MG4: Drag, Various Conditions

[Fogey]

I thought I'd have a look at the power levels needed to combat aerodynamic drag at constant road speeds, in various conditions. At high speed, these will dominate in economy terms, particularly if those conditions are adverse. To get the overall power level requirement, there'll be an added (constant) figure to overcome rolling resistance, and a slightly (probably!) variable figure for motor/transmission losses.

Assumptions are: altitude: sea-level, temperature: 15C, atmospheric pressure: 1000mbar, MG4 estimated frontal area: 2.3 sq metres, cd 0.28. Values in kWh/100miles.

At 70mph: 17.1
At 60mph: 12.7
At 50mph: 8.8

For a ground-level headwind of 20mph:

At 70mph: 28.4
At 60mph: 22.5
At 50mph: 17.1

At temperature of -10C, add about 10%; for pressure 1100mbar, add about 10%; for altitude 1000ft, subtract about 10%.

So, comparing 100 miles at 50mph in calm conditions versus 70mph against a 20mph headwind, there's about a 20kWh difference. Do note also that there'll always be a penalty when doing a round-trip in windy conditions relative to calm conditions.

 

[siteguru]

What formula are you using to derive those figures?

 

[Fogey]

What formula are you using to derive those figures?

I used this calculator:

Drag Equation Calculator

The Drag equation calculator helps you compute a force exerted on a moving object immersed in a fluid.

www.omnicalculator.com

I was just going to compare power figures, but ended up calculating energy specific to a distance travelled (100 miles), as it's more useful.

 

[Deleted member 9909]

I will be honest to me they don't matter a jot. If you have to make a journey you have to make a journey irrespective of drag and weather conditions. I am not going to hold off until the wind dies down or the temp warms up and you can't easily change the vehicle aero dynamics so we have what we have. I already know more speed = less efficiency and tend to do 50 - 60 where appropriate and viable. But on motorways sometimes it's just not a good thing if you value your safety when an overtaking lorry driver is giving you a cheery two fingered greeting.

Good info for debate and in depth technical discussion but for me ?‍ sorry

Not criticising as forums are a good place to share and discuss exactly this stuff its just me.

Alan

 

[tsedge]

I thought I'd have a look at the power levels needed to combat aerodynamic drag at constant road speeds, in various conditions. At high speed, these will dominate in economy terms, particularly if those conditions are adverse. To get the overall power level requirement, there'll be an added (constant) figure to overcome rolling resistance, and a slightly (probably!) variable figure for motor/transmission losses.

Assumptions are: altitude: sea-level, temperature: 15C, atmospheric pressure: 1000mbar, MG4 estimated frontal area: 2.3 sq metres, cd 0.28. Values in kWh/100miles.

At 70mph: 17.1
At 60mph: 12.7
At 50mph: 8.8

For a ground-level headwind of 20mph:

At 70mph: 28.4
At 60mph: 22.5
At 50mph: 17.1

At temperature of -10C, add about 10%; for pressure 1100mbar, add about 10%; for altitude 1000ft, subtract about 10%.

So, comparing 100 miles at 50mph in calm conditions versus 70mph against a 20mph headwind, there's about a 20kWh difference. Do note also that there'll always be a penalty when doing a round-trip in windy conditions relative to calm conditions.

Shows the value of driving slowly. Interesting analysis, thanks.

 

[Macadoodle]

This fellow on the u tube claims to be getting over 5miles kWh including motorway driving from his SE SR. The GOM does seem to show well over 200 miles, I'll be well pleased with anywhere near that, expecting 180 and will be happy with that .

 

[Fogey]

I should point out that the figures apply to any car with the same aerodynamics as the MG4's stated cd and my estimated frontal area. And powered by any motor - electric, petrol, clockwork etc. etc. Weight is not, of course, a factor either.

 

[tsedge]

I should point out that the figures apply to any car with the same aerodynamics as the MG4's stated cd and my estimated frontal area. And powered by any motor - electric, petrol, clockwork etc. etc. Weight is not, of course, a factor either.

I assume you mean weight is not a factor affecting your calculations above (which are only considering aerodynamics) - clearly it does take energy to accelerate mass, so a heavier car will also take more energy to get up to speed, reducing range.

 

[Fogey]

I assume you mean weight is not a factor affecting your calculations above (which are only considering aerodynamics) - clearly it does take energy to accelerate mass, so a heavier car will also take more energy to get up to speed, reducing range.

The calculation is for steady speed only. Weight will affect normal day-to-day driving during acceleration and will affect rolling resistance too.

 

[MartinSEsr]

very interesting data, thanks

 

[wattmatters]

Weight also matters when gradient is non-zero. Which is often the case in many locations.

Interesting estimated CdA figure of 0.28 x 2.8m² = 0.784m².

It is testable if one could do some roll down testing with an accurate data logger and a suitable location - it's possible to tease out the CdA and Crr/drivetrain components.

Wind direction is rarely aligned with vehicle direction as well and so there can be yaw angle variances with CdA. I've no idea if we can assume Reynolds number is fixed over the typical velocity range for a car. It probably is but aerodynamics is a pretty weird beast at times.

While EVs recover energy during regen braking, it's never as much as the energy expended to increase kinetic or gravitational potential energy in the first place.

Weight also adds to rolling resistance, and due to the way the maths works, each 0.01 in Crr is equivalent to adding 1% to the gradient. Crr for a car is in the 0.008 - 0.015 range, while going straight at least (cornering is a different matter).

Road surface matters a lot. IOW a dead flat but rough road surface adding 0.004 to Crr compared with smooth asphalt results in an equivalent power demand to going up a very slight incline (gradient 0.4%) on smooth asphalt.

 

[Fogey]

What I was saying was that weight isn't a factor which affects drag - and the figures are for aerodynamic drag only.

Energy required to overcome rolling resistance (which is pretty much constant for a given weight over a given distance/route) and in changing speed are of course extra.

 

[wattmatters]

What I was saying was that weight isn't a factor which affects drag - and the figures are for aerodynamic drag only.

I see what you're saying - I was thinking more about the scenario where available power is limited, and that's when weight and its demand on the available power matters.

Increase those demands (e.g. increase rolling resistance, climbing a gradient, increasing kinetic energy) and less power (or total energy budget if you like) is available for overcoming drag. But for most motoring power is not a limitation.

 

[Fogey]

But for most motoring power is not a limitation.

Yes, as I said, the energy per 100 miles is a more useful figure (as it has a more direct relationship to range - which very much is a limitation).

By the way, the energy figure for lifting an MG4 1000 feet is about 1.5 kWh. (Not a lot of people know that!)

 

[wattmatters]

By the way, the energy figure for lifting an MG4 1000 feet is about 1.5 kWh. (Not a lot of people know that!)

Gravitational potential energy, U = mgh

m, mass: car is close enough to 2,000kg with passengers/luggage.
g, acceleration due to gravity, close enough to 9.8 ms⁻²
h, height in metres, 1000 feet is ~ 305 metres

U = 2000 kg * 9.8 ms⁻² * 305 m = 5.98 kJ. Divide by 3.6kJ/kWh = 1.66 kWh

My local climb is close to 730 m, so looking at burning 4 kWh just to lift the car and occupants/cargo. But I get some energy back on the way down. Will be heading up and down it a couple of times this coming weekend.

 

[Fogey]

No it's not - I think you've mixed up hours and seconds.

Edit: in response to @wattmatters' comment.

 

[wattmatters]

No it's not - I think you've mixed up hours and seconds.

Yeah, I realised as soon as I posted and corrected my units. I was thinking kJ, not kW•h.

 

[Fogey]

OK, as at least one FM is now feeling confused, I'll repeat, the correct figure is:

~ 1.5kWh per 1000 feet.

 

[wattmatters]

OK, as at least one FM is now feeling confused, I'll repeat, the correct figure is:

~ 1.5kWh per 1000 feet.

Which is in line with the 1.66kWh I posted.

But the exact number depends on what assumption is made for the mass of the car.

My Essence 64 has a curb weight of 1,672 kg and a GVM rating of 2,123kg, so I just added passengers and luggage and rounded it to 2 tonnes for the sake of estimates.

Using 305 metres for 1000 feet...

(2000 kg * 9.8 ms⁻² * 305 metres) / 3600 seconds/hour = 1.66 kWh

 

[Fogey]

Which is in line with the 1.66kWh I posted.

The point is that you contradicted my post by saying that the figure should have been 5.98 kWh. Which was INCORRECT - hence my response.

You then admitted your mistake, so I repeated the the figure of 1.5kWh as being the correct one, and not 5.98kWh.

But it's apparent that you edited your post to change it to kJ (and then to convert it to 1.66kWh). And since the fact that it was edited was not recorded (why isn't it flagged by the software?), the thread has now became even more confusing to anyone reading it.

I'm NOT arguing against your corrected figure of 1.66 (it's realistic for a loaded car).

By the way, can I suggest in future you might not leap to contradict someone until you've double-checked your calculations?