That's Mow Like It!

[Title shamelessly stolen from The Simpsons.]

Yesterday, I mowed the lawn with our fancy electric mower, which has a nice speed control for its self-propelling feature. It got me wondering though: What's the most efficient speed to mow the lawn? The blade is constantly running, so it seems like finishing the lawn as quickly as possible would be best, but then the drive is using more power. Is it better to use a lower drive speed to save energy?

I couldn't find many details on the battery/motor specifications, so we'll have to make some assumptions: I'm not sure how the variable drive speed/constant blade speed would work with a single motor, so I'm betting there's two. That means we can split up the power use like this:

The manual gives a couple different cases for battery life, but since it applies to both the self-propelled and meatbag-propelled models, I'm guessing it's assuming the drive is off, so that only tells us about the blade power. For the drive power, I decided the closest analogy would be an electric golf cart. We can look at its power use, then scale it by the ratio of weights and speeds. This model uses 2.6 kW to haul 270 kg at 15 mph. The mower weighs 28.4 kg with a top speed of 3.1 mph, so a rough approximation for power would be 57 Watts.

Now there's the question of efficiency: When we put power into an electric motor, how well does it put that power back out? Luckily, I found a great DoE paper on how the efficiency changes with percent of maximum power. I grabbed a couple points from Figure 1 and did a rough curve fit:

where e is the output efficiency and L is the fractional load.

For the "light load" case, the manual says the battery, which holds 7.5 Amp-hours of charge and runs at 56 Volts, will last for 60 minutes. After a bit of computation, that means the blade is using 420 Watts. Now we need to know how long it takes to mow the lawn, depending on the speed. Using a tool like this, we can find the total area of the property, 1065 m^2, and the total area of the house, 246 m^2, to find the approximate area of the lawn, 819 m^2. The mower is 0.53 m wide, so the minimum total distance it would have to travel is 1545 m.

We now have a long chain of relationships we can put together. The total energy used is

where P is the power, and t is the total time. The time it takes to mow is

where d is the distance, and v the velocity, which in turn is

The power, from earlier, is

The equation we get from putting all these together is a bit involved, so I just made a plot instead:

The battery's total capacity is about 1,500 kJ, so even at the slowest speed we're well within a single charge. I was hoping using the motor efficiency curve would give a more interesting result than "go as fast as you can," but since the maximum load is most efficient, that's the result. Maybe my next step will be figuring out how to balance the exertion of following the mower with the time out in the hot sun!