What is the accelerational efficiency of your car(s)?

What is the accelerational efficiency of your car(s)?

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TurboHatchback

Original Poster:

4,162 posts

154 months

Thursday 20th February 2020
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Warning: nerd content below.

So I was thinking about the improvements made in the 0-60 times cars can manage with surprisingly modest power and it led me on to thinking about how close we are to the maximum achievable exploitation of available power and how it has changed over the years. We can calculate this very simply:

E = 1/2MV^2 and of course P=E/T thus theoretical accelerational efficiency = (1/2MV^2)/PT

V is fixed at 60mph and the units are SI (kW and M/S) so if we substitute the constant and convert to mph and bhp this becomes:

Efficiency = 0.482M/PT where M=Mass in kg, P = power in BHP and T = 0-60 time in seconds.

Obviously the discrepancy is made up by rolling resistance, wind resistance, tyre slip, clutch or torque converter slip, power curves and gear ratios that don't allow maximum power at all times, electrical drain, energy stored rotationally in the wheels and drivetrain etc etc. I thought it would be interesting to see how far manufacturers have got in minimising these factors, I predict the best results from a modern DSG diesel with relatively low power of some kind.

My car (E90 330i manual) does quite poorly, 258bhp, 1490kg and 6.4s = 43.5% efficiency.

I tried a B9 A4 Quattro TDI 272 and it does much better, 268bhp, 1665kg and 5.1s = 58.8% efficiency.

How do your cars compare and what examples can you think of that fare particularly well or poorly?

TurboHatchback

Original Poster:

4,162 posts

154 months

Thursday 20th February 2020
quotequote all
300bhp/ton said:
0-60mph can be drastically reduced with just the right tyres and maybe a gearing change. Same weight, power and torque. I’m no mathematician, how does your equation cope for this?

Part of the reason we see such good 0-60mph times these days are down to tyres and launch control systems. Rather than outright power or low weight.
The equation is very simply the linear kinetic energy at 60mph divided by the maximum possible energy output of the engine in the time taken to get to 60mph.

TurboHatchback

Original Poster:

4,162 posts

154 months

Friday 21st February 2020
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FunkyNige said:
Agree, just look at mine -
Disco Sport - 54%
Street Triple bike (with a 70kg rider) - 39%

So a 2 tonne 4x4 is much more efficient than a bike.
That's exactly what I'd expect, the figure is a measurement of how much of the engines potential output is converted into kinetic energy in the time taken to reach 60mph. A diesel 4x4 with a flat power curve, lots of gear ratios and perfect traction should do reasonably well, limited by high wind and rolling resistance mostly.


Edited by TurboHatchback on Friday 21st February 10:11