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NEGLECT FRICTION |
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In this section, we set out to solve the equation of motion for a car if we neglect all frictional forces and aerodynamic drag. While this might not seem a very realistic thing to do, it does enable us to deduce some interesting results. |
Solution without Friction
In the case where we neglect all frictional forces, the equation of motion becomes
Let us now further assume that the torque is a constant value t0 across the rev range. Evaluating this integral then gives the following expression for the time required to accelerate from an initial velocity vi to a final velocity vf:
Application of the Formula
For my SEAT Toledo Tdi, with Superchip, a suitable test for this expression is the time required to accelerate from 30 to 50 mph in third gear, as the torque generated by the engine is approximately constant in the rev range 1800 to 3000 RPM, with a value of 208 lb-ft. For this car, we have:
M = 1400 kg (including weight of me and diesel)
vf = 50 mph = 22.4 m/sec
vi = 30 mph = 13.4 m/sec
d = 0.3 m
g = 15.5 mph / 1000 RPM = 4.55
f = 0.85 (i.e. assume that a 15% loss occurs between engine and wheels)
t0 = 208 lb-ft = 283 Nm
Substituting into our formula gives a time of 3.5 seconds. For the standard car, for which the torque is around 173 lb-ft, the time is found to be 4.2 seconds.
To get a more realistic estimate, we need of course to take account of aerodynamic drag and other frictional forces. The mean aerodynamic drag that arises in the 30 mph to 50 mph speed range will be, very approximately, that which occurs at 40 mph:
Mean aerodynamic drag ~ 0.5 ´ 1 ´ 1.7 ´1.4 ´ 0.3 ´ 17.92 ~ 115 N
A crude estimate of the other frictional forces can be made by driving the car to, say, 40 mph on a straight flat road and dropping the clutch, and timing how long the car takes to drop from 40 mph to some lower speed, calculating the deceleration, and noting that the resistive force equals the car mass times the deceleration. The frictional forces then equal the decelerative force minus the aerodynamic drag (i.e. the 115 N obtained above). For my Toledo, the frictional force was about 150 N.
The expression for acceleration time is modified to
where the first term in this expression is what we had before, and the friction force = 115 + 150 = 265 N.
Using this expression gives the following amended acceleration times:
Standard car: 4.6 sec
Superchipped car: 3.7 sec
The advantage afforded by the Superchip is clear. It should be noted that the acceleration time for the standard car (i.e. 173 lb-ft and 110 BHP) is close to that obtained by Volkswagen Car Magazine for the Volkswagen Bora Tdi, namely 4.9 seconds. The difference between this calculated value and that obtained by Volkswagen Car can be attributed to various factors, including different test conditions, slightly different mass of the Bora, our crude estimates of frictional forces, etc.
Optimised Gearing
It is now interesting to re-examine the problem by asking the following question: how can the acceleration times be minimised? It is clear that the acceleration time depends on a number of factors, not the least of which is the gear ratio g. Let us therefore choose the gear ratio to be as low as possible, while still allowing completion of the acceleration from speed vi to vf. Let the maximum engine revs (in revs per second) be Rmax. Recalling the expression that relates car velocity to gear ratio,
,
it is clear that this gear ratio will be given by
where we are simply requiring that the engine hits its maximum revolutions at the final velocity vf. Now lets substitute this expression for g back into our original expression for the acceleration time:
Now for the interesting bit. We observe that the denominator is equal to
which is nothing more than the maximum power generated by the engine, as measured at the wheels. We therefore see that the following important rule holds:
the acceleration time is inversely proportional to the power-to-weight ratio, provided that the optimum gearing is chosen
The last clause is important – it is often said that the ultimate determinant for the performance of a car is the power-to-weight ratio, but this only strictly applies if the gearing is chosen in the optimum way. Note also that the optimum gearing will change, depending on the initial and final velocities chosen. So, choosing the gearing to provide optimum acceleration over the whole speed range of the car will not be easy. For example, many petrol cars are geared to hit 60 mph in second gear – thus reducing the time required for the all important 0-60 mph sprint and making the car look fast on paper. The payoff is that the car can feel underpowered during acceleration in second gear at lower speeds, as the gearing is now too high for optimum acceleration from, say, 20 mph to 40 mph. In contrast, a turbodiesel will trade some ability to sprint from 0 to 60 mph, with the benefit of enhanced acceleration from 20 to 40 mph.
For my SEAT Toledo, the peak power occurs at about 4000 RPM (=67 revs per second), and the mean torque in the range 2000 RPM to 4000 RPM is about 180 lb-ft (=244.5 Nm). Using the formulae above gives the optimum gear ratio for this increment of about 11.2 mph per 1000 RPM, and a minimum acceleration time of about 3.2 sec (neglecting friction).