1st gear ratio ... the key to better numbers?
#81
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Sheesh, I wondered how people got those high Post numbers.
Who cares? Any gains (if there are any) from better gearing are "negligible" meaning any gains would be lost to varying testing methodologies and drivers. I mean you have car rags publishing 0-60 times that vary by more than 1/2 sec so a gain of .1 or .2 sec is meanigless, it's within the error of the measurement. Now if it made you "feel" better (faster) that's another matter.
Who cares? Any gains (if there are any) from better gearing are "negligible" meaning any gains would be lost to varying testing methodologies and drivers. I mean you have car rags publishing 0-60 times that vary by more than 1/2 sec so a gain of .1 or .2 sec is meanigless, it's within the error of the measurement. Now if it made you "feel" better (faster) that's another matter.
#82
Originally posted by calbear2k1
Someone (Buff-daddy?) suggested that Germans compared to Japanese tend to get more out of the same HP.
Someone (Buff-daddy?) suggested that Germans compared to Japanese tend to get more out of the same HP.
#83
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Originally posted by biker
Sheesh, I wondered how people got those high Post numbers.
Who cares? Any gains (if there are any) from better gearing are "negligible" meaning any gains would be lost to varying testing methodologies and drivers. I mean you have car rags publishing 0-60 times that vary by more than 1/2 sec so a gain of .1 or .2 sec is meanigless, it's within the error of the measurement. Now if it made you "feel" better (faster) that's another matter.
Sheesh, I wondered how people got those high Post numbers.
Who cares? Any gains (if there are any) from better gearing are "negligible" meaning any gains would be lost to varying testing methodologies and drivers. I mean you have car rags publishing 0-60 times that vary by more than 1/2 sec so a gain of .1 or .2 sec is meanigless, it's within the error of the measurement. Now if it made you "feel" better (faster) that's another matter.
Gearing makes all the difference. Stock tsx first gear multiplies engine torque about 15.5 times to the groud..say you made first gear more similar to that in the auto which is just under a 12 times multiplication. ARe you saying that 500bft+ of torque being delivered to the ground is negligible? 25% less drive power through the entire gear has a minmal effect?
#84
Originally posted by TinkySD
Gearing makes all the difference. Stock tsx first gear multiplies engine torque about 15.5 times to the groud..say you made first gear more similar to that in the auto which is just under a 12 times multiplication. ARe you saying that 500bft+ of torque being delivered to the ground is negligible? 25% less drive power through the entire gear has a minmal effect?
Gearing makes all the difference. Stock tsx first gear multiplies engine torque about 15.5 times to the groud..say you made first gear more similar to that in the auto which is just under a 12 times multiplication. ARe you saying that 500bft+ of torque being delivered to the ground is negligible? 25% less drive power through the entire gear has a minmal effect?
Still though, I'd tend to agree that shorter gearing makes a car faster overall as it increases the average delivered horsepower per the earlier discussions.
#85
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Originally posted by rb1
I think you still might have to simulate this. If you did what you say above, you'd get to stay in 1st gear about 30% longer, while the stock TSX has shifted to 2nd and is now getting a (say) 7.5 multiplier.
Still though, I'd tend to agree that shorter gearing makes a car faster overall as it increases the average delivered horsepower per the earlier discussions.
I think you still might have to simulate this. If you did what you say above, you'd get to stay in 1st gear about 30% longer, while the stock TSX has shifted to 2nd and is now getting a (say) 7.5 multiplier.
Still though, I'd tend to agree that shorter gearing makes a car faster overall as it increases the average delivered horsepower per the earlier discussions.
#86
Originally posted by TinkySD
Actually i find a good way to think about this whole arguement is a CVT transmission; which work buy infinitely varying the gear ratio to keep the engine at peak horsepower. In the case of the tsx a CVT equipped auto would be quicker than a 6mt in a straight line.
Actually i find a good way to think about this whole arguement is a CVT transmission; which work buy infinitely varying the gear ratio to keep the engine at peak horsepower. In the case of the tsx a CVT equipped auto would be quicker than a 6mt in a straight line.
VW/Audi, for example, claims identical 0-60 and fuel economy numbers for the A4 CVT and A4 MT.
#87
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Originally posted by rb1
While true in theory, losses due to friction are somewhat higher in current CVTs than in an MT and this offsets the advantage somewhat (basically, at least in today's CVTs, you have a belt-driven car ).
VW/Audi, for example, claims identical 0-60 and fuel economy numbers for the A4 CVT and A4 MT.
While true in theory, losses due to friction are somewhat higher in current CVTs than in an MT and this offsets the advantage somewhat (basically, at least in today's CVTs, you have a belt-driven car ).
VW/Audi, for example, claims identical 0-60 and fuel economy numbers for the A4 CVT and A4 MT.
#88
"d^2E/dt^2 = F dv/dt
which would indicate that v^2 is proportional to the double integral of F (torque)."
that's torque to the ground, which you maximize by having maximum average HP and proper gearing. it's not engine torque.
which would indicate that v^2 is proportional to the double integral of F (torque)."
that's torque to the ground, which you maximize by having maximum average HP and proper gearing. it's not engine torque.
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Originally posted by drchaos
that's torque to the ground, which you maximize by having maximum average HP and proper gearing. it's not engine torque.
that's torque to the ground, which you maximize by having maximum average HP and proper gearing. it's not engine torque.
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Originally posted by drchaos
"d^2E/dt^2 = F dv/dt
which would indicate that v^2 is proportional to the double integral of F (torque)."
that's torque to the ground, which you maximize by having maximum average HP and proper gearing. it's not engine torque.
"d^2E/dt^2 = F dv/dt
which would indicate that v^2 is proportional to the double integral of F (torque)."
that's torque to the ground, which you maximize by having maximum average HP and proper gearing. it's not engine torque.
You derive something with "power" at the end and you conclude with "maximum HP."
I derive something with torque at the end and you conclude with "maximum HP."
My question for you now is this: Forget about the first gear ratio. Will adding a "Type R" sticker increase horsepower, or will it increase torque?
#92
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Originally posted by dnb
Oh, sure ...
You derive something with "power" at the end and you conclude with "maximum HP."
I derive something with torque at the end and you conclude with "maximum HP."
My question for you now is this: Forget about the first gear ratio. Will adding a "Type R" sticker increase horsepower, or will it increase torque?
Oh, sure ...
You derive something with "power" at the end and you conclude with "maximum HP."
I derive something with torque at the end and you conclude with "maximum HP."
My question for you now is this: Forget about the first gear ratio. Will adding a "Type R" sticker increase horsepower, or will it increase torque?
the key term he used was average horsepower which has nothing to do with peak horsepower but instead is a measurement of area under the curve. It's why I thought it really didn't tell us much as I think most of us know you ant maximum area under the hp curve...but there are several ways to get there.
#93
You want to maximize the integral of the power output over time.
Since there is a local maximum in power with RPM, then ideally you would want to upshift at RPMS a little bit above this so it ends up a little bit below this.
Now since you maximize power with time, you have to be able to figure out what P(t) to use. If you had a CVT at always optimal ratio, this would be a constant: set the engine speed at maximum HP. With fixed gears there is a bit more complicated optimization problem and the shape of the curve with RPM will come into play.
First gear ratios of course will control how fast you can get up to the power peak as well and therefore that will also change the integral of the power with time.
Since there is a local maximum in power with RPM, then ideally you would want to upshift at RPMS a little bit above this so it ends up a little bit below this.
Now since you maximize power with time, you have to be able to figure out what P(t) to use. If you had a CVT at always optimal ratio, this would be a constant: set the engine speed at maximum HP. With fixed gears there is a bit more complicated optimization problem and the shape of the curve with RPM will come into play.
First gear ratios of course will control how fast you can get up to the power peak as well and therefore that will also change the integral of the power with time.
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Originally posted by drchaos
You want to maximize the integral of the power output over time.
You want to maximize the integral of the power output over time.
t(0-60) = integral_from_0_to_60 of F(Power, v,t) dv
where F is a can of worms.
Note to mods: vBulletin should be able to process TeX so we can have nice mathematical notation in our discussions. What's a nerd to do without his partial derivatives and integrals?
dnb, who used to write in TeX (LaTeX is for wimps)
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