I started drawing up what the robot lawn mower would look like if we didn’t care about separating the compass from the electric motors. Removing this constraint allows several design efficiencies, some of which I was not expecting.
I decided to use two battery bays on version 5 because I had to mount a mast smack dab in the middle of the chassis. I didn’t want to mount it on a removable lid because it would be cumbersome to remove to get access to the batteries. Instead, I put hinged doors on both bays.
It looks neat in the picture above, but what you don’t see is all the wires running through my chassis tubes between bays to connect the batteries and all the signal wires run through the mast weldment up to the control enclosure. It started getting ridiculous drawing all of that up.
A separate design constraint I’ve been trying to achieve is to keep the wheel base of the robot to a minimum for handling reasons. Unfortunately, the mower deck design I settled on has a motor in the middle of the deck that is a pain to locate such that it doesn’t interfere with the battery bays.
Because I was splitting the battery bays anyway, I positioned the mower deck the way you see in the version 5 picture above. One downside to doing this is that the mower deck is pointing backward from what you see on virtually every riding mower.
Combining the battery bays let me rotate the mower deck 180º. In the back of my mind I have been worrying the backward orientation of the mower deck might cause performance issues. Now we won’t have to find out.
Additionally, both power and control enclosures can be mounted directly to the battery bay, which will drastically shorten the wire runs I’ll have to make. I’m actually excited to start drawing wires again. Things aren’t so claustrophobic anymore.
And on top of all the benefits above, the chassis weldment went from having 28 total parts to 15. Not too bad!
Batteries are often advertised with a nominal voltage and nominal charge. It’s tempting to take those numbers at face value and assume that regardless of operating circumstances, a battery will sit at its nominal voltage and run until its nominal charge has been depleted. Reality is more complicated than that.
Christopher Milner reminded me of this in a brief conversation this week. He is using 13 3.2V, 100Ah LiFePo4 batteries in series to power the deck motor on his mowing rig. Stringing them together in series gets you close to 48V in their charged state. He reports getting about 3 hours of runtime while powering a motor that consumes 1300W.
This surprises me because my own calculations suggested four 12V, 35Ah lead acid batteries should be sufficient. Christopher obviously knows what he is doing and I trust his experimental results much more than I do my own back of the envelope calculations. The discrepancy means I need to reevaluate my numbers.
When making my calculations I did not take battery discharge rates into account. This is a big mistake, as I’ll show below. But first, for my own educational benefit, I’d like to introduce you to the C-rate, a way to quantify how fast you pull current out of a battery.
What is a battery’s C-rate? Our good friends at Wikipedia define it as:
The C-rate is defined as the charge or discharge current divided by the battery’s capacity to store an electrical charge.
In layman’s terms, the C-rate is just the ratio of an arbitrary discharge rate to the battery’s charge capacity. It’s a simple way to describe how much current you’re demanding from a battery relative to the battery’s total stored charge. If you’ve got a battery rated for 10Ah and you discharge it at 5A, the C-rate would be:
It’s kind of a janky way of defining things in my opinion. A C is really an inverse hour, or stated differently, the unit of measure for a C-rate is h−1. Also, don’t confuse the C-rate with Coulombs, the unit of measure for charge. Clear as mud?
The reason I write all of this is so we’re all on the same page about what a C-rate is. If I’m wrong, please comment below, because I am going to proceed with this understanding of the C-rate going forward.
The C-rate is a useful measure because the total amount of charge you can extract from a battery depends on how fast you take it out. If you try to pull 100A out of a 10Ah battery, you’re not going to get nearly as much charge out of the battery as you would if you discharged it at 1A.
Additionally, when you demand a large amount of current from a battery, your voltage starts dropping fast. This is important because electrical power, which is ultimately what we’re after, is voltage times current. If your voltage drops you get less power, even if you’re withdrawing the same amount of current from the battery.
Your batteries are going to be much happier and live much longer if you stick with a reasonable C-rate. Having said all of that, how do the four 12V, 35Ah batteries I selected stack up against the expected current draw they’ll experience?
Our SLA Batteries Reevaluated
Previously I estimated the total current consumption of our robot at 197A. I think this is a conservative number, but I’m going to roll with it anyway. The number reflects current consumed by the three deck motors, two drive motors, and various electronics at their worst case scenarios.
I’m using two sets of two 12V, 35Ah batteries wired in parallel, then in series to get an equivalent battery that’s 24V, 70Ah. The discharge rate for one battery in this configuration is half the total current consumption because we have two sets of two batteries wired in parallel. This means that one battery is discharged at a rate of 98.5A.
The datasheet for one of these batteries shows the following chart, with various C-rates:
At a discharge rate of 98.5A, our C-rate is 2.81C. Looking at the chart above, that would mean our batteries will last for ~8 minutes. Yikes. And realistically, the 2.81C curve may extend to about 8 minutes, but the voltage drops off so fast after about 4 minutes that you’ll probably start noticing performance problems very quickly.
Also interesting to note from the chart above is that the one hour discharge rate is in fact not 35A as you’d expect with a 35Ah battery. It’s actually 0.628 times 35Ah, or 21.98Ah. To truly extract the 35Ah charge from the battery, you can only discharge it at a C-rate of 0.05C, or 1.75A.
Having written all of this, I wonder to myself why manufacturers don’t just list curves with specific discharge rates in Amps. The chart above would be completely unambiguous if they just showed a curve for 105A, 70A, 35A, 21.98A, etc. The conversion is tedious and honestly, if you don’t truly understand C-rates you leave with the impression that these SLA batteries are much more capable than they truly are.
Looks like we will need to use some lithium batteries after all. Thanks for saving me $400 on some batteries that wouldn’t have worked Christopher!
What would a composite lithium ion battery look like if we were to use one for the mower?
We need a nominal voltage of 24V, so if you string 7 18650 batteries with a nominal voltage of 3.6V you’d end up with 25.2V, which should work fine.
String 15 sets of those 7 rows of batteries connected in series together and you get a battery with 45Ah of charge. Not too shabby. This would be a 7s15p battery in lithium ion parlance, I think.
Two of those 7s15p batteries would fit in each of our two battery bays. You’d need a total of four to achieve the numbers shown above. They’d fit in our robot something like this:
The weight savings here are significant: 70lb. This is to be expected, but I’m surprised it’s this high. I don’t have a truck to haul this robot around in, so the lighter I can make it, the easier it will be to take it out for field testing. The reduced weight should also decrease the power consumption from the robot’s drive motors, making it more efficient, and possibly more agile.
The cost is still going to be over $1,000 though. Talking with some suppliers, I think the 18650 cells cost a little less than what Amazon will quote you, probably closer to $4.50 per cell. But you still need some fancy charging equipment, and there will be labor and material cost for building up the battery and coming up with a way to secure them in the battery bay.
I may start out using SLA batteries because the monetary risk is pretty low, about $300. They may perform better than I’m expecting. If they don’t, the lithium ion batteries appear to be a good plan B if we need more run time.