about gtr's

[BlueSkunk]

hi, a question to those who might know: i really like the fxr best of all the gtr's but have to accept the fact that it's not a necessarily competitive car (compared to the xr and fz). unless, of course, i would be an above average driver....

so my question is whether there are any plans to even out the odds somewhat. if not then i'd consider making an effort and starting to learn to drive with the xr straight away.

much appreciated

[tristancliffe]

The FXR will always be at a disadvantage because of the weight of it's 4WD system - the other two are RWD and thus ultimately faster, just like it would be in real life. It also means that the FXR is the easiest car to drive because it's nigh on impossible to spin it.
The XRR is next 'best', but it'll struggle around some corners for having a front engined layout and less power than the FZR. Not too hard to drive, but not the fastest.
The FZR is mid/rear engined, and the most powerful, and as such can really hit you if you make an error. You need to be much smoother to get it to work. The high nose bug also effects this car the most as the decreased rear travel and peculiar roll rates make keeping the car pointing the right way difficult at times (in slower corners mostly).

I'd say practice the XRR at first and get good at that, then move up to the FZR. The FXR (edit: typo, thanks Android) (edit2: I have lost the ability to press the keys in the right order, so had to fix it again...) is only useful at South City, especially the shorter tighter courses.

[bobvanvliet]

In races of about an hour, the FXR is actually quite competative at Blackwood GP also.

And as for the XRR and FZR, I get into low laptimes a lot faster with the FZR, but it does suffer the most from tyre wear.

[Bob Smith]

What are you on Tristan?

The FXR will always be at a disadvantage because of the weight of it's 4WD system
The measly 30kg increase in mass is unlikely to much of a factor in the FXRs sloweness. The massive additional power loss in the drivetrain, however, would be.

The XRR is next 'best', but it'll struggle around some corners for having a front engined layout and less power than the FZR. Not too hard to drive, but not the fastest.
1 - how exactly does enginer power output affect cornering speed?
2 - I think you'll find all three GTRs have matching peak power figures.


The FZR is mid/rear engined, and the most powerful, and as such can really hit you if you make an error.
Again, no difference in absolute power. However, due to the lack of turbo lag, it does offer the most immediate power under foot, hence the increased fuel consumption.

Additionaly, the FZRs weight distribution allows it to full take advantage of that rear wing, whereas it would just be wasted on the other two GTRs.


Remember what they told you at school, quality over quantity. ;)

[Hyperactive]

The fzr can really beat the other two when accelerating out of corners, I think XRR and FXR lose here the most. FXR is slow on straights and it burns the fronts pretty quickly making it even slower in corners.

I would never suggest starting with the XRR. It's the hardest and least forgiving of the GTRs and also least rewarding to start with. Start with FXR and learn it how it handles and you can give some good opposition to the other players with not-so-much racing experience.

Then move to the FZR. It bites a lot in the beginning but just keep grinning and shouting evil things, in the end you get it drivable. After you get fast and competitive with this car you should move to the XRR. The turbo lag feels horrible and the whole car feels like brick on wheels at first but keep driving it and you'll love it! I was few months ago complaining about the XRR and its turbo lag etc. but now after few hundred laps I'm totally in love with. I have a poster of XRR and I sleep with it, I see dreams only about XRR when sleeping and I have painted my real life car to look exactly like an XRR and ... well you got the point ;)

Patience is the right word with the GTRs.

EDIT: And the power/torque curves are quite different in every GTR. The FZR has defenately the widest power/torque curve and thus it can light up its rear wheels at will. XRR is the worst. Just to get it off the line you need the floor the throttle and sit in the smoke until it starts moving. While doing this you can admire the beatiful skins of the passing FZRs and FXRs. :)

[severin_schoepke]

I'd like to learn the XRR, but after a few laps I always miss my FZR. It's just a superb car with my favourite layout: RR.

If you remember that, it's actually an easy car to drive:

1. slow in, fast out: The rear-heaviness offers superb traction when accelerating out of a corner, use it!


2. Be aware of lift-off oversteer: Because the weight is in the back, the rear will come round when suddenly lifting off the throttle. This makes some corners hard to drive, I especially hate the final chicane in Aston.

3. The same with trailbraking


If you remember these points you'll love the FZR, after a few laps I always want to go out and buy a 911, unfortunately, I don't have the money ;-)

[Gunn]

I have become very good friends with the FZR and now find it a pleasure to drive. It's a real beast that requires a gentle touch. I love the way it looks on the track too, in the replays the FZR always seems to be absolutely hammering along.
It gets my vote for handling and speed over its two rivals.

[Jakg]

please be joking, anyway, i drive AS Nat quite a lot in the XFR, but ill soon try the big GTR's on it!

[Hyperactive]

Btw, this is not the right forum ;)

When I first tried XRR I though that the car was pure sh**. If I now had a time machine I would "travel" back to that moment and slap myself :D

[Primoz]

FZR looking good on the track? Sorry, no, it looks like a woman with a fat ass. My 2 cents.

[bLaCk VaMpIrE]

on my opinion the FZR is much easier to control than the XRR, because of his heavy arse pressing the tyres much more on the ground than the XRR does.
additionaly i think, that the XRR has quite a little bit too much disadvantages for its difficulty.

[Bob Smith]

EDIT: And the power/torque curves are quite different in every GTR. The FZR has defenately the widest power/torque curve and thus it can light up its rear wheels at will. XRR is the worst. Just to get it off the line you need the floor the throttle and sit in the smoke until it starts moving. While doing this you can admire the beatiful skins of the passing FZRs and FXRs. :)
Oh yeah the FZR probably does have a flatter torque curve, although I've never gone through the hassle of actually measuring it. The XRR and FXR have, as far as I can tell though, identical engines. The only reason the XRR is poor off the line is traction of the powered wheels - front heavy and RWD versus either front heavy and AWD (slightly front heavy is best for AWD) or rear heavy and RWD (with most of the weight where it's useful).

Also you have to nail the throttle in neutral in the FXR/XRR to build up some boost so it actually spins the wheels at all (obviously a worse problem in FXR). Damn turbo lag!

[bbman]

on my opinion the FZR is much easier to control than the XRR, because of his heavy arse pressing the tyres much more on the ground than the XRR does.
Exactly... I first learned to drive XRR (more or less... :D) and when I switch to FZR from time to time now, it is much easier to drive... :nod:

Personally, I like the XRR more though (maybe because of it's underdog factor :D)...

2. Be aware of lift-off oversteer: Because the weight is in the back, the rear will come round when suddenly lifting off the throttle. This makes some corners hard to drive, I especially hate the final chicane in Aston.
Yeah, I hate that too... :D

3. The same with trailbraking
I can't agree with that one... When I drove the FZR, I always got the impression that it really needs the trailbraking... :shrug:

[Hankstar]

They all have their good points, but for some reason I prefer the XRR - maybe something to do with my favourite touring car series (Oz V8s) being front/rear layout - it's also quicker than the FXR and kinder to tyres/fuel than the FZR.

Having said that I drive the XFR more than the big ones as I think it suffers less from the low-speed grip issue (which keeps me away from the FZR to an extent). I'm not all that competetive in the big GTRs as I can't be bothered compensating for the low-speed no-traction thing (I get enough of that in GPL and one game with tyre bugs is enough)...I'm waiting for the patch that fixes it :)

[Gunn]

FZR looking good on the track? Sorry, no, it looks like a woman with a fat ass. My 2 cents.I know a few fat-arsed women who also look great on the track.

[Hankstar]

I like big butts and I cannot lie :nod:

[BlueSkunk]

never mind women's behinds, but all that has been said here means that the only competitive cars in the gtr class are the fz and xr, no? unless your name is flotch...

[richy]

maybe one day there will be weather effects that might give it a bit more of an edge?

[Hankstar]

BlueSkunk, in an enduro, the FXR often has the edge as it's kinder to tyres and fuel and doesn't need to stop as often as the 2WD GTRs. In sprints though, the FXR usually needs to be driven by aliens to beat quick guys in 2WDs :)

[Jonesy_]

^^ the FXR actually eats tires the most from my experinces, at least when speaking the R2 compound.
And think again about the fuel consumption, if the FXR and XRR has the same engine, but the FXr weights more, which one will eat more fuel?
at least thats how it is supposed to go, and I think I´m not totally wrong if I say that the 4wd somehow makes it to use more fuel to get the same distance than XRR, some otherway than just the extra weight.

[Hankstar]

Well, that's from my limited experience anyway. As I said I tend to drive the XFR more than the big ones due to the unrealistic low-speed no-grip issue :)

[Hyperactive]

About the front tyres heating and wearing. I was few minutes ago driving the FXR (10 lappers) and I had R3 in front and R2 in rear. The R3 in front were almost worn out but the rears had lots of rubber left after the race. No locking/braking, some slides but nothing major.

And I hate pushy cars soo much :)

[Funnybear]

maybe one day there will be weather effects that might give it a bit more of an edge?

Oh god. The FZR in the wet would be a nightmare. Imagine if they put a drying track in as well, get that big 'ol arse hanging out anywhere off the racing line thats it, your gone. The FXR would nefinatly be the car to drive in the wet. But I am a FZR racer and I would do my best to get a FZR competative in the wet. It's a car that requires patience and understanding. You need to whisper sweet nothings to it and feed it sugarcubes. But if it's in the right frame of mind it will give you the best ride of your life. It's more than likely to give you a kick up the behind on the next lap though . . .

[bbman]

You're partly right... The FXR eats the tires if you want to be anywhere near the laptimes of the two 2WD-cars, but for endurance, you don't have to be fast primarily, thus, you could go way longer than the 2WDs...

And it actually uses less fuel than the 2WDs also, don't ask me why... :shrug:

@hyper: How about setting power more to the rear? 36 % to front is very good to drive, 25 % is perfect for tire wear... :)

[AndroidXP]

And it actually uses less fuel than the 2WDs also, don't ask me why... :shrug:
FuelControl online speaks otherwise :p
On short laps it has the same and on longer laps up to 0.4l more per lap than the XRR. I guess it loses most on long straights where it's really struggling to keep up.

Also like bbman said, you need to put the power way to the rear to get even wear and a good driving behaviour. 2/3 to 3/4 to the back is optimal, everything else makes it very pushy and tyre eating.

[Hyperactive]

@hyper: How about setting power more to the rear? 36 % to front is very good to drive, 25 % is perfect for tire wear... :)

Might work though the biggest problem was that the tires heat up after few laps to over 100 degrees and stay there quite long. But I guess I won't be driving the FXR too soon again ;)

[Funnybear]

Try keeping the FZR's rears on the straight and narrow. It really needs pampering if you wanna make them last for any amount of time.

[Bob Smith]

@hyper: How about setting power more to the rear? 36 % to front is very good to drive, 25 % is perfect for tire wear... :)
I like it at 22% to the front myself. A touch of power oversteer in there somewhere, feels so nice, easily controlled by playing with the front and rear power locking. Plus you'll always keep both tyres until everything on the F9 display is back to blue again (well, with sensible camber anyway).

[DasBoeseC]

Mahlzeit.....

About the FXR being competitiv in endurance racing.....take a look at the results of the Masters of Endurance.


The FXR is NOT gentle with its tyres.....we almost always had to pit due to tyre wear and only once due to fuel consumption. And whoever said it is near impossible to spin the FXR has never driven it fast enough over long enough stints.....trust me ;)

CU, Sebastian

[Moonclaw]

The XRR is next 'best', but it'll struggle around some corners for having a front engined layout and less power than the FZR.
XRR
365kW (490bhp) @ 6278 rpm
627Nm (462 lbft) @ 4782 rpm
332 W/kg (453 bhp/ton)

FZR
365kW (490bhp) @ 8106 rpm
503 Nm (371 lbft) @ 5267 rpm
332 W/kg (452 bhp/ton)

So where's this less power? :razz:

[Hyperactive]

@moonclaw: You are just looking at the peak values. The XRR may have higher peak torque but the torque curve itself is quite sharp, unlike the FZR which has more torque around the peak value.

Also the turbo lag means that there is a lag when yopu push the throttle in XRR/FXR.

[Moonclaw]

But to shag the , XRR still has more power.

[tristancliffe]

More power at some specific point. Remember the peak power figure of a car is the most useless figure bandied about by anyone. Peak torque is much better, but still fairly useless. What you want is a torque curve - the only bit that matters.

Do a test - run the FZR and the XRR on the same tyres, the same gearing etc, and watch which one is fastest in terms of speed and acceleration. Sure the FZR might be a bit more aerodynamic (who can really tell if that's modelled accurately or not), but the FZR is a LOT faster than aero differences would suggest, meaning the XRR has less tractive effort, which means less torque, which means less useable 'power'.

[bbman]

Plus, the FZR is way better in getting the power on the tarmac, as the XRR has a very light rear (compared to its direct competitor)...

[Ball Bearing Turbo]

Peak torque is much better

No.


What you want is a torque curve - the only bit that matters.


The only reason you want that is so you can derive a power curve from it :razz:


Measuring the engines ability to do work (POWER output) is FAR more useful than knowing an instantaneous measure of force it's capable of generating. One is a measure of force over time (work) and the other is instantaneous force. More force over time is what moves cars over larger distances in less time. All a peak torque figure is good for is giving you a rough idea of what the power curve is going to look like!

UhOh everyone duck

[tristancliffe]

How do you use power to calculate the performance of the car? That's right, you can't.

You use the torque curve and the gear ratios to generate the tractive effort curves, and from that you can calculate acceleration, top speed (if you know rolling and aerodynamic drag), everything you need.

Power curve is only useful for deriving a torque curve. Unfortunately people can only cope with 'power' without realising that in terms of performance it's a rubbish thing to quote.

If you can tell me when a power curve is useful, other than using it to calculate torque, then I'm all ears... I know a good estimate of an engine is if the lbft and hp figures of an engine are similar then it'll probably be a well balanced engine, and a high hp low lbft (peak figures) means a peaky engine, but thats about it.

Last time we had a disagreement we ended up with a 3 page conversation. Gulp.

[Bob Smith]

You use the torque curve and the gear ratios to generate the tractive effort curves, and from that you can calculate acceleration, top speed (if you know rolling and aerodynamic drag), everything you need.
Having programmed just that thing in GRC, I can assure you all that Tristan is correct. The power curve is just some pretty you can draw once you're done with the performance estimations. Power is only really useful for a quick top speed analysis.

[Ball Bearing Turbo]

@ Tristan (edit2: & Bob too)

As you're well aware, torque at the drive wheels is what matters (which you alluded to above). Due to gearing, the vast majority of the time wheel torque is higher at the engines peak power output than it is at the engines peak torque output.

As far is power calculations being useful:

P = net horsepower
F = the sum of the longitudinal forces at the driven tires.
V = velocity

P = F * V
or
F = P / V

Therefore we achieve maximum torque at the wheels if you maximize your POWER at any given velocity.

Yes this is simplified since I am at work and could not resist replying, but the math still holds true.

I enjoy this topic despite the inherant futility of debating whether two pieces of data derived from each other are relevant or not... It's still just ... fun.

EDIT: Bah, beaten by BOB.

[tristancliffe]

Actually, if you look at a tractive effort curve (the force or torque at the wheels) it will peak always at peak engine torque, but at varying speeds due to the torque multiplication of the gears.

Power doesn't really come into. Your equations are wrong - peak torque occurs at peak torque, peak power occurs at peak power. The peaks are not at the same time.

[Hyperactive]

It would be nice if someone actually would measure the the torque curves of some of the cars. Would it be this simple:

1) create a simple autocross layout on BL car park with possibility to do more than just one "laps". All it needs is a long straight.
2) take a car, put the tire pressures to full and take all the aero away from the setups. *
3) Make simple acceleration&engine braking tests so you could get the acceleration&speed values (curves) between the clutch bite rpms and the red line.
4) Save these "laps" from the replay in raf and open it in F1perfview. Also check the tires size and transmission ratios.
5) calculate&draw
6) publish the result
7) put flamesuit on
8) do it again just to prove it works!
9) well...

So it would a very simple job actually. Comparing the acceleration/time curve to the rpm/time curve and taking the right values... And vice versa for the engine braking to get the power train "friction" values, rolling resistances and air resistances.

Of course I can't do it myself because I need to learn some 3d-modelling ;)

*I think LFs shows the forces (drag) in the setup menu, the speed slider thingy? Just write down the values in 5m/s steps and it should be fine

[tristancliffe]

You might be able to do this with the g-meter in LFS, or in F1 perf view. If you know the mass of the car and the acceleration of it you can derive the torque at the wheels (ignoring drag of any sort, so keep the speeds low). And if you know the gear ratios, the dynamic wheel circumference (pi*r^2 will do though) and the torque you can easily work out engine torque at certain rpms.

I'll have a lookie at the weekend (work commitments til then :() but maybe someone will beat me to it...

[Ball Bearing Turbo]

I knew I should've waited till I got home to reply :doh:

:D

Your equations are wrong

Why? P=Fv is a common formula! :schwitz:

Let me ask you this - when is the most energy being produced by an engine - at peak torque or peak power output?

The more energy produced, the more energy expended on moving the car therefore the faster it will accelerate.

The very definition of power speaks for itself: the ability to do work. Doing work on the car is what we're concerned with. I don't care how much force you apply to the crankshaft, rather I am concerned with how quickly you can apply the given force, with the end result being work which takes power to accomplish, of which force (torque) is only a component.

If this is wrong (which would seem to contradict a lot of physics, but I'm open to learning) then someone needs to tell me exactly why time is not important in this debate, since acceleration occurs over time - and force is an instantaneous measurement, whereas power takes this into consideration providing what would seem to me to be a much more relevant indication of the potential of an engine.

[RMachucaA]

Also, dont forget that the FZR responds nicely to the EXPLOIT, so once the patch is out it will level things that little bit more.

[tristancliffe]

Okay, do we agree that the acceleration will be greatest when the force at the wheels is highest. It's a simple F = M x a problem. For a fixed mass, the greatest force will result in the greatest acceleration.

So below is a graph of tractive effort at the wheels (force at the wheels, in Newtons) against engine speed for 5 gears. As you can clearly see the tractive effort curves are squashed torque curves, and the peak occurs at peak torque in any gear, where the most acceleration is to be had. Also the top speed of the car is shown by where the drag line (red) intersects the tractive effort curves.
It also shows when to change gear - when there is more force in the next gear than the current gear, and in this graph it is where each line crosses the next.

The second graph is the same as the first, but I have also put on the power derived forces using your equation. Power in watts = Newton meters per second, and I converted the speed for the calculation to meters per second, so the end result is in Newtons. As you can see, it shows peak accelerations at the point of peak torque NOT peak power. Quite why the two lines don't coincide perfectly I'm not sure, and I can't be bothered to inpect it, but it's probably just a poor conversion between units or something.

Hopefully this will answer your questions. Peak acceleration ALWAYS occurs at the point of peak torque. The car in question in this graph has a peak torque of 100lbft at 5700rpm, and a peak power of 117hp at 6700rpm.

Feel free to ask more questions, but you won't win this one :p :D

[Hyperactive]

http://auto.howstuffworks.com/framed.htm?parent=horsepower.htm&url=http://www.revsearch.com/

[Bob Smith]

You can get the torque curves of the engine much easier than that, I did it in S1 days. Just drive the car from rest at full throttle (no wheelspin) until you hit the rev limiter, output the RAF data from the replay, use F1perfview to output to the longitundal force on the drive wheels to CSV, sum this in Excel, divide by total gearing reduction (look at 1st/final drive ratios in setup used) and tyre diameter, then adjust drivetrain losses until peak torque matches up. Voila and you can plot the torque curve. A little more work and you've the power curve too.

OK so it's not a one step process. :D

-------------------------------

To the debate:
"Therefore we achieve maximum torque at the wheels if you maximize your POWER at any given velocity."
Yes, but that is coincidence - I mean, you don't need to know anything about the power curve. Just look at the wheel torque in every gear and you'll end up straddling the rpm peak power anyway.

Tristan: peak acceleration IN ANY ONE GEAR certainly happens at peak torque. By making use of multiple gears, acceleration will be a constantly dwindling number once you hit peak torque in first. The point is you are only at the rpm of peak torque, unless you gearing is over spaced, changing down will decrease engine torque output but increase torque at the wheels.

It is best to only think about torque and not power though - by optimising torque you'll be optimising power anyway.

[Hyperactive]

I think it's the best to use a slightly taller 2nd gear. Accelerate from the low clutch bite rpms to the redline and lift off the throttle and let the car slow down by engine braking and air resistance + all the other "resistances".

Then just separate the raf output into two parts. The acceleration and the deacceleration part. The latter gives the info you need to reduce the resistances from the acceleration curve and therefore you don't need to calculate them by yourself.

The result is a torque curve from the axle of engine (or how do the london people call it :D )

[Hatemaker]

I prefer the XRR, mainly because I haven't taken the time to learn the FZR better. Even then, I still like the XRR the best, and the XFR second best.
And, would it be a bad thing if I've beaten FXRs with the XFR? (Allthough I'm definately sure he was a shoddy driver.

[Ball Bearing Turbo]

To the debate:
Ball Bearing Turbo said:
"Therefore we achieve maximum torque at the wheels if you maximize your POWER at any given velocity."

Yes, but that is coincidence - I mean, you don't need to know anything about the power curve. Just look at the wheel torque in every gear and you'll end up straddling the rpm peak power anyway.

Thank you! :D

Since most lamen off the street can't relate to "wheel torque in each gear", and it's not published in magazines (hehe maybe it should be, can you imagine seeing tractive effort curves in auto rags ROFL, people would be like "WTF?") Ahem... Right then, since they're NOT, peak hp/tq values are both relevant IMO.

[tristancliffe]

I still disagreee. I've done a number of programs and spreadsheets for predicting perfomance of various race cars we've owned over the years, and not once has the power curve or any power figure been useful. All you ever need for any performance calculation is the torque figures.

And Bob - I don't get what you mean by straddling the rpm peak power anyway... it's perfectly possible that the optimum gear shift point of a car is BEFORE peak power. It just depends on your gearing. I don't see what that sentence means, although BBT ^ seems to think it proves him right. Maybe it does, but can you clarify your sentence.

[Ball Bearing Turbo]

it's perfectly possible that the optimum gear shift point of a car is BEFORE peak power. It just depends on your gearing.

I agree - and that point can also come after your peak power output just as it can come before (granted not as likely as before depending on what gear your're in, but still...) It's about maximizing area under the curve.

But: Which curve? :D

It's a simple F = M x a problem. For a fixed mass, the greatest force will result in the greatest acceleration.

Only for that moment in time, which is somewhat not relevant, or at least doesn't relate to what my argument is.

You still need to explain to me why you maintain that absolute force has more to do with acceleration over time than the rate that the force is applied does. Again, I feel that it's about work - it takes more work to move an object from here to there in less time, and that takes more power, not necessarily more force. Take any given force. Apply that same force twice to the same body under the same conditions, but at different rates. The quicker you apply the force, the more work has been done, and more POWER has been utilized. That example is a bit fantastic and impossible but it illustrates my point about the factor of time, and the potential to do work (which is power not force).

I could be dense, but it still seems like you're comparing apples to oranges. I am speaking about getting from A to B as fast as possible - not about any specific moment within the time it takes to get from A to B.

:tilt:

Edit: here's a question, and I've seen it before on here: Will a CVT equipped car accelerate most rapidly from A to B by holding RPM at peak power or peak torque? :D

Think about top fuel drag cars..... The whole objective of such beasts is to perform the most work on that car as possible to get it from here to there like a whirlwind on crack. They want the most force in the least TIME as possible, which takes power.... I think around 6000 horses of it or so.....

edited for clarity again.

[Funnybear]

I am taking bets ladies and gentlemen. I have my favorite . . .

[tristancliffe]

I agree - and that point can also come after your peak power output just as it can come before (granted not as likely as before depending on what gear your're in, but still...) It's about maximizing area under the curve.

But: Which curve? :D


Indeed, maximising the area under the tractive effort chart (torque) will yeild stronger performance, either through tuning, more revs (longer in lower gears) or less transmission losses. And maximising the area under the tractive effort chart will result in more area under the power curve, as the power curve is derived from the torque curve.
Only for that moment in time, which is somewhat not relevant, or at least doesn't relate to what my argument is.
Eh? Okay, let mass change as fuel changes or whatever. You still know the force and the mass at any one time and can therefore workout the speed, acceleration and velocity if you know the starting conditions. Thus starting from rest and assuming no wheel spin or clutch slip it's possible to work out how long it'll take to do a 1/4 mile or reach 200mph. Track work is different as braking and turning come into it, but for pure acceleration tractive effort and mass is all you need.
You still need to explain to me why you maintain that absolute force has more to do with acceleration over time than the rate that the force is applied does. Again, I feel that it's about work - it takes work to move an object from here to there, and that takes more power, not more force. Take any given force. Apply that same force twice to the same body under the same conditions, but at different rates. The quicker you apply the force, the more work has been done, ore POWER has been utilized. I could be dense, but it still seems like you're comparing apples to oranges. I am speaking about getting from A to B as fast as possible - not about any specific moment within the time it takes to get from A to B. Power ISN'T the rate at which force is applied. 200N applied is 200N whether you do it for 1ms or 1 year. And the rate of acceleration will be the same if the mass doesn't vary. Power is the measure of work done on an object.
At it's simplest, lots of work doesn't necessarily mean lots of acceleration, just as lots of acceleration doesn't necessarily mean lots of power. The two are connected, but not interchangable. Torque is what drives a vehicle. Torque is what you (should) tune in an engine. You might tune for high end torque or low end torque, or mid range torque, but it's that that you want to alter. I'm not quite sure what you're saying still. It's probably me being Mr Thicko, but I'm sure we'll sort it out sooner or later...
Edit: here's a question, and I've seen it before on here: Will a CVT equipped car accelerate most rapidly from A to B by holding RPM at peak power or peak torque? :D

Think about top fuel drag cars..... The whole objective of such beasts is to perform the most work on that car as possible to get it from here to there like a whirlwind on crack. They want the most force in the least TIME as possible, which takes power.... I think around 6000 horses of it or so.....
A CVT car will accelerate quickest with the engine at peak torque. edit: hmmm, I have a feeling the extra variable (the gear ratio) means this is wrong, but for fixed geared cars it's 100% correct. *goes to read about cvt's and do some maths*

The drag car does indeed what the most force in the least time. Clearly the faster it accelerates the quicker it's 1/4 mile time will be. And to get the quickest acceleration at any given point or differentiated over the whole time period, it needs the most force at the wheels it can acheive. This is, for any given gear ratio, varying or not, at the point of peak torque, when the brake efficiency is at it highest.

Edit2: phew, my edit was before Colcobs edit, so I correctly deduced the CVT trick question just in time :D

[colcob]

OMG, KILL ME NOW.

Not this again, please for the love of all that is holy.

*col goes searching for the TORQUE VS HORSEPOWER THREAD OF DOOM.....*

(..and you're still wrong Tristan, a CVT car will accelerate quickest at peak power because by definition, the CVT will adjust the transmission to give greatest torque AT THE WHEELS. You're confusing engine torque with wheel torque again)

[Ball Bearing Turbo]

:nana::monkey:

[Funnybear]

Hey keep it going . . . I never caught the last Power v torque thread of doom so this is really interesting . . . . . You guys know your stuff.

[skiingman]

The drag car does indeed what the most force in the least time. Clearly the faster it accelerates the quicker it's 1/4 mile time will be. And to get the quickest acceleration at any given point or differentiated over the whole time period, it needs the most force at the wheels it can acheive. This is, for any given gear ratio, varying or not, at the point of peak torque, when the brake efficiency is at it highest.

Cars that want to get from point a to point b in the minimum amount of time always do so with the engine running at its maximum power, not its maximum torque. These cars have appropriate gearing.

This is commonly accomplished in drag cars by the use of various forms of loose gearing. Drag racing torque converters for automatic transmissions seek to bring the engine to peak power as quickly as possible. In true drag only applications, the point is to have flash stall at or very near peak power, not peak torque. The drag racers then further make use of the change in tire diameter that comes along with thin sidewall slicks, minimizing the number of ratio changes necessary and thereby maximizing time spent at peak power.

Really fast drag applications simply run the engine up to peak power and then slip the clutches just enough to keep it there all the way down the track. The power lost via this inefficient method of delivering energy to the road is far less than that lost by having fixed gearing that requires time consuming shifting and running the engine at points other than peak power.

BBT is correct that power is what determines the time from a to b, and given the appropriate gearing torque is completely meaningless. This is why all rudimentary estimates of time to distance and top speeds utilize power figures, not torque figures or curves.

[Bob Smith]

To the CVT - think about it, for any given speed, the extra wheel torque given from the increased gearing reduction at peak power rpm outweighs the lower torque output at the engine.

The whole cofusion is caused because you keep thinking about ONE gear. You have to remember, you'll never have the as revs low as peak torque, because you'll have changed down, reduced engine output but increased wheel torque. It's something that took me a while to click from the last thread but it finally makes this weird kind of clarity in your head.

See attached GRC sceeny that the maximum wheel torque can be kept by keeping the revs above peak torque (ignoring first gear of course). OK it's not always the case but this proves that peak wheel torque coincides with peak power.

[Ball Bearing Turbo]

I hear Tristan getting his towel out....

And warming up his throwing arm....

:D:D:D

[tristancliffe]

I'm going to ignore CVT's for the time being until I can come up with or be shown a graph that actually proves anything... I'm reading a few things, and I'll see what I can come up with.

But for manual transmissions, I still can't see the peak wheel torque coninciding with maximum power. That screenshot of the GRC shows the the wheel torque is falling away after peak engine torque. A gear is a torque multiplier, and as such the peak wheel torque (or force at the wheels) will have to be at the peak engine revs in any given gear, although the speed at which this peak arrives will obviously change.

I have taken the GRC sceenshot and added the points where peak wheel torque in any of the gears occurs. In this example it is just after the shift point because of the gearing. To the left of the red blob you would be better to change down a gear due to torque multiplication, and anything to the right of the blobs is less that optimum for that gear, but better than the next gear until it crosses over. Thus peak torque in this case isn't at the engines peak torque because the gearing suggests you would change differently.

Perhaps we are trying to explain different things?

Edit: forgot screenshot :S

Edit2: Right, I've done some calculations and reading, and I am 100% happy that CVT's will run at peak power for maximum tractive effort at the wheels. It took a bit of thinking, but I understand it now... As for Manual cars, you still get peak torque at the wheels when the engine is at maximum torque. The gearing you have defines when the crossover point is, which defines the gear change point. The more gears you have the closer you can get to the maximum area under the graph being a smooth hyperbola (i think it is), but as most cars only have 4, 5 or 6 gears this bit of the argument isn't brilliant. You either have (mostly) 6 or less, or continuously variable. For a manual transmission with definite, well spaced gearing for all conditions (acceleration, top speed, economy etc) as on road cars then the maximum wheel torque, as I have said, WILL occur at maximum engine torque NOT maximum power.

[Ball Bearing Turbo]

I am back :D

Mr. Tristan:
Well, you conceded that with a CVT - peak power determines maximum acceleration, but in doing so you also condeded the whole debate, because the fact that a CVT is infinitely variable is not relevant. If you consider this carefully you will see that if it's true for a CVT, it's true for any GIVEN gear ratio, if for no other reason than simply on the basis that CVTs "use" every gear ratio possible - including the ratios you would use in a regular manual. Put differently, if it's true for a CVT, it's true for ANY transmission known to man because a CVT is effectively incorporating every ratio you could ever have in any feasible transmission.

Which brings me to my second point, the death blow.

Recall how I've been trying to convey the importance of time in this whole debate? It FINALLY occured to me to define power in terms of it's components (yes, it took a long time for that revelation....) being force and time, and therefore I submit to you that 1HP = 550 lb/ft per second.

This is exactly why more power yeilds greater acceleration, not necessarily more force alone, but more force over time is the key factor. Thus, even an engine that has long past it's torque peak, but is at it's POWER peak has more actual output. Surely I don't even have to post the numbers for everyone to realize that HP*550 gives you the "amount of force per second" (vulgarized term for convenience) being generated.

However I will give an example:
in the attached graph, Note the following:

Peaks:
Torque: ~300lb/ft at ~4000RPM.
HP: ~290HP @ ~5500RPM

Note the force output at both peaks:
5500RPM = 290HP = 159500 Pound Feet of Torque per Second
4000RPM = ~200HP = 110000 Pound Feet of Torque per Second

Thus, 45% more work is being done at peak power than at peak torque.

Your need for force is now satified in the proper venue: time.

[Ball Bearing Turbo]

OMG, KILL ME NOW.

Not this again, please for the love of all that is holy.


Forgot to mention: ROFL LMAO

:thumb:

[AndroidXP]

:munching_

[colcob]

What is happening here is a clash of definitions.

There are two ways of defining acceleration in this debate.
The strictly scientifically correct definition, which is the rate of change of velocity at an instantaneous moment in time; and the more practical but semantically innacurate definition of 'how quickly can we go from point A to point B'.

In the first case, the highest instantaneous acceleration will always occur when the longitudinal force is highest, which is at the torque peak in any given gear. Of that there is no doubt.

But when people talk about the 'acceleration' of cars, they tend to be actually thinking of how long does it take to get from point A to point B, or even how long does it take to get from velocity A to velocity B.

In these cases we get back to thinking about the area underneath graphs, rather than the maxima of them.
If you take a graph of acceleration over time, the total area under that graph is equal to the average velocity over that time. So anything you do to increase the overall area under the tractive effort curve gives a faster average velocity over the course of the acceleration. Which means that your final velocity will be higher and your overall distance travelled will be greater.
Ergo, in laymens terms, the 'acceleration' is better.

So basically, the performance potential of an engine is dictated by the area under the torque curve. Now before someone jumps in and says 'but what about peak power', I should say that peak power is dictated indirectly by the area under the torque curve, the longer the torque curve stays above a certain point, the higher the peak power will be.

An excellent illustration of all these points is the restricted V10's that Torro Rosso are using in F1 this year. They have been rev and inlet restricted to give supposedly the same peak power as a V8, but they have a greater area under the torque curve due to those 10 cylinders pulling before the restrictor kicks in, which means they have a better performance potential than they should.

So in summary, acceleration performance is not dictated by peak power only, or peak torque only, but by the total area underneath the torque curve (within the rev range used by the current gearing setup).

[Funnybear]

I love mixing in enlightened company. Makes me feel clever.

I must say that I am reading this with interest and with only a slight fuzzing around the edges.

Hopefully, when you have come to some kind of consensus I might know whether it's better to put the gear change of my FZR on the power peak or the Torque peak. TBH I'm going for the Torque atm. From what I know of engines and cars, torque does seem to be the more important of the two. But I do realise that you can't have one without the other . . .

[colcob]

No. You change gear when the red light comes on. Simple as that.

LFS calculates when the next gear will give you more acceleration than the current one, and thats when the light comes on.

[Funnybear]

Yea, ok. Smartarse. Maybe what I really meant was to . . Maybe what I meant was that if you . . . . Maybe . . . Damnit. I'll just shut up and read.

[AndroidXP]

Yeah, the ideal shift point is very shortly after the red light comes on. ("Shortly after" because you lose a bit of speed in the shift process)

[tristancliffe]

Let's return to the original question - why is the FZR faster. Someone said the XRR has more power, and I said that didn't matter one iota. Neither did the max torque figure. I stated that the torque curve was the important thing. This is because of the area beneath it.

More power at some specific point. Remember the peak power figure of a car is the most useless figure bandied about by anyone. Peak torque is much better, but still fairly useless. What you want is a torque curve - the only bit that matters.

For a given fixed gear ratio the acceleration in that gear WILL be greatest at peak torque output. For a CVT you can't state the same thing as you don't have a fixed gear to judge when 'best' acceleration takes place. The CVT alters it's ratio to ensure that the area under the tractive effort curve is at it's greatest.
With a manual it is neccessary to choose the gear ratios for a number of factors - low speed control (i.e. acceleration with wheelspin or boggin down), high speed atainability (reaching max speed on the straights), and drivability in between. For best performance you will, of course (and I've never stated differently) have to change gear around the peak power figure so that you 'use' as much of the area under the tractive effort curve as you can.
That's what the red light in LFS tells you. If you change when it comes on you use the area under the graph more effectively. Shift too early and you waste a gear, whift too late and you'll waste a gear. In wasting the gear you lose some potential (except when purposely stretching a gear before a corner, but thats different).

[Bob Smith]

Hence power is only good for judging top speed, where you are at a fixed rpm (eventually) and only in one gear. I don't think any shortcut for estimating acceleration will ever be particularly accurate, there are just too many factors to consider - you need to sit down and simulate it. That's why I don't really like power-to-weight ratios, they don't tell you enough about acceleration to be that meaningful, and because the traction of the drive wheels needs to be known also, you don't know how difficult they'll be to get off the line either.

[skiingman]

Remember the peak power figure of a car is the most useless figure bandied about by anyone. Peak torque is much better, but still fairly useless.

This is just woefully untrue.

Some rather obvious (tired, even) examples come to mind:

-I have an electric motor here capable of peak torque greater than an F1 engine. It produces less than 5HP.
-I have seen hydraulic wheel motors capable of similar feats.
-In terms of more plausable stuff in the automotive realm, the peak torque of the 3.0 liter Vulcan in my van is 165 ft-lbs. Same as the motor in my Mazda. The motor in my Mazda produces an extra 70 horsepower. It gets from A to B much quicker. Yes, in the same vehicle with the same gearing, the Vulcan is capable of the same peak instantaneous acceleration figure. Speaking of dumb figures, there is one right there. The only thing that matters is the average acceleration over time. High peak accelerations just break things and break loose tires.
-Importantly, the reverse of the non-auto examples above do not demonstrate a situation in which a power figure is "worthless". The torque output of the power turbine of a PT6A is rather tiny, but gear the thing from 20,000+ rpms to 2100 rpms at a propeller, and you've got more than a 1000ft-lbs and hundreds of horsepower. No matter what speed the engine moves at, you can find a way to make use of the power it creates.

In summary, a 300ft-lb torque electric motor might chirp the tires, but might not get you anywhere quickly, while a 40ft-lb turbine engine might get you there faster than the 200ft-lb car engine.

There are some really good reasons why you do, as you say, need a torque curve to look at. However, for a cursory inspection or simple matchup, the peak power figure will be vastly more meaningful than the peak torque figure. This is why nearly everything on earth used to provide motivation to twist anything has a power output specified and printed on it, not a torque output.

Maybe the reason why this isn't so obvious to you is that you haven't driven a 1970's Cadillac equipped with a emissions regged big block. Uh, 400ft-lbs of torque, check. Dead slow, check. Less power than a new Camry, check.


I don't think any shortcut for estimating acceleration will ever be particularly accurate

I think you are quite correct, Bob, although the inaccurate shortcuts are quite useful when trying to determine what size motor you need to run the lawn mower, or roughly how many horsepower you'll need to drop in the 2800lb auto to run the quarter mile in 13's. For relatively slow cars, the simple three/four variable quarter mile calculators can be accurate within +/- 10%, which is damned impressive considering the small number of variables used. Stuff gets twisted when you get to cars with enough power (meh, torque over a bunch of time) to spin the rubber bits in bigger gears.

[Ball Bearing Turbo]

What is happening here is a clash of definitions.

Ok...

There are two ways of defining acceleration in this debate.
The strictly scientifically correct definition, which is the rate of change of velocity at an instantaneous moment in time; and the more practical but semantically innacurate definition of 'how quickly can we go from point A to point B'.

How can there be measure of rate of change at an instant? Change is measured over time, if no time elapses it's impossible for change to take place!

In the first case, the highest instantaneous acceleration will always occur when the longitudinal force is highest, which is at the torque peak in any given gear. Of that there is no doubt.


I beleive I proved above that the the delivery over time of force is actually highest at peak power output.

But when people talk about the 'acceleration' of cars, they tend to be actually thinking of how long does it take to get from point A to point B, or even how long does it take to get from velocity A to velocity B.

Again with my density, but I fail to see how one can be true but not the other. If "peak acceleration" occurs at "peak torque", then how can holding a CVT at the "peak torque" not yeild "peak acceleration" from point A to B? That's paradoxical to me! If the "rate of change for that instant" is the greatest it can be, how could it (the torque peak!) NOT produce the greatest rate of change over time? But it doesn't!


We've already established that CVTs will have maximum acceleration at peak engine POWER output (Power; BTW I've clearly also defined above, in terms of "force output" for lack of better ability to articulate.) That being true, and it is, how could the same possibly NOT be true for any given gear ratio - since by saying it's true for a CVT, we have said it's true for an infinite number of ratios! That includes the 3, 4, 5, 6 ratios in any automatic or manal transmission on the earth!


In these cases we get back to thinking about the area underneath graphs, rather than the maxima of them.
If you take a graph of acceleration over time, the total area under that graph is equal to the average velocity over that time. So anything you do to increase the overall area under the tractive effort curve gives a faster average velocity over the course of the acceleration. Which means that your final velocity will be higher and your overall distance travelled will be greater.

Yes. We've all been saying this.

Ergo, in laymens terms, the 'acceleration' is better.

The problem is if a particular condition (which some are saying is the peak torque output of the engine) is capable of producing the greatest acceleration even for a "moment", then holding that condition constant (as in the case of a CVT) would yeild the greatest acceleration over time as well (because as stated, "for any given gear ratio peak acceleration happens at peak torque" right?) , but it does not. Why? Because the overall output is still less than it is at peak power, and less tractive effort is available.

So basically, the performance potential of an engine is dictated by the area under the torque curve. Now before someone jumps in and says 'but what about peak power', I should say that peak power is dictated indirectly by the area under the torque curve, the longer the torque curve stays above a certain point, the higher the peak power will be.

Peak power is not dictated indirectly, it's DIRECTLY influnced by the torque curve!

Analogy:
What's the point in measuring a room in 2 dimentions? It tells you nothing about the "potential" of the room (what you can fit) Torque = 2D, Power = 3D

So in summary, acceleration performance is not dictated by peak power only, or peak torque only, but by the total area underneath the torque curve (within the rev range used by the current gearing setup).

Both curves are relevant. NOT just the torque curve!

[tristancliffe]

How can there be measure of rate of change at an instant? Change is measured over time, if no time elapses it's impossible for change to take place! How can you state the momentary acceleration of an object if it's at a moment. Acceleration is a rate of change, yet you can state the acceleration at time x.
I beleive I proved above that the the delivery over time of force is actually highest at peak power output.Not really, you just stated some stuff. Some is correct, some isn't.
Again with my density, but I fail to see how one can be true but not the other. If "peak acceleration" occurs at "peak torque", then how can holding a CVT at the "peak torque" not yeild "peak acceleration" from point A to B? That's paradoxical to me! If the "rate of change for that instant" is the greatest it can be, how could it (the torque peak!) NOT produce the greatest rate of change over time? But it doesn't!
This is what I was struggling with last night - how come in CVT's you keep them at peak power for best performance but with a MT you want max torque for best performance in any gear. This is because of several factors.
If you had a manual gearbox with an infinite number of gears then maximum acceleration in any one of those gears is still at peak torque. If the ratio is constant, best acceleration is at peak torque. Always. But the ratio in a CVT is an extra variable. What I did to understand it is take a tractive effort graph for a 6 speed gearbox, and then plot curves with the same torque curve for a gearbox with 100 ratios. The area under the tractive effort curves is maximised when you run the engine on the right hand side of the torque curve, which happens to be around peak power. I'm not sure if it is actually AT peak power though, but it certainly is close.

We've already established that CVTs will have maximum acceleration at peak engine POWER output (Power; BTW I've clearly also defined above, in terms of "force output" for lack of better ability to articulate.) That being true, and it is, how could the same possibly NOT be true for any given gear ratio - since by saying it's true for a CVT, we have said it's true for an infinite number of ratios! That includes the 3, 4, 5, 6 ratios in any automatic or manal transmission on the earth!Dealt with that above. You are trying a compare two systems with a different number of variables - one has fixed ratios the other are variable.
The problem is if a particular condition (which some are saying is the peak torque output of the engine) is capable of producing the greatest acceleration even for a "moment", then holding that condition constant (as in the case of a CVT) would yeild the greatest acceleration over time as well (because as stated, "for any given gear ratio peak acceleration happens at peak torque" right?) , but it does not. Why? Because the overall output is still less than it is at peak power, and less tractive effort is available. Still, comparing systems with different variables. If you could 'lock' a CVT at a given ratio (and you can on some modern CVT's) you will find the peak acceleration in that gear will occur at peak torque. If you are in your car and you want to overtake a tractor it's best to hold the car in a gear so that it's just below peak torque - because acceleration will be greatest whilst you overtake.
Another test - use LFS. On a flat road (dragstrip) accelerate in 4th gear and watch the g-meter. It should peak at peak torque (although I suspect that air resistance will skew the results too much anyway).
Peak power is not dictated indirectly, it's DIRECTLY influnced by the torque curve! Yes, but indirectly dictated by the AREA UNDER THE torque curve (sorry for caps, using it to exmphasise).
Analogy:
What's the point in measuring a room in 2 dimentions? It tells you nothing about the "potential" of the room (what you can fit) Torque = 2D, Power = 3D Sorry, thats rubbish. Torque is more relevant. We all, apart from you, say so. Read a few books and you'll see :D
Both curves are relevant. NOT just the torque curve!Yes, power is relevent, but a LOT less than the torque curve.

[Ball Bearing Turbo]

How can you state the momentary acceleration of an object if it's at a moment. Acceleration is a rate of change, yet you can state the acceleration at time x.

An instant is not relevant when you're concerned with time, over which acceleration happens.

Not really, you just stated some stuff. Some is correct, some isn't.

Which is correct and which isn't and why? I noticed you conveniently skipped that post! :razz: I did in fact prove completely that the most force is delivered in the least time at peak power when compared with peak torque, and I even did it with numbers. You cannot deny that that is true, because it would be ludicrous. Plain and simple, and I don't even need a book! :x :D

If you had a manual gearbox with an infinite number of gears then maximum acceleration in any one of those gears is still at peak torque.


Are you being serious? You just basically pulled a 180! A CVT IS a transmission with an infinite number of ratios!, and the max acceleration you stated yourself, using graphs and all, occurs at peak power. I don't know how you could say that with a straight face! :shrug: :tilt:

If you could 'lock' a CVT at a given ratio (and you can on some modern CVT's) you will find the peak acceleration in that gear will occur at peak torque.

Then why wouldn't it happen (peak acc @ peak tq) if you didn't lock it? It still has to "pass" that particular ratio as it modifies the ratio! There are no more variables, all there is is a greater resolution of ratios, which if broken down in into any subset of ratios be it 1000, 100, or 6, all yield the same result I'm afraid.

If you are in your car and you want to overtake a tractor it's best to hold the car in a gear so that it's just below peak torque - because acceleration will be greatest whilst you overtake.

That's not true for that reason. Ideally, at any given speed, in any given ratio, you would want peak power output to perform the most work possible in the least time possible, regardless of available force. You gear there so that you can move into the max area under the curve, which moves you towards peak power, and that's how they relate. Passing a tractor takes time, not an instant.

Yes, but indirectly dictated by the AREA UNDER THE torque curve (sorry for caps, using it to exmphasise).
same with all my caps, just for emphasis. Not yelling :)

Sorry, thats rubbish. Torque is more relevant. We all, apart from you, say so. Read a few books and you'll see :D

Actually, not everyone agrees with you, I'm sure you've read all the posts. I think I have at least one supporter and a few wishwashers lol.

And I don't read. Too much bogus information out there.

[colcob]

Oh I give up.

BBO, I was with you up to a point, but obviously you havent studied maths or physics to a high enough level because although your instincts are good, you're struggling with the maths.

The concept of rate of change in an instant is fundamental to the entirety of calculus. In a sense you've been quite astute because the instant is actually defined as an infintestimal quantity of time. But if you take a graph, the gradient of a graph at any point is its rate of change.

Clearly you can point to a discrete spot on a curve and say that it has a gradient. This is what calculus is all about in a way, its a mathematical method of determining what the gradient is at a given point on a curve (and loads of other things as well, but lets not complicate it).

In the question of the CVT transmission, you are making the same mistake as Tristan in confusing the force from the engine with the force applied at the tyres.
A car accelerates ONLY by applying force to it. Acceleration is directly proportional to force (a = F/mass). So the torque applied at the wheels is the only determining factor (ignoring mass) as to how fast the car will accelerate at a given moment.

The torque generated by the engine is multiplied by the gearing factor to give the torque at the wheels. So if an engine puts 100Nm at 1000RPM into a gearbox with a ratio of 1:2 (or 2:1, I get confused) the wheels will apply 200Nm of torque and turn at 500RPM.

So your CVT gearbox basically calculates the correct ratio to ensure that for the current road speed, the engine will be turning at maximum power RPM, and this will result in a higher torque at the wheels than would be the case at peak torque, because the gear is lower.

An example.

the FXO gives 225lb ft at 4338RPM, and 234bhp @ 6365.

Using the formula BHP = torque * rpm / 5252 (yes, torque and bhp are directly related)

Torque at peak BHP = 234 * 5252 / 6365 = 193 lb ft.

Lets say we are travelling at 25m/s (about 50mph) and the circumference of the wheel is 1.7m, that means the wheel is rotating at 15rps = 900 RPM.

If the CVT matches our speed to peak power revs, that means the gear ratio will be 6365/900 = 1: 7.07.

So if we multiply our torque at peak power (193 lb ft) by our gear ratio, we get:

torque at wheels = 1365 lb ft

If our CVT matches the speed to our peak torque rpm, the gear ration will be 4338/900 = 4.82.

So the torque at the wheels will be 4.82 * 225 = 1084 lb ft

Voila. The reason a CVT produces greater acceleration at peak power is because it lets you use a lower gear which gives greater torque at the wheels. This is fundamental even ignoring CVT's. The reason more power gives you better acceleration over time is because it lets you stay in a lower gear for longer, thus giving you more torque at the wheels for longer.


And finally, as the astute will have notices above, torque and power are directly mathematically related to eachother. power = torque * rpm*constant depending on units used.

So all this nonsense about how power is more relavant than the torque curve, or peak torque is more relavant than peak power, is all just tripe.

In the beginning there was the torque curve. A dyno measures torque. From the torque at each given RPM we mathematically derive the power at each given RPM using the simple formula above. The point at which peak power occurs is a product of the torque curve, or the other way round depending on how you look at it.

The point is that the two curves are not independent, their shapes are inexorably bound to eachother by maths.

[colcob]

Actually, having read through both your posts again, I think you just both fundamentally dont get it, but in different ways.

[skiingman]

What I did to understand it is take a tractive effort graph for a 6 speed gearbox, and then plot curves with the same torque curve for a gearbox with 100 ratios. The area under the tractive effort curves is maximised when you run the engine on the right hand side of the torque curve, which happens to be around peak power. I'm not sure if it is actually AT peak power though, but it certainly is close.

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.
one has fixed ratios the other are variable.
Still, comparing systems with different variables. If you could 'lock' a CVT at a given ratio (and you can on some modern CVT's) you will find the peak acceleration in that gear will occur at peak torque. If you are in your car and you want to overtake a tractor it's best to hold the car in a gear so that it's just below peak torque - because acceleration will be greatest whilst you overtake.

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.

Another test - use LFS. On a flat road (dragstrip) accelerate in 4th gear and watch the g-meter. It should peak at peak torque (although I suspect that air resistance will skew the results too much anyway).

Duh. Now downshift to a gear right at peak power and gun it. Bingo---bigger g-meter reading.

Sorry, thats rubbish. Torque is more relevant. We all, apart from you, say so. Read a few books and you'll see :D

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.

Yes, power is relevent, but a LOT less than the torque curve.
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.

[tinvek]

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

[tristancliffe]

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...

Which is correct and which isn't and why? I noticed you conveniently skipped that post! :razz: I did in fact prove completely that the most force is delivered in the least time at peak power when compared with peak torque, and I even did it with numbers. You cannot deny that that is true, because it would be ludicrous. Plain and simple, and I don't even need a book! :x :D 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...
Are you being serious? You just basically pulled a 180! A CVT IS a transmission with an infinite number of ratios!, and the max acceleration you stated yourself, using graphs and all, occurs at peak power. I don't know how you could say that with a straight face! :shrug: :tilt:
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.
Then why wouldn't it happen (peak acc @ peak tq) if you didn't lock it? It still has to "pass" that particular ratio as it modifies the ratio! There are no more variables, all there is is a greater resolution of ratios, which if broken down in into any subset of ratios be it 1000, 100, or 6, all yield the same result I'm afraid.
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).
That's not true for that reason. Ideally, at any given speed, in any given ratio, you would want peak power output to perform the most work possible in the least time possible, regardless of available force. You gear there so that you can move into the max area under the curve, which moves you towards peak power, and that's how they relate. Passing a tractor takes time, not an instant. 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.
Actually, not everyone agrees with you, I'm sure you've read all the posts. I think I have at least one supporter and a few wishwashers lol. Well, all my imaginary friends are nodding, and they don't like you. They keep whispering nasty rumours about you and everything :Looking_a
And I don't read. Too much bogus information out there.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.

[skiingman]

So all this nonsense about how power is more relavant than the torque curve, or peak torque is more relavant than peak power, is all just tripe.

In the beginning there was the torque curve. A dyno measures torque. From the torque at each given RPM we mathematically derive the power at each given RPM using the simple formula above. The point at which peak power occurs is a product of the torque curve, or the other way round depending on how you look at it.

The point is that the two curves are not independent, their shapes are inexorably bound to eachother by maths.

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.

[skiingman]

If you have an infinite number of fixed ratios it's not the same as an infinitely variable ratio.

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.

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).

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.

[tristancliffe]

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.

[Ball Bearing Turbo]

BBO, I was with you up to a point, but obviously you havent studied maths or physics to a high enough level because although your instincts are good, you're struggling with the maths.

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!

The concept of rate of change in an instant is fundamental to the entirety of calculus. In a sense you've been quite astute because the instant is actually defined as an infintestimal quantity of time. But if you take a graph, the gradient of a graph at any point is its rate of change.

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.

A car accelerates ONLY by applying force to it. Acceleration is directly proportional to force (a = F/mass). So the torque applied at the wheels is the only determining factor (ignoring mass) as to how fast the car will accelerate at a given moment.

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?:scratchch

torque at the wheels than would be the case at peak torque, because the gear is lower.

Is that really the only reason?

The reason more power gives you better acceleration over time is because it lets you stay in a lower gear for longer

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?


And finally, as the astute will have notices above, torque and power are directly mathematically related to eachother. power = torque * rpm*constant depending on units used.

Indeed, in my VERY first post, I noted the futility of debating the relevance of two things derived from each other.

The point is that the two curves are not independent, their shapes are inexorably bound to eachother by maths.

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. :thumb:

[skiingman]

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.
.

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.


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?

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.

Thus the 4th gear test I defined above is correct.

Depends on what you are trying to prove with it. It does prove what you stated above. It does NOT prove:

If you are in your car and you want to overtake a tractor it's best to hold the car in a gear so that it's just below peak torque - because acceleration will be greatest whilst you overtake.

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.

[colcob]

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 :) ).

[tristancliffe]

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

[Hyperactive]

What is this? Who writes the longest post -competition :D.

So when do I see the power and torque curves of the GTRs? :)

[tristancliffe]

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...

[Bob Smith]

OMG! What happened to this thread? - it went beserk! Will read on my lunch break, don't have time now. :D

[colcob]

What is this? Who writes the longest post -competition :D.

So when do I see the power and torque curves of the GTRs? :)

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.

[Funnybear]

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.

[colcob]

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.

[tristancliffe]

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!

[Funnybear]

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.

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?

[Ball Bearing Turbo]

Another kick at the cat!


Power ISN'T the rate at which force is applied. 200N applied is 200N whether you do it for 1ms or 1 year

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?

[Ball Bearing Turbo]

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.

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.

Would you have to just take a point in that time frame and just examine that just to get any usable data from it.

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.

[tristancliffe]

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.
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.

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.
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).

In either case, over that second, where have I exerted more force folks?!

Apparantly at the power peak of course! 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.

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? 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.
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?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.
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 :)). 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.

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.

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.

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?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.

[skiingman]

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.

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 a CVT 'box then you run at peak power because the gear ratio gives you the best ratio possible at every rpm.

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.

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.

No, but if you let go of it, it goes further. :smileypul

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).

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.

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.

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.

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.

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.

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)

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've had enough, you'll never learn.

I would hope everyone involved would learn something. I learned my roommate has a copy of Excel to let me borrow. Amongst other goodies.

[Ball Bearing Turbo]

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! :tilt: , 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.:)

I have not said you want to run at peak torque all the time for best A-B.

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... :shy:


I have known for over 15 years (a large proportion of my life to date) about torque multiplication.

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!

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.

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.

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.

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?

I don't see what you're going an about with force over time all the time. .....

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. :shrug: 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.

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.

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. :tilt: :shy:

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.

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 :)

:schwitz:

[colcob]

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.

[maczo]

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.

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.

[Bob Smith]

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

[tristancliffe]

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.No, you obviously misunderstand me, which I will get to in a second.
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.ypu, that's what I've been saying all the long, well done.
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. 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.
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. :smileypulThats got nothing to do with force/time...
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.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).

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)
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.
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.
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.
Road and racecars alike almost never have those gears, because other concerns are more important than a particular performance from a to b.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...
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.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...

Yes, the torque curve is the useful bit of info. Sometimes they are hard to come by.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.
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.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.

[tristancliffe]

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... :shy: 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.
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)? 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.

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. :shrug: 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.If you maximise one you maximise the other. But the torque curve is what you use when working out performance.
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.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.
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. :tilt: :shy: 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.

Anything else I need to do?

[colcob]

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.

So, y'know, let it lie.

[tristancliffe]

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'.

[tristancliffe]

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.

[maczo]

That is really great, Tristan, I think these two posts could be made a Sticky (although I do not see in which part of the forum :scratchch ).
Thanks a lot :thumb:

[Vain]

I think these two posts could be made a Sticky I know which forum! The physics-section on RSC!

Anyway, long but nice discussion. :thumb:

Vain

[Ball Bearing Turbo]

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.


ACK, This is what I've been saying for 3 pages now! :schwitz:

I am glad Tristan made those graphs, they are excellent work.

I'm not totally done yet, because I need someone to explain what I perceive to be contradictions in some things stated in the last few posts.

[Ball Bearing Turbo]

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.

:scratchch 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....

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.

I realize you're talking tractive effort (wheel torque).

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...

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.

ypu, that's what I've been saying all the long, well done.

When you previously stated:
If you have an infinite number of fixed ratios it's not the same as an infinitely variable ratio.
it appeared to contradict what skiingman said there... :)

If you have any transmission, you maximize the area under the power curve for the time spent going from A to B.
(read: engine's power curve) this has been my argument from the get go!

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.

Again this is has been my point from my first post!


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.

This is all I've ever been trying to say!:shrug:


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.

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.

[colcob]

This is all I've ever been trying to say!:shrug:


To quote my dear tutors at uni "So why didn't you then?" :)

[Ball Bearing Turbo]

I was er... just trying to use as many words as possible to say it :razz:

[lrdbsi]

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

[tristancliffe]

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...

Bet Leo's more mature than you... *runs*

[skiingman]

I still stand by the Riemann sum comment, because its one of the only actual practical uses I've ever seen for Riemann sums.

[Ball Bearing Turbo]

Rofl

:D

[spankmeyer]

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? :shrug:

[Hyperactive]

Do the XRR/FZR/FXR have different tyre sizes?

[Tweaker]

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? :shrug:

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.

[Hyperactive]

Let me guess. The fzr has the widest tires? / XRR has the thinnest?

[spankmeyer]

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. :D

[skiingman]

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.

[Ball Bearing Turbo]

Indeed, check out my "Boost Modelling Questions" Thread from a while back....

edit: here's a link for your convenience!
http://www.lfsforum.net/showthread.php?t=4324

[tinvek]

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

[spankmeyer]

Hypothetical question: In real world racing, would it be anywhere near reasonable to build and maintain a race car powered by a tiny engine with massive turbo if bigger engines were allowed in the first place? (XRR engine vs. FZR engine)

[colcob]

I dunno about today, but I'm fairly sure there was an era when formula one engines could be either 3 litres or 1.5 litres with a turbo, and plenty of teams went with the crazy turbos instead of the 3 litres.

[Hyperactive]

Imho they should have exactly the same tire size and engine size. The different sizes in tires are very hard to balance, as one could think of. Having a 2 litre turbo engine against 3.6 litre NA with 500hp is just not realistic as these cars aren't exactly high-tech anyway.

A twin turbo might make it more interesting but the power isn't the main problem in this class. It's the tire/fuel consumption which are totally different in every model.

[spankmeyer]

Imho they should have exactly the same tire size and engine size. The different sizes in tires are very hard to balance, as one could think of. Having a 2 litre turbo engine against 3.6 litre NA with 500hp is just not realistic as these cars aren't exactly high-tech anyway.

A twin turbo might make it more interesting but the power isn't the main problem in this class. It's the tire/fuel consumption which are totally different in every model.

Dingdingding we have a winner!

[Primoz]

BMW made a 1,5 4 cyl. turbo engine that produced 1500 bhp for qualification. After the fast lap the engine blew up :D

[Hyperactive]

Dingdingding we have a winner!

Please put the doorbell back to your pocket. ;)

How are the educational movies progressing, btw? :)