Quote:
Originally Posted by Shortsguy1
That top thick black line is the max output on the engine as a function of RPM. The road load is the necessary output of the engine on a flat road to maintain speed. When those intersect at the far right side of the graph, you have the top speed of the car. Using your logic, with a headwind lowering the road load curve, the vehicle would have a higher top speed (as the intersection would shift right to higher rpm and vehicle speed).
Pretty sure that a headwind would raise the road load.
I think you may be getting "fuel use", and "fuel use per power produced" mixed up.
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Hmm, let me think.... If we take two situations represented by say a steady state change in the road load, one increasing and one decreasing it might look something like this.
So at point 1 and 2 (and the original point) respectively we would have
1- The car needs 15 kW and uses 330 g/kWh so that's 4.9 kg fuel or 6.2 l/hr or about 1.6 gallon per hour or 37 mpg
x- The car needs 20 kW and uses 300 g/kWh so that's 6.0 kg fuel or 7.5 l/hr or about 2.0 gallons per hour or 30 mpg
2- The car needs 27 kW and uses 275 g/kWh so that's 7.4 kg fuel or 9.2 l/hr or about 2.4 gal/hr or about 25 mpg
So yes, I have it backwards. I was confusing fuel use with fuel use per power produced. I stand corrected
03-19-2013, 09:53 PM
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