Hi All,

Did some calculations as follows

Hi Girls, This imaginary vehicle going along a theoretical horizontal 100 kM road, what will the power burn be? So let us look at the theory!

The power burn will be mostly to overcome wind resistance (drag) and rolling resistance. If we take an approximately standard car, work out drag, make an allowance for rolling resistance (not a lot) we can estimate a value to maintain the desired speed.

The standard car that I have chosen is 2M wide by 1.5M high making 3 sq M, and fudge a factor of 0.85. A modern reasonably aerodynamic 4 door saloon has about a 0.3 Cd value. Using these in the following calculations will give a drag force, from which we can determine the power ‘burn', and energy for the 100 kM drive.

Power = Force X Velocity. You want to know the power, you have selected the velocity at 100 km/hr so we need to know the force required.

At a constant speed that force is mostly drag. Drag Force F = Cd . p . ½ V . V .

A Cd factor ranges from 0.25 low, 0.3 mid modern 4 door, 0.4 SUV.

Density of air p = 1.225 Kg/ M3

Area of vehicle normal to direction A = 0.85 x Width x Height (an approximation).

100 Km/hr is 27.78 m/sec.

So we have might have: Drag = 0.3 x 1.225 x 0.5 x 27.78 x 27.78 x 0.85 x 2 x 1.5 Drag force = 361.6 Newtons

Power = F x V = 361.6 x 27.78 = 10045 Watts .... ????

10.045 kW Or 13.5 Hp or consumption per hour, nominally 10 kWhr

This does not take into consideration rolling resistance which is small, but exists. This is perhaps a guide for an ideal constant speed running. You need to consider a factor for head winds, and the power to accelerate the mass of your vehicle up to the 100 kM. I was surprised how much power was needed, unless you accept a very long acceleration time. In my previous calculations a decent hill climbing rate also was a surprise.

If we do an out and back against a 15 kM head wind in one direction and a 15 kM tail wind in the other we can see what happens to consumption. I will be interested! If I put all the common factors into a constant K, we get:- 0.4686

So burn against the head wind = k x (27.78 +4.17)squared = 0.4686 x 31.95 x 31.95 = 478.35 N The velocity is still the same so power is F x V = 478.35 x 27.78 = 13,288.6 W (13.3 kWhr)

The return journey will be : K x (27.78 – 4.17) squared = 261.2 N.

As above Power = 261.2 x 27.78 = 7,256.5 W (7.26 kWhr)

The average = 20545.1 /2= 10272.5 W. (10.27 kWhr) Interesting a surprisingly small increase. Note that the weight of the vehicle is not included in the calculation. This would potentially effect rolling resistance. I suppose adding a nominal 10% or less would cover that.

Adjusting for other variables would be a linear calculation. Cd, and cross-sectional area would be to divide the power by the existing variable and multiply by the new.

For example if the Cd was say 0.4 then new power burn = 10 x 0.4/0.3 = 13,333 W

If CS Area also increased from 3 sq mtr to 4 sq mtr then new power = 13,333 x 4/3 = 17,777 W Now I am rusty on my maths and all the above may be cobblers, but hey Guys, trying to help. Cheers, Joe

P.S. Hill climbing and acceleration are different sums, and weight (mass) do come into the picture! I take it that someone lives on the flat lands!