. If they don't explain I can do so later, but I'm at work at the moment!
Ah, thanks Blank, I'll recheck the links.There is a link to the maths earlier in this thread or a link at the start of the Parkers thread. If they don't explain I can do so later, but I'm at work at the moment!
Alfanatic said:
blank said:
Power at wheels is equal to power at engine (ignoring transmission loss which will be near enough a set percentage anyway) due to conservation of energy.
There is a link to the maths earlier in this thread or a link at the start of the Parkers thread. If they don't explain I can do so later, but I'm at work at the moment!
Ah, thanks Blank, I'll recheck the links.There is a link to the maths earlier in this thread or a link at the start of the Parkers thread. If they don't explain I can do so later, but I'm at work at the moment!
Power of engine = enginetorque*enginerpm/5252
torque at wheels = (enginepower/wheelrpm)*5252
With those equations the poster gets the same results as he does calculating through torque instead of power. In those two equations I can see that both engine rpm and wheel rpm are used, and this would be where doing it by power is taking the gear ratio into account.
So, I no longer dispute the Parker Forum method, I just need to figure out why they get to a slightly different result when looking at the 4th / 5th change than I did doing it through torque
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