To push a vehicle at low speeds where rolling resistance predominates, the force required is proportional to speed. More speed -> more force for the given distance --> more work/mile. Have you ever pushed a vehicle any distance? Tried to push it faster? I believe that you will find it takes more force to push it faster.
The required power to meet the rolling force needs is a constant * V^2. Aero force is a constant*A*Cd*density*V^2. Aero power is proportional to V^3. So total required road load horsepower is rolling constant * V^2 + Aero constant * V^3.
Work done/mi = power * Time. That divides the velocity out for a given mile getting back to work done = ~ (rolling resistance force + Aero force)* distance.
So one measures the coast down time. The vehicle mass (ignoring the energy in rotation of wheels/tires/brakes/driveline (you do do the coastdown in neutral? right?) times V^2 gives the energy at the starting speed and the ending speed. The difference gives you the energy expended. Divide by the time and you get average horsepower over that time. Multiply the required horsepower times the expected engine BSFC (at the appropriate engine speed and torque) * gal/lb (~1/6.15 for gasoline) * driveline efficiency gives you gallons/hr. Take mph and divide by the above and you should have an estimate of mi/gal.
I was trying to simplify the problem to make it easy for people to understand. Technically speaking the force of rolling resistance has two components: one is constant with velocity, the other part tends to be proportional to velocity; I ignored the second part to make the analysis easier for people to understand the effect of aero drag.
Have you ever pushed a vehicle any distance? Tried to push it faster? I believe that you will find it takes more force to push it faster.
Why, yes I have; 180 yards at a time in the low 20 second range, to be very specific about it. And I've done it enough to understand that the increase in force is required not because the rolling resistance increases by any significant amount, but because Newton says you have to add force to accelerate the mass of the vehicle to a higher velocity. And once at the higher speed. the power required remains higher at a higher speed because you are doing the same amount of work (force over distance) in a shorter period of time. P=Fv.
The fact remains that, as I said originally, to a first order approximation, the force you exert on the pushbar with your hand is essentially the same whether you're going 10 mph or 20 mph, as long as the speed is constant and the pitch of the hill is neglible. That's a fact I have empirically verified countless times. That's why the rolling resistance equation, as it applies to a particular wheel/tire setup, is usually stated in the form
You'll notice the conspicuous absence of a velocity term in the equation.