Quote:
In which case, why bring up the analysis of acceleration in the context of how much more power is needed to go at twice the top speed then?
Graham
I suspect Graham, that you might have missed the point! For any given speed, no matter how slow or fast, one needs twice as much reserve power to obtain the same degree of acceleration, when travelling at twice that given speed.
As a vehicle’s road-speed increases, the maximum power that can be developed by the engine at a given corresponding engine speed, is dictated by gear ratios, final-drive ratio and tyre’s external circumference. At a given road speed, varying proportions of that power, are each requiring to overcome (a) rolling resistance, (b) aerodynamic drag, and (c) hill-climbing resistance. Any remaining unallocated power capability, can be used for acceleration (i.e. increasing a vehicle’s road-speed and its associated kinetic energy).
However as the road-speed (m/s) increases, so the extent of unallocated power available for acceleration [m/s² = (m/s)/s] tends to diminish, because the demands of (a), (b) & (c) increase with road-speed (m/s) and this probably increases more rapidly than the total power output of the engine as engine speed increases. Not only has the available power diminished as road-speed (m/s) increased, but the amount of power needed to achieve the same degree of acceleration [m/s²] increases in proportion to the road-speed (m/s).
Recall that the power for acceleration, P = speed x acceleration x mass; most conveniently expressed in the metric S.I. system of units, with power P in W or kW, speed v in m/s, mass m in kg and acceleration a in m/s². Recall that for a skydiver in free-fall, the initial acceleration due to gravity is approximately 9•81 m/s² (also a car’s theoretical maximum braking deceleration), gradually diminishing to zero at “terminal-velocity” as descent-speed increases, owing to increasing aerodynamic drag.
Hence for a car of 1 tonne = 1000 kg mass, accelerating at 1 m/s², at a speed of 5 m/s (i.e. 18 km/h = 11•25 mph), P = 1000 x 1 x 5 = 5 kW
Hence for a car of 1 tonne = 1000 kg mass, accelerating at 1 m/s², at a speed of 10 m/s (i.e. 36 km/h = 22•5 mph), P = 1000 x 1 x 10 = 10 kW
Hence for a car of 1 tonne = 1000 kg mass, accelerating at 1 m/s², at a speed of 15 m/s (i.e. 54 km/h = 33•75 mph), P = 1000 x 1 x 15 = 15 kW
Hence for a car of 1 tonne = 1000 kg mass, accelerating at 1 m/s², at a speed of 20 m/s (i.e. 72 km/h = 45•0 mph), P = 1000 x 1 x 20 = 20 kW
Hence for a car of 1 tonne = 1000 kg mass, accelerating at 1 m/s², at a speed of 25 m/s (i.e. 90 km/h = 56•25 mph – fast enough for New Zealand roads! – ideal cruising speed for Triumph Toledo 13/1500!), P = 1000 x 1 x 25 = 25 kW
Hence for a car of 1 tonne = 1000 kg mass, accelerating at 1 m/s², at a speed of 30 m/s (i.e. 108 km/h = 67•5 mph – too fast for New Zealand roads!), P = 1000 x 1 x 30 = 30 kW
Hence for a car of 1 tonne = 1000 kg mass, accelerating at 1 m/s², at a speed of 35 m/s (i.e. 116 km/h = 78•75 mph – more than fast enough for British roads!), P = 1000 x 1 x 35 = 35 kW
Hence for a car of 1 tonne = 1000 kg mass, accelerating at 1 m/s², at a speed of 40 m/s (i.e. 144 km/h = 90•0 mph), P = 1000 x 1 x 40 = 40 kW
Assume a Dolomite Sprint of circa 900 kg mass (would be much more than this, with a driver & three passengers, averaging 65 kg each, plus luggage) were able to maintain a constant acceleration (which no car can do!) from rest to 60 mph = 96 km/h = 26•66 … m/s in 8•5 seconds.
The average acceleration would equal (26•66 … – 0) / 8•5 = 3•137 m/s²
Average Power, Pav = mechanical work done to increase kinetic energy, E = ½mv² / time taken
Hence, Pav = 0•5 x 900 x 26•66 … x 26•66 … / 8•5 = 37•65 kW
Peak power at peak-power engine speed is often only relevant in the lower gears, and usually for top gear is a separate issue, that forms part of the discussion re “under-gearing”, “optimal-gearing” & “over-gearing”, which will affect the vehicle’s greatest attainable maximum road-speed.
To reliably predict a car’s maximum attainable road-speed even under non-windy conditions, and its acceleration & hill-climbing abilities, one would need precise, accurate information about the following things:
• Overall mass, including mass of driver, passengers, luggage, fuel & oil etc;
• Aerodynamic drag coefficient
• Projected frontal area
• Rolling resistance characteristics
• Power & torque versus rpm characteristics of the engine
• Efficiency re mechanical losses of the drive train
• Gear ratios, final-drive ratio & tyre external circumference
_________________
Regards.
Nigel A. Skeet
Independent tutor of mathematics, physics, technology & engineering, for secondary, tertiary, further & higher education.
https://www.linkedin.com/profile/view?id=308177758
Upgraded 1974 Triumph Toledo 1300 (Toledo / Dolomite HL / Sprint hybrid)
Onetime member + magazine editor & technical editor of Volkswagen Type 2 Owners' Club
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They certainly proved their worth driving around northern, eastern, southern & western Europe, as well as Great Britain & Ireland. I've heard that one can commonly get about 60,000+ miles of use out of them. 
