Note:
I have no theoretical background except for what I've read on forums such as this one. Don't laugh, please have patience. Thank you.
Can you please please please explain why the difference in these two situations? I would expect that in point 1 it would also be max power. Why is it so?
Perhaps I can also ask a question with an example. Lets say we have two identical bicycles, with only one gear. I find someone my weight and we both have a seat on our bikes. Someone pushes us both so that we achieve the same velocity. Now:
Let's assume the pedals are on 1meter rods so that Nm=N numerically (right?). My friend applies 20N of force on his right pedal every time it is in the same position (that is under his foot), but isnt interested in his other pedal. I apply 10N of force on each pedal. That is, he is applying 20N once per rotation, while I apply two times 10N per rotation. Will my friend be accelerating faster? This is what has been bugging me, and I think it might also be what BBT meant with the "force over time" thing. Thanks in advance.
If the engine is producing maximum torque at a certain rpm, then for that gear (no changing allowed) that means this is also the point where the most longitudinal force is applied through the tyres, so ignoring drag, this is the point of most acceleration. Or at least, most acceleration potential.
The only reason peak acceleration can be found in situation 2 is because you can alter the gearing, which is what confuses people, since the additional gearing reduction (i.e. torque multiplication) outweighs the reduced torque output from the engine.
To the bike scenario, and assuming they are both applying the force for the same amount of time, and ignoring resitant forces, I think it should be equal. :S
No, you obviously misunderstand me, which I will get to in a second.
ypu, that's what I've been saying all the long, well done.
But this is where you have completely failed to see my example. Maybe you have it in your head that I MUST be wrong, so you can't see that I'm not in fact incorrect.
I never said (uisng you example) that you would change gear between 2 and 6000rpm. There is every chance that you will have more tractive effort (and thus utilise the area under the graph more completely) if you change at 7000rpm. But it might also be that the gearing used causes the curves to overlap at ANY rpm, so it might be best to change at 4500rpm. You don't know, as you haven't told anyone the gearing.
[quote=skiingman]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.[/quote[ You are still missing the point, either because you can't understand my point or becuase you want me to be wrong and refuse any statement by me.
[quote=skiingman]No, but if you let go of it, it goes further. [/quote]Thats got nothing to do with force/time...
[quote=skiingman]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.[/quote]Who said anything about optimising TC systems... Stop invented extra scenarioes where the previous examples fail to hold true. And besides, if you maximise the intantaneous acceleration for every timestep then you will be a perfect TC system, so there).
[quote=skiingman]
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)
[/quote] I'm not saying that at all. With my manual transmission examples I have tended to use one gear, with a fixed ratio, and the peak acceleration in that gear doesn't occur at peak power but at peak torque. But a CVT is interested in maximising the area under the curve (THE most important thing) and NOT the maximum accerlation in any one ratio. It's what I have been saying all the long (well, it took me a night with the CVT stuff, but even before that I knew maximum acceleration in a single gear occurs (as Colcob has confirmed more than once) at the engines peak torque. If you won't accept that now then you will be wrong.
[quote=skiingman]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.
[/quote]Perfect peak power gearing - that sentence doesn't make any sense... Trying to confuse the issue with Riemann sums isn't a good tactic, as people will just lack respect. A teacher doesn't bring quantum theory into year 5 lesson when discussing why apples fall of trees.
[quote=skiingman]Road and racecars alike almost never have those gears, because other concerns are more important than a particular performance from a to b.[/quote]Well, if A is the race start and B is the finish line, then all race cars want to get from A-B in the shortest time. Road cars aren't, as noise, economy, adaptability etc come into play). You're trying to make me sound like I don't know this...
[quote=skiingman]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.[/quote]The area under the wheel torque curves (for there are more than one) is a copy of the engine torque curve. Thus making the engine torque curve contain more area means that for the same gearing performance is increased. BUT, as you say, it's often possible to increase overall performance just by altering the gearing to increase the area under the wheel torque curve. However, you can only optimise one thing. If you go for top speed then your acceleration MUST suffer, and if you go for maximum acceleration (ignoring wheelspin for now) then your top speed will be low. Nice simple stuff...
[quote=skiingman]
Yes, the torque curve is the useful bit of info. Sometimes they are hard to come by.[/quote]No they're not. Most cars have them published, and any car used on in race will have a torque curve of some sort, either manufacturer published or obtained from a dynamometer. Admittedly they're notas prevailant as power curves, but thats because the general public is under the misapprehension that power curves at the thing to have.
[quote=skiingman]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.[/quote]Not sure I agree with that - I think both peaks on their own are useless, and I'd rather have a torque curve than a power curve when estimating performance.
In my accelerate or get shot example I was trying to state that (as Colcob confirmed, but it's what I've been saying all the long) is that maximum acceleration in any one fixed gear will occur at the engines peak torque. Thus if you were told to accelerate hardest in 4th gear you would want the engine to cross the peak torque of the engine. It by no means states that that peak acceleration will continue after the peak, and thus for A-B performance somehow trying to keep the engine at peak torque wouldn'tr work.
I will try to come up with a series of graphs that explain what I mean, and I will add them later when I've done them.
If you maximise one you maximise the other. But the torque curve is what you use when working out performance.
No, if you short shift you have not used all the area under the torque curve. I will try to explain this in the stuff I add later in a graphical form with annotations.
For best performance you have to stop thinking of the engine torque curve or the power curve. You are interested in maximising the area under the wheel torque curve, which takes gearing into account. Hopefully this will become clear after my excel session...
I accept your apology (indeed it has proved to be one of the best apologies I've ever seen without resorting to grovelling or other demeaining stuff), and I just felt that people were becoming too angry to continue a sane discussion. If we can continue as we were then I am more than happy to continue. I will do a bit of work to define what I mean. I will try to cover:
1. Why acceleration in a given gear is maximum at the engines peak torque.
2. Why maximising the area under the wheel torque graph gives the best performance
3. When gear changes have to be done, and why the peak torque rpm or the peak power rpm isn't of consideration here
4. Why CVT's are run at peak power for best performance.
I wouldnt bother really, I thought we'd already covered those things.
I would just say that maximising the area under a wheel torque/time curve, and maximising the area under a power/time curve (assuming freedom to gear appropriately), are effectively the same thing.
So really, some of you are arguing that the answer is 10+10, and the others that the answer is 40/2.
Okay, I've done some initial graphs. Hopefully they show that peak torque produces the highest acceleration in a given gear, that gear changes should not necessarily be made at peak torque, peak power or peak revs but where the wheel torque line (tractive effort if you like, which is what my graph is) crosses the next gear. If they don't cross then you use all the revs you can. They might cross below peak torque, at peak torque, above peak torque, below peak power at peak power above peak power or any revs you choose really. Does that make sense?
In all of these cases the maximum speed of the car is the same (118mph with air reistance, 162 withoutmph without air resistance i.e. rev limited), but the time taken to reach that will be less if you don't change at the optimum revs and use the most area possible under the graph.
I will now try to add CVT to that in two ways. One, I will make a 30 gear gearbox and shade the optimum, and second to draw the tractive effort curve a true CVT would give at WOT. I haven't done this yet, and it might take a little while whilst I copy equations, and graph lots of graph series'.
Sorry Colcob, I've already done the 'work' before I read your post...
Here is the graphs showing a 30 gearbox behaviour. The maximum and minimum gear ratios are the same as before, as are the torque and power curves, the final drive ratio and the wheel diameter.
You can see that if you shift at maximum torque or maximum revs you are getting a lot less wheel torque than at maximum power becuase the gear ratios are so close. Note that in any one of those gears maximum acceleration still occurs at peak torque.
The last one is a CVT. Obviously it's one curve as the gear ratios change keeping the revs constant (at WOT). This you can't use the argument that peak acceleration occurs at peak torque, becuase the gear box would just keep you at peak power, give you a numerically higher ratio and therefore more tractive effort. The ratio, rather than the revs becomes the variable which is why it's not as straightforward to see, and why it took me an evening of background reading to understand (actually it only took me 30 seconds to understand once I'd found a source that explained it well).
I hope this helps, and you can all now see what I'm saying. It's all been said now, and I think it's coming to the end of it. I hope no one can now misunderstand what I am trying to say, nor argue that I am wrong. You might be coming from a different angle to me, and the hard part is explaining your angle so that everyone else can understand. This is my angle, and it's been VERY accurate when performance predicting racing cars. Brunel University use a program I made for performance predicting the Formula SAE car, so it can't be that wrong.
Ok..... So then are you saying then that in that example you wouldn't necessarily have the most rapid increase in velocity over time (in any given gear) even though you crossed the torque peak?
Previously, until I saw your graphs, I've read what you said above as a contradiction, because they don't "sound" like two different scenarios. When you say "accelerate the hardest by crossing the torque peak of the engine".... it tends to imply that if you COULD hold it there then you would continue to accelerate at a maximum rate over time, thus maximizing A-B performance, but I now understand you're saying this is not the case, which is what I've been saying too.
Is there any way that you could, on the 5 spd graph, include indications of where we are in the engine's POWER curve as well? That would really help me visulize this whole mess in a way that I can comprehend! I see that you indicated the engine's power peak....
I realize you're talking tractive effort (wheel torque).
Ok.... Well that puts everything into perspective doesn't it...
Originally Posted by skiingman 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.
When you previously stated:
it appeared to contradict what skiingman said there...
(read: engine's power curve) this has been my argument from the get go!
Again this is has been my point from my first post!
This is all I've ever been trying to say!
This is part of my point, yes. Total work done is the same, but pretend now that the first example was happening at twice the rate (double the "engine RPM" ), you've done more work right?! Thank you for bringing that up, you've applied the same amount of overall torque in half the time, thus done more work and accelerated faster. Which as why, as stated, horsepower and torque are related by RPM.
AHAHAHAH , funny reading first couple of comments then skipping few pages, your all pathetic think u all know best , now im rebering why i left this community your all hoity-toity noobs washed with knowledge from games, go out there get some real expereince, start by leaving school first
Sorry Irdbsi! I forgot that you know best, weren't you banned once or twice from here?
And I've NEVER EVER seen you say anything remotely interesting or informative, whereas pretty much every poster in this thread has. And for your information some of us DO have real race cars, and most of us aren't in school. In fact, judging by your 6 year olds response I'd say you didn't have a race car and were still in school...
There are some weird unbalances in car classes that I cannot comprehend. FXO has wider tyres than XRT? Why give an obvious advantage to FZR by making the other two GTRs turbocharged?
Well they are turbocharged because they are only little 4-bangers . FZ50/FZR is a 6 cylinder engine. That should create enough power output already. I think it is all because of the idea for creating a more powerful 'brother' to the little roadcar sibliing. Only way to make those other 2 specific GTR cars to keep up is with some boost of power.
And yes, tire sizes are different size. Just looking at them shows it.
I do see the logic behind road cars and their GTR versions (beefing them up without changing the concept of the car), but can you honestly deny that turbo is nothing more than a handicap at the moment? Sure you can hit the gas pedal earlier when exiting turn, but that's just because you aren't putting any power to the wheels - you are just waiting for boost to raise and hopefully the horsepower comes back from lunchbreak before next braking.
The question I haven't seen asked is why the turbo car has as lousy of transient response as it does.
There are plenty of real-life vehicles producing similar power with far more tractable torque curves and throttle response...at least it seems so to me.
what gets me about the turbo modeling is the xrgt turbo and the complaints.
i use to own one of the original mitsi lancer 2000 turbo's which were the same running gear as the starion (see where im going here ?)
the turbo lag / power curve, especially after boost had been raised to produce 210 bhp was very similar to the xrgt turbos and yes it was a nightmare to drive, especially in the wet, on one occasion i ended up, going up a hill in the wet from a 40mph bend, in fifth showing 90 mph on speedo on 1/4 throttle and road speed was falling below 30 mph
to be honest it was always very optomistic of mitsubishi to try and even get the standard 160 bhp through a live axle with no lsd and 185 section tyres (or even 175s on some). i owned a garage at time and insurance allowed 20 year old working for us to drive a porsche but he was specifically prohibited from driving the lancer
[/quote]Thats got nothing to do with force/time...
).
. FZ50/FZR is a 6 cylinder engine. That should create enough power output already. I think it is all because of the idea for creating a more powerful 'brother' to the little roadcar sibliing. Only way to make those other 2 specific GTR cars to keep up is with some boost of power.