Thanks @dlang! I appreciate the overview - I’m not very familiar with motor loading, especially as they approach stalling.
So the voltage is regulated by the system power supply brick and limited by the PWM, and the current is (non-linearly) controlled by the motor load… so the net power is not measurable by any of the versions of Motor control systems (without modification) - all we can really know it’s relative output on the scale of 0V-to-Vmax. Is that correct?
I would say that sums it up pretty well. In addition, there are three motors sharing that power source and possibly browning it out during high demand.
we know we hit 100% duty cycle when cutting along the top edge (especially if
the the top beam is a bit low or the sled is a bit heavy)
yep, that’s right.
David Lang
The first time I read this, I was not thinking of it as non-rotating - What you’re saying is clear to me now- thanks for your persistence!
In other words, It doesn’t matter if the inputs are relative or different between machines, we’re simply finding out what PWM for a given machine produces 40lb of pull at speed=0.
I think that this is a reasonable way to identify the safety cutoff (as measured by Duty Cycle) for a given machine.
The other thing to figure out to make this work is the weight cut-off point. (the 40lb you use as an example) With the example that @dlang provided (with the sled at the top/centre) the acute angles mean that it would take more than 40lbs on each chain to lift the 25lb sled vertically. But this is probably calculable. Maybe we’d be aiming for something more like 100lb per motor.
Edit: (after some estimating, more like 60lb per motor at 14 degree chain angle)
Good to know. I had assumed that the dynamic chain tension while cutting would be close to the static numbers in the spreadsheet. That may actually still be true. But 100% duty cycle while moving generates much less chain tension than 100% duty cycle stalled. Thus revealing a major flaw in my solution. Oops, that wont work.
I’m having trouble with the ‘motor speed zero’ part. The motor speed is only zero if the power is off or the motor is stalled. What am I missing?
Our assumption is that, with a given load (say 100lbs), the motor would stay stalled up to a minimum pwm level, before it start moving.
the motors are rated at a max of 66 pounds/motor. that would be at 100% power
(full power, not PWM)
David Lang
While doing the chain measuring, the motor is stalled. If it isn’t stalled, the gears are being stripped. The small bit of movement in a spring scale can be neglected once it reaches steady state.
But it doesn’t matter, because apparently we need to operate on the right side of the motor curve for at least part of the time, which isn’t handled by this limiting strategy.
The max PWM setting is full power, on 100%
So I think that the only problem with PWM limiting to protect the gearbox & structure, as @slomobile suggested, is the power margin of the motors. If we limit their output to protect the machine, there will be modes of operation were the limiting will stop function where it would have worked before (although possibly as it approaches structural limits).
So this strategy should be reserved for when we upgrade power supplies, H-bridge shields, & gearboxen.
Not always. The max PWM setting means full power is available, not necessarily used. If the motor is still turning, it isn’t using max power yet.
At 100% PWM, same as connected straight to power supply, the motor will turn at some no load RPM and consume some amount of power. As the load increases, rpm will decrease and power will increase until it reaches stall, the point of o rpm and max power.
Ok, got it now.
I do a 1 hour burn in on the Blue Smoke Herder with no heat sinks installed. - FWW
The wide temp range in the chips seems to play very well.
Thank you
That is good to know. Since the figure is not a torque, I assume that rating is with the 10 tooth #25 sprocket? Any other qualifications to the rating?
Just to be clear, does that mean 66 pounds hanging on the chain will just stall the motor at 12v and no current limiting?
Or does it mean 66 pounds is the max that a motor will lift given Maslow driver/power supply?
Just trying to understand why it is different from the 50kg cm on the spec page.
but we don’t have any way of measuring the current, all we have it motor
position (from the encoder) and the power we are applying via PWM
so there’s no way to limit the power without making it so that the motor has
less power when it needs it for system operation.
David Lang
do a stress test of a couple motors pulling against each other at full power,
see if you can manage to damage them (I’ll get you a pair of the better motor
mounts to use)
That is good to know. Since the figure is not a torque, I assume that rating
is with the 10 tooth #25 sprocket? Any other qualifications to the rating?
no, the rating is 30 kg/cm of torque, but it so happens that the 10 tooth #25
socket is just over 1 cm, so the 30 kg (66 pounds) per cm translates into almost
66 pounds of force (close enough that we just use that value)
Just to be clear, does that mean 66 pounds hanging on the chain will just stall the motor at 12v and no current limiting?
Or does it mean 66 pounds is the max that a motor will lift given Maslow driver/power supply?Just trying to understand why it is different from the 50kg cm on the spec page.
If it’s really 50Kg/cm that it’s rated for, that would be 110 pounds/motor.
what spec page are you looking at?
but since we know we have trouble across the top of the machine, I have trouble
believing that it can actually apply that much force.
David Lang
The one copied into this thread. Another left motor gear failure, need a solution - #16 by Gero
same as
ET-WGM58A-dc gear motor manufacturer/Etonm Motor Co., Limited
Its hard to be sure which line refers to our motors, but mine are marked ET-WGM58A-E
E is the 5th letter, so I looked at the 5th line down. Same line that @Gero highlighted.
