Calculate uphill efficiency on different gears

Question

I was interested to find whether it would be more efficient to climb a hill on a lower gear, so I did some calculations.

I have the following data of a custom vehicle:

Gear	Drive force (N)	Max speed (km/h)
1	591	4.87
2	284	10.12

I have chosen to calculate for a slope of 8 degrees, so opposing force will be equal to

$$ F_s = m * g * sin(\alpha) $$

Where m is the mass of the vehicle (200kg) and g is gravity, so

$$ F_s = 200 * 9.81 * sin(8) = 273 N $$

I'm ignoring other resistive forces to keep it simple as we can consider them constants at such low speeds.

The drive force is provided from an electric motor with rated power of 1kW, with 80% efficiency, so 800W of usable mechanical power.

I'm calculating the time needed for climbing the distance of 1km with a constant 8 degree slope, assuming we're already at the top speed for the respective gear:

$$ t = 60 / V $$

Gear	Time (minutes)
1	12.32
2	5.91

Here comes the part I'm not very certain about. If already at maximum speed for the respective gear, the electric motor should only be loaded with the resistive force, thus to calculate the amount of power that will be drawn for both gears I came up with this formula:

$$ P = P_m * F_s / F_d $$

Where $P_m$ is the motor's power, $F_s$ is the resistive force from the slope and $F_d$ is the drive force.

Gear	Power (W)
1	369
2	769

And with the resulting powers I also calculate the total amount of energy that will be consumed like this:

$$ E = P * t $$

Gear	Energy (Wmin)
1	4544
2	4546

I'm not entirely sure about the unit on the energy there but it's not really relevant as I only need to compare the two values and it is identical for both.

Looking at the results it turns out that in both cases the same amount of energy will be spent, which seems a little counter intuitive to me, as if we imagine the same situation on a bike, we'll clearly be more exhausted climbing the hill on a higher gear.

I'm suspicious the method I came up with for calculating the amount of power drawn from the motor on the respective gears might be incorrect.

Answer 1

You found that the total energy needed to climb the hill is the same despite using two different gear ratios. This is in fact correct because of conservation of energy.

Look at is from another point of view. To get to the top of the hill energy is needed because we are working against gravity. The amount of gravitational potential energy a object contains is given by: $$E=mgh$$ With $h$ being the elevation with respect to some datum. Notice that power is not involved and neither is any sort of applied force. For a car climbing a hill $m$, $g$, and $h$ will not change whether it climbs fast or slow. Increasing power will only mean you get to the top of the hill faster. The energy is still the same because, for example, doubling the power will halve the time meaning energy will be unchanged. Gears don't change energy, they just trade between force and velocity. This equation might help: $$F*V=P=E/t$$ In you analogy with the bike, how exhausted you would be is actually not totally correct. While it is definitely easier to climb on a lower gear because of the lower force, you will need to pedal much more to reach the top. If you climb a steep hill on a bike with even a very low gear you will find yourself quite tired by the time you reach the top, even though the pedals are still quite easy to turn.

Just to be sure, imagine that it did take less energy to climb a distance in a lower gear. Well if we climbed in a low gear but then descended in a high gear driving a generator, we would expect that we generated more energy going down than we used going up. This is impossible though because energy must be conserved.

All this, of course, only applies in an idealized situation with no friction, wind resistance, and an assumption that our car is equally efficient at all speeds. In real life the actual motor efficiency curve would determine the most efficient speed with wind resistance skewing the answer slightly slower.