View Poll Results: HP -> Acceleration... not Torque!
Voters: 203. You may not vote on this poll
HP -> Acceleration... not Torque!
Registered User
Joined: Jan 2002
Posts: 250
Likes: 0
From: Redford, MI
Zoran, you seem to be confusing acceleration and speed, the two are very different. Using your own equations,
WheelTorque = EngineHP/WheelRPM
(EngineHP = EngineRPM*EngineTorque)
WheelTorque = EngineRPM*EngineTorque/WheelRPM
(Gearing = EngineRPM/WheelRPM = constant)
WheelTorque = k*EngineTorque
You get max WheelTorque at max EngineTorque, according to your equation.
WheelTorque = EngineHP/WheelRPM
(EngineHP = EngineRPM*EngineTorque)
WheelTorque = EngineRPM*EngineTorque/WheelRPM
(Gearing = EngineRPM/WheelRPM = constant)
WheelTorque = k*EngineTorque
You get max WheelTorque at max EngineTorque, according to your equation.
Registered User
Joined: Jan 2002
Posts: 250
Likes: 0
From: Redford, MI
Originally posted by KnightRider
You're assuming the car has only one gear. Zoran was assuming the car had infinte gear ratios.
You're assuming the car has only one gear. Zoran was assuming the car had infinte gear ratios.
Registered User
Joined: May 2003
Posts: 8
Likes: 0
From: Worthington
I think HP and TQ have their place in this discussion. Leaning on either one is inane and pointless.....
Torque moves your car, no argument, hp is not a physical force and CANNOT be directly measured, only calculated....
However the car with greater torque dosen't always win.
Further more looking at HP and TQ peak numbers van be very helpful! In the end i would rather have a plot fo TQ and TQ info over HP info....
Ben
Torque moves your car, no argument, hp is not a physical force and CANNOT be directly measured, only calculated....
However the car with greater torque dosen't always win.
Further more looking at HP and TQ peak numbers van be very helpful! In the end i would rather have a plot fo TQ and TQ info over HP info....
Ben
Registered User
Joined: Jun 2003
Posts: 24
Likes: 0
From: Portland
[QUOTE]Originally posted by FYRHWK1
[B]
correct, but the only reason to rev higher is more wheel RPM, so why not just choose a wider ratio and use the point of higher torque? less engine stress and wear, same torque at the wheels and less noise.
[B]
correct, but the only reason to rev higher is more wheel RPM, so why not just choose a wider ratio and use the point of higher torque? less engine stress and wear, same torque at the wheels and less noise.
Registered User
Joined: Jun 2003
Posts: 24
Likes: 0
From: Portland
I just searched for "CVT hp torque" and found this:
http://www.v8914.com/Horsepower-v-torque.htm
Check the tables at the bottom to see what I'm talking about.
Edit: Nevermind, saw that it was posted earlier but if anyone missed it, it covers exactly what I was talking about.
http://www.v8914.com/Horsepower-v-torque.htm
Check the tables at the bottom to see what I'm talking about.
Edit: Nevermind, saw that it was posted earlier but if anyone missed it, it covers exactly what I was talking about.
Registered User
Joined: Jul 2002
Posts: 163
Likes: 0
From: NY
Originally posted by KnightRider
Good observations. I'll try my best to explain this detail.
The reason you don't want a wider ratio is because that would actually lower the torque at the wheels with a CVT. HP measures the potential for acceleration at that engine RPM. But since the HP changes with RPMs you have to keep the engine RPMs fixed to make use of that potential. Imagine that for every possible wheel RPM you could choose whatever gear ratio you want. So basically you are picking whatever engine RPM you want. Now believe it or not the wheel torque will be greatest at the engine RPM that produces peak HP not the torque peak. It goes back to those equations, the easiest one for me to visualize is [WheelTorque = EngineTorque * gearing]. Seems like engine torque is directly related to the wheel torque right? True if the gear ratio does not change. So why not run the engine at it's peak torque? Because to run at that lower rpm the gear ratio would have to be lower and the wheel torque would be lower as a result. Funny thing is if you slowed down using that same gear ratio you would actually make more wheel torque when the engine slows down to it's torque peak. BUT... at that speed with a CVT you could again change the gear ratio higher to get the engine back up to the HP peak and make more torque at the wheels. Probably the best way to understand it is to try calculating wheel torque with actual values from a magazine or something. Pick an arbitrary wheel rpm and calculate the gear ratios needed for that wheel rpm at the torque peak and at the hp peak, then calculate the torque at the wheels. For example with the s2000:
240hp@8300rpm (~151.9lb-ft) 240lb-ft*5252/8300rpm
153lb-ft@7500rpm (218.5 HP) 153hp*7500rpm/5252
say we pick 500rpm as the wheel rpm, our gear ratios will be 16.6 (8300/500) and 15 (7500/500).
So the torque at the wheels is torque * gearing:
151.8 * 16.6 = 2519.9 (at the HP peak)
153 * 15 = 2295 (at the engine's torque peak)
This will work for any car and any wheel rpm. And once again this is only for a continously variable transmission which lets you choose any gear ratio you like. With a standard transmission the peak acceleration will be at the torque peak. With the CVT the peak acceleration will be at the lowest wheel rpms and will smoothly drop off as the wheel rpms increase.
Good observations. I'll try my best to explain this detail.
The reason you don't want a wider ratio is because that would actually lower the torque at the wheels with a CVT. HP measures the potential for acceleration at that engine RPM. But since the HP changes with RPMs you have to keep the engine RPMs fixed to make use of that potential. Imagine that for every possible wheel RPM you could choose whatever gear ratio you want. So basically you are picking whatever engine RPM you want. Now believe it or not the wheel torque will be greatest at the engine RPM that produces peak HP not the torque peak. It goes back to those equations, the easiest one for me to visualize is [WheelTorque = EngineTorque * gearing]. Seems like engine torque is directly related to the wheel torque right? True if the gear ratio does not change. So why not run the engine at it's peak torque? Because to run at that lower rpm the gear ratio would have to be lower and the wheel torque would be lower as a result. Funny thing is if you slowed down using that same gear ratio you would actually make more wheel torque when the engine slows down to it's torque peak. BUT... at that speed with a CVT you could again change the gear ratio higher to get the engine back up to the HP peak and make more torque at the wheels. Probably the best way to understand it is to try calculating wheel torque with actual values from a magazine or something. Pick an arbitrary wheel rpm and calculate the gear ratios needed for that wheel rpm at the torque peak and at the hp peak, then calculate the torque at the wheels. For example with the s2000:
240hp@8300rpm (~151.9lb-ft) 240lb-ft*5252/8300rpm
153lb-ft@7500rpm (218.5 HP) 153hp*7500rpm/5252
say we pick 500rpm as the wheel rpm, our gear ratios will be 16.6 (8300/500) and 15 (7500/500).
So the torque at the wheels is torque * gearing:
151.8 * 16.6 = 2519.9 (at the HP peak)
153 * 15 = 2295 (at the engine's torque peak)
This will work for any car and any wheel rpm. And once again this is only for a continously variable transmission which lets you choose any gear ratio you like. With a standard transmission the peak acceleration will be at the torque peak. With the CVT the peak acceleration will be at the lowest wheel rpms and will smoothly drop off as the wheel rpms increase.
Registered User
Joined: Apr 2001
Posts: 854
Likes: 0
From: Houston
Originally posted by FYRHWK1
right right right, i should've worded what I said better, all i'm saying is that in a CVT, if you're looking for economy (which the fiesta is) then you wold only need to spin up to your torque peak as that gives you the most torque to multiply, which means less gearing needed, less engine speed and all the benefits that go with that. After plugging around some numbers using a 14:1 ratio an the CVT stopping the engine at 4000 RPM, with 81 ft lb i go 1134 RWTQ and 285.71 WRPM, having it shift at HP peak 5200 with 75 ft lb available and using an 18:1 ratio (I tried to get the wheel RPM close as i could without going nuts over decimals) you get 288.9 WRPM and 1350 ft lb, you'd not only accelerate harder but you'd be able to reach higher speeds, albeit a small difference in that. Obviously this is optimal for a performance car, I just wonder when it would be a case of diminishing returns, as where you're simply adding gearing and only making up the torque lost through engine output dropping and not actually gaining RWTQ.
right right right, i should've worded what I said better, all i'm saying is that in a CVT, if you're looking for economy (which the fiesta is) then you wold only need to spin up to your torque peak as that gives you the most torque to multiply, which means less gearing needed, less engine speed and all the benefits that go with that. After plugging around some numbers using a 14:1 ratio an the CVT stopping the engine at 4000 RPM, with 81 ft lb i go 1134 RWTQ and 285.71 WRPM, having it shift at HP peak 5200 with 75 ft lb available and using an 18:1 ratio (I tried to get the wheel RPM close as i could without going nuts over decimals) you get 288.9 WRPM and 1350 ft lb, you'd not only accelerate harder but you'd be able to reach higher speeds, albeit a small difference in that. Obviously this is optimal for a performance car, I just wonder when it would be a case of diminishing returns, as where you're simply adding gearing and only making up the torque lost through engine output dropping and not actually gaining RWTQ.
It took a very long time, but you finally seem to be getting it...
There IS a point of diminishing returns, and that's the HP peak. Revving past the HP peak in a car with a CVT doesn't make any sense, since you're losing wheel torque that way, and just revving the car up past the HP peak won't do any good: the car will accelerate slower than at the HP peak, and it will burn more fuel and stress the engine.
But I am glad that you finally understand. The equation:
WheelTorque = EngineHP/WheelRPM
means that to maximize wheel torque (assuming you can regulate EngineRPM and gearing at will) you need to maximize EngineHP. Maximum horsepower occurs at HP peak, that's why all the cars with CVT transmission accelerate the hardest at their HP peak.