Quote from Ball Bearing Turbo :

I realize that, which is why I noted originially that there is no "next gear" so the fact remains the max acceleration for the given gear occurs at peak torque, and the acceleration in top gear will follow the torque curve. I can't stress this enough - for the GIVEN ratio this is true. Changing the final gearing changes the torque multiplication just as you said.

Yes - and if we change the final gearing so that it is lower, as I am suggesting, we accelerate faster

Yes, but there will come point where torque multiplication is moot because of the drastic drop in torque at the engine - and this happens very quickly after peak power.

depends entirely on the engine, my 2 stroke motorbike revs out to scary levels. It's easily worked out roughly though, work out how many rpm you drop in a gear change, grab your power graph, get the biggest chunk you can over the rpm spread. Most area under the curve.

Agreed, it must depend in the engine's characteristics... I guess you can't generalize it too much.

Well, the simple logic I go by (even tho I don't really sit around making setups but I'm learning ) is to adjust the final drive (basically the general top speed of the car) to make it so that the car reaches maximum revs down the longest straight in top gear on the course and then I just adjust the gear ratios in between for acceleration or traction. I learned that bit from the lovely GPAEDIA which comes with Grand Prix 4 which has some vids on making setups. So, I got that from a guy called Mark Hemsworth whom I think was an engineer in the Arrows F1 team so he does have credibility.

As for the whole peak torque/peak power thing, from what I read from a book called "Ayrton Senna's Principles of Race Driving", he stated that the best time to shift is a bit after the peak power so that you maximise the speed you get from that gear and when you shift gear, the next gear is at the peak power of the engine, thus giving you much better acceleration. Thats what I've read and seen, don't blame me if they're wrong but I do think they know what they're talking about and it is logical.

Sadly, a lot of people that should have a clue... don't.

If you've ever seen a NASCAR broadcast, you'll know exactly what I mean

Quote from poirqc :

I'm starting to understand what all everybody's adding, but i'd like to go a little off-topic, since good on topic things have been told.

What's are the pro/cons of a close ratio transmission.

I think you stay closer to peak power around most of your shifts but you could lose some time shifting?

Pros
More likely to keep a car in the powerband, especially on tuned engines with narrower powerbands, thus giving quicker lap times

Cons
Longer first gear (worse starts usually, and harder to drive slowly on the clutch)
More shifts required, which might start to lose you time (though unlikely except at high level motorsport)
More likely you'll need to shift in the middle of a corner, which can unsettle the car
Cost


Despite the longer lists of cons, any racing driver/team will use them if suitable/legal, assuming they can afford it. Any self respecting racing driver would sell their grandmothers for additional performance on the track.

it seems like the torque vs power thing confuses people a lot.

as others have already stated, torque and power aren't really two different measurements. they are mathematically related to each other, and that's why on all dyno graphs the torque and power curves cross at the same point.

maybe it would help those who find the theory hard to understand if we used some examples of how this all plays out.

take two motors that are both capable of producing the same amount of torque, but one revs higher than the other. the higher reving engine will have more power. even as you go past the torque peak, the power will continue to climb for a while because the extra revs make up for the lower torque. it's only after the torque starts to really drop off a lot that the extra revs don't help, and then the power drops off as well.

another way to think about it is this:

torque = how hard someone can punch you in the face
revs = how many times someone can punch you in the face in a minute
power = how beat up you are after a minute of taking punches

so if somebody can punch you really hard once or twice a minute, they will probably break your face less than someone who can punch a little less hard, but can do it 30 times a minute, but someone who can't punch hard at all can hit you as many times as they want and it wont hurt.

so coming back to gear ratios, it's power that counts, and you'll want to shift just before power starts to drop off, and your gears should be spaced so that you get enough revs in the next gear to keep the power up, which means you will never shift into a gear that puts you below the torque peak. (remember, you want a large number of medium hits, not a small number of hard hits)

hope that helps!

ROFL

Interesting analagy...

I'm pretty sure torque at the wheels is all that matters my friend, and the only two factors involved are torque at the engine, and gearing.

The only reason you're enamoured with being around the power peak is because in so doing you're employing gearing that enables you to have more torque at the wheels, despite producing less torque at the engine .

It still bothers me that this is correct, but until someone shows me MATH that proves otherwise (which would also be proving Todd Wasson wrong BTW) then it's just life.

Quote from Ball Bearing Turbo :

ROFL

Interesting analagy...

I'm pretty sure torque at the wheels is all that matters my friend, and the only two factors involved are torque at the engine, and gearing.

The only reason you're enamoured with being around the power peak is because in so doing you're employing gearing that enables you to have more torque at the wheels, despite producing less torque at the engine .

It still bothers me that this is correct, but until someone shows me MATH that proves otherwise (which would also be proving Todd Wasson wrong BTW) then it's just life.

indeed. the engine makes torque, revs turn it into power, and gears turn it back into torque. so, as you say, we maximize torque at the wheels by chosing gears that put the revs where the max power is. my point was only to show why max power is always higher than max engine torque, as well as why max engine torque doesn't translate into max wheel torque.

Quote from Ball Bearing Turbo :

Really the only thing that still perplexes me about the whole issue is the fact that the definitions for power and torque cause contradictions... F=MA is so simple, and as pointed out in the thread applies to a theoretical instant. However, acceleration takes place over time. You said in the thread that the greatest rate of instantaneous acceleration (in any GIVEN gear) takes places at the torque peak (simple F=MA, makes perfect sense to me). However if that was completely true in practice - then how is it possible that greater acceleration over time (a whole bunch of "instants" of which the previous theory is comprised) does not cooincide with higer torque values, but rather higher power values?

i think the problem your having is that youre not looking at the full picture of many torque curves laid on top of each other

but on a side note youre mistaken about acceleration ... theres no such thing as instantaneous acceleration theres only acceleration which is what you mean by instantaneous ie the tangent to speed over time ... what you mean by acceleration is just the net change in speed over a longer period of time



but back to the actual problem which is why on the whole youre using the engine past its peak torque

obviously youre right about f = ma or rather a = f/m so youll get the most acceleration at peak torque in any gear
however this is only true for cars which only have a single gear

what gearing does is basically it streches the torque curves out

now lets take a look at a few pics from bobs (bens or whatever he likes to call himself today) setup tool
(out of pure lazyness ive taken the xfg but it applies to any car in lfs ... note that lfs has very smooth torque curves which arent necessarily correct but help with understanding the basic concept)

the first pic shows 2 gears rather close together
the y axis is torque at the wheels and the x axis is speed of the car (discounting slip but for all that matters in this discussion its speed)

note how the 2nd higher gear only overcomes the first lower one well past its peak

2nd pic is if we push the 2nd gear a little further out

the situation hasnt changed much and the higher gear still offers more wheel torque only well past both gears individual peaks

now heres an extreme example of 2 gears which are spaced stupidly far from each other

only in such a situation youll find that the 2nd gears torque curve is streched out far enough for it to overcome the first gears before its peak

looking at the full picture with all 5 gears in place

you can see how for each individual torque peak theres a part on the "waterfall" region of the lower gears curve that offers more wheel torque


the problem here is the fact that gearing doesnt shift the torque curve
each gears wheel torque curve will allways start at 0m/s and extend to "revLimit * gearing"
so what it really does is it streches the engines torque curve and much like a rubber band the curves gets thinner while being streched ie its peak is lower
once you look at more than a single gear youll find that on the whole the new lower peaks will almost always be below the drop off of the lower gears wheel torque

Thanks Shotglass, yeah - I already understand what you're saying, although I appreciate the reinforcement. Anything to help pound it into my skull helps

Still, it's the definitions and the single gear issue that still irritates me. Let's forget about gears all together - direct drive

Suppose you had a fluid coupling between the engine and the wheels - that's it. Or a slipper clutch, whatever.

Graphwise, if you use the definition of horsepower: 33000 * x will give you how many "foot pounds of force per minute" is being applied through the couple / clutch device to the wheels, where x is of course the engines horsepower output at whatever RPM you choose. Obviously, the engines max power RPM yeilds the largest value.

Yet, if you use the engines torque peak, the horsepower output is less, thus less "foot pounds of force per minute" is being transferred - but if you use the A=F/M method, more torque is available at the wheels at the engines peak torque value.

Anyone see what I am getting at here? This does not make sense, puting gearing aside.

could you put that in si units so it makes sense for someone on the right side of the globe ?


youve made a total mess of the definitions swirling around in my head ^^

my current hunch is that you have a misunderstanding of what power is
power is not force/time its work/time
it only works out because torque is of the same dimension as work

Well 1 horsepower = 33000 ft.lbs per minute (44742 n.M :razz

That means it takes 1 horsepower to move 33000 lbs one foot in one minute

Clearly, (at least to me, please tell me if I am wrong but explain why) the ability to do work should have some effect on how fast I can move my direct drive slipper clutch mobile down the track - since it takes work to do just that.

I need to go get food, since this is hurting my head all over again.

DAMN THIS TOPIC

haha, i was going to mention direct drive electric motors, but i decided not too add to the confusion.

btw another fundamental flaw in your reasoning is that power is not force/time
power is force*speed

Quote from Ball Bearing Turbo :

Clearly, (at least to me, please tell me if I am wrong but explain why) the ability to do work should have some effect on how fast I can move my direct drive slipper clutch mobile down the track - since it takes work to do just that.

exactly that is why power plays a role in a vehicle max speed
its the maximum amount of force (drag N) with (against) which the engine can move a car a certain amount (m) in a given time (s / speed)

Check your PMs, if I post here Colcob might put a bounty on my head... :hide:

i prefer to keep this public to possibly get input from trist col and others if you dont mind

i think ive made up my mind about cvt now and imho the best acceleration would be neither at peak power nor at peak torque

now the reason why its not at peak torque (in any normal car that is) is explained rather well by col in this thread:
http://www.lfsforum.net/showthread.php?t=5754&page=3

but (not having finished reading the thread yet) i havent read this interpretation yet which is imho the correct one:
1) by varying the gear ratio with a cvt while accelerating at any given rpm youre constantly reducing the wheel torque with a function of gradient -1
2) therefore if the torque curve is relatively flat past peak (less drop off than -1) following the torque curve further (past peak) will yield a better acceleration than reducing the gear ratio (ie yield more wheel torque)
3) so if im not mistaken the best acceleration with a cvt should be achieved by keeping the engine at the exact rpm at which the torque curve (first) has a gradient of -1 past peak --- this bit im very unsure about as of yet


unsure or not what this means is that its very important for the maximum acceleration with a normal gearbox or a cvt how flat the torque curve is past peak

relating to the original discussion
basically power is an indirect method to express the flatness of the torque curve or in other words how much torque is available at high rpm which enables you to use a higher gearing ratio for more wheel torque




(now that i think about it peak power is exactly the spot at which the torque curve falls with gradient -1)

ack

BBT:

I think the fundamental question you are asking is: "Why doesn't constant power cause constant acceleration?"

And the answer to this question can be found easily by looking at it from an energetic point of view. The amount of energy that you need to accelerate an object from v1 to v2 is 1/2*m*(v2^2 - v1^2). This means that the amount of energy that is needed to cause a given difference in velocity (v2 - v1) is not constant. The faster v1 is, the more energy you will need to get to v2.

In english: it's harder to accellerate when you are already going fast. That means your single gear vehicle will accelerate worse at peak power than at peak torque but this is because it's going a lot faster at peak power than it was at peak torque.

OK look here´s the deal with CVTs.

I think we can easily agree on that this is correct:
wheelTorque = engineTorque * gearRatio

Now as you can see there is a linear relation between torque at the wheels and the gearing ratio.
So if we reduce the gear ratio at any given engine/vehicle speed the wheel torque will fall of linearly (gradient -1).

Now let´s rewirte the relation with this dependency:
wheeTorque = engineTorque * gearRatio(speed)

To fully understand that assume a CVT that keeps the RPMs at a constant (ie. engineTorque is constant) so while you adjust the gear ratio with speed the wheel torque will drop of in a linear fashion.

Let´s look at the situation with a normal gearbox in any given gear. With that kind of situation the dependency to speed would look like this:
wheeTorque = engineTorque(speed) * gearRatio
The torque at the wheels obviously varys with the point you´re at on the engine´s torque curve.

Now if we combine these two things the following happens.
- If you are at peak torque and reduce the gear ratio the torque at the wheels will drop off in a linear fashion
- If you stay in the current gear the torque at the wheels will drop of less than linear since the torque curve is relatively flat
- Therefore the point at which you have to reduce the gear ratio is exactly where the engine torque drops faster than linear
- That point is exactly at peak power

Assume it was a slipper clutch that held the engine at a constant RPM, all the way from zero velocity to max velocity (which would just be the speed at which the wheel speed was able to match the chosen engine RPM). Or better yet some magic lossless coupling or something...

edit: woops, this was directed at JB

edit2: That really nails the CVT issue Shotglass. Pretty eloquently put I think:

Quote :

Therefore the point at which you have to reduce the gear ratio is exactly where the engine torque drops faster than linear

That should be nailed to a wall somewhere

That slipper clutch is either in reality a CVT or it's converting lots of power to heat. Or it's magic.

i think i should restate that:
Therefore the point at which you have to reduce the gear ratio is exactly where the engine torque drops faster than linear

as:
Therefore the point at which the impact, on acceleration, of reducing the gear ratio is less than the impact of staying in gear, is where the torque curve drops faster than linear

Quote from evilgeek :

as you go past the torque peak, the power will continue to climb for a while because the extra revs make up for the lower torque. it's only after the torque starts to really drop off a lot that the extra revs don't help, and then the power drops off as well.

Quote from Shotglass :

i think i should restate that:
Therefore the point at which you have to reduce the gear ratio is exactly where the engine torque drops faster than linear

as:
Therefore the point at which the impact, on acceleration, of reducing the gear ratio is less than the impact of staying in gear, is where the torque curve drops faster than linear

if i'd thought about it a little more i would have replaced "a lot" with "at a slope of -1". so i think we agree, yes?