||The cable drive for the
DIY motion platform II also makes use of
the cable-pulley system. To have some idea about the accelerations that can
expected, I have done some calculations and verification measurements.
If you don't know the the stall torque of your motor, you can measure it
in the following way:
The DC motor shaft rotation is geared down via two sprocket wheels with radius R1 and R2. The torque of the second shaft will increase with ratio R2/R1. R2/R1 is basically the teeth ratio of the two sprocket wheels. In my case this is 39/11= 3.55 The torque in the second shaft is therefore 0.24*3.55=0.85Nm/A
With the rotation direction as shown, the chain that is connected to the
platform side will be pulled up via sprocket wheel with radius R3. R3 is the
arm-length of this torque that pulls the chain. As the torque is specified
at 1 meter arm length, the smaller arm-length R3 will increase the force by
ratio 1m / R3[m] which in my case is 1000mm/19.5mm= 51.3 The
force in the chain will thus be 0.85*51.3 = 43.6 N/A
I did some measurements to verify these results:
Now the force by which the platform is pushed and pulled is known, it is possible to calculate the accelerations of the platform. Since the platform will pitch and roll around a fixed rotation point, the accelerations are angular. For ease of calculations, I have assumed the weight that needs to be accelerated is mostly balanced. The platform can thus be seen as a big flywheel, that needs to be accelerated or decelerated.
Platform drive torque T = 100N*0.61m = 61 Nm
As can be seen, the tacho shows a ramp (sort-of) response during the constant 3Amp current pulse. Since the tacho voltage is a measure for motor rpm, it is also a measure for the cable speed that pulls the side of the platform. Therefore you can derive the acceleration from the tacho signal, being the derivative of the speed: dv/dt. Since we are interested in the angular acceleration, we have to make some conversions.
Tacho signal (motor spec) 14V = 1000 rounds per minute = 16.6 rounds per second. 1V corresponds with 1.19 rps. From tacho ramp, we see that the tacho peak is 6.4V . so this corresponds with 7.6 rps. From the motor gearing, one motor turn = 14* 11/39 = 3.94 teeth of sprocket wheel. Teeth distance = 0.95cm, so 1 motor turn is 3.74cm platform side movement. 7.6rps equals 0.28m/s platform side speed. The total circle that the side of the platform would make equals 2*p*Rp = 3.86m .Angular speed w = 2*p* 0.28/3.86 = 0.46 rad/sec Angular acceleration is the difference in angular speed over time dw/dt from beginning to end of the step pulse. dt can be found from the tacho graph: 102msec. Angular acceleration a = dw/dt = 0.46/0.102 = 4.5 rad/sec2
Seems the actual angular acceleration is about 33%
lower than the calculated value. So not completely accurate, but not too bad
either, taking into regard that I made lots of assumptions and
simplifications. Possible reasons: Inaccurate inertia value, motor torque
not constant, or other stuff I don't know about.
It should be noted that I have deliberately used the motor in a constant current drive. Therefore my motor torque stays relatively constant, as long as the power supply can deliver sufficient voltage. In end application, the gearing should be such that the motor stays in the low rpm range to be able to deliver the required torque. When too much gearing is used, the motor rpm will be so high that the torque becomes very low. In that case the platform will no longer accelerate, but move with constant speed.
More measurements will be done with all gear (including me) installed on the platform.