Actually, having read through both your posts again, I think you just both fundamentally dont get it, but in different ways.
Wow. Forest for the trees.
It takes a really cursory understanding of physics to understand that running the prime mover at peak power is going to get the car down the track the quickest, fastest, whatever. There is no need for spreadsheets and tractive effort curves to make this simple statement. Peak torque, indeed, has NOTHING TO DO WITH IT.
Jeebus...how many times do I have to point that out? Its like I'm writing into a void. Electric motors (usually) have peak torque at 0 rpm. The power peak is anywhere from a few to several tens of thousands of rpm later. The torque peak has zero to do with the power output of the motor. It has zero to do with how fast the car moves from any to point to any other point.
You are making this all up in your .confused head. Colcob already gave you the hint, and you didn't take it. Torque at the wheels, guy.
If you actually believe running the motor at peak torque is the best way to pass someone, you need to take a step back and look at some simple logic.
-At 40mph, the engine running at peak torque with a total gear ratio of 5:1 makes 5x peak torque at the wheels.
-If you downshift to nearer peak power and change the ratio to 7.5:1, you need to be making less than 2/3rds peak torque at your power peak for the car to accelerate slower than it would at peak engine torque. Very few cars make less than two thirds peak torque at the power peak. Reference widely available dyno evidence. If you make 80% of peak torque at the power peak, you now have 6x the peak torque at the wheels. Gee, that might be quicker to get around the "lorry" or whatever you call them.
Duh. Now downshift to a gear right at peak power and gun it. Bingo---bigger g-meter reading.
Who's books? Seriously, what the heck are you talking about? I don't know of a single quality race/automotive engineering book that claims that a torque figure is more relevant. Several of them extol the virtues of a smooth and tractable torque curve, but no right minded racer is ever going to sacrifice power for torque. If you can increase specific power output at the cost of the specific torque, you do it.
The torque curve is solely relevant for the purpose of developing the appropriate gearing, and making the driver happy with transient responses. The only thing relevant to getting from a to b faster than the other guy's is the peak power.
If your racing series limits what you use for gearing, the shape of the torque curve may gain more importance. That is an arbitrary bit of importance, however. As noted, airplanes fly perfectly well with internal combustion engines making 500ft-lbs of torque, but even better with a PT6 developing 50.
why does every one get fixated with horspower?
horsepower is never ever measured it is calculated from the torque figure and rpm figure
horsepower = torque X rpm
--------------
5252
dynos measure twisting effort, i.e. torque and then derive the hp figure
going off topic if you ever see power and torque curves for an engine that are not of equal value at 5252 rpm then they are wrong.
as stated by others its the total area of torque that determins acceleration, gearboxes and final drives are torque multipliers. this is where hp comes in' the torque output and its rpm gives the power entering the gearbox' say 100 hp at 6000rpm, if you have a 1.50:1 ratio then the power (ignoring mechanical losses) is the same but the rpm has fallen to 4000 rpm so for the power to be the same the torque must rise
therefore you get
100 hp =( 6000 rpm / 5252 ) * 87.52 lbft = ( 4000 rpm /5252) * 131 lbft
so an increase of 31 % in torque, add in a 3.5 : 1 final drive and then you get 459 lbft at the wheels.
all figures aprox and ignoring mechanical losses etc.
was going to carry on about cvt boxes but head hurst too much
edited for punctuation
horsepower is never ever measured it is calculated from the torque figure and rpm figure
horsepower = torque X rpm
--------------
5252
dynos measure twisting effort, i.e. torque and then derive the hp figure
going off topic if you ever see power and torque curves for an engine that are not of equal value at 5252 rpm then they are wrong.
as stated by others its the total area of torque that determins acceleration, gearboxes and final drives are torque multipliers. this is where hp comes in' the torque output and its rpm gives the power entering the gearbox' say 100 hp at 6000rpm, if you have a 1.50:1 ratio then the power (ignoring mechanical losses) is the same but the rpm has fallen to 4000 rpm so for the power to be the same the torque must rise
therefore you get
100 hp =( 6000 rpm / 5252 ) * 87.52 lbft = ( 4000 rpm /5252) * 131 lbft
so an increase of 31 % in torque, add in a 3.5 : 1 final drive and then you get 459 lbft at the wheels.
all figures aprox and ignoring mechanical losses etc.
was going to carry on about cvt boxes but head hurst too much
edited for punctuation
Colcob - no no, I now fully agree with you. I see where I was misinformed yesterday, and I've revised where I was wrong. I fully understand what you're saying now. I just wish I was as good at explaining my thoughts as you. Anyway...
I clearly don't understand what you are trying to state. You have the most force at the wheels with a fixed ratio at peak torque. I showed that with the tractive effort graph derrived from both power and torque. I think I need to re-read your posts to see quite what you're saying now...
No, read it again. If you had a MT with an infinite amount of gears you would be able to use the whole of one ratio, and in that ratio peak acceleration occurs at peak torque. But with a constantly variable infinite number of gears (CVT) then the gear ratio is automatically 'chosen' to give the most performance by fixing the revs at peak power.
Because the ratio varies, meaning the torque multiplication varies to give the most wheel torque (what I call tractive effort) at peak power...
If you have an infinite number of fixed ratios it's not the same as an infinitely variable ratio. With a CVT you can't run at peak engine torque because the ratio changes to give maximum wheel torque (and maximum area under the curve), and makes the engine run at peak power (at WOT).
Yes, area under the graph is the important thing. And maximising that is different to stating when, in a given fixed gear, maximum acceleration occurs at an instant.
Well, all my imaginary friends are nodding, and they don't like you. They keep whispering nasty rumours about you and everything
This skill in research (which I have yet to perfect, but I'm better than many, and getting better) is to sort the wheat from the chaff. Knowing what is misinformation is difficult, which is why proper, worked proof is a must.
I clearly don't understand what you are trying to state. You have the most force at the wheels with a fixed ratio at peak torque. I showed that with the tractive effort graph derrived from both power and torque. I think I need to re-read your posts to see quite what you're saying now...
No, read it again. If you had a MT with an infinite amount of gears you would be able to use the whole of one ratio, and in that ratio peak acceleration occurs at peak torque. But with a constantly variable infinite number of gears (CVT) then the gear ratio is automatically 'chosen' to give the most performance by fixing the revs at peak power.
Because the ratio varies, meaning the torque multiplication varies to give the most wheel torque (what I call tractive effort) at peak power...
If you have an infinite number of fixed ratios it's not the same as an infinitely variable ratio. With a CVT you can't run at peak engine torque because the ratio changes to give maximum wheel torque (and maximum area under the curve), and makes the engine run at peak power (at WOT).
Yes, area under the graph is the important thing. And maximising that is different to stating when, in a given fixed gear, maximum acceleration occurs at an instant.
Well, all my imaginary friends are nodding, and they don't like you. They keep whispering nasty rumours about you and everything
This skill in research (which I have yet to perfect, but I'm better than many, and getting better) is to sort the wheat from the chaff. Knowing what is misinformation is difficult, which is why proper, worked proof is a must.
Thank you very much for providing the solid math that I've been avoiding typing up because it shouldn't be necessary really to illustrate a simple concept.
I do believe, however, that while its obvious that torque and power are directly related when one has a dyno sheet to work with and an application in mind, power figures are vastly more useful for making cursory comparisons than any simple torque figure can be. It is my contention that this is why we have the unit power, because it offers a convenient and effective means of measuring work done over time.
When it comes down to it, no one really cares how much torque you have at any given moment when looking at the big picture. This is why AFAIK every thing under the sun that produces power is rated in units of power, not units of torque or force.
When you calculate the resistance of a hull in water at 15 knots, you go to the engine supplier with an estimate of how much horsepower you need, not how much torque you need. Same for building your train, plane, automobile, lawn mower, and everything else on earth. The torque figures and curves are needed for more subtle reasons, like making parts with appropriate duty cycles (re: speed of parts) and gearing that is efficient and won't break. For the race car...those are all secondary or tertiary considerations.
I know you've studied the calculus, so I don't understand how you can make such an obviously false statement.
Point of the calculus is to make an infinitely large number of infinitely small things the same as something singular and variable. Area under a curve, for instance, or volume of a solid of revolution.
You made this up. You can run a CVT at peak torque. Sometimes, you do. Particularly when trying to save fuel, as BSFC usually hits its lowest point around the peak torque figure.
When trying to get someplace in a hurry, you run the CVT at peak torque. For exactly the same reasons you run the infinite-number-of-ratios MT at peak power.
There are lots of people stating different things here now, and it's getting a touch hot.
I am going to state a few things I believe to be correct.
1. In a fixed gear, driving through the engine rev range, the car will be accelerating hardest at the point of peak torque. Forget about changing gear for the moment, just one gear. That much we know, right? Thus the 4th gear test I defined above is correct.
2. Changing down a gear into a numerically higher gear will result in more wheel torque at the higher engine rpm than the original gear at peak torque (unless the torque curve is a crazy shape or the gearing is weird, but ignore those possiblities)
3. A CVT is constantly changing to give you this highest numerical gear at any given moment to maximise your wheel torque.
4. @skiingman, I said at WOT, implying your not trying to save fuel (incidentally the best bsfc does occur at peak torque because the overall efficiency of the engine is highest at this point). I know you can run a CVT an any revs you want, but for WOT work you will run at the maximum revs you can get away with to give the most wheel torque.
Does all of that make sense? I sincerly hope so, cos I'm getting bored of this discussion now. It's becoming to crowded and too argumentative.
I am going to state a few things I believe to be correct.
1. In a fixed gear, driving through the engine rev range, the car will be accelerating hardest at the point of peak torque. Forget about changing gear for the moment, just one gear. That much we know, right? Thus the 4th gear test I defined above is correct.
2. Changing down a gear into a numerically higher gear will result in more wheel torque at the higher engine rpm than the original gear at peak torque (unless the torque curve is a crazy shape or the gearing is weird, but ignore those possiblities)
3. A CVT is constantly changing to give you this highest numerical gear at any given moment to maximise your wheel torque.
4. @skiingman, I said at WOT, implying your not trying to save fuel (incidentally the best bsfc does occur at peak torque because the overall efficiency of the engine is highest at this point). I know you can run a CVT an any revs you want, but for WOT work you will run at the maximum revs you can get away with to give the most wheel torque.
Does all of that make sense? I sincerly hope so, cos I'm getting bored of this discussion now. It's becoming to crowded and too argumentative.
Truthfully I have never studied either subjcet, all I have is a highschool education. I do however have a Mensa certificate which is probably the only reason I can at least semi-respectably participate in this conversation without looking (totally) like a jacka... er... you get it!
I have all the respect in the world for those on here with higher education, and the grey matter necessary to apply it!
I know it didn't seem like it, but I do understand that. I was looking for a way to articulate my questioning of the relevance of it in this debate. Even highschool math teaches calculus, at least in Canada. The crux of my poorly worded argument hinges on the importance of time in this whole debate.
But why be concerned about that moment? It seems logical to me in my uneducated brain that a single moment is not relevant to the performance of a car. If you're measuring ANYTHING that takes TIME to happen, that moment is a waste of math. Isn't it? This is why in the first post of this page I brought in the idea of force over time, because that's what's relevant in moving a car from here to there. Knowing how much force can be applied per unit of time will tell me how much work I can perform on the body in question. Prove me wrong if you want though.
Perhaps you can explain to me why the ability to do work is less relevant than instantaneous force?
Is that really the only reason?
It's not because performing more work on the car move it from here to there faster???
Then why wouldn't a drag car with a direct drive not seek to launch at peak torque, but rather peak power?
Indeed, in my VERY first post, I noted the futility of debating the relevance of two things derived from each other.
Absolutely, but they tell us different things. Which is why power needs to be derived from the torque curve The Power curve tells you MORE about how the engine will perform. I mean in terms of motive ability, not perform in the sense of it's personality - for that you need both curves. Otherwise we wouldn't do it.
At least this is a heck of a thread.
The point I was making is that it matters not if you are using a CVT or MT. If you are running at WOT and accelerating, you seek to run the engine at peak power. Hopefully the dudes that designed your gearing made that easy.
Yes. Thankfully, we have gears so we can run the engine at peak power rather than peak torque, maximizing the area under the curve of acceleration vs. time.
Depends on what you are trying to prove with it. It does prove what you stated above. It does NOT prove:
Which is from experience quite wrong...I'd have to have awfully wide ratios for me to settle with being in a gear lugging the motor that far below the power peak. In practice, that wouldn't happen in fourth gear. Second gear, yes, as the ratio is very wide between first and second in street cars.
Wow, this is getting confusing. BBO, you seem think that I've 'gone over to the torque side'. Not at all, i'm just trying to explain the way the two things interellate.
Your expression of it in terms of work is just as correct as my expression of it in terms of the aggregate of instantaneous forces over time, the two methods of explanation are not mutually exclusive.
I agree with Skiing man that peak power is a much more useful and relevant figure than peak torque in estimating the performance potential of an engine, but I'll repeat till I'm blue in the face that the torque/power curve is the be all and end all descriptor of the performance of an engine, and once you have that, all other relevant information can be derived (in theoretical performance terms at least, it wont tell you if the thing drinks oil, or blows up if you dont get it serviced every 1000 miles ).
Your expression of it in terms of work is just as correct as my expression of it in terms of the aggregate of instantaneous forces over time, the two methods of explanation are not mutually exclusive.
I agree with Skiing man that peak power is a much more useful and relevant figure than peak torque in estimating the performance potential of an engine, but I'll repeat till I'm blue in the face that the torque/power curve is the be all and end all descriptor of the performance of an engine, and once you have that, all other relevant information can be derived (in theoretical performance terms at least, it wont tell you if the thing drinks oil, or blows up if you dont get it serviced every 1000 miles
).With regard to the passing a tractor bit, which is where I believe the only contention lies now, I will use the picture below to explain what I mean.
Scenario - you are driving along at 60mph in top (5th)gear, just cruising. You come across a tractor doing 35mph. You slow down to wait for an opportunity to pass, and thus you need to select a new gear to do so safely and in good time. Being normal road car gearing the change will take you from peak revs (and around peak power) to a bit below peak torque, so that the maximum area under the graph is obtained. You now have a choice. You could use 1st gear (i know, I know, but for the sake of this example bear with me) and be at peak power. You would have lots of tractive effort. Great! But ! The thing about peak power is that it occurs near the end of the torque curve (i.e. high up the rev range) and thus you would instantly have to change gear. Therefore it would be better, overall, to stay in the higher gear (numerically lower) at below peak torque to make the bet pass.
If the tractor was at 30mph 1st gear would put you at peak torque (peak torque in 1st gear is the quickest acceleration the car is capable of. You will accelerate slower at peak power in 1st gear because you have less tractive effort, or (in the other term that people might find less confusing) less wheel torque.) whilst 2nd would be someway below peak torque. So, 1st gear at peak torque, or 2nd gear at much less than peak torque. If speed is the ultimate aim, then yes you would always use the lowest gear you can get away with as the wheel torque (I will try to use this term in this conversation but it's the same thing without the wheel diameter), but in this example the revs would increase very rapidly in 1st gear, meaning you'd still need to change gear. This it's still 'better' to use 2nd gear and perform the overtake with the minimum fuss and the minimum overall risk.
Of course you will always WANT to use the lowest gear (numerically high) possible, but quite often that doesn't make any sense for a variety of reasons.
The only other thing I can really grasp in this bitty conversation is perhaps there is still misunderstanding over when the maximum acceleration will occur. Lets say (using the same graph) you are stuck for some reason in 3rd gear, and you will be (keep with me) shot if you don't accelerate at the greatest rate possible. At what revs (or speed) do you slow to first to get the greatest possible acceleration? 70mph (peak torque of the engine in this gear) or 80mph (peak power of the engine in this gear)?
For clarities sake I must point out the dotted line on the graph. This is the ideal curve that a CVT would follow (or as closely to it as is possible), and at any point on that curve the engine would perform best at maximum power revs (in this case 6500rpm, peak torque is at 5500rpm). Ignore this bit for the time being as it only applies to constantly variable ratios.
Edit: Picture added
Scenario - you are driving along at 60mph in top (5th)gear, just cruising. You come across a tractor doing 35mph. You slow down to wait for an opportunity to pass, and thus you need to select a new gear to do so safely and in good time. Being normal road car gearing the change will take you from peak revs (and around peak power) to a bit below peak torque, so that the maximum area under the graph is obtained. You now have a choice. You could use 1st gear (i know, I know, but for the sake of this example bear with me) and be at peak power. You would have lots of tractive effort. Great! But ! The thing about peak power is that it occurs near the end of the torque curve (i.e. high up the rev range) and thus you would instantly have to change gear. Therefore it would be better, overall, to stay in the higher gear (numerically lower) at below peak torque to make the bet pass.
If the tractor was at 30mph 1st gear would put you at peak torque (peak torque in 1st gear is the quickest acceleration the car is capable of. You will accelerate slower at peak power in 1st gear because you have less tractive effort, or (in the other term that people might find less confusing) less wheel torque.) whilst 2nd would be someway below peak torque. So, 1st gear at peak torque, or 2nd gear at much less than peak torque. If speed is the ultimate aim, then yes you would always use the lowest gear you can get away with as the wheel torque (I will try to use this term in this conversation but it's the same thing without the wheel diameter), but in this example the revs would increase very rapidly in 1st gear, meaning you'd still need to change gear. This it's still 'better' to use 2nd gear and perform the overtake with the minimum fuss and the minimum overall risk.
Of course you will always WANT to use the lowest gear (numerically high) possible, but quite often that doesn't make any sense for a variety of reasons.
The only other thing I can really grasp in this bitty conversation is perhaps there is still misunderstanding over when the maximum acceleration will occur. Lets say (using the same graph) you are stuck for some reason in 3rd gear, and you will be (keep with me) shot if you don't accelerate at the greatest rate possible. At what revs (or speed) do you slow to first to get the greatest possible acceleration? 70mph (peak torque of the engine in this gear) or 80mph (peak power of the engine in this gear)?
For clarities sake I must point out the dotted line on the graph. This is the ideal curve that a CVT would follow (or as closely to it as is possible), and at any point on that curve the engine would perform best at maximum power revs (in this case 6500rpm, peak torque is at 5500rpm). Ignore this bit for the time being as it only applies to constantly variable ratios.
Edit: Picture added
(thisnameistaken)
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by thisnameistaken
What is this? Who writes the longest post -competition 
.
So when do I see the power and torque curves of the GTRs?
. So when do I see the power and torque curves of the GTRs?
When I've finished my dissertation which has to be in, in duplicate, by midday tomorrow. I will then play some LFS, have a nap (not going to sleep much tonight), get rather drunk in an un-Tristanlike fashion just for the hell of it, sleep some more, then have a look at stuff like that on Friday...
OMG! What happened to this thread? - it went beserk! Will read on my lunch break, don't have time now.
If you download Bobs GRC2 you can see some estimated torque curves. But we dont really need them.
Basically, the XRR acheives its peak power by producing more torque at lower revs (6278 rpm), and its peak torque point is only 1500 revs lower. So the torque curve is quite peaky, as you'd expect from a turbo.
The FZR acheives peak power by producing less torque but at higher revs (8000+ ), but its torque peak is a full 3000 revs lower, which means basically that it has a wider powerband. So, because the power comes on earlier in the rev range, over the course of an acceleration, the area under the graph is greater, so the car covers the distance in less time.
Another factor to think about is because the FZR has a higher rev limit, it can always be in a slightly lower gear than the XRR and FXR, which will compensate for its slightly lower torque output.
I think a couple of big hitters waded in Bob. Not to take anything away from Tristan or BBT, they might be big guys too.
TBH I'm reading Colcobs with interest. Actually, scrap that, I'm reading it all with interest. It's fascintating stuff.
But one thing I am getting (being strictly a layman) is that possible the engine size doesn't matter a huge amount. If you have geared it right then the Wheel Torque would be suffeicent . . . . I think I've just answered my question. Would it be that you would have to have so many gear changes to make the most from a small engine that the acceleration would be destroyed by the constant changing of gears . . . Right?
So the more power an engine has the higher the ability to pull in any gear and give you the maximum acceleration for that gear.
Er. And just one thing I picked up on. You guys are talking about acceleration times and argueing whether it describes a moment in time or a period of time. Other than the obvious 'It takes 4.5 seconds for this car to travel to 60 MPH' what are the units of measurement used to describe an objects acceleration. Because if you describe a point at which a car is accelerating at 60m /s/s (A purely notional figure. Not intented for accuracy) then you are saying that that car will continue to accelerate at 60m /s/s to infinity. Only when many measured points are plotted will the Acceleration curve be generated. So surely a time lapsed over distance travelled factor must be incorporated into any given unit of acceleration. (Which I might have just done using the per second per second approach. God, I wish I'd gone to uni)
P.S. I have a sneaky feeling I must just be confusing myself here. So feel free to pass over this post.
TBH I'm reading Colcobs with interest. Actually, scrap that, I'm reading it all with interest. It's fascintating stuff.
But one thing I am getting (being strictly a layman) is that possible the engine size doesn't matter a huge amount. If you have geared it right then the Wheel Torque would be suffeicent . . . . I think I've just answered my question. Would it be that you would have to have so many gear changes to make the most from a small engine that the acceleration would be destroyed by the constant changing of gears . . . Right?
So the more power an engine has the higher the ability to pull in any gear and give you the maximum acceleration for that gear.
Er. And just one thing I picked up on. You guys are talking about acceleration times and argueing whether it describes a moment in time or a period of time. Other than the obvious 'It takes 4.5 seconds for this car to travel to 60 MPH' what are the units of measurement used to describe an objects acceleration. Because if you describe a point at which a car is accelerating at 60m /s/s (A purely notional figure. Not intented for accuracy) then you are saying that that car will continue to accelerate at 60m /s/s to infinity. Only when many measured points are plotted will the Acceleration curve be generated. So surely a time lapsed over distance travelled factor must be incorporated into any given unit of acceleration. (Which I might have just done using the per second per second approach. God, I wish I'd gone to uni)
P.S. I have a sneaky feeling I must just be confusing myself here. So feel free to pass over this post.
Well when you say an object is accelerating at 10 m/s/s you are only referring to its acceleration at that moment in time (or to put it another way, the gradient of the velocity/time graph at a given point).
You arent saying that it will continue to accelerate at the same rate at any point in the future. But i suppose the units suggest a kind of prediction. An acceleration rate says "at this rate of acceleration, in 1 seconds time, this object's velocity will have changed by this much".
Imagine, a graph of speed/time. If your acceleration rate is constant, the graph will simply be a straight line going off at an angle to the top right. The gradient of that straight line is your constant acceleration.
Now imagine a real life speed/time graph like you'd see in F1 Perfview, it starts off kind of steep, when your acceleration is high, but the faster you go, the more it flattens off, because your acceleration is lower.
At any point on that graph, the gradient of that point is the current rate of acceleration at that moment in time.
This is interesting stuff actually, you're kind of discovering the edges of calculus by yourself.
You arent saying that it will continue to accelerate at the same rate at any point in the future. But i suppose the units suggest a kind of prediction. An acceleration rate says "at this rate of acceleration, in 1 seconds time, this object's velocity will have changed by this much".
Imagine, a graph of speed/time. If your acceleration rate is constant, the graph will simply be a straight line going off at an angle to the top right. The gradient of that straight line is your constant acceleration.
Now imagine a real life speed/time graph like you'd see in F1 Perfview, it starts off kind of steep, when your acceleration is high, but the faster you go, the more it flattens off, because your acceleration is lower.
At any point on that graph, the gradient of that point is the current rate of acceleration at that moment in time.
This is interesting stuff actually, you're kind of discovering the edges of calculus by yourself.
Indeed, then you get into the realms of rate of change of acceleration.
And as acceleration is rate of change of speed, and speed is rate of change of position (if you like), then the rate of change of acceleration (how acceleration varies over time) is the rate of change of the rate of change of the rate of change of position... Yikes!
And as acceleration is rate of change of speed, and speed is rate of change of position (if you like), then the rate of change of acceleration (how acceleration varies over time) is the rate of change of the rate of change of the rate of change of position... Yikes!
Thats what I figured. So what is the correct termanology to use when dealing with a cars acceleration. I feel awfully crude just saying a 0-60 time of such and such. Surely there is a better way of presenting the acceleration over time. But is it that because it is a curved graph then to actually define it in any practical sense would be impossible due to the non linear nature of said graph. Would you have to just take a point in that time frame and just examine that just to get any usable data from it.
How do they do it in Racing? What engineering terms do they use when changing gear ratios, for example, and applying them to the engine. How can they show the change, or do they just generate the graphs and leave it at that?
Another kick at the cat!
Actually, yes power IS the rate at which force is applied. The most basic definition of Horsepower is 33000 foot pounds per minute. That is not debatable! It also sounds suspicously akin to force over time. That is of course linear motion, but who cares, for the scope of this debate, force is force. This is why I pointed out that 1HP = 550 foot pounds per second in an earlier post. If it takes one second to apply 550 foot pounds of force, you have spent one horsepower, therefore your statement is errant. This is the whole basis for the relationship between Horsepower and Torque! Hence, the other factor that relates them firmly to each other is engine RPM, which is how fast the engine is turning!
Here is a real life example. Observe my attachment now. It's the torque / power curve for great little engine. Colcob stated above that turbo engines generally have very peaky torque curves... Well then this well designed engine is a great exception, because it peaks at 250 pound feet of torque all the way from 2400 to 4400 RPM. Horsepower peaks at 230hp at 5300RPM.
According to Tristan, we want to run at the torque peak as much as possible to get from here to there ASAP. Just like passing a tractor. Therefore with that logic, I am best to operate in any given gear, all the way down to 2400RPM. Let's look at the force over time principle again, and actually calculate the total amount of force capable of being generated... At 2400RPM, the power output of this engine can be defined in terms of force over time as 120HP*550 = 66000LB/FT/S. Assuming wide open throttle, if one second of time passes at 2400RPM (obviously the value will change as RPM increases... Hence a POWER curve, but let's pretend the engine is held against a load that holds rpm there... Guess what load that would be...) we will have exerted a total of 66000 pound feet of force over that second of time. Climb to 5300RPM, at 230HP*550 = 126500 LB/FT/S. Again, in the same situtation, holding the engine at 5300RPM for one second (at WOT obviously) yeilds a total force output of 126500lb/ft over the duration of that second.
In either case, over that second, where have I exerted more force folks?!
Apparantly at the power peak of course!
Taking out resistances etc just to keep it simple, do you think a 10km/h increase in speed will occur at WOT FASTER if we floor that car at 2000RPM compared to say 5000RPM?
Over a second of time in each case, which one will I feel more on my back? Which situation has dumped more energy into my car over that second of time?
Even LFS shows this to be true. LFS does not tell you to shift in a way that straddles the torque peak in ANY given gear, and as scared as I am to say it, the area under the POWER curve determines how fast you get from A to B, not the torque curve. That's why we HAVE power curves! See attached graph again. Anyone dragging this car (which I have had the pleasure of experiencing!) runs the thing from 4800-5800 RPM in every gear, which they can because it's a close ratio unit (especially for a road car). This is because the total force exerted over the TIME it takes is greater at peak power than it is at peak torque. So, to answer Funnybear's question: beyond doubt, you want to operate as close to your power peak as you can at all times, period. This will ensure your car's greatest average velocity over time!
This whole debate started when Tristan exclaimed how useless power figures are in terms of estimating vehicle performance... Which clearly can not be viewed as the case any longer. An inexperienceable moment means nothing in the context of getting a car around a track in the least time possible. For that, you need power, not JUST torque. Torque needs to be applied at a rate sufficient move me from here to there in less time. In the automotive world, this is a much, much more relevant (practical as Col put it) description of acceleration. To describe it otherwise is terribly misleading to the average enthusiast, because you're giving a false impression of what results will be for any given vehicle/engine.
Should someone decide that this is wrong, please provide detailed analysis and math to back up your argument. Don't just give me "that's wrong".. or "your formula is wrong" and dodge the point. If I am wrong, the I need to know why, and if you say I am wrong, then you must know why, right?
Actually, yes power IS the rate at which force is applied. The most basic definition of Horsepower is 33000 foot pounds per minute. That is not debatable! It also sounds suspicously akin to force over time. That is of course linear motion, but who cares, for the scope of this debate, force is force. This is why I pointed out that 1HP = 550 foot pounds per second in an earlier post. If it takes one second to apply 550 foot pounds of force, you have spent one horsepower, therefore your statement is errant. This is the whole basis for the relationship between Horsepower and Torque! Hence, the other factor that relates them firmly to each other is engine RPM, which is how fast the engine is turning!
Here is a real life example. Observe my attachment now. It's the torque / power curve for great little engine. Colcob stated above that turbo engines generally have very peaky torque curves... Well then this well designed engine is a great exception, because it peaks at 250 pound feet of torque all the way from 2400 to 4400 RPM. Horsepower peaks at 230hp at 5300RPM.
According to Tristan, we want to run at the torque peak as much as possible to get from here to there ASAP. Just like passing a tractor. Therefore with that logic, I am best to operate in any given gear, all the way down to 2400RPM. Let's look at the force over time principle again, and actually calculate the total amount of force capable of being generated... At 2400RPM, the power output of this engine can be defined in terms of force over time as 120HP*550 = 66000LB/FT/S. Assuming wide open throttle, if one second of time passes at 2400RPM (obviously the value will change as RPM increases... Hence a POWER curve, but let's pretend the engine is held against a load that holds rpm there... Guess what load that would be...) we will have exerted a total of 66000 pound feet of force over that second of time. Climb to 5300RPM, at 230HP*550 = 126500 LB/FT/S. Again, in the same situtation, holding the engine at 5300RPM for one second (at WOT obviously) yeilds a total force output of 126500lb/ft over the duration of that second.
In either case, over that second, where have I exerted more force folks?!
Apparantly at the power peak of course!
Taking out resistances etc just to keep it simple, do you think a 10km/h increase in speed will occur at WOT FASTER if we floor that car at 2000RPM compared to say 5000RPM?
Over a second of time in each case, which one will I feel more on my back? Which situation has dumped more energy into my car over that second of time?
Even LFS shows this to be true. LFS does not tell you to shift in a way that straddles the torque peak in ANY given gear, and as scared as I am to say it, the area under the POWER curve determines how fast you get from A to B, not the torque curve. That's why we HAVE power curves! See attached graph again. Anyone dragging this car (which I have had the pleasure of experiencing!) runs the thing from 4800-5800 RPM in every gear, which they can because it's a close ratio unit (especially for a road car
). This is because the total force exerted over the TIME it takes is greater at peak power than it is at peak torque. So, to answer Funnybear's question: beyond doubt, you want to operate as close to your power peak as you can at all times, period. This will ensure your car's greatest average velocity over time!This whole debate started when Tristan exclaimed how useless power figures are in terms of estimating vehicle performance... Which clearly can not be viewed as the case any longer. An inexperienceable moment means nothing in the context of getting a car around a track in the least time possible. For that, you need power, not JUST torque. Torque needs to be applied at a rate sufficient move me from here to there in less time. In the automotive world, this is a much, much more relevant (practical as Col put it) description of acceleration. To describe it otherwise is terribly misleading to the average enthusiast, because you're giving a false impression of what results will be for any given vehicle/engine.
Should someone decide that this is wrong, please provide detailed analysis and math to back up your argument. Don't just give me "that's wrong".. or "your formula is wrong" and dodge the point. If I am wrong, the I need to know why, and if you say I am wrong, then you must know why, right?
Correct. This is why a POWER curve is useful, because it gives you an idea how much work will be performed during the time it takes to travel through the RPM band, which could be calculated based on mass/resistance/gearing etc.
Not if you want any truly relevant / accurate data. It may give you a premise for prediction as Col stated, but it's only a glimpse of a whole picture as you alluded to.
Look matey, don't quote me out of context. I have not said you want to run at peak torque all the time for best A-B. I have known for over 15 years (a large proportion of my life to date) about torque multiplication. I am well aware that you will probably have more wheel torque in a lower gear at a higher rpm than a higher gear at low RPM. I am well aware that for the best A-B time would want to maximise the area under the wheel torque curves, either by maximising the area under the torque curve or by increaseing the working range (revs) of the engine.
Don't piss on me by saying I didn't know that. What I stated (or was trying to before you got your bee in your bonnet) was that maximum acceleration will occur at peak torque in a given gear. Thus in 1st gear maximum acceleration will be at 4000rpm if the peak torque is at 4000rpm. In 5th gear the peak acceleration (of that gear) will be at the peak torque of the engine, i.e. 4000rpm. If you have a CVT 'box then you run at peak power because the gear ratio gives you the best ratio possible at every rpm.
I don't see what you're going an about with force over time all the time. A force is a force. An apple doesn't weigh more if you hold it for longer. I'm not interesting in your force over time (maybe I am, but I don't understand your blatherings). I am interested, in terms of acceleration and performance, of the instantaneous force, and thus the instantaneous potential acceleration at every time step, be it 100 times per second (LFS physics engine) or infinite times per second (calculus).
I already did a graph for you using your equations of power = force / time, and it quite clearly showed that the peak wheel torque and the peak wheel force (and thus the peak acceleration) occurs at peak torque for a fixed ratio when derrived from both torque AND power.
If there is more tractive effort (i.e. area under the graph) between 2000rpm and your final engine speed than area between 5000 and the final engine speed then yes, 2000 would be quicker. You will have more total force exerted at the wheels. But it might well be that the engine does produce more torque (more tractive effort between 5000 and 5000+x rpm, in which case that would be better.
Assuming the same gear for the 2000 and 5000 problem, then yes the 5000 will put more energy into the car as kinetic energy is proportional to speed squared.
Yes, we've known this all along. You will get more acceleration by utilising more of the area under the wheel torque curve, thus you run at those rpms. Of course you don't change gear at peak torque, that would be silly. You use the rest of the gear to the redline (or where the wheel torque curves cross for each gear) to get the most performance.
I still maintain that a power curve is useless. The shape of the wheel torque figure, which is what moves your car, is the same shape as a torque curve, albeit flattened due to torque multiplication. Therefore I'd much rather see a torque curve and gearing info that a power curve (cos it means I don't have to convert it into torque). Power is derrived from torque, and torque is derrived from power. They are interchangable (with engine speed and gearing known), but the torque curve is the useful bit of info.
Normally these threads with you are quite nice and interesting, but this last one has quite a strong air of rudeness to it, and I don't like it. Therefore, whilst I am not admitting defeat, I am not going to post in this thread. You can believe what you like, and I will belive what I like. I am the one (out of use two) using the equations and theries on actually tuning race cars (of sorts), and I know for a fact that my estimates of acceleration and performance for a given cars torque curve/gear ratios/wheel size/rev range is pretty much spot on. I'm not going to argue with you, as you've tuned nasty. Thanks for the interesting bits at the start, and I learnt a lot about CVT's. I just hope you sort out your ideas and present them in an understandable fashion rather than bithing and doing all the 'apparently, according to Tristan' and 'Tristan says this which we all know is shit' etc etc, and then quoting me out of context.
So yeah, thanks, and bye.
This is what you keep doing. You haven't yet looked at what I've posted and picked away at it bit by bit to show where I'm wrong, you just keep charging in with your force over time whatnots and your poor maths and incomplete examples. I've had enough, you'll never learn.
This is a confusing statement. If you want to make the gearing work best to get from a to b, you gear the vehicle to have the maximum area under the engine's power curve for the event...maximizing the area under the engine's torque curve might come close to coinciding with this, but in many cases won't.
If you have any transmission, you maximize the area under the power curve for the time spent going from A to B. You are confused by thinking that a CVT is anything more than an optimized multi-speed transmission.
You believe that optimizing the area under the torque curve is optimal. I don't. I'd like to provide an example as proof.
Lets say we have a modern engine tuned with electronic turbo wastegate to have a flat torque peak from 2000 to 6000 rpms. After 6000 rpms, the torque decreases 10 percent by 7000 rpms.
Your theory would show the car to accelerate fastest from A to B if shifted between 6000 and 2000 rpms, where the area under the torque curve is greatest. However, the car will actually get from A to B faster if you don't upshift until some point past 6000 rpms. Also, if you have enough ratios, it will get there much quicker if you don't make it shift all the way back to 2000 rpms. A gearbox spending time on a torque plateau between 5000-6000rpms will get there faster than one on the same plateau between 2000-6000, regardless of "area under torque curve". The important quantity thus becomes more obviously the power generated over time.
I'm going to leave the actual math as an exercise to the reader (mainly because I don't have Excel on this computer, and my Excel macros aren't playing nice with OpenOffice), but there are numerous programs on the web that will allow you to simulate this and bear this out. Importantly, by shifting after the torque peak of an engine with a wide, flat torque curve, you don't maximize the area under the torque curve. You also don't necessarily go slower, you may go faster.
No, but if you let go of it, it goes further.
The danger here is that you risk missing the forest for the trees. You don't actually care that much about instantaneous accelerations unless you are trying to write a traction control system. You care about the average acceleration over time, and optimizing it. You do this by optimizing the torque delivered to the wheels over time. Your stumbling block is understanding that this is different than the torque delivered by the engine over time.
I already did a graph for you using your equations of power = force / time, and it quite clearly showed that the peak wheel torque and the peak wheel force (and thus the peak acceleration) occurs at peak torque for a fixed ratio when derrived from both torque AND power.
Thankfully, we don't have single speed car transmissions. We are offered a number of ratios to use. You admit that the CVT goes from A to B fastest when used at peak power, but refuse to recognize the same relationship in a 6MT. (a lower resolution CVT, as Ball Bearing Turbo correctly put it)
Since you bring up the calculus, if you determined the perfect peak power gearing from A to B for a CVT, the optimal gear ratios for a 5MT would be the interval points on the CVT ratio curve for a midpoint Riemann sum where n was the number of ratios.
Road and racecars alike almost never have those gears, because other concerns are more important than a particular performance from a to b.
Quite correct. You need to admit to yourself (hey, publically if you want, I don't care) that optimizing the area under the wheel torque curve doesn't necessarily mean optimizing the area under the engine torque curve. Particularly if you have something with a broad and flat torque curve, such as a modern turbocharged gas engine or an electric motor or a gas turbine.
Yes, the torque curve is the useful bit of info. Sometimes they are hard to come by.
This is why people are being confused when they state that "peak power" is worthless. On the contrary, peak power tells you more than peak torque. If you have the opportunity to grab a valid torque curve, you are completely set. The confusion sets in when you misinterpret the use of that curve and make dangerous assumptions about areas under curves.
Oh, race cars eh? Well, I turbo'd a track day car once, crashed it, now I'm going to do so again this summer. Unless I had a lot more money, this discussion becomes rather moot because I don't have the money for bespoke gear clusters.
I would hope everyone involved would learn something. I learned my roommate has a copy of Excel to let me borrow. Amongst other goodies.
Er.. Wow... 
edit: skiingman beat me, this part is @ Tristan:
Before I present any further questions, rebuttals, theories or "blatherings", I would like to sincerely apologize if I have (and clearly I have) offended you, or anyone else reading this thread. This was never my intent, and perhaps I should make better use of smileys at appropriate locations to indicate my intent. I don't always have the time for that, so I apologize. I am
surprized at your response, it seems very uncharacteristic for you. Nonetheless, please accept my humblest apologies. I didn't set out to frustrate you or anyone else, that I can promise you. And I truely never intended to appear aggressive, arrogant or any of the other things you indicated I was displaying. If I appear terse in any thread, especially those with long posts, please understand it's just me trying to articulate something as quickly as possible. And I never meant to quote you out of context, which I will address in my response. And I never said what you say is poo or any other form of waste, human or animal!
, Quite the contrary I respect everyone on this forum. We might as well wrap this up though pretty swiftly since it's clear in this circumstance we're not receiving properly what the other person is saying. Indeed I am sure there is a plethora of misunderstandings, probably mostly on my part. So in closing, please forgive me and I hope we can once again have an interesting debate in the future.
What confuses me is that you say in your "accelerate or get shot" example that you want to be at peak engine torque, which by extention says to me that the fastest way from A to B is to hang out there. If the rate of acceleration in any fixed gear ratio is truly the greatest at the engines torque peak rather than the power peak (which I undestand F=MA is supposed to say) then how could it also not be the fastest way to get from A to B? I hope I am wording that in a way that articulates my question accurately, and maybe you'll undestand why I mistakenly quoted you out of context in your above quote...
I am aware of your background, and having the experience you do at an age a few years younger than me is indeed something to be proud of. I would be too if I had it!
I never said you didn't know about torque multiplication, or at least I didn't mean to express or imply that at all. That's a Gr9 topic and I would never insult you like that, I'm not out to insult anyone. If anyone needs teaching, it me and I already know that LOL! As for max/acc per gear I addressed my question above.
Forgive me, I still just can't see that. You stated in an earlier post that the variable ratio "is an extra variable in the system". Could you show me in numbers an example of each type of system (Reg Manual and CVT)?
Maybe then I would get it. The graph you posted didn't tell me much. It still seems logical so me that since a CVT is "infinitely variable", you basically are just having the transmission change (for lack of better interpretation) through an infinite number of fixed ratios very rapidly. I truly fail to understand why the principles of a CVT would be different than any other transmission, since at any instant in time you can pinpoint what ratio the transmission is using right?
Clearly this is the case.... And perhaps I am full of crap. It seems logical to me that it's paramount, but I am starting to think we are describing two totally separate things and calling them the same thing.
FTR, my math about force over time shows why maximizing the area under the engines power curve (not the torque curve) yields the results we are all looking for at the track.
This is also my only point of contention, and the point I've been trying to make all along.
The reason I used that SRT-4 Graph, is because it is a great example of what I have been trying to convey this whole time. When you said above: "area under the graph"... I agree if you're talking the power curve, I don't agree if you're talking the torque curve, which is the fundamental point of our "debate". One does not short shift that car and hang out between 2400-4400RPM for best performance, which would be maximizing area under the torque curve. People hang out in the space that maximizes the area under the power curve for best performance, the quickest way around the track, up the dragstrip, or any time you want to be quick. Even though the torque curve has started to dwindle, we are just about to maximize the area under the power curve, which I still think is more important, and practical experience in that car seems to prove that.
Again, I am sorry you feel that way! I never meant to be rude. I am sorry you feel my "maths are poor and examples incomplete", perhaps someone with a better understanding can at least articulate properly what I've been trying to say, I regret that I couldn't be more precise in my presentation. I do wish though that you could've shown me where my math was wrong. When I first put math into the (don't hit me) "force over time" concept, you stated it was "some stuff, some correct and some not" but you never showed me where I was wrong! Contrary to your current beleif, I am indeed willing to learn
edit: skiingman beat me, this part is @ Tristan:
Before I present any further questions, rebuttals, theories or "blatherings", I would like to sincerely apologize if I have (and clearly I have) offended you, or anyone else reading this thread. This was never my intent, and perhaps I should make better use of smileys at appropriate locations to indicate my intent. I don't always have the time for that, so I apologize. I am
surprized at your response, it seems very uncharacteristic for you. Nonetheless, please accept my humblest apologies. I didn't set out to frustrate you or anyone else, that I can promise you. And I truely never intended to appear aggressive, arrogant or any of the other things you indicated I was displaying. If I appear terse in any thread, especially those with long posts, please understand it's just me trying to articulate something as quickly as possible. And I never meant to quote you out of context, which I will address in my response. And I never said what you say is poo or any other form of waste, human or animal!
, Quite the contrary I respect everyone on this forum. We might as well wrap this up though pretty swiftly since it's clear in this circumstance we're not receiving properly what the other person is saying. Indeed I am sure there is a plethora of misunderstandings, probably mostly on my part. So in closing, please forgive me and I hope we can once again have an interesting debate in the future.What confuses me is that you say in your "accelerate or get shot" example that you want to be at peak engine torque, which by extention says to me that the fastest way from A to B is to hang out there. If the rate of acceleration in any fixed gear ratio is truly the greatest at the engines torque peak rather than the power peak (which I undestand F=MA is supposed to say) then how could it also not be the fastest way to get from A to B? I hope I am wording that in a way that articulates my question accurately, and maybe you'll undestand why I mistakenly quoted you out of context in your above quote...
I am aware of your background, and having the experience you do at an age a few years younger than me is indeed something to be proud of. I would be too if I had it!
I never said you didn't know about torque multiplication, or at least I didn't mean to express or imply that at all. That's a Gr9 topic and I would never insult you like that, I'm not out to insult anyone. If anyone needs teaching, it me and I already know that LOL! As for max/acc per gear I addressed my question above.
Forgive me, I still just can't see that. You stated in an earlier post that the variable ratio "is an extra variable in the system". Could you show me in numbers an example of each type of system (Reg Manual and CVT)?
Maybe then I would get it. The graph you posted didn't tell me much. It still seems logical so me that since a CVT is "infinitely variable", you basically are just having the transmission change (for lack of better interpretation) through an infinite number of fixed ratios very rapidly. I truly fail to understand why the principles of a CVT would be different than any other transmission, since at any instant in time you can pinpoint what ratio the transmission is using right?
Clearly this is the case.... And perhaps I am full of crap. It seems logical to me that it's paramount, but I am starting to think we are describing two totally separate things and calling them the same thing.
FTR, my math about force over time shows why maximizing the area under the engines power curve (not the torque curve) yields the results we are all looking for at the track.This is also my only point of contention, and the point I've been trying to make all along.
The reason I used that SRT-4 Graph, is because it is a great example of what I have been trying to convey this whole time. When you said above: "area under the graph"... I agree if you're talking the power curve, I don't agree if you're talking the torque curve, which is the fundamental point of our "debate". One does not short shift that car and hang out between 2400-4400RPM for best performance, which would be maximizing area under the torque curve. People hang out in the space that maximizes the area under the power curve for best performance, the quickest way around the track, up the dragstrip, or any time you want to be quick. Even though the torque curve has started to dwindle, we are just about to maximize the area under the power curve, which I still think is more important, and practical experience in that car seems to prove that.
Again, I am sorry you feel that way!
I never meant to be rude. I am sorry you feel my "maths are poor and examples incomplete", perhaps someone with a better understanding can at least articulate properly what I've been trying to say, I regret that I couldn't be more precise in my presentation. I do wish though that you could've shown me where my math was wrong. When I first put math into the (don't hit me) "force over time" concept, you stated it was "some stuff, some correct and some not" but you never showed me where I was wrong! Contrary to your current beleif, I am indeed willing to learn Blimey. Here we go again. I'll be honest and say I didnt have the patience to read all those posts in detail, but it just seems to me that everyone is coming to a roughly equivalent understanding, but are still arguing about minutae that got posted ages back.
I also think its difficult to have a meaningful debate because the definitions haven't been sorted. Also, I think that for myself as well as others, existing pieces of knowledge are preventing us from properly assimilating other bits of knowledge, and also we are making statements that apply to a particular context without necessarily qualifying that context.
When I step back from cars and engines and gearings for a moment, and just think about raw physics, it all seems blindingly obvious.
If over period of time t, an average power of p is expended, the amount of kinetic energy added to the object will be p*t.
So any change in the average power over the period of time will result in a proportional change in the amount of kinetic energy added.
So lets ignore transmissions or drivetrain losses etc. and say that our CVT produces a dead flat power curve (line?) at max power, so the average power for the time t = max power.
In any manual transmission, the power curve will rise and fall depending on gear changes, but will always be below or equal to max power, so average power over time t < maxpower.
So that means that after time t, less kinetic energy has been added in the second example, which means that any time our engine is below maximum power, we are adding less kinetic energy.
Now i'm starting to state the obvious now, but if we have initial velocity u, our final velocity v will always be higher when more kinetic energy has been added. And if our final velocity is higher, we have 'accelerated better'.
So, for a given engine, with complete freedom to gear appropriately, maximum final velocity will be acheived when the engine spends as much time as possible as close as possible to maximum power.
Just for completeness, I going to make a few statements that have come up in the debate, together with the context needed to make them correct.
1. For a given fixed gear ratio, the highest instantaneous acceleration acheivable in that gear will occur at the moment when the engine is producing maximum torque.
2. For a given road speed, the highest instantaneous acceleration acheivable at that speed will occur if the gearing is such that the engine is producing maximum power.
3. Over a given period of time, ignoring any gearing limitations, the greatest overall increase in velocity will occur when the area under the power/time graph is maximised.
And thats it from me.
I also think its difficult to have a meaningful debate because the definitions haven't been sorted. Also, I think that for myself as well as others, existing pieces of knowledge are preventing us from properly assimilating other bits of knowledge, and also we are making statements that apply to a particular context without necessarily qualifying that context.
When I step back from cars and engines and gearings for a moment, and just think about raw physics, it all seems blindingly obvious.
If over period of time t, an average power of p is expended, the amount of kinetic energy added to the object will be p*t.
So any change in the average power over the period of time will result in a proportional change in the amount of kinetic energy added.
So lets ignore transmissions or drivetrain losses etc. and say that our CVT produces a dead flat power curve (line?) at max power, so the average power for the time t = max power.
In any manual transmission, the power curve will rise and fall depending on gear changes, but will always be below or equal to max power, so average power over time t < maxpower.
So that means that after time t, less kinetic energy has been added in the second example, which means that any time our engine is below maximum power, we are adding less kinetic energy.
Now i'm starting to state the obvious now, but if we have initial velocity u, our final velocity v will always be higher when more kinetic energy has been added. And if our final velocity is higher, we have 'accelerated better'.
So, for a given engine, with complete freedom to gear appropriately, maximum final velocity will be acheived when the engine spends as much time as possible as close as possible to maximum power.
Just for completeness, I going to make a few statements that have come up in the debate, together with the context needed to make them correct.
1. For a given fixed gear ratio, the highest instantaneous acceleration acheivable in that gear will occur at the moment when the engine is producing maximum torque.
2. For a given road speed, the highest instantaneous acceleration acheivable at that speed will occur if the gearing is such that the engine is producing maximum power.
3. Over a given period of time, ignoring any gearing limitations, the greatest overall increase in velocity will occur when the area under the power/time graph is maximised.
And thats it from me.
about gtr's
(131 posts, started )
FGED GREDG RDFGDR GSFDG