I need to verify something: If I have an electric cylinder pushing against a horizontal load, say a box or pallet, which is not physically attached to my cylinder's tooling, it will increase in the torque requirement, but not factor into inertial calculations? The additional load does impact the system inertia if the load is dropped onto a tooling fixture, the load is vertical, or it is a pick-&-place operation, correct? -- Mike Little Application Engineer Motion Science 864-647-5454 x318

It must be counted during acceleration. If the programmed deceleration rate is greater than the deceleration rate of the box alone when decelerated by the friction torque only, then they will separate. The cylinder will decelerate based on the programmed rate and the box will decelerate based on the friction torque. Torque equals inertia X acceleration. If you're looking at it from a servo point of view, you've got a real problem because it will be a non linear function based on the speed you're moving and the frequency of any excitation. You don't want to go there. Tom Thomas B. Bullock, President Bull's Eye Marketing Incorporated Industrial Controls Consulting Division 104 S. Main Street, Suite 320 Fond du Lac, WI 54935 PH: 920: 929-6544 FX: 920: 929-9344 E-mail: [email protected]

The mass of the load is still contributing to the system inertia. What you do not factor in on a horizontal application is the weight of the load. The mass is weight divided by gravity constant. Gravity constant varies with altitude, but is considered to be around 32.14 feet/sec^2 or 386 in/sec^2. Depending on whether you are using ft/sec or in/sec for your velocity. This is why you can push a car but not lift it. You are accelerating the mass but you are not lifting against gravity. Bill Sturm

----- Original Message ----- > I need to verify something: > If I have an electric cylinder pushing against a horizontal load, say a > box or pallet, which is not physically attached to my cylinder's > tooling, it will increase in the torque requirement, but not factor into > inertial calculations? The inertia always factors into the situation, but it may not adversely affect your situation. TRUE: The standard servo motor sizing calculations don't account for your situation, as described. For sizing think of it this way: The starting inertia is both the cylinder and the pallet. (assuing they are in contact) The stopping inertia depends on whether or not the pallet is still in contact with the cylinder. By the way, the contact point is also a coupling. Think of is as a spring. Both inertia (kinetic energy) and spring (potential energy) are energy storage. Everything we do in controlling motion is relative to controlling energy storage. That spring (contact point) could cause as much problem as inertia. > The additional load does impact the system inertia if the load is > dropped onto a tooling fixture, the load is vertical, or it is a > pick-&-place operation, correct? Inertia in a vertical plane is no different than a horizontal plane. Inertia is a way of mathematically representing energy caused by a change in speed, in the direction of motion. Gravity (vertical plane) works as potential energy. EXAMPLE: A pendulum: kinetic and potential energy A pendulum in a zero friction system would swing (oscillate) forever Constantly transforming energy from potential (maximum potential is where is reverses at each side) to kinetic (inertia) energy (maximum at center-the highest speed) Assuming minimum oscillation is desired, we keep spring low by using stiff components. And we manage the inertia by planning for how the system is going to react. Typically this is done with inertia matching guide lines. Hopefully this helps Dave Kane [email protected]

This isn't that complicated, nor a big deal unless you're doing something besides shoving it out of the way. I'm assuming that's the case because if you were trying to position the box it would be attached to the cylinder. As previously stated, the load will contribute to the system inertia regardless of whether or not it is physically attached. As you indicate it will also increase the torque requirement. That increase will be due to inertial acceleration (added mass) & frictional torque (added weight). The torque required to decelerate may actually decrease depending on your deceleration rate. If the box decelerates more quickly (due to friction) than the cylinder, it will act as an aid to your system such as a damper. I'll give you an example calculation. I'll assume that the electric cylinder uses a ball screw for transmission. Also, for clarification "pitch" means revolutions per inch in my equations. Formulas (I'm used to working in English units): Load inertia = (Weight (lbs) * 16) / ((2*pi*pitch)^2) units are oz*in^2 Cylinder inertia = constant out of the catalog (if it's not given, you'll have to calculate it based on the ball screw's rotational inertia & any carriage weight/mass) units are oz*in^2 Motor inertia = constant out of the catalog units are oz*in^2 System inertia (Jsys) = Load inertia + Cylinder inertia Note that the load's inertia is all of the weight that the cylinder is moving regardless of attachment & orientation / angle. Those factors are used to determine the frictional component only....they are irrelevant for inertia calculations. Assume a move comprised of distance moved in a certain amount of time w/ a trapezoidal move profile: Vmax = [Distance (inches) / time (seconds)] * pitch * (3/2) units are rev/sec "3/2" is a constant based on the trap move. Time to accelerate load (taccl) = time (seconds) / 3 Torque due to acceleration (Taccl) = (1/386)*[{(Jsys)/(cylinder efficiency*system efficiency)} + motor inertia]*[(2*pi*Vmax)/taccl)] units are oz*in Torque due to friction (Tf) = [(Weight of load (lbs)*16)*coefficient of friction] / (2*pi*pitch*cylinder efficiency*system efficiency) units are oz*in The coefficient of friction is actually a useful term in this equation. For instance, assume that the actual coefficient is 5%. Also assume that the load is on a 30 degree angle. Sin(30) = 0.5. Therefore, your new coefficient is actually 0.55. You can always check yourself because a vertical load is the same as using a 90 degree angle: sin(90) = 1, which means you see 100% of the load's weight. In this example, the peak torque (Tpeak) = Taccl + Tf units are oz*in The continuous torque would be based on your duty cycle. It's also a good idea to keep an eye on your inertia ratio (system inertia to motor inertia), but that's another discussion all together. Let me know if you any other questions. Regards, Guy Guy H. Looney Motion Control Engineer A.C.E. Systems, LLC 170 Medearis Drive Old Hickory, TN 37138 email: [email protected] website: www.acesystemsllc.com work: (615) 754-2378 fax: (425) 944-5017 cell: (615) 330-0044