On a hill terrain scenario where a vehicle is going downhill from the top of one hill and then continue without stopping uphill to the top of another hill of equal height, the following hypermiling technique can be used to save fuel/energy used regardless of what type of energy the vehicle is using.
From the top of the hill apply energy gently and make no attempt to increase velocity until the vehicle is moving downward the top of the hill. With the vehicle is pointed downhill, apply additional energy to increase the velocity of the vehicle so that when it is at the bottom of the two hills its velocity is 1.5x the speed that desired at the middle of the uphill climb approaching. At the bottom of hill pull back on the energy being applies so that vehicle gradually decelerates to the desired velocity as it passes the middle of the uphill climb. Allow the the vehicle to further decelerate to about 3/4 to the speed when it was in the middle of the uphill climb so that when the vehicle is on the top of the next hill its velocity is 2/3 of when it was in the middle of the uphill climb.
Why does is this method more energy efficient? ================================== One word - gravity.
At any given velocity V, it takes less power to propell a vehicle downhill than it does uphill. If a vehicle is on cruise control and going a fixed velocity on a hilly terrain where all hills are of equal height then the uphill energy savings will approximately equals the downhill energy extra cost. Hence, if the downhill energy cost requirement, E1, is about 20% less ( 0.80*E ) THEN the downhill energy cost requirement, E2, is about 20% more (1.20*E) where E is the energy requirement to move the vehicle at a velocity V if the vehicle was moving on a flat terrain. Hence the total estimated energy requirement, ET, for going downhill and then uphill using cruise control look like this:
ET = E1 + E2 = .80 * E(v) + 1.20 E(v) = 2 * E(v)
We also know that it takes more energy to move a vehicle the higher the velocity. Suppose that at the velocity range we are using the vehicle power plant has a linear power/velocity performance (air resistance is considered neligible if V < 50 mph) then if V is increased in by 20% then E is increase by
20% and that if V is decreased by 33% then E is decreased by 33%. Then E(v') = 1.20 *E(v) and E(v'")=0.66*E(v)ET(hypermiling hill) = .80* E(v') +1.20(v") = .80* (1.20*E(v) + 1.20(.66*E(v)) = .96 *E(v) + .79* E(v) = 1.75 E(v)
The would result in a hypothetical results of a 12% savings in energy over using a constant velocity regardless of what type of power the vehicle is using. Of course if there is a stop sign or a red light at the bottom of the hill then this technique does not work (9_9)
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