L = integral(dL/dx with respect to x from 0 to X)

it follows that

FE = 1/X*integral(100*dL/dx with respect to x from 0 to X)

but this is just the distance-weighted average value of 100*dL/dx, and 100*dL/dx is the instantaneous FE in L/100 km. Hence my statement that to optimize FE one needs to minimize this average. Put another way, it's the distance travelled at a given instantaneous FE that is important for the trip's overall FE, and not directly the time spent at each instantaneous FE. (Of course, the time spent will depend on the vehicle's speed and hence indirectly on its instantaneous FE, but won't in and of itself determine the overall trip FE.) One needs to drive so as to minimize the distance travelled at high instantaneous L/100 km, and maximize the distance travelled at low instantaneous L/100 km. [One can make a corresponding statement in terms of distance in miles and instantaneous FE in gallons/mile, but since no gauge I know of reads instantaneous FE in gallons per mile, you're stuck.]

Droid13 — I don't think you can truly conclude that time spent at a given inatantaneous FE can be used instead of distance travelled, since the car's fuel efficiency is not the same at all accelerations. Were the efficiency the same at all accelerations, then your statement would be correct. Suppose that we start from rest at a constant acceleration 'a' and accelerate up to a given speed 'v' in a time 't,' covering a distance 'x.' Then these quantities are related by the formulas

x = v^2 / (2*a)

and

t = v/a

so that the distance x travelled and the time t taken both depend inversely on the assumed constant acceleration a. Changing a changes both x and t in the same proportion. Eliminating 'a' shows this:

x = v*t/2

Were the car's instantaneous FE the same independent of the acceleration 'a' chosen, then it would be true that one would get exactly the same trip FE no matter how you drive the route, since the work done would be the same in all cases (ignoring varying air resistance), and hence the fuel used would then be the same too.

Stan

Originally Posted by SPL turk — Let me see if I can be clearer. The trip FE in L/100 km is 100*(liters L used)/(kilometers X travelled). Since L = integral(dL/dx with respect to x from 0 to X) it follows that FE = 1/X*integral(100*dL/dx with respect to x from 0 to X)

Originally Posted by SPL ymiheere — The TCH's air conditioning is electric, powered by the high-voltage NiMH battery through a 3-phase ac compressor motor. Only if the NiMH battery drops too low will the ICE need to come on. Stan

Originally Posted by n8kwx Short 5 minute trips? Live in Fairbanks, AK? Either can effect city mileage significantly. (Unless you are jackrabbiting and slamming on the brakes every block). Trip length and temperature are much bigger factors.