How much bhp required at a steady 70mph cruise?

Discussion

kambites said:

No idea, I just nicked a figure of wikipedia as an example.

Ta. E36 325 apparently.Using your calculation, if you put in its maximum speed of 160mph (71.5m/s) then you get a required power of 146kW. The real car supposedly has 141kW.

What do you make of that? No other drag accounted for in that, obviously.

trashbat said:

I might be being dim, but for an approximation, can you take a car's BHP and maximum speed (assuming no limiter) and work back from there? Power required = velocity ^ 3.

Obviously your results are only good for a car of that weight and aero.

For example: my car can supposedly do 137mph with 165bhp (I should be so lucky on either count), so were that idea valid and my maths correct, it'd need a mere 22bhp to do 70mph.

Yes that's a valid comparison. You really don't need much power for steady-state. You've already overcome the inertia so it's just rolling resistance and aero drag to slow you.Obviously your results are only good for a car of that weight and aero.

For example: my car can supposedly do 137mph with 165bhp (I should be so lucky on either count), so were that idea valid and my maths correct, it'd need a mere 22bhp to do 70mph.

Edited by trashbat on Monday 22 September 10:35

james_gt3rs said:

Mr E said:

I had a Chinq Sporting with 54 rampant horsepower. It would show about 95mph absolutely flat out, and was not aerodynamic...

A 1.0l Micra will do an indicated 100mph...TBH, the fiat wasn't geared for any more, that was pretty much redline in top.

We have to assume the car is optimally geared, ie. ratios chosen such that the engine is producing its peak power at the car's top speed. This isn't always the case, but I do my sums on BMWs and they always seem to be set up like this!

With this assumption we can use the peak power and the top speed to extract an approximation of the aerodynamic drag. Let's take my E36 328i - it's got 193PS and a top speed of 150mph. You need to work in kW and m/s, so 142kW and 67.0m/s.

Rolling resistance = Crr * mass * g

Approximation for road tyre on tarmac = 0.03 * 1400 * 9.81

Rolling resistance = 412N

As power P = force * velocity, we need 412 * 67 = 27.6kW to overcome this drag. This is subtracted from peak power below.

As aero drag force F = 1/2 * air density * CdA * velocity^2

power to overcome aero drag = 1/2 * air density * CdA * velocity^3, so

CdA = power / (1/2 * air density * velocity^3).

For this example then, CdA = (142,000 - 27,600) / (1/2 * 1.225 * 67^3)

CdA = 0.621

Pleasingly, this is very close to the book figure of 0.627 so we're on the right lines.

70mph = 31.3m/s

power to overcome aero drag = 1/2 * 1.225 * 0.621 * 31.3^3

power to overcome aero drag = 11.7kW = 15.8PS

Rolling resistance is still 412N as above, but at 70mph the power needed = 412 * 31.3 = 12.9kW = 17.5PS

Finally, we get total power at 70mph = 17.5 + 15.8 = 33.3PS

So as it turns out, the E36 only needs around 33 horsepower to maintain 70mph on level ground without headwind. Pretty impressive.

I hope it's not completely opaque how this is calculated you can do the same for any other car, all you need is the top speed and the peak power. If the car's gearing is really far from optimal, eg. a mega-long 5-speed box, it'll be a little out but still close enough at 70mph.

McSam said:

We have to assume the car is optimally geared, ie. ratios chosen such that the engine is producing its peak power at the car's top speed. This isn't always the case, but I do my sums on BMWs and they always seem to be set up like this!

With this assumption we can use the peak power and the top speed to extract an approximation of the aerodynamic drag. Let's take my E36 328i - it's got 193PS and a top speed of 150mph. You need to work in kW and m/s, so 142kW and 67.0m/s.

Rolling resistance = Crr * mass * g

Approximation for road tyre on tarmac = 0.03 * 1400 * 9.81

Rolling resistance = 412N

As power P = force * velocity, we need 412 * 67 = 27.6kW to overcome this drag. This is subtracted from peak power below.

As aero drag force F = 1/2 * air density * CdA * velocity^2

power to overcome aero drag = 1/2 * air density * CdA * velocity^3, so

CdA = power / (1/2 * air density * velocity^3).

For this example then, CdA = (142,000 - 27,600) / (1/2 * 1.225 * 67^3)

CdA = 0.621

Pleasingly, this is very close to the book figure of 0.627 so we're on the right lines.

70mph = 31.3m/s

power to overcome aero drag = 1/2 * 1.225 * 0.621 * 31.3^3

power to overcome aero drag = 11.7kW = 15.8PS

Rolling resistance is still 412N as above, but at 70mph the power needed = 412 * 31.3 = 12.9kW = 17.5PS

Finally, we get total power at 70mph = 17.5 + 15.8 = 33.3PS

So as it turns out, the E36 only needs around 33 horsepower to maintain 70mph on level ground without headwind. Pretty impressive.

I hope it's not completely opaque how this is calculated you can do the same for any other car, all you need is the top speed and the peak power. If the car's gearing is really far from optimal, eg. a mega-long 5-speed box, it'll be a little out but still close enough at 70mph.

Statically, their power to weight values are not too far apart:

Mono: 285 bhp / 550 kg = 520 bhp/tonne

P1: 903 bhp / 1490 kg = 606 bhp/tonne

Now, lets see what happens at 130mph, where both cars are using around 150 bhp to push through the air:

Mono: 185 / 550 kg = 245 bhp/tonne

P1: 803 / 1490 kg = 505 bhp/tonne

This^^ is why the P1 disappears off down the straight, leaving the mono looking like the driver has just found reverse gear by mistake!! ;-)

kambites said:

trashbat said:

What do you make of that? No other drag accounted for in that, obviously.

I don't think it's particularly surprising. Mechanical drag is non-trivial. kambites said:

Max_Torque said:

Rolling friction & mechanical drag is non trivial at low speeds, but is "swamped" by aero drag as speed increases.

By the looks of the calculations above, for a 3-series the drag is about half aero and half mechanical at 70mph. kambites said:

Max_Torque said:

Rolling friction & mechanical drag is non trivial at low speeds, but is "swamped" by aero drag as speed increases.

By the looks of the calculations above, for a 3-series the drag is about half aero and half mechanical at 70mph. Max Torque, that use of subtracting the drag power to demonstrate high power high weight vs low power low weight is very telling, nice one!

Oh, and I realise that the above calculation takes your quoted engine power, which is at the flywheel, as if it was at the wheels. Seeing as we're talking about comparing the "power" required for a given speed, and the layman's understanding of "power" is at the flywheel, I didn't worry about transmission loss but if you want truly accurate figures you'd need to account for that.

(Static drag co-efficient) + (Rolling drag co-eff x speed) + (Aero drag co-eff x (Speed ^ 2))

However, in order to simulate the precise shape of the road load curve for a modern car, that includes significant non linearities, typically 5 or 6 term polynomials are required to define a more precisely matched characteristic.

The EU publish "standard" road load drag curves for passenger cars, that can be used for emissions and fuel economy testing, but as they are old now, they are easily beaten by modern "low drag" vehicles and so pretty much all manufacturers use real "coastdown" testing to determine the road load for their particular vehicle.

mrfunex said:

My Lexus has a power gauge, in kW, rather than a rev counter. Assuming it's accurate, 70mph uses about 15-20kW. 100mph is somewhere in the region of 50-60kW.

I'll let someone else convert that to bhp!

15-27hp at 70mph, 67-80 at 100mph. Seems a little low, but I suppose it could well be quite aerodynamic.I'll let someone else convert that to bhp!

If you drive a typical electric car over the EUDC driving cycle, surprisingly, the energy lost in the tyres of the vehicle is more than that lost in the rest of the powertrain put together! (battery, power distribution, inverter, motor, gearbox and shafting, wheel bearings etc). This is why there is now a huge push to develop and market "eco" tyres!

I find it slightly interesting that the drag factor is also why cars with the same power to weight at rest but different power outputs feel different once rolling. For example say you have an Elise which is 120hp/750kg and then you have a Civic Type R which is 197hp/1230kg. Both are 160hp/tonne at rest (I hope!). We will ignore the effect of driver weight as I can't be bothered.

You'd expect them to accelerate at the same rate and in first/second gear in the Elise I imagine it would be a traction off the line race and the Elise would win. However at 70mph the Type R has a better power to weight ratio than the Elise, despite them being the same at rest. Going up in speeds obviously this change grows exponentially with drag until the Elise stops and the Civic Type R keeps going. On the road power to weight is arguably more important, but on a race track it's more complicated.

Gassing Station | General Gassing | Top of Page | What's New | My Stuff