[bwrlane]

Originally Posted by NISFAN

It is an interesting topic.

The basic equation is of course Newton's own: Force = Mass x Acceleration

Force being the driving force at the wheel, specifically at the contact point of the tyre with the road.

With this in mind, regardless of how important I made engine torque sound, it is largely irrelevant. That is why engine power is a more meaningful metric.

If you build the picture, the engine applies it's torque to a gearbox, which in the first few gears at least amplifies the torque.

Lets take a hypothetical 500lb.ft engine.

First gear might typically have a ratio of 3:1 which amplifies the 500lb.ft to 1500lb.ft. Ignoring any losses of course in this whole example.

This 1500lb.ft then goes through the rear diff and might see another amplification in the order of 3:1. So now we have 4500lb.ft travelling towards the wheels.

The wheels then reduce the torque, due to the diameter difference between the drive shaft and the outer circumference of the tyre. Quite massively as you might imagine as there is something like a 26:1 difference.

The equation to arrive at what actually moves the car forward is a fairly simple one, albeit long as it has many equations to get from engine to forward movement.

The real clue about power being more important comes when we consider what it actually means on a practical level. As you know torque and power are mathematically linked with this equation: HP = Torque x RPM / 5252. So a higher horsepower engine basically means it has the ability to do more work.

Note that with gears we can trade torque for revs and vice versa. And that is how a 300lb.ft Petrol engine can out accelerate a 500lb.ft Diesel (assuming the petrol has a higher peak horsepower figure). In effect the gearing nullifies the actual crank torque figure, making horsepower a more meaningful parameter.

In your example of an engine making the same torque at double the revs, if torque was a completely flat parameter, so 500lb.ft all the way through the rev range, both your examples could run the same first gear ratio, and acceleration would be exactly the same. Of course the one that stops at 2000rpm will require a gear change at half the road speed, where it will drop into a 'less amplified torque' gear ratio, meaning it's rate of acceleration drops, whilst the high revver continues at 1st gear torque amplification gearing. When it gets to needing a second gear, that could be a much 'closer ratio' than on the low revver, meaning it gains an advantage with every gear change.

Hope that makes sense.

Thank you. Yes, I think the key to understand is actually the role of gearing. And I've a feeling that the second thing to understand is kinetic energy, which is proportional to the square of speed. This means that double the power is needed to achieve the same rate of acceleration at double the speed.

Think I'm getting there... thanks

[pdk42]

Originally Posted by bwrlane

And I've a feeling that the second thing to understand is kinetic energy, which is proportional to the square of speed. This means that double the power is needed to achieve the same rate of acceleration at double the speed.

No, that's not right. The same force will produce the same acceleration irrespective of the speed the object is moving at - assuming no effect from friction or air drag. In the real world, drag is the big issue. Drag also goes up as the square of the speed, so you need progressively more power as the speed increases just to maintain that speed due to the drag. So effectively, the power available for acceleration decreases with speed. But, it's got nothing to do with kinetic energy.

[JustChris]

Originally Posted by pdk42

No, that's not right. The same force will produce the same acceleration irrespective of the speed the object is moving at - assuming no effect from friction or air drag. In the real world, drag is the big issue. Drag also goes up as the square of the speed, so you need progressively more power as the speed increases just to maintain that speed due to the drag. So effectively, the power available for acceleration decreases with speed. But, it's got nothing to do with kinetic energy.

Agreed. Love a bit of GCSE physics.

Kinetic energy = 1/2 mv2 and isn't that relevant here unless just wondering how many joules your car is using at any given speed.

F=MA

Therefore A = F/M

Thus assuming the car has the same mass it still takes the same force to accelerate it. Now as stated above at higher speeds (it's not really velocity were talking about here as a car is not travelling in a constant vector) there is more force required due to higher wind resistance for the same rate of acceleration.

[pdk42]

Originally Posted by ChesterM2

Kinetic energy = 1/2 mv2 and isn't that relevant here unless just wondering how many joules your car is using at any given speed.

Well, it's not "using" that energy (joules) per se. The kinetic energy is the energy inherent in the object due to its speed (I'll substitute speed for velocity here and ignore direction vectors). You only get that energy back if you decelerate it (it's the kinetic energy that heats the brakes up), and it only gets there in the first place due to the work put in accelerating the object's mass (Force x Distance). Ignoring friction or drag, you don't need any energy to keep an object moving at constant velocity (which is why spaceships can go a long way once away from earth's gravity without using any fuel).

I'm sounding like a physics teacher now...

[nas80]

Also depends on the driver you know.

[jumperjohn]

Originally Posted by nas80

Also depends on the driver you know.

I agree, all the figures in the preceding 4 pages are suited to lab conditions only. Stick different drivers in the different cars and you will get different 'irrelevant' 0-60 times etc.

If you own a 335, or similar, you may well be smug that in theory your car is quick but may easily become humbled by a well driven Ford Focus 'd' on the twisties.

It's a nice feeling to own a car like these but I think the willy waving is a bit embarrassing.