Power versus Torque - Part 2

Last week we looked how power and torque are defined and related. This time we apply some of that knowledge to seeing how hard cars accelerate when endowed with engines with either lots of peak torque, or lots of peak power. Which is required to give the best acceleration?

Let's say that we have three engines:

• Engine A, which is representative of the old Ford 4.1 carby engine.
• Engine B, which has the same torque curve, only it is spread out, so that the torque numbers are at one and a half times the revs of A.
• Engine C, which has the same power curve as A, but spread so that the power figures are one and a half times the revs of engine A, but it only has two-thirds of the torque.

Revs	Torque (A)	Power (A)	Torque (B)	Power (B)	Torque (C)	Power (C)
1000	190	20	170	18	113	12
1500	255	40	190	30	127	20
2000	280	59	245	51	163	34
2500	300	79	265	69	177	46
3000	280	88	280	88	187	59
3500	255	93	295	108	197	72
4000	225	94	295	124	197	82
4500	180	85	280	132	187	88
5000	120	63	260	136	173	91
5500	Past Redline		235	135	157	90
6000			225	141	150	94
6500			195	132	130	88
7000			160	117	107	78
7500			120	94	80	63

To shows how this data looks graphically, here are the power curves of the three engines:

and the torque curves of the same three engines:

Okay, so now we have some numbers to work with to determine which of these engines will give the most performance in terms of acceleration. If peak torque were the most important characteristic, you would anticipate that engines A and B would perform similarly (they have near identical peak torque), and that engine C would trail.

If peak power were the important factor, you would expect that engine B would be quickest, with engines A and C level pegging.

For argument sake, let's assume that each of the cars has a gearbox with a 2:1 first ratio, and a direct second, and that in first the car does 10 km/h per 1000rpm, and 20 km/h per 1000rpm in second. Also assume for argument's sake that the diff ratio is 1:1. Use the basic equation handed down from old Isaac Newton, F =ma, where F is the force (related in this case to the torque at the driven wheels), m is the mass and a is the acceleration. The force at the driven wheels is related to the torque by considering factors such as wheel radius, but these are constant so we will leave them out, and basically say that the torque at the tyres is directly proportional to the engine torque. So, in turn, the acceleration is directly proportional to the torque at those driven wheels.

Let's have a look at what happens when we compare the cars.

Speed	Torque at the Wheels (A)	Torque at the Wheels (B)	Torque at the Wheels (C)	Torque at the Wheels (D)
10	380	340	254	380
20	560	490	314	560
30	560	560	394	560
40	450	590	374	450
50	240	520	346	240
60	280	450	300	280
70	255	320	214	255
80	225	295	197	225
90	180	280	187	180
100	120	260	173	120

Now suddenly things look very different!

Notice that car B pretty much always has the acceleration advantage, but that car C trades blows, so to speak, with car A. Car C has a lot torque at the wheels to 40 km/h, more from 50 km/h to 60 km/h, less between 70 and 80, and more from 90-100 km/h. On the whole, car C would accelerate roughly equally to car A, except over the first part of the rev range, where car A would have a decisive advantage.

But here is where gearing plays its part. If we put in a diff of ratio 1.5 in car C (now call it car D), you will have the same torque figures at the wheels throughout the range as car A (which is what you would expect, given that the torque of engine C was only 2/3 of that of engine A). So, it is clear that you can compensate for a lack of torque if you have power. Simply change the final drive ratio. The question is, can we change gearing to compensate in the case of an engine that has a relative lack of power?

Let's have a look by comparing engines A and B. The first problem that you are stuck with is that if you attempt to lower the final drive ratio, the car will no longer reach its top speed (it runs out of revs). The second point to realise is that you could quite easily change the gearing in engine B (use a 1.5 diff ratio again) to get significantly more torque at the wheels than engine A. So, in short, you can compensate for a lack of torque in a powerful engine by shortening gearing. But you cannot compensate for a lack of power by gearing.

(An interesting example - by appropriate gearing, you can get a human being to generate a lot of torque at the driven wheels at extremely low (like about 1 km/h!) speeds. The ability to hold this torque at higher revs is power, and the fact that a human being does not generate much power is the reason that a human-powered vehicle are not able to accelerate quickly at any but the very lowest speeds.)

Earlier I mentioned that, by knowing engine power at a certain speed and mass alone, you can determine the acceleration. The equation is:

P = Fv

Where P is the engine power, F is the force and v is the velocity. Now, remembering that F = ma,

P = mav

	P
a =
	mv

So all you have to do at a certain speed is look at the tacho, look at the power graph and determine how much power you develop at those revs, and you can determine the acceleration at that speed (assuming no drivetrain losses). You will note that you cannot do the same for engine torque without knowing the gearing to determine the actual torque at the wheels, rather than the torque at the engine, which is modified by the gearing. As you'll now realise, gearing is a torque multiplier, not a power multiplier.

It is therefore clear that in order to maximise your acceleration, it is very important to have high power, rather than high torque. You can get high torque at the wheels by using appropriate gearing, but you cannot get high power from gearing. In short:

• for best acceleration, maximise power
• for best driveability, a high torque, non-peaky engine is better

An engine with high torque throughout the rev range will have good low-rev and part throttle response; this is the reason that cars like 4.1-litre Cortinas feel that they are very quick, even though they are not in absolute terms. That 4.1 has a very flat torque curve, with a maximum torque similar to the Commodore Ecotec V6 - but I know which engine I would prefer in outright acceleration terms (the V6 has about 50% more power)!

Conclusion

So power is the critical determining factor for maximum acceleration, and torque is necessary for driveability. This can quite clearly be seen in looking at the types of engines used for certain applications. Trucks and industrial vehicles use engines with large amounts of torque with a fairly flat torque curve, but relatively little power. Racing and sporting vehicles use high revving engines, with high power and relatively little torque in comparison to their power outputs. That is why you have F1 having engines revving to over 18000 rpm. And there's no use doing that if to maximise acceleration all you needed to do was maximise the amount of torque you had!

Top Speed The top speed of a vehicle has almost nothing to do with the mass of the vehicle (apart from bearing friction and tyre losses which, in terms of power consumed at top speed, are minimal). The two major factors are power and aerodynamic drag. The equation to get a pretty good approximation of top speed of a vehicle is that relating power to drag:- where P is the power in Watts, r is air density (1.3), CD is the coefficient of drag, A is the frontal area in square metres, and v is the velocity in m/s. To see how well this works, lets take, for example, a modern Falcon XR6 and an old 4.1 Cortina. The XR6 produces about 110kW at the wheels, has a CD of 0.31, and a frontal area of about 2.4 square metres. In the case of the Cortina, the numbers are 60kW, 0.48 and 2.2. To get top speed: For the XR6, this equates to 66.6m/s or 239km/h. The XR6 has been tested at over 230km/h, so this seems pretty reasonable. For the Cortina, the speed is 48.4m/s, or 174km/h, which is pretty lineball with the tests for the car. An interesting exercise would be to test the numbers for your own favourite car, remembering to use power at the driven wheels, not at the flywheel. And again, note here that it is peak power, not peak torque, which is the important engine performance variable.

Did you enjoy this article?

Please consider supporting AutoSpeed with a small contribution. More Info...

Share this Article:

Tweet