- Which of the two will have a higher top speed ?
- Which of the two will accelerate faster ?
The answer?….. they will be as near to the same as makes no difference….the reason? They both have the same horsepower….but that is with one proviso… and that is that we can optimise the gearing for both applications to make sure of peak performance from each engine installation.. The important point being illustrated is that performance between the two will in no way be in proportion to the massive difference in torque.
Now for those of you who wish to do a calculation, you will find that the ‘Chevy’, due to max torque happening at low rpm and well below the mid-point of max engine speed (in this case 1800 rpm & 3900 Rpm respectively) , will probably only need a three speed gearbox and a moderately “tall” final drive (axle) ratio (a guess around 2.9:1) to optimise acceleration and top speed. The Ford on the other hand develops max torque way up the rev range, relatively close to max (7500Rpm and 9000Rpm), requiring at least a 5 speed gearbox with a very, very “short” final drive to achieve the same (a guess at around 6,5:1)…. Once that is done I can assure you that the two engines operating in their best individual operating ranges will result in a very similar vehicle performance as installed.. So how do we get this to happen.?
Well to start with, the Engineers involved will know (from known data on the vehicle design), the absolute max speed of the Fortuna driven by 120Bhp. (this is irrespective of the power source… whether that 120 bhp is for example generated by a large or small IC engine, gas turbine or electric). Let’s say that figure for this exercise…120 Bhp can get that Fortuna to 160 Km/h…max. We then gear the car to achieve 160Km/h. In the case of the Chev, we do so to achieve that speed at 3900 engine rpm and the Ford at 9000 engine Rpm.
So why don’t we just change the gearing to get more speed? …Well we can deviate from the optimised gearing, but because max power dictates the ultimate max speed, changing the gearing, either way from ‘ideal’, will result in the vehicle being slower because it will not be running at max power.
Next, we optimise the acceleration portion of the exercise and to do this we try to keep the engine rpm between the max torque point and max power, right up to the max speed we can achieve. As noted earlier, for the Chevy engine this requires only a few gears because of the relatively wide spread of power. The Ford may require 5 or more gears to keep it in the narrow power band… but once we have done this, the acceleration of both vehicles will be pretty similar, if not the same.
The point here is that we have taken an engine with just 75 ft lbs of torque and made it compete with an engine having almost 3 times that number and done so on equal terms…with both doing the best they can….because they generate the same horsepower, or work…and more significantly, we have pretty well equalised the ultimate Tractive Effort available at the driving wheels between the two engines.
We can now go into the realm of… “OK so why don’t we fit small high revving engines to big 4X4s then?”. The answer is because they are noisy, inefficient, busy (frequent gear changes), wear out more quickly and in the spec noted, probably more expensive. The converse, of why not fit a big engine to that small race car is self-explanatory, it is just too big and heavy…besides, the homologation police would have a collective heart attack.
So why do we not talk about tractive effort in the advertising blurb then? That’s a very good question because it is, after all, more relevant to the performance of a car/vehicle than quoting the torque number. The practical explanation is that we would have to quote tractive effort in all the gears and at differing engine speeds in order to have some form of comparison between vehicles ….it gets complicated and is only really needed by engineers to optimise performance.
Why is it complicated?…well to give an idea of just one calculation of available tractive effort at the wheels using the torque available from the engine at just one engine speed, it goes like this:
(I stick to my decision on not using actual calculations but show the following just to illustrate the point.)
Tractive Effort calculation from Torque to Tractive Effort at the wheels
This is the force available at the contact between the drive wheel tyres and road…. The tractive effort relates engine power and torque as follows: