# Simple arm pivot scenario questions - gear ratio, load ratings, etc

Hi all,

On this separate thread, I asked & received some help on a wrist mechanical design. Thanks to those who shared their experience and recommendations there!

Here, I have a related but different discussion. I think I understand the major considerations for choosing a sufficient motor - gearbox combination to power a generic arm pivot, BUT there are some little things I’m unsure about, and I’m wondering if some folks might talk through this.

Here’s the scenario:

• The joint needs to handle 5 ft-lbs
• I use Ari’s AMB mechanism ratio calculator to determine a Neo550 current limited to 20A + 50:1 reduction should work. Here’ s a link

Questions:

1. In my brain, I interpret the information from the calculator as… The motor will require 16.7A to hold the 5 ft-lb arm straight out parallel to the ground. Is that right?
2. I assume less current will be needed to lower the arm and more may (will?) be needed to start raising the arm from the straight out position. If so, then maybe a higher reduction should be used, to be safe?
3. To that point, is it better to use the highest reduction that will still allow the arm to rotate fast enough, even if it is far more reduction than needed? Seems wise - why not?
4. This is where the load ratings question comes in… I was originally interested in using a Neo550 (20A limited) + Ultraplanetary, and I wanted to use 100:1 ratio (or 50:1 with another 2:1 from pulleys or sprockets) because it would be fast enough. BUT looking at the UP’s load ratings guide, the max ratio recommended for Neo550 is around 30:1. What I don’t truly understand is… How conservative is that number? If I’m current limiting to 20A (that alone) does the load ratings guide change for the Neo550? (If so, that might be a good one to just add to the ratings since so many people seem to be using that limit).
5. And so overall, if I’m current limiting the Neo550 and if the actual task I’m asking of it + the UP is not large, can I go ahead and just use the 100:1 ratio and be fine? What’s the best way to figure that out? Calculate the max torque the outer stage would face (maybe with a buffer), and if it is less than rated max for the stage, then it will be fine?

Thanks!

1 Like

The 16.7 Amps is the current the motor will draw as the arm passes through horizontal spinning at its loaded speed, 3.1 rev/s in your case.

The current the motor will draw while holding the arm still horizontally is the current associated with the stall voltage. This is approximately the motor’s stall current times the stall voltage divided by 12, or 15.5 Amps in your case.

As always, it’s a tradeoff. The higher your reduction, the better acceleration your system will have and the lower current it will draw under equivalent loads. But it will also be a heavier gearbox, have lower efficiency, and a lower maximum speed. What’s always true is if you go too low you exceed the motor’s maximum power (on the wrong side of the motor curve), and if you go too high you’ll exceed its maximum efficiency point (where the motor is spinning really fast and barely generating any torque). You can see these points as the default min and max ratios on my graph, so you’re good as long as the dotted red line stays on the graph.

1 Like

Thanks. It seems clear the higher reduction than absolutely needed is good from the motor’s perspective, though some extra weight & inefficiency are incurred. It would be helpful to also confirm how to think about the UP’s load rating (~30:1 max with Neo550). If the actual load is relatively small, it seems like one can safely go higher than the 30:1 on the Neo550+UP combo… just looking to confirm I’m not missing something there. Thanks again.

UPDATE: Analyzing more:

• 5 ft-lbs = 23.1 Nm
• The UP overview says, “During testing, the output gear of the cartridge fails at 40 N-m. All load ratings are based on a safety factor of 1.2 to accommodate manufacturing tolerances.”
• Logically, if the torque the system faces is significantly lower than the max the output gear can handle, then using 100:1 with Neo550 should be ok.
• The other thing I notice is… at 100:1, even the relatively weak HD Hex motor (seems similar in profile to AM Neverest) that comes with the UPs could work for this scenario. Half a second to pivot 90 degrees isn’t too bad, but the Neo550’s more than 3 times faster and is also a bit lighter.

One thing that we often fail to take in to concideration is the acceleration/deaceleration of the robot, and the forces that the arm experiences during that time.

last year my team moved our intake 4bar using a NEO 550 with a large planetary reduction(100:1 and another belt reduction after IIRC)
And everytime our robot spun around our intake would open immideatly.

My point is that while in static(ish) conditions the motor seems fine, I would definetly recomend taking a large safety factor on your output torque, as well as thinking about ways to lock the arm in its folded position.

2 Likes

If your wrist weighs 5.5 lbs and is 16 inches (from your other thread), your arm must be very short to get 5 ft-lbs at the pivot. I suspect there’s something missing in the calculation of the torque required.

I chose to use roundish numbers for the hypothetical scenario in this thread… it’s a similar but different torque than we’re designing for. After our weight reduction last meeting, we’re probably sitting at around 5.2 lbs. With the motor and more material being located on the back of the gripper vs the front, I guess the distance from pivot to com may be between 6 and 7 inches rather than the midpoint of 8. I have a couple identical kitchen scales; I should bring them in and measure. Thanks

The above concerns the wrist… our main pivot is powered by a couple Neos, maxp’s, chain. Total reduction in the neighborhood of 130:1.

Yes. It is best to allow a large margin to account for shock loads. This was very evident in the 2016 game where we were blowing up the planetary gearbox at least once per event due to bouncing when going over the Defenses. Our arm design had “sufficient margin” for the static case.

Afterwards, I asked a really good mechanical engineer who has done a lot of work on mechanisms that have to operate in very rough conditions. He said it is difficult to predict how much margin over the static load numbers would be needed. He said he typically starts with 5x or 6x, then does alot of worst-case field testing.

1 Like