Is it possible to calcuate this?

[peteinsg]

Hello there... I am not an engineer so please excuse my ignorance if I am missing some important information but I'm wondering if its even possible to calculate this... What would the ideal size of motor be (speed/torque) to elevate a 20mm OD threaded rod 1100mm in approx. 30 second while lifting a 30KG load. The thread rod passes through two 25mm bushes and can connect directly (inline) to the shaft of a 12/24v electric motor. The speed of the motor and the thread pitch of the rod would ideally be commonly available parts. Any assistance would be greatly appreciated. Apologies if this sound complex. Thx. Pete

[kjoiner]

Hello there... I am not an engineer so please excuse my ignorance if I am missing some important information but I'm wondering if its even possible to calculate this... What would the ideal size of motor be (speed/torque) to elevate a 20mm OD threaded rod 1100mm in approx. 30 second while lifting a 30KG load. The thread rod passes through two 25mm bushes and can connect directly (inline) to the shaft of a 12/24v electric motor. The speed of the motor and the thread pitch of the rod would ideally be commonly available parts. Any assistance would be greatly appreciated. Apologies if this sound complex. Thx. Pete

Hi Pete,

Your example can be calculated, but some more information will be needed. As far as the basic power requirement, you are lifting a 30KG load 1100mm in 30 seconds. I'm going to switch to english units for a moment here to get the power - 30Kg=66.138lb=294.196N. Power (watts) is N-m/sec so you have (294.196N x 1.1m)/30sec = 10.78Nm/s which equals 10.78 watts which is around .014hp. This value is the basic power requirement - a factor of safety needs to be added to the value. You need to multiply this number by 2-4 times to account for unknowns. One very important issue - if this mechanism is to be used in a way that a failure could lead to injury or worse, you need to be very careful and possibly consult with someone who can certify your design. My input is for general information only.


To get a torque requirement, you will need to establish what rpm your motor or screw will be running. To get an rpm, you will need to select the lead (linear travel per turn) of the screw. This brings up screw selection. If you want the load to stay put once you turn the motor off (non backdriving), you will need to select a lead screw with a short lead and lower efficiency. This does not mean the screw is "bad" but rather a term for lead screws and nuts that relates to the backdriving ability. In other words, you definitely don't want to use a ball screw. In addition to being expensive, not matter how short the lead, they can backdrive on you and once you shut off the motor, the load can drift back down or come down quite rapidly depending on your lead.

As far as screw diameter, the rpm and the length of the screw between supports will determine what diameter you need. If you are running at high rpm with a long distance between supports, the screw can hit its critical speed and start to whip. I don't think you will have that problem but it's good to keep in mind. Here are some companies that make lead screws. They have very good documentation for designing a lead screw system:

Links removed....

Design guides will have examples of different support options for the screw and how it affects the operation of the screw.

Hope this helps.

Kyle

[peteinsg]

Hi Kyle. Many thanks for your advise. The screw will be mounted vertically secured to a bulkhead by the two 25mm long bushes - through which the screw passes. This there a way to calculate the amount of friction (and therefore the amount of additional torque required to overcome it) caused by the two bushes? The 30KG load bearing down on the threads within the bushes is my main concern. If the overall amount of torque can be calculated approx, I can then select the screw thread pitch & motor rpm to match the 30 seconds elevation time - assuming the rpm/pitch doesn't affect the calculation. There are other aspects to the design which I didn't go into to prevent the risk of the screw reversing (unintentionally) or causing injury. Cheers. Pete

[kjoiner]

Hi Kyle. Many thanks for your advise. The screw will be mounted vertically secured to a bulkhead by the two 25mm long bushes - through which the screw passes. This there a way to calculate the amount of friction (and therefore the amount of additional torque required to overcome it) caused by the two bushes? The 30KG load bearing down on the threads within the bushes is my main concern. If the overall amount of torque can be calculated approx, I can then select the screw thread pitch & motor rpm to match the 30 seconds elevation time - assuming the rpm/pitch doesn't affect the calculation. There are other aspects to the design which I didn't go into to prevent the risk of the screw reversing (unintentionally) or causing injury. Cheers. Pete

This is a test post. My previous ones failed to show up.

[kjoiner]

Hi Pete,

Just to clarify, the bushings are at either end of the screw and you are looking for the friction to turn the screw within them? It depends on the type of bushings you are using. For applications such as yours, an oil impregnated bronze bearing (SAE 841 also called "oillite") would probably work. You do need to consider the axial load on the screw from the weight of the object your are lifting. You will need some type of thrust bearing to take the load. You don't want the shaft of the motor to be supporting the weight. Another option is to use a ball bearing at the bottom with a shoulder on the screw bearing against it. The ball bearing would need to be the deep groove type so it can support an axial load. Better yet is an angular contact bearing that is designed to take the load. Without knowing all of details that will evolve during the design, the friction of the bearings is not going to be very large. Will you be turning a smooth section of the screw to fit into the bearing?

To determine the amount of torque required, you will need to look at your motor selections and screw selections together. The pitch of the screw will determine two things - torque to raise the load and the rpm it must turn. This will guide you to which motors you can use. When looking at the motor, the rpm and torque will determine the current draw and power supply size you will need.

Going back a step, when you refer to the "bushes" are you talking about the lead nut that travels along the screw? Just to make sure we are talking about the same things, you will have a 30KG load mounted in some fashion to a lead nut. The lead nut travels along the lead screw which is driven by a motor. The lead screw is vertical and supported at the bottom by some type of thrust bearing and at the top by some type of plain bearing.

As an example of something I did a while back, I designed a small linear actuator that generated a nominal 65lb of force but capable of 92lb peak. The load was horizontal, but force is force. The lead screw I used is a 3/8" diameter with a 3/8 pitch. The mechanism travels about 2" per second and I used a small motor that produces 175 oz-in of torque and turns at around 180-200 rpm. I also have a timing belt between the motor and screw since some shock loading is involved. It pulls around 1.6A. The lead nut is plastic since I cannot run the screw with grease or oil for the application. The lead nut can take around 300lb of static load so it has ample reserve strength. Normally I would support the screw at both ends, but constraints allowed me to only support the screw at the back end and I used a small ball bearing for radial/axial support. The screw is only about 6" long. A length of 1100mm like yours will definitely need support at both ends.

The parts in the above example came from Roton. We buy the screw stock and then have it machined with a shoulder for the ball bearing and a groove for a snap ring to retain the bearing.



Kyle