So I characterized an elevator single motor but two chains using sysId.
I know that kG / kA should give 9.8 m/s^2 but for some reason I’m getting double that value.
Would using a single motor but with two chains cause the kA value to halve and the therefore kG/kA to double.
If so would this be equivalent to running two motors and one chain?
Can you share diagnostic plots?
I can do you better and send the sysID file
ElevatorNEO.json (341.5 KB)
Now a note on this. I collected the data in straight up motor rotations. No gearing or drumRadius was used in sysId. (I was still learning about its features at the time).
Though not factored into the sysid setup the gearing for the motor was 16 : 1. Each sprocket had a diameter of 1.76 inches
And the diagnostics image
Your data look fine, and you’re probably seeing that your system loses about half of its available torque to frictional losses in the transmission.
Try using ReCalc to obtain some theoretical gains to compare to.
Thank you. That seems incredibly weird that it is almost a factor of 2. Since I did it in rotations and attempted to see that kG/kA equals 9.8 I kept thinking I wasn’t converting to meters correctly or that I had the wrong sprocket diameter. I’ve checked that by the way. I’ll look into ReCalc.
So could one argue that g = kG / (e * kA) where e is efficiency factoring in frictional loses?
It is probably actually a torque efficiency of 1/sqrt(2) - the gearing loss gets double-counted in the ratio you calculated.
Wouldn’t kG and kA both be affected the same by torque efficiency losses?
I also tried recreating your conversion of listed kA of .001991 V/(motor rotation/s^2) to V/(m/s^2) and got kA = .001991 / (1.76*.0254*pi/16) = .2268 V/(m/s^2), which gives kG/kA = 1.15, which is far off from 2x 9.81. Could you elaborate?
Your right. I don’t know where I got that value. They were all back of the envelopes and I don’t have the envelopes. They were also done a week ago. Seeing it again and with someones else’s input makes me wonder what I was thinking last week.
Anyways whats the verdict. Is KG and kA both affected. or just kA
Could you elaborate on that?
Actually now that I think some more about it, wouldn’t kG/kA be considered effective gravity. The gravity of the system given external frictional forces. This would make sense (barring whatever bad math I was doing last week) as to why it is 1.15 m/s^2
When torque efficiency of the transmission decreases, kG gets smaller but kA gets bigger. So, their ratio shrinks by the square of the efficiency. Remember that the motor and the gravity loading are on opposite ends of the transmission.
If Rafi’s calculation is correct, we have that the efficiency is about sqrt(1.15/9.8), or approximately 34%. That’s a very lossy transmission.
Thank you for the clarification. Am I right about the effective gravity bit?
That should be correct, disregarding kS.
of course 
Thank you.
I agree that the static holding voltage decreases as the efficiency decreases, because in that case the driving force that has to overcome friction is the back-driving force due to gravity.
That said, while in motion/accelerating upwards wouldn’t the voltage required to overcome gravity be increased by reduced efficiency? I’m thinking of it as the net force in the cable is (accel + g)*m, but I don’t see how the motor would “know” to distinguish between the accel and the g terms. With this interpretation, kG = C*m*g/e, where C is some constant involving kT, gearing, drum diameter, etc and e is the efficiency, while kA = C*m/e.
Regardless, the 34% efficiency seems extraordinarily low, so unless this elevator is very mechanically lossy due to overtightened chain/cables, tight bearings, or steep angle (which would also reduce the 9.81 gravity figure), it’s possible that the kA estimate is just not that good or that the gearing or sprocket diameter are incorrect.
I tried manually estimating kA and just from eyeballing it looks like it could be as high as .0036 in native units or .411 in SI units, yielding an effective gravity of .639 m/s^2 (which is even worse than before).
I’m not suggesting that this is worth doing, but I would be interested in seeing if repeat trials or alternate procedures could yield values that make more sense. Realistically, I don’t think any of this is particularly impactful to on-field performance.
To get a more-accurate kA estimate, you can trim your step voltage test data to exclude the long steady-state “tail”, which contains redundant information and dilutes the fit.
