My team has our 2-stage cascade elevator built, but we are still hammering out the details of the gearing. We want to use a Cim on a 9:1 versaplanetary together with a 22:15 chain reduction, which is powering a 2.375" diameter PVC drum. The reason we would like to use this specific gearing is that we already have the required parts. When we input the above ratio into JVN’s design calculator, the current draw on the Cim is 19 amps.
My questions are the following: does the calculation factor in the fact that the force to lift the 2 stage is double that of the first? And is 19 amps an acceptable current draw for a Cim?
For a cascade elevator, the force is the same throughout. For a continuous elevator, the force needed will increase based on the number of stages engaged.
The JVN calculator does not calculate the load for you. You have to supply it with the load placed on the motors, you have to create this estimate. The default value may or may not be appropriate for your application.
Can you post a screenshot of how you are using the calculator? This would help.
To answer your question, no the calculator does not natively account for that. By using a cascading elevator, you’re essentially adding a 2:1 “reduction” in the opposite direction of normal gear ratios. So the elevator travels twice as fast with half the force. With this additional anti-reduction, you may need to reconsider your hearing choice.
I’d also suggest using my design spreadsheet to help with these calculations. It does the same thing JVN’s calculator does with a number of extra features that will be helpful (like stall voltage calculation, max power and efficiency, and various characteristic inputs)
I wanted to make sure you understood it. The reason I say that is that by asking the question, you did not understand how powerful a CIM motor is. You can calculate the torque required by your values and find out where on the CIM power curve you are. This will give you your powered RPM, your efficiency, power and Amps.
This kind of work is what he is doing with the calculator already.
OP: See above comments about how cascade gearing is essentially twice as fast / half as much reduction as you’d expect. Basically you use the calculator with about the load you estimated, but with the travel distance equal to what you would expect one stage to move - knowing that both stages will actually move at once. You are not lifting the load twice, however…
A loaded current draw of about 20 amps is a nice, safe value for your elevator. You could gear a little bit lower to be safe (and to have a slower, more controllable elevator) which would also reduce current draw.
Many times I see folks use this sheet without any real understanding of what they are doing. I was simply pointing out that OP is trying to apply the process to a motor that the OP does not have good data for. I compare all these values back to the actual tested motor curves to see where my operating point is on the curve. To choose a motor you must understand torque and the speed you need. This sheet is great but its not the final answer. I did not say the OP did anything wrong.
I think you’re misunderstanding the OP. He is referring (as far as I can tell) to the amplifying effect the cascade rigging system has on the weight of elevator stages 2+.
If stage 1 weighs 10 pounds you have to pull on the cable with 10 pounds of force. If stage 2 weighs 10 pounds, you have to pull on the cable with 20 pounds of force PLUS the 10 pounds to lift stage 1. This is because of the 2:1 mechanical disadvantage between your winch and stage 2 caused by the cascade rigging.