My team is hoping to utilize an elevator to both raise and lower our intake mechanism and climb at the end of the match. To keep things simple, we’d like to use just one motor/gearbox to achieve both. However, we know that these two applications have massively different loads. We hope to keep our intake mechanism as light as possible.
If we design our motor/gearbox combination to achieve the climb (so lots of torque to achieve a climb in ~1 second), what will happen if we run that same motor/gearbox combination with a much smaller load?
Should we expect speed to remain constant and current draw to drop?
Should we expect speed to increase and current draw to remain the same?
Should we expect both of them to change?
Would something else happen entirely?
We are hoping to avoid requiring a 2-gear shifter or requiring two separate motor/gearbox systems for both weight and space savings.
Most people would point you towards JVN’s mechanical design calculator, which you can find here.
I also made a spreadsheet with the same tools and a few more, which you can find here. Under the Mechanism Ratio tab, you can figure out what motors and ratios you need to get certain speeds with certain loads. There is even an example for a linear elevator that tells you exactly how to use the calculator.
EDIT - to answer your questions generally, with decreased load (weight lifted) you will see:
• faster speed
• lower current
• higher efficiency
• lower power
In different cases, these effects may not be very noticeable, and in rare cases they may actually work the opposite way The play around with the calculators to see for yourself.
Here is a spreadsheet by JVN (148 mentor) where you can plug in different motor combinations and loads to different gearboxes and see what the result is. It’s been extremely useful for me when designing any mechanism for FRC.