  HomeUp 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. The above drawing shows the basics of the drive: The power source is the DC motor. The important spec of this motor is the (stall) torque. This is listed on the motor label: Torque constant = 0.24 Nm/A This means that the motor shaft can produce a force of 0.24 Newton over an arm length of 1 meter at 1 Ampere of drive current. If you don't know the the stall torque of your motor, you can measure it in the following way: Mount a lever to the shaft of your motor. Let the end of the lever push a kitchen scale. When driving your motor with current I, you can measure the number of kilograms the end of the lever pushes, and torque in Nm/A can be calculated with the formula as shown. 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 So I should get 43.6 Newton pulling force in the chain at 1Amp of motor drive. I did some measurements to verify these results: I connected the motor to a constant current supply (bench lab supply in constant current mode)  I measured the force in the chain by means of home-made spring-balance. (You could also make the side of the platform push bathroom scale, but there was not much room for that setup).   The plot shows the results of the force per Amp drive. (Force based on 1kgf = 9.8N). Of course there are lots of inaccuracies due to friction, etc, but the results at least show that I'm in the ball-park.    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. The balanced platform load can be simplified as two weights, with equal distance Rm from rotation point For checking the above calculation method, I did a test with 2x16kg load on the roll axis of the platform. The motor was driven with 3 Amp constant current. (platform drive pulling force ≈ 100N)  Rp was 0.61m, Rm was 0.51m, Total platform mass 35kg, (I added 3kg for the platform mass). Platform drive torque T = 100N*0.61m = 61 Nm Platform total Inertia I = 35*(0.51)2  = 9.1 kgm2 Angular acceleration a = 61/9.1= 6.7 rad/sec2   (= 381 degrees/sec2,  which is not such a bad value for a motion platform)  To check these results, I used the motor tacho signal (14V/1000rpm) for measuring the acceleration. The above graph shows the measurement result (tacho and motor drive current during the short 3Amp current step). 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. But the model is good enough to know what I can expect with the full simpit on the platform. Inertia will increase about 4 times due to higher mass, but motor torque can also be increased by higher current. Values around 15Amp should give me comparable accelerations as measured now. From the inertia formula it can be seen that one should try to place most of the weight around the platform rotation point, as inertia increases with the square of the distance from center. 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.