Plowperson is simply wrong, because he doesn't understand what a gearbox does: The dimesnions of acceleration at a given velocity are in fact power, not torque, which only governs instantaneous acceleration from rest. Torque that is not sustained at high RPM will not accelerate a car as well as lower torque that is, suitably geared down.
The type of gearbox makes no difference to the principle.
But if I remember the last CVT I drove, it tended to keep the engine around the peak torque engine speed with your foot hard down (until the car speed got above that in the highest ratio). It certainly didn't go as high as peak bhp initially. Would have been a pretty noisy drive if it did.
Most conventional autos change up when the engine reaches peak bhp. Or thereabouts.
I can't see anyone arguing with that.
But I'm damned sure that when Vir Campestris wrote "peak torque", he meant torque at the engine, not at the wheels. If you think he meant torque at the wheels, that's your mistake right there.
I was attempting to get to the basic principle. I chose CVT because that makes it theoretically possible to run the engine at the "best" speed for a given road speed. I'm not talking about a production CVT, I'm talking about a theoretical one optimised for maximum acceleration rather than economy, noise or anything else.
So I'll ask again, do you agree that with an appropriate CVT, the maximum acceleration will be at peak BHP for the engine?
That gets my vote as the biggest load of s**te from you to date.
Thanks for confirming it's yet something else you don't understand.
I'll give you a clue for learning up about it. Just how an engine's power (BHP) is calculated.
I'm thinking aloud here and I may be wrong - in which case tell me where...
Torque is a rotational force - well, a turning moment: the product of a force and a radius from the centre of rotation.
F=ma, so peak rotational acceleration will be at peak torque.
That's torque at the wheels.
A gearbox increases torque by the same factor that it reduces the rotational speed. Hence the lower the gear, the higher the multiplication factor for torque.
So I'd expect maximum acceleration to be at maximum engine torque in the lowest gear.
But I may be missing something crucial, since some people are talking about maximum power.
Now where have I gone wrong?
I've already explained that and the answer is no. Unless the peak BHP coincided with peak torque, which would be unusual, to say the least.
Have you ever driven a CVT? Did you have the figures for the RPM that peak torque and peak bhp occurred on that engine?
On most car engines, peak BHP occurs quite close to the maximum rpm of that engine. Basically, there is no point in it revving very much higher.
If a CVT went immediately to near peak revs from rest when you wanted best acceleration, it would be a very fussy vehicle indeed. Most describe it as a constant drone - more akin to the lower revs, but peak torque figure.
But peak torque at the flywheel will translate to even greater torque at the road wheels
In article , "Dave Plowman (News)" writes
Friction causes waste of power
I agree with you Dave. Power is important at the stable state e.g. top speed when acceleration is actually zero. And has been stated before poser and torque are linked. You can't have one without the other, to coin a phrase.
You haven't. Gear box ratios are then selected to keep the engine around its max torque point as road speed increases. However all engineering requires compromises and those for an F1 car or Le Mans will be different to those for a city run about such as a Fart 500.
Well exactly. By that token an infinite reduction gearbox will give infinite acceleration at zero speed.
The point is, once above zero speed there is no acceleration at all. :-)
Ok, you are wrong then. Imagine the engine has peak torque at 2000 rpm and peak power at 4000 rpm. The torque at 4000 rpm could be anything down to just over 1/2 that at 2000 rpm for that to happen.
With the gearbox optimising for maximising acceleration by running the engine at maximum BHP, it will run at 1/2 the ratio of one trying to run the engine at maximum torque. So if our engine torque at 4000 rpm is
0.55 that of it at 2000 rpm, the wheel torque will be 1.1x that of what it would be if geared to run at the lower rpm. That means it will accelerate the car faster.I shall repeat : "I'm not talking about a production CVT".
This much is true. My proposed CVT runs the engine at peak BHP, no higher, no lower, while it's trying for maximum acceleration.
And once again : "optimised for maximum acceleration rather than economy, noise or anything else."
You've already agreed that autos try and do what I describe - peak power is normally pretty close to the red line, and autos will aim for that when being asked to go as fast as possible. Obviously those which aren't CVT do have to drop below maximum power due to the gaps in the gears.
So what? if it spins the wheels its no damned good.
Of COURSE in a given FIXED gear peak torque is peak acceleration. at that speed, the point is that we actually have gearboxes with many ratios.
And rate of acceleration at speed is in fact a function of power, given the perfect gearbox.
You will always have better acceleration by dropping a gear to exploit peak power, than changing up to peak torque with less power
Power is torque times speed. The more power you have the more torque at any given speed, given a suitable gear ratio in between
Another total plonker.
You are right however that 'poser' and 'torque' are inextricably linked.
No, at around its max POWER point as road speed increases,
Power = acceleration times velocity times mass.
F = ma P=Fv
therefore P=mav.
ergo maximum acceleration demands maximum power, and the torque is irrelevant since you select the right gear to be at maximum *power* which will result in maximum torque at the road wheels *BUT NOT AT THE FLYWHEEL*.
The time you change up a gear is well past peak torque, when the power has also peaked such that the power you are at is at the same value as it will be in the next gear up, with te engine on the lower rev side of the peak power point .
However all engineering
It doesn't make any difference. Torque from the engine ends up at the wheels. And the maximum torque from a car engine is never at maximum BHP.
All you really need to do is find some acceleration curves for any car. The maximum rate of change of speed in any gear always occurs at peak torque - not peak bhp.
I'm really quite surprised so many don't know this.
Correct.
Correct
Correct. Which always bears a relationship to torque from the engine.
Correct.
Correct. Assuming the tyres can transmit that torque without slippage.;-)
You haven't. As I keep on saying, there is a direct relationship between torque and bhp. BHP is merely a product of torque and engine speed - via a couple of constants.
Vir Campestris wrote "given suitable gearing", but you've assumed a limited choice of gear ratios. Take your scenario, change the gearing to give the same wheel speed at maximum engine power, and the torque (and therefore acceleration) will be higher.
Nobody is disagreeing with that. What they are saying that the maximum rate of change of speed would be achieved by selecting the gear such that the engine is at peak power for a given road speed.
They do.
If you change up at peak torque, you will now be (well) below it in a higher gear. With a well designed engine and box, changing up at about peak BHP will bring you near the peak torque rpm.
And you have that exactly the wrong way round emphasis wise. BHP is a function of torque and engine speed. So the more torque, the higher the BHP. Not that I'd expect you to understand the difference.
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