The design for our bot calls for 6 Falcon 500 motors running at 4000 RPM each. If I’m reading the Falcon datasheet correctly, the nominal current draw at 4000 rpm is 100 A. Therefore, our total current draw will be 600 A. This means that with the standard 18 Ah battery, we’ll only get 1.8 minutes of operating time. Is this a correct analysis?

Everything goes through a 150A breaker on a competition legal robot. On top of that, each motor is capped to 20A, 30A, or for drive motors usually 40A. This will increase your operating time significantly. If you go over this limit, you will trip breakers and fuses.

Your analysis isn’t quite right. Do you mind if I ask what the 6 Falcons will be used for? Drivetrain?

yes drive train

OK so a Falcon 500, if you’re providing 12 V to it and it’s running at 4000 RPM, will output 0.5 Newton-meters of torque. In a normal situation (if you’re just driving across the field), that torque only has two places to go: overcoming friction in drivetrain gearboxes and accelerating the robot. Assuming you don’t have a ton of friction, the result is that the robot will start to go faster. But you would get less torque from the motors and you would also draw less current (moving to the right on the motor curves). This would continue until the torque output matches the gearbox friction.

If you don’t want the motors to go above 4000 RPM, you would need to provide less than 12 V (I don’t know exactly how much because it depends on friction). For example, if you only sent 8 V to the motor (sending arbitrary voltage is what motor controllers do), then the effective free speed, stall torque, and stall current would all be multiplied by 8/12. The effect: you’d be getting very little torque and drawing very little current at the same 4000 RPM. Your top speed would be lower if you sent less voltage, because torque output would match gearbox friction at a lower speed.

Human drivers naturally do this, since usually pushing less on the joysticks corresponds to sending less voltage, and results in the robot going not as fast. If you did velocity control, what that’s doing is adjusting the voltage to get the requested speed.

So in steady-state (when the robot isn’t accelerating), the current draw of the motors will correspond to whatever torque it takes to cancel out gearbox friction.

Does that make sense?

I just came back to this to try to explain what I said better; this is kind of what I was trying to get at. Great explanation. Much better than I could have ever worded it at this point.

I’m gonna say this again, but a 6 falcon drivetrain is 200% overkill. You will not need that much power and you are traction limited anyways so you won’t be able to use all that power. It’s better to stick with a 4 falcon drivetrain and use the other PDP slots for mechanisms and such. If you want the top tier analogy, most modern 254 robots only use 4 motors on the drivetrain.

And if you’re plan is to be a defense bot, just load your bot up with weights and that’ll increase your pushing force.

@HowDoIRobot , you’re reading the datasheet correctly, but not understanding it.

The graph you’re referring to has one poorly labeled feature. That all values are at 100% duty cycle, for 12 V.

If you’re really running your falcons at 4000 RPM and at 100% duty cycle, then yes you would draw that many amps. Except you very likely aren’t going to use 100% duty cycle and only run at 4000 rpm.

The curves for torque and current scale linearly with duty cycle. So if your user input is 10% and at 4000 RPM, you would draw only 10 amps each. In your case, it’s very seldom for it to actually be at 100% duty and not a higher RPM where the current is lower.

If you pull out the data sheet for your battery (I’ll use MK as an example) you’ll see that the maximum output current is 270 amperes. Exceed that at your own risk.

That’s also at peak for approximately 5 seconds (which will probably trip your main breaker). Continuous operation is about 80 amps according to this datasheet.

So here’s a little devil’s advocate moment. Although I would generally agree 6 Falcons on a drivetrain is a little much, there are some configurations/gearboxes that allow for the swap between 6 and 4 motors for driving. Whether or not this is what is going on, we’ll have to see come competition time, but having competed against 230 Gaelhawks for years, I’m sure they have some pretty good reasons for this design!

Not in certain cases. A fast full-field cycle requires a top speed somewhere between 15 & 18 ft/s. For a robot to make that speed useful while also having enough acceleration power to efficiently do shorter sprints, it needs either a shifting transmission or more power. Ergo, the solution for playing every strategy “as well as possible” in 2020 is either a 6-motor BLDC drive train, or a 4-motor drive train that shifts (BLDC’s help, but not quite as advantageous).

Full disclosure: ILITE is doing a 4-NEO drive because ~~we’re playing a specific strategy~~ we’re out of PDP slots anyways .

Math should inform the situation. [Pardon me if I’m physicsplaining… Someone might find the below useful].

If you know the CoF of the “best” tires/wheels and the maximum weight of the heaviest possible robot, you can find the maximum traction possible under that arrangement, which then gives you the maximum theoretical acceleration of your robot. (F_traction = mu * weight = m * a)

Look up or calculate the maximum speed of your robot. v_max = rpm_motor * 2 * pi/60 * wheel_radius/gear_ratio.

Use kinematics to figure out how long it will take to reach the maximum theoretical speed of your robot. (v_max = a * t)

Also, calculate the kinetic energy of the robot at that speed. (K = 1/2 * m * v_max^2)

Calculate the power. (P = K/t)

Divide this by your estimated drivetrain efficiency (in decimal form–this is probably around .90 per stage of gear or chain, so a toughbox mini has two gear stages and then a belt system, so e = .90^3 = .73, or 73%). Note that because you’re dividing by a number less than one, you’re going to get a larger number than you started with – this reflects the fact that less efficient systems need more power in to get the same power out.

If the input power of your motors running at the point on the torque-power curve where you want them running (call it 40A for maximum oomph, which means 400 W per Falcon in this case) is bigger than this number, then you have more motor than you can use on the field, unless you’re using the six-motor arrangement for, say, some ridiculous PTO to lift three robots 18" off the ground in a half of a second or something.

Hmmm… Doing my own math here for a kitbot drivetrain and wheels with a CoF of 1.2 (which is a bit of a guess), I’m getting an input power of 2600 W, a little bit higher than the output of four falcons at 40A, if you want to yeet yourself from 0 to 15.5 fps in 0.4 seconds (that is, from stopped to maximum speed within the length of one typical robot).

I think the flaw here is that the 40A number of 400W is what the Falcons would come down to, from their peak power of 783 W, so the average power would be closer to 600W upon yeetceleration, making two Falcons just about perfect (within 10% of the theoretical maximum), and six Falcons massive, needless overkill (35% above the theoretical maximum useful power).

There’s a couple things here.

First - I don’t know why someone picked the number 6 for shooter motors, but I highly doubt that was based on any kind of analysis whatsoever. Start with one and add one motor at a time until your returns diminish.

Second, you’re reading the spec sheet somewhat incorrectly. What it is saying is, when the Falcon is under enough load such that at 100% throttle it is spinning at 4000 RPM, it will draw 100A of current. When the motor is not under full load, in order to get it to spin less than its free speed, you will limit the throttle - this does not raise the current level of the motor. Otherwise, wouldn’t every motor be drawing its stall current when turned off?

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