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)|
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:
$$ t = 60 / V $$
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.
And with the resulting powers I also calculate the total amount of energy that will be consumed like this:
$$ E = P * t $$
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.