matt0941 wrote:I thought you wanted to shift at torque peak and ideally switch gears so that RPM's fall flat in the middle of the torque curve.
You want to optimize the amount of torque goes to the wheels. The more the better.
Low End Torque
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No Matt, shifting at 4000 or so would bring you to the bottom of the torque curve. By shifting at, lets say 6500, you drop right back down to where the motor is already putting out max torque. Don't let the HP vs. "Q" battle confuse you. Just cause your car's torque starts to taper off after 4300 or so, doesn't mean the car will quit pulling.
WD
WD
When you reach 100% volumentric efficiency, you hit your peak torque, which is 4300-4500rpm in KA-de. When you VE go down, you torque curve diminish too.
Changing the VE characteristic will change your torque curve. VE also has direct relationship on how fast a turbo will spool. Intake, exhaust, port & polish, runner length, size and shape affect your VE. ( which is torque curve )
A good size turbo should hit max psi right after your peak torque, and support enough CFM with psi desired to all the way your RPM will reached. Anything too small is choking your engine, and too big will cause too much lag. Of course, is not easy to find a turbo like this.
Good work C-kwik. That's the best torque/hp explanation I seen in this forum.
Changing the VE characteristic will change your torque curve. VE also has direct relationship on how fast a turbo will spool. Intake, exhaust, port & polish, runner length, size and shape affect your VE. ( which is torque curve )
A good size turbo should hit max psi right after your peak torque, and support enough CFM with psi desired to all the way your RPM will reached. Anything too small is choking your engine, and too big will cause too much lag. Of course, is not easy to find a turbo like this.
Good work C-kwik. That's the best torque/hp explanation I seen in this forum.
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C-Kwik
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erich wrote:I was under the impression the ideal place to shift is when the torque divided by the gear ratio is greater in the next gear then the current gear.
erich
Torque is multiplied by the gear ratio, but aside from that, you are correct. But with OEM gearboxes, you can not always achieve that result since gear spacing may be too far apart, particularly in the lower gears.
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C-Kwik
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ADAMHU wrote:the ideal place to shift is at the redline of the engine....WHY you ask..cause all of our cars have a little thing in them called......
gearing......
This is usually the case, but not always. With the KA and the manual transmission, I plotted the torque figures at 6000 RPM shifts in each gear, and in the higher gears, 6000 RPM seemed optimal. I was little limited by the fact that the dyno chart only went up to 6000 RPM, but with the way the torque was falling at that point, lmore torque could be made at the wheels by shifting around 6000 RPM.
here is a good article...someone plug in the nubers for a 240sx....and lets see....
Unfortunately, *engine* torque does not tell you the full story. What matters is the torque *delivered to the tires*, including the effects of the transmission. We all know a car does not accelerate as hard in second gear at peak torque RPM as it does in first gear. The transmission amplifies or multiplies the torque coming from the engine by a factor equal to the gear ratio. So to determine how much the car is accelerating at a particular instant, you have to know both the torque output of the engine as well as the gear ratio.
To figure out your shift points knowing only torque, generate tables of transmission output torque vs. RPM for each gear. To get transmission output torque, multiply the engine torque by the gear ratio. You are simply comparing gear to gear, so the final drive ratio can be ignored. You may also need to know the relationship between RPM in one gear and RPM in another gear (which is RPM * (gear2ratio/gear1ratio) at any particular vehicle speed.) Then it's easy to see what shift points to choose to maximize your transmission output torque at all times.
Here's an example for the 1999 Neon DOHC engine with a five-speed manual transmission. Before you flame, understand that I do not have an accurate torque curve for this motor. I'm estimating visually from the curve printed in the '99 brochure, which is seriously flawed (it makes a lot more sense if the torque curve is shifted to the right 1000 RPM). I get:
Engine Transmission output torque (ft-lb):
Torque 1st 2nd 3rd 4th 5th
RPM (ft-lb) 3.54 2.13 1.36 1.03 0.72 <- gear ratio
---- ------- ---- ---- ---- ---- ----
1000 50 177 107 68 52 36
1500 65 230 138 88 67 47
2000 80 283 170 109 82 58
2500 92 326 196 125 95 66
3000 104 368 222 141 107 75
3500 114 404 243 155 117 82
4000 120 425 256 163 124 86
4500 125 443 266 170 129 90
5000 130 460 277 177 134 94
5500* 133 471 283 181 137 96
6000 130 460 277 177 134 94
6500* 122 432 260 166 126 88
7000 110 389 234 150 113 79
(note: peak torque is at 5500 RPM, peak horsepower is at 6500 RPM)
Without graphing, there's something immediately apparent: in any gear, at 7000 RPM, the transmission torque output is always higher than at any RPM in the next gear up. What this means is, for this car:
* Shift at the redline, not at the torque peak!
Walk through an example. You're hammering down the track in 1st gear. Engine RPM is 6000, just past the engine's torque peak. Do you shift? Well, if you do, the engine will be pulled down to 3600 RPM, and 2nd gear will send 246 ft-lb of torque to the wheels (actually, to the differential first, which amplifies the torque by a constant factor and sends it to the wheels). Don't you think it would be better to hold it in first gear? Torque is dropping off, but it's still 389 ft-lb at 7000 RPM, right before the 7200 RPM redline. So, for this powertrain, first gear is *always* the best deal for acceleration, at any speed, except that you can't accelerate past the redline.
The 1-2 shift at 7200 RPM pulls the engine down to 4400 RPM, where 2nd will deliver 265 ft-lb of torque. Not only did you win by maintaining the high torque of 1st all the way to 7200 RPM, you are now better off in second gear.
Same thing goes for the 2-3 shift. 2nd gear output torque at the redline is still greater than 3rd gear output torque at any engine speed, so you wind her out as far as she'll go before you shift to 3rd. Same for the 3-4, same for the 4-5.
But, you ask, isn't your acceleration greatest at the torque peak? Yes, it is! But only within that gear. The next gear down will give you even greater acceleration at the same speed, unless the vehicle speed is too high for that gear.
To use engine torque to understand how your car performs, you MUST include the effects of the transmission.Maximum Acceleration Vs. Power
OK, so what about power? As has been noted by a previous contributor, Power (hp) = Torque (ft-lb) * RPM / 5252. Note that power is also force * velocity, specifically:
Power (hp) = Force (lb) * Velocity (MPH) / 374
That's net horsepower, which is engine power minus losses like transmission and tire friction. The force is the sum of the longitudinal forces at the contact patches of the two driven tires.
Hmmm... P = F * V ...rearrange to get F = P / V ...
that means that you get the maximum force pushing the car if you maximize your *Power* at any given velocity. This gives us another useful rule:
* Shift to maximize engine POWER, not engine torque!
This is *exactly* the same as saying "shift to maximize transmission output torque". But it's a little easier to apply. Here's how.
Using the torque information above, I get the following power curve:
RPM HP
1000 10
1500 19
2000 30
2500 44
3000 59
3500 76
4000 91
4500 107
5000 124
5500 139 (peak torque)
6000 149
6500 151 (peak power)
7000 147
The tires don't see quite these numbers due to [friction and aerodynamic] losses, but I'm going to assume that the losses are comparable from gear to gear and that the overall shape of the power curve remains the same.
Applying the maximum power rule, we'd like to race down the 1/4 mile with the engine always as close to 6500 RPM as possible. If we had a continuously variable transmission, the lowest E.T. would be achieved by keeping the engine dead on 6500 RPM. 5500 is not the best; at any vehicle speed, the engine would put out more torque but the transmission will have a less advantageous gear ratio, so you get a net loss of force to the tires. Apply P = F * V or P = T * RPM to prove this.
Since the Neon doesn't have a CVT, we have to shift. The shift points are pretty easy to determine. In fact, you don't really need to know anything about the gear ratios of the different gears, which is why power is sometimes easier to understand than torque.
I'm going to assume that the DOHC puts out at least 145 horsepower at the redline (7200 RPM). Shifting at the redline in each gear should drag the engine down as follows:
shift RPM drop Horsepower change
- ----- ---------- ------------------
1->2 7200->4700 145->114
2->3 7200->4600 145->110
3->4 7200->5500 145->139
4->5 7200->5000 145->124
(I derived this, but all you really need to do is drive the car, shift, and find out where the motor lands)
Note - and this is important - the transmission does not amplify power.
Power in = power out, minus losses(which are low for a manual transmission).
This is predicted by the law of conservation of energy.
Is 7200 the correct shift point? It would *not* be the correct shift point if the engine was making more power in the new gear than the old gear. That would mean that you should have shifted earlier. But in this case, the power output at redline is always greater than the power output after the shift. So it's the best performance you can get.
A more rigorous way of doing this is to graph horsepower vs. velocity in each of the gears. If power in one gear drops below the horsepower of the next gear at a particular MPH, then that MPH is where you should shift, otherwise shift at the redline.
I leave as an exercise for the reader the following: predicting shift points based on engine torque, RPM, and gear ratio gives the same results as predicting shift points based on power and vehicle velocity.Exceptions
There are no exceptions; a car running at its (net) power peak can accelerate no harder at that same vehicle speed. There is no better gear to choose, even if another gear would place the engine closer to its torque peak. You'll find that a car running at peak power at a given vehicle speed is delivering the maximum possible torque to the tires (although the engine may not be spinning at its torque peak). This derives immediately from first principles in physics.
However, note the following: - Transmission losses are not shown on engine power curves. The net power curve (power delivered to the ground) may have a different shape or even a different peak RPM as a result. This would result in different shift point. Best results are obtained from a power curve measured by a chassis dynamometer. - The discussion above assumes negligible tire slip. If you exceed the maximum traction available from the tires, then additional power doesn't help. That's why it's sometimes no loss at all to shift early when the tires break loose, and in fact it can be a benefit.To the Point
Torque and power are (almost) flip sides of the same coin. Increasing the torque of an engine at a particular RPM is the same as increasing the power output at the same RPM.
Power is just as useful and relevant in determining vehicle performance as is torque. In some situations it's more useful, because you may not have to play with gear ratios and a calculator to understand what's going on.
A car accelerates hardest with gearing selected to stay as close as possible to the engine *power* peak, subject to the traction capability of the tires.
Not all cars should be shifted at the redline for maximum performance. But it's true for many cars. You can determine optimal shift points by graphing horsepower vs. velocity or transmission torque vs. RPM. Engine torque alone will not determine shift points.
Unfortunately, *engine* torque does not tell you the full story. What matters is the torque *delivered to the tires*, including the effects of the transmission. We all know a car does not accelerate as hard in second gear at peak torque RPM as it does in first gear. The transmission amplifies or multiplies the torque coming from the engine by a factor equal to the gear ratio. So to determine how much the car is accelerating at a particular instant, you have to know both the torque output of the engine as well as the gear ratio.
To figure out your shift points knowing only torque, generate tables of transmission output torque vs. RPM for each gear. To get transmission output torque, multiply the engine torque by the gear ratio. You are simply comparing gear to gear, so the final drive ratio can be ignored. You may also need to know the relationship between RPM in one gear and RPM in another gear (which is RPM * (gear2ratio/gear1ratio) at any particular vehicle speed.) Then it's easy to see what shift points to choose to maximize your transmission output torque at all times.
Here's an example for the 1999 Neon DOHC engine with a five-speed manual transmission. Before you flame, understand that I do not have an accurate torque curve for this motor. I'm estimating visually from the curve printed in the '99 brochure, which is seriously flawed (it makes a lot more sense if the torque curve is shifted to the right 1000 RPM). I get:
Engine Transmission output torque (ft-lb):
Torque 1st 2nd 3rd 4th 5th
RPM (ft-lb) 3.54 2.13 1.36 1.03 0.72 <- gear ratio
---- ------- ---- ---- ---- ---- ----
1000 50 177 107 68 52 36
1500 65 230 138 88 67 47
2000 80 283 170 109 82 58
2500 92 326 196 125 95 66
3000 104 368 222 141 107 75
3500 114 404 243 155 117 82
4000 120 425 256 163 124 86
4500 125 443 266 170 129 90
5000 130 460 277 177 134 94
5500* 133 471 283 181 137 96
6000 130 460 277 177 134 94
6500* 122 432 260 166 126 88
7000 110 389 234 150 113 79
(note: peak torque is at 5500 RPM, peak horsepower is at 6500 RPM)
Without graphing, there's something immediately apparent: in any gear, at 7000 RPM, the transmission torque output is always higher than at any RPM in the next gear up. What this means is, for this car:
* Shift at the redline, not at the torque peak!
Walk through an example. You're hammering down the track in 1st gear. Engine RPM is 6000, just past the engine's torque peak. Do you shift? Well, if you do, the engine will be pulled down to 3600 RPM, and 2nd gear will send 246 ft-lb of torque to the wheels (actually, to the differential first, which amplifies the torque by a constant factor and sends it to the wheels). Don't you think it would be better to hold it in first gear? Torque is dropping off, but it's still 389 ft-lb at 7000 RPM, right before the 7200 RPM redline. So, for this powertrain, first gear is *always* the best deal for acceleration, at any speed, except that you can't accelerate past the redline.
The 1-2 shift at 7200 RPM pulls the engine down to 4400 RPM, where 2nd will deliver 265 ft-lb of torque. Not only did you win by maintaining the high torque of 1st all the way to 7200 RPM, you are now better off in second gear.
Same thing goes for the 2-3 shift. 2nd gear output torque at the redline is still greater than 3rd gear output torque at any engine speed, so you wind her out as far as she'll go before you shift to 3rd. Same for the 3-4, same for the 4-5.
But, you ask, isn't your acceleration greatest at the torque peak? Yes, it is! But only within that gear. The next gear down will give you even greater acceleration at the same speed, unless the vehicle speed is too high for that gear.
To use engine torque to understand how your car performs, you MUST include the effects of the transmission.Maximum Acceleration Vs. Power
OK, so what about power? As has been noted by a previous contributor, Power (hp) = Torque (ft-lb) * RPM / 5252. Note that power is also force * velocity, specifically:
Power (hp) = Force (lb) * Velocity (MPH) / 374
That's net horsepower, which is engine power minus losses like transmission and tire friction. The force is the sum of the longitudinal forces at the contact patches of the two driven tires.
Hmmm... P = F * V ...rearrange to get F = P / V ...
that means that you get the maximum force pushing the car if you maximize your *Power* at any given velocity. This gives us another useful rule:
* Shift to maximize engine POWER, not engine torque!
This is *exactly* the same as saying "shift to maximize transmission output torque". But it's a little easier to apply. Here's how.
Using the torque information above, I get the following power curve:
RPM HP
1000 10
1500 19
2000 30
2500 44
3000 59
3500 76
4000 91
4500 107
5000 124
5500 139 (peak torque)
6000 149
6500 151 (peak power)
7000 147
The tires don't see quite these numbers due to [friction and aerodynamic] losses, but I'm going to assume that the losses are comparable from gear to gear and that the overall shape of the power curve remains the same.
Applying the maximum power rule, we'd like to race down the 1/4 mile with the engine always as close to 6500 RPM as possible. If we had a continuously variable transmission, the lowest E.T. would be achieved by keeping the engine dead on 6500 RPM. 5500 is not the best; at any vehicle speed, the engine would put out more torque but the transmission will have a less advantageous gear ratio, so you get a net loss of force to the tires. Apply P = F * V or P = T * RPM to prove this.
Since the Neon doesn't have a CVT, we have to shift. The shift points are pretty easy to determine. In fact, you don't really need to know anything about the gear ratios of the different gears, which is why power is sometimes easier to understand than torque.
I'm going to assume that the DOHC puts out at least 145 horsepower at the redline (7200 RPM). Shifting at the redline in each gear should drag the engine down as follows:
shift RPM drop Horsepower change
- ----- ---------- ------------------
1->2 7200->4700 145->114
2->3 7200->4600 145->110
3->4 7200->5500 145->139
4->5 7200->5000 145->124
(I derived this, but all you really need to do is drive the car, shift, and find out where the motor lands)
Note - and this is important - the transmission does not amplify power.
Power in = power out, minus losses(which are low for a manual transmission).
This is predicted by the law of conservation of energy.
Is 7200 the correct shift point? It would *not* be the correct shift point if the engine was making more power in the new gear than the old gear. That would mean that you should have shifted earlier. But in this case, the power output at redline is always greater than the power output after the shift. So it's the best performance you can get.
A more rigorous way of doing this is to graph horsepower vs. velocity in each of the gears. If power in one gear drops below the horsepower of the next gear at a particular MPH, then that MPH is where you should shift, otherwise shift at the redline.
I leave as an exercise for the reader the following: predicting shift points based on engine torque, RPM, and gear ratio gives the same results as predicting shift points based on power and vehicle velocity.Exceptions
There are no exceptions; a car running at its (net) power peak can accelerate no harder at that same vehicle speed. There is no better gear to choose, even if another gear would place the engine closer to its torque peak. You'll find that a car running at peak power at a given vehicle speed is delivering the maximum possible torque to the tires (although the engine may not be spinning at its torque peak). This derives immediately from first principles in physics.
However, note the following: - Transmission losses are not shown on engine power curves. The net power curve (power delivered to the ground) may have a different shape or even a different peak RPM as a result. This would result in different shift point. Best results are obtained from a power curve measured by a chassis dynamometer. - The discussion above assumes negligible tire slip. If you exceed the maximum traction available from the tires, then additional power doesn't help. That's why it's sometimes no loss at all to shift early when the tires break loose, and in fact it can be a benefit.To the Point
Torque and power are (almost) flip sides of the same coin. Increasing the torque of an engine at a particular RPM is the same as increasing the power output at the same RPM.
Power is just as useful and relevant in determining vehicle performance as is torque. In some situations it's more useful, because you may not have to play with gear ratios and a calculator to understand what's going on.
A car accelerates hardest with gearing selected to stay as close as possible to the engine *power* peak, subject to the traction capability of the tires.
Not all cars should be shifted at the redline for maximum performance. But it's true for many cars. You can determine optimal shift points by graphing horsepower vs. velocity or transmission torque vs. RPM. Engine torque alone will not determine shift points.
Its not that I am stupid, just lazy.-
C-Kwik
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matt0941 wrote:If it is optimal to shift at 6000 why do people remove the rev limiter? I am not being facisious (sp?) just wondering.
Because higher revs generally mean more HP as long as the engine is able to breathe well enough to produce a decent amount of torque up to the new redline. The KA is quite lacking in torque in any revs above redline so raising the rev limit will probably net a vey small gain if any. But if you were to add modification or change parts so that it is able to breathe better above the OEM redline, then it is possible that it can make more HP. And this is true from what I had mentioned about torque and HP previously and what Adam has posted as well.
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Structure240sx
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C-Kwik
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Structure240sx wrote:one thought that opoed into my head today. if you had a built ka with lowered compression say like 8.0 or 8.5:1 you would lose most of that low end torque anyway like the sr's low compression
You would lose torque everywhere. But, you'ld be able to boost more and get more overall power, assuming your fuel system can keep up.