Ze wrote:yeah I get that the external power required to overcome grade will be proportional to grade, but didn't know if engine efficiency is also affected by grade, which would make the internal energy (fuel) relationship to grade nonlinear.
On a grade, the power-to-fuel efficiency would actually be higher, assuming speed and gear ratio (thus engine speed) remains constant. This is because of the relationship between fuel and power (roughly y=mx+b) that I mentioned before, where y is the fuel usage and x is the engine's load, at a given rpm. It's not proportional (b>0), but it is roughly linear (x^1) for a typical engine. So at a given engine speed, at higher power output, the engine is more efficient, basically because it takes a certain amount of power to just spin the engine, under no load. So the higher the load, the higher the proportion of fuel energy is going to the load.
So compare level ground to a grade, given the same vehicle speed on both. If the grade is such that you can remain in the same gear, the comparison can be made. Otherwise, it's more complicated because the engine speed will be different between the two cases.
On level ground, the vehicle is using fuel at a certain rate. Part of that fuel energy goes to power at the drive wheels, and part goes to losses within the drivetrain. On an uphill grade, at the same vehicle speed and same engine speed, the drivetrain losses don't change much, but there is a lot more power going to the wheels, specifically the amount of power required to increase the car's potential energy at a given rate.
On an uphill grade, the total power-to-fuel efficiency will be higher. The vehicle will obviously be using more fuel on the grade, but the efficiency is higher because a smaller proportion is going to drivetrain losses. And by "efficiency," I don't mean fuel per distance; I mean fuel per work output, which includes the increase of potential energy of the car as it goes up the grade.