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03262009, 06:00 PM  #23 
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In any single gear, there is a proportional relationship between engine torque and acceleration. If engine torque goes up, so does acceleration and vice versa. Engine torque times transmission and differential gearing multiplied by the lever arm length of the tire will result in the linear force on the car. Divide by the mass of the car to get acceleration. There is no such correspondence between power and acceleration with fixed gearing. When you say the car "pulls" harder at 8K than 4K I'm interpreting that to mean you feel the car accelerates faster. If you feel greater pull at 8K than at 4K then you are either in a different gear, it's illusionary or I don't understand what you mean by "pull".
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03272009, 07:59 AM  #26  
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For the rest of it, I think you made your case. But I suppose in a way it is all moot since anyway, since you cannot change the torque curve of the motor, you can only shift gears.
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03272009, 09:20 AM  #27  
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Since there's no such thing as a perfect transmission, horsepower is actually fairly meaningless. The only relevant measure of power is the average torque between the shift point (RPMs where you end up after shifting to the next highest gear) and redline. If you add those averages up over all the gears, take into account the multiplication, mass, driveline friction and aerodynamics, you can calculate the exact acceleration of the car. So the average torque between about 5500 and 8300 is "power." 

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03272009, 09:30 AM  #28  
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Horsepower can be measured on an engine dyno, where there is no gearing or wheels. Therefore horsepower, in the proper sense, has nothing to do with gearing nor wheels. Horsepower is simply a unit of power, which is defined as torque x rotational velocity (RPM commonly but other unit could be used). I'll see if I can dig up Bruce Augstein's explanation, since it is arguable the best out there. Or maybe he'll just chime in.
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03272009, 10:04 AM  #29  
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Well if we are going to dig into details, we should get them right  horsepower is actually (torque * RPM) / 5252. And you can't *measure* horsepower on a dyno. Dyno's measure torque. Horsepower is a mathematical calculation derived from torque based on RPMs and a constant. RPMs by themselves are only relevant if you've got a tranmission that can take advantage of higher RPMs via gearing. You can't *do* anything with horsepower without gears. I, and several other, posters may be misreading you, but your original post seemed to imply that you thought horsepower somehow "took over" from torque at a certain RPM to provide acceleration. If you only had one gear, more RPMs would give you a higher top speed but not any more acceleration. torque at higher RPMs (horsepower) only helps you if you have gears that can take advantage of it. To talk about horsepower, in the context of acceleration, independent of gearing is completely, utterly, totally, and without exception, meaningless. More details: Peak horsepower is actually the mathematical inflection point where the theoretical incremental wheel torque that you could achieve with increased gear multiplication (gearing) is less than the incremental instantaneous torque lost due to increasing RPMs. In other words, it's the point where the torque curve falls faster than the additional wheel torque you could get by increasing gearing in a perfect transmission. Said yet another way, if you had a perfect transmission, below the horsepower peak (and even though torque is falling), you could get more wheel torque because you could increase gearing faster than the torque is falling. Above the horsepower peak, torque falls faster than RPMs increase. More simply, the horsepower peak occurs at the inflection point where each incremental RPM (8000 to 8001) is a lower % increase than the % decrease in torque available at 8001 vs 8000. Gearing is absolutely implicit in the meaning of horsepower. Horsepower and gearing are as joined at the hip as Bernanke and Inflation. 

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03272009, 10:31 AM  #30  
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Another way to look at it is, suppose you put a CVT in an M3 (a stong capable CVT, say). The car would perform better if the engine were brought to 4k RPM and held there, than if the engine were brought to 8k RPM and held there. However, since we don't have a CVT, we must rev the motor past the RPM where peak torque is made in order to get best acceleration. The reason is, we want to have the torque remain near peak once RPMs fall back down to pick up the next gear ratio. The beauty of the M3 engine, then, is that it allows you to keep increasing RPM way past peak (torque curve stays relatively flat)  in preparation for the next gear, as I say  without giving up too much in acceleration vs. if you could instead keep the RPM at 4k and simply change the ratio instead. (And before someone says it, I realize 4k and 8k RPM are arbitrary and don't necessarily represent exact optimal/leastoptimal figures, I was just using the previously established examples.)
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03272009, 10:50 AM  #31  
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The truth is exactly the reverse of what you are saying. You can accelerate at a greater rate IN ONE GEAR at 4k than 8k, but you can accelerate at a faster rate overall in a lower gear at 8k than 4k. The torque you *get* from 2x the gearing (8000/4000) is greater than the torque you lose because the engine can't make as much torque at 8000 as it can at 4000. That's the whole point of horsepower. Cars get torque from 2 different things  either gears or the engine. Horsepower gives you a measure of how the two are related. Horsepower tells you, in a simplified way, the point where you are better off using gears than torque. Horsepower gives you a measure of "net torque after taking into account gearing." 

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03272009, 11:36 AM  #32  
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03272009, 11:57 AM  #33 
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hard math might help. As you move up the RPM curve, the rate at which gearing can make up for engine torque falls. It's just a function of math and %.
If you had an engine with a 1000RPM redline and 100ftlbs and a 10x gear ratio, you would only need 50ftlbs on a 2000RPM engine to make the same wheel torque because you could use a 20x gear ratio. (100x10 = 50x20 = 1000) 3000/2000 = 1.5x relative torque multiplication from gearing, but 5000/4000 is only 1.2x and 8000/7000 is only 1.14x. So from 1k to 2k, gearing can "make up" for a 50% fall in engine torque and you'd have the same wheel torque, but by the time you go from 7000 to 8000RPMs, you could only make up for a 14% fall in engine torque with gearing. As you increase RPMs, it gets mathematically harder and harder to *replace* engine torque with gear torque. Peak horsepower provides the mathematical inflection point where you can no longer replace engine torque with gear torque because the engine torque is falling too fast. 
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03272009, 12:06 PM  #34  
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If the engine is spinning at 8,300 rpms in 1st gear that would correspond to 42 mph (with a properly adjusted CVT or 6MT). That is ideal. If the car were going 21 mph, then with a properly adjusted CVT the engine would still be at 8,300 rpms but with the 6MT in first gear you would be at 4180 rpm. Because the engine is spinning so much faster with the CVT you would need to roughly double the gear ratio (otherwise the wheels wouldn't be going 21 mph) and that will double the wheel torque. If the CVT were calibrated to spin the engine at max torque then at 21 mph the engine would be at 4,000 rpm. In that case the CVT would have roughly the same gear multiplication as first gear and thus no more rear wheel torque and/or acceleration. Accelerating this max torque CVT equipped car to 42 mph the CVT would have half the gear ratio of the 6MT (because the engine is spinning at half the speed, 4,000 vs. 8,3000) in order to get the wheels to go 42 mph. This would give the CVT equipped car half the rear wheel torque as the 6MT car. Another way to look at this is with the graph I posted earlier. If you connected the shift points (the place to the right in each gear's curve where the torque goes to zero) with a straight line that would roughly look like the rear wheel torque vs. speed curve for a properly adjusted CVT equipped car. If you adjusted the CVT to max torque then you would be connecting the peaks from each individual curve (far left of the flat portion of each curve). For this engine, that would be a very slow thing. Of course, driving around at 8,300 rpms will cause people to give you funny looks. 

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03272009, 01:07 PM  #35  
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Suppose we take whay you've said, and increase RPM more  make it much more  say 20000 RPM (lets assume our engine is built to this spec). At the same time, torque descreases at well. Are we still better off? And do we continue to be better off as RPM approaches infinity, and torque approaches zero? If no, then I don't get it. Well, I should say I don't automatically get it. Maybe if I had the data in front of me, and a way to see the results as different parameters were varied, then I'd get it. In fact I am sure I would, so perhaps I should seek out some software that does this. Quote:
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03272009, 01:13 PM  #36 
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I like this explanation
http://www.allpar.com/eek/hpvstorque.html 
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03272009, 01:41 PM  #37  
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I fully understand torque and the fact that gearing directly multiplies torque (increases mechanical advantage). I also fully understand that power is torque multiplied by RPM. I know how to read your graph above, and what it says. I know how to look at dyno graph and read it. And I know most of the other elementary things about what makes a car go. None of this knowledge is helping me understand why your assertion should be true. Unfortunately. I am happy to continue the discussion, but I don't want to waste your time, or earlyapex time, trying to explain something that I could probably resolve myself with a little bit of reading, math, and experimentation. Quote:
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03272009, 02:02 PM  #38  
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If you double the gear ratio (from say 10:1 to 20:1), you double the torque multiplication. Example: Torque = 100ft lbs at 4000 2nd gear = 2:1 final drive = 3:1 torque at wheels = 100ftlbs * 2 (2nd gear) * 3 (final drive) = 600ftlbs. Assume the car is going 20mph. Now, take torque of 100ft lbs at 8000 rpms. 1st gear = 4:1 (4:1 instead of 2:1 in 2nd gear) final drive = 3:1 torque at wheels = 100ftlbs * 4 (1st gear gives more torque multiplication) * 3 (final drive) = 1200 ftlbs at the road. Car is still going 20mph. 100ft lbs at 8000 gives you 2x the torque of 100ftlbs at 4000 because of the gearing difference. Let's say you only had 75ftlbs at 8000, that would give 75*4*3= 900ftlbs. So 75 ftlbs of torque at 8000 is still 300ftlbs more (900 vs 600) than 100ftlbs at 4000 due to being able to take advantage of lower gearing. because of the gearing, 75ftlbs at 8000 provides more road torque than 100ftlbs at 4000 even though the engine is making less total torque. The torque from the gears more than makes up for the drop in actual engine output. Horsepower provides us with the point where the tradeoff no longer works (theoretical gear torque stops making up for drop in engine torque at HP peak). The only *problem* with horsepower is that it implicitly assumes a perfect CVT. That doesn't exist, so a broader/fatter torque curve is better in most real world applications. 

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03272009, 02:41 PM  #39 
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Here we go again.
It might help to think of gearing like a lever. A longer lever gives you more force on the one end but you have to pull the other end farther. So it is with car gearing  the engine turns more revolutions to move the car over one mile in lower gears, but there is more force (torque) at the wheels. Lower gear: faster engine but less car speed. Higher gear: less force but more speed. Power is how fast you're using energy. Energy is force times distance, so power is force times speed. More power means more torque and more speed. For the same power you can trade off one for the other using different gearing. The upshot here is that the horsepower figure takes into account the gearing and torque. When manufacturers cite torque numbers they're only giving you the maximum torque at one rpm. What is more important is the torque over the whole rev range. Power can give you an overall feel for how fast a car is. Peak torque at one rpm cannot.
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03272009, 03:17 PM  #40  
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But I still don't understand why running a car at peak torque and varying the ratio will make it accelerate worse than running it at peak power and varying the gear ratio. (i.e the CVT example) Your examples (which I've clipped to keep post size down) and the math therein are very simple, but I don't readily see how they apply to the issue of accelerating at peak torque vs. peak power. Quote:
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03272009, 03:36 PM  #41  
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If you are going 20 mph at 4000 rpms you HAVE to use a numerically taller gear (by a factor of 2x) than going 20mph at 8000 rpms. When you start accelerating at 20mph at 4000RPMs with a CVT, you are beginning in a gear that has one half the torque multiplication because you are beginning at 4000. Assuming an infinitely variable CVT, at any speed, the gear used at 8000 will be 2x the gear used at 4000 producing 2x the torque. With a CVT, the STARTING torque multiplication at 20 mph and 4000 rpms is exactly 1/2 the torque multiplication at 20mph and 8000rpms. 

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03272009, 03:39 PM  #42 
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Here's a better example. Put your car in 1st gear at 20mph and accelerate. Now put it in 6th gear at 20mph and accelerate. You get a lot more acceleration in 1st, right?
With a CVT at 8000 it's like starting in 1st gear rather than 6th gear. You get the advantage of a CVT either way, you just start off with much, much better gearing. 
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03272009, 04:50 PM  #43  
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"Here's an example for the 1999 Neon DOHC engine with a fivespeed 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 1999 brochure, which is seriously flawed (it makes a lot more sense if the torque curve is shifted to the right 1000 RPM). I get: Transmission output torque (ftlb): XXX XX 1st2nd3rd4th5th RPM TQ 3.542.131.361.030.72 < gear ratio  1000 050 177 107 068 052 036 1500 065 230 138 088 067 047 2000 080 283 170 109 082 058 2500 092 326 196 125 095 066 3000 104 368 222 141 107 075 3500 114 404 243 155 117 082 4000 120 425 256 163 124 086 4500 125 443 266 170 129 090 5000 130 460 277 177 134 094 5500 133 471 283 181 137 096 6000 130 460 277 177 134 094 6500 122 432 260 166 126 088 7000 110 389 234 150 113 079 (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!" I've copied a section from the link provided by Paddy. It seems to me the easiest to follow and seems consistent with much of what is being discussed here. Note that the 2nd column is engine torque, while the remaining columns to the right are transmission output torque. I realize a similar chart for the M3 may look a little different, but I suspect the implications would be very similar, if not the same. The attached image is easier to read. I had to add some dashes and leading zeros to get things somewhat lined up above. 

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03272009, 04:56 PM  #44 
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Here's a visual. Sometimes a picture helps. Everyone else is explaining it as good as can be done so I won't waste anyones time repeating.
The solid red line is rear wheel torque for a normal 6MT car. The dotted red line is the engine rpm for the same normal 6MT car. All shifts done at redline. The engine RPMs for the two CVTs would be at their respective values (8300 and 4000) regardless of vehicle speed. I don't plot those lines because they would be trivial. 
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