New lower price for Axon II ($78) and Axon Mote ($58).

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Force = Torque/Radius = 2.5/6 kg(f) <- Kilograms of forceThen (and this is what I'm not entirely sure about) I found a calculator which said that 1kg(f) = 1*g N (g = 9.81m/(s^2))So: Force = (2.5/6)*g N

Now (and this is also something I'm not sure about) I said that the force acting at the center of the wheel will act against the mass of the object causing the acceleration, such that:a=f/ma=(2.5/6)*(g/m)assuming m=2.5kg:a=g/6 m/(s^2)a=1.635 m/(s^2)

v = r*c (r = rotations per second, c = circumference of wheels)w = 250/60 = 4.16 /sc = pi*.12mv = 4.16*.12*piv = 1.56m/s

So the final values from the calc's are that I'd have:a=1.635m/(s^2)v=1.56m/s

Also, the final consideration I have is the following: In the article I linked to above (with the torque/rotation curve) they said that the optimal efficiency is at 250rpm and 2.5kg-m torque. But how does one "get" those figures? I assume it's with appropriately setting the l

factors I need to think about when I say that I want to operate in the optimal operation point of the motor (250rpm 2.5kg-cm)?

So double the motors = double the torque which will double the acceleration.

From what I understand now from you response nottooolly, ideally (not taking friction into consideration) the motor will accelerate the robot until it reaches a point where it can no longer be accelerated anymore (reaching top RPM). As it gets towards this point, the output power will decrease. From what I understand, the output torque of the motor when the robot is at a standstill (0 RPM) would be about the same as the stall torque, and would then decrease linearly until the max RPM is reached. Is this the correct assumption?

max RPM is dependent on the friction acting on the robot at that speed.Assuming that the total friction isn't a HUGE amount, it's not likely that the max RPM of the motor on the robot will be much less than the rated

(from standstill) in a linearly decreasing amount until the max speed is if the stall and no-load current are both given, and the frictional forces etc. are all known?

I'm guessing that most of my assumptions are incorrect here, but if they're wrong could you please tell me how I should look at this whole thing?

-Slamming into reverse. Not sure how motors behave in this condition, hopefully still linear!

-Acceleration and top speed for turning on the spot. Probably no big deal unless your robot is much longer than the wheel spacing.

Nope, not linear. But you never want to slam into reverse. It wreaks massive havoc not only in the electrical system (giant voltage/current spikes), but also mechanically (gear damage) as well.

Did you write the RMF calculator? Good stuff! But why the bizarre RMF instead of just calling it power and using conventional units?

That said, I added 'watts' to the tutorial just as an FYI.

I just notice that now! But the units aren't actually watts, that would be Nm * radians/second

Yea another thing I noticed is if you try to use the calculator for a constant speed on a flat surface you get RMF = 0 ! I thought the efficiency value would allow for power lost between the motor output and the movement of the robot, but it doesn't seem to.

You mean acceleration of 0? Of course, if your robot never accelerates, it'll go nowhere Efficiency is what'll factor in your robot adding extra output to keep pace under friction.

Sure, it can move at constant speed! This seems to be the most tricky part in selecting a motor because depending on the friction, the power requirement can be anywhere between 0 and loads.

The thing with efficiency/friction is that its incredibly hard to predict. Without experimental data (ie the robot was already built), rule of thumbs are generally the best way to go about it.