Power Consumption Revisited

I think I may have incorrectly estimated my power needs for the mower. A key assumption I’ve been making is that the motor will generally need to be capable of generating ~5ft-lbf of torque during maximum operation. I’m not sure this is really true though.

Do We Really Need 5ft-lbf of Torque?

The 5ft-lbf of torque figure comes from taking a typical gasoline push mower engine and looking at the gross torque output of the engine. But one variable I forgot to consider is that the torque curves I looked at are associated with an engine typically used with an 18in to 21in blade. Our mower uses a 12in blade.

Intuitively, the torque we need to cut through grass is going to be positively correlated to the amount of grass we’re trying to cut at once. So a smaller cutting blade should require less torque than a larger blade. There’s less grass for the blade to run into, sapping momentum from the rotating blade.

I have no idea what the relationship between blade length to required torque looks like. I am going to assume it is linear for simplicity, but I have no clue if this is a good assumption. The torque you need is also going to be related to the quantity of grass clippings circulating around under the deck impacting the blade. Good luck modeling that.

Given the smaller blade size, let’s say you only need 60% of that 5ft-lbf torque value, so 3ft-lbf or 2.2N-m of torque. That’s the ratio between a 20in blade and a 12in blade.

How Much Current Does the Motor Draw at 3ft-lbf of Torque?

The performance curves for the E30-400 motor say that the motor consumes 56A of current at 3ft-lbf of torque. I think this is a more accurate number for current draw from the motor.

E30-400_Chart
Performance curves for the AmpFlow E30-400 DC motor.

How Much Power Does the Motor Consume at 3ft-lbf of Torque?

Another mistake I made was pulling power numbers off this chart thinking they were power supplied to the motor, not shaft power output by the motor.

This is an important distinction, because no motor is 100% efficient. The input power should be the power supply voltage of 24V times the current consumed at a given point on the curves. At 3ft-lbf of torque, it’s (56A)(24V) = 1344W.

This jives with the chart above, because shaft output power at 3ft-lbf or 2.2N-m of torque is about 1040W. That would imply an efficiency of (1040W)/(1344W) = 77%. The chart says the motor is about 75% efficient at this torque, pretty close to this estimation.

So under maximum operating conditions, each motor should consume 56A of current and 1344W of power. The three motors collectively consume 168A of current and 4062W of power.

Is That a Good or a Bad Number?

The 168A number is acceptable because it is right at the limit of what the Mauch current sensor can handle. It’s rated for 200A of current and that leaves us 32A of current for drive motors and miscellaneous control electronics, which should be enough.

So assuming our three mower deck motors consume 56A, our two drive motors consume 12A and our control electronics consume 5A of current, you could have a maximum of 197A drawn from the batteries. Very little margin, but I think it should be okay because…

Maximum Versus Typical

One additional thing I’d like to mention is that I think these are maximum power consumption numbers. Previously I referred to them as typical power consumption numbers.

Do you need 3ft-lbf of torque while mowing the entire time? I doubt it. The calculations above prove that if our electric motors need to operate at 3ft-lbf of torque, they should be able to do it. Operating at 3ft-lbf of torque drops the rotation speed down to 4500RPM which results in a blade tip speed of 14100ft/s, which is a little lower than I’d like but should work.

Run Time Recalculated

Turns out I also miscalculated how battery charge adds when batteries are connected in parallel versus series. In series, battery voltage adds. In parallel, charge (your amp hours) add. Previously I assumed your total charge is the sum of each individual battery charge.

Since we have two sets of batteries connected in series, and then in parallel, our equivalent battery is 24V, 70Ah. This makes sense because I think the Ryobi lawn mower is advertised at 24V, 70Ah too. It’s the same battery set up, apparently.

If we were to run all three deck mowers with a load of 3ft-lbf torque on them, it would take (70Ah)/(197A) = 21 minutes to completely drain our batteries (again, assuming that’s even possible to do, in reality it isn’t).

At half this torque value, total current consumption would be 28A for each deck motor, resulting in 113A total. That results in (70Ah)/(113A) = 37 minutes of run time. The E30-400 motor consumes 29A of current at peak efficiency, so I’m hoping that I’ve sized these motors for the sweet spot of their performance.

If you were to bump up the battery size used on the mower to four 50Ah batteries, run time would be (100Ah)/(113A) = 53 minutes. Doing this would add 34lb to the mower, which would show up in the current consumed by the drive motors.

Thoughts

Even though I have more confidence in these numbers being correct, they’re still disappointing. I would like to shoot for a minimum 2 hours of run time. The only two ways I can think of to get there:

  1. More efficient motors and electronics.
  2. Larger batteries.

Using BLDC motors would increase our efficiency, but they cost 4 times the brushed DC motors I intend to use. Reduced run time is an acceptable trade off to save $700 in my opinion.

Larger SLA batteries start getting pretty ridiculous beyond the four 35Ah’s I’m using currently. The battery bay has to grow to accommodate the larger batteries, and that pushes the wheels out, increasing the wheel base and negatively affecting vehicle performance.

Additionally, the added weight makes me wonder if the 0.125in sheet metal battery bays are sufficient to support the weight of the batteries. Two 50Ah SLA batteries weigh 64lb. I’d probably want to reinforce it just to make sure.

We could switch to some Lithium Ion batteries, but here the cost is at least as bad as switching to BLDC motors.

Lithium Lead Acid Comparison
A brief cost and performance comparison of a composite lithium ion battery versus our current 12V, 35Ah SLA battery.

If you were to make a composite battery out of 18650 cells equivalent to the four 35Ah SLAs I’m using currently, it would cost just shy of $1,000 in 18650 cells alone. And that doesn’t even include labor to build the battery and a fancy charging system to go with it.

I found some guys that make custom 18650 batteries, and maybe they can do it for cheaper. I’m starting to understand why Tesla’s use lithium ion technology. If you need a boat load of power and have any kind of space or weight constraint, you kind of have to. Unfortunately, I drive a 2003 Honda Accord, not a Tesla Model X and so the mower project can’t afford some legit lithium ions.

I may have to get used to about 30 minutes of run time.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s