# Arm rotation speed?

Thank you.

This makes sense.

I will sneak in another question.

Was 233 arm made with 2, 3 and 4 inch square tube or larger sizes?

Without knowing the details of the system itâs hard to give a firm yes or no. But, you can calculate this.

If you know the load on the arm, you can get its rough mass moment of inertia about the axis of rotation (or estimate it conservatively by assuming a point mass at a distance equal to the length of the arm or something like that).

The moment of inertia multiplied by the desired angular acceleration of the arm (not the speed of the arm), is equal to the net torque applied to the arm.

The net, or total, torque applied to the arm is the sum of the torque applied by the motor and the moment caused by the gravitation fore acting on the arm. For a worst case, assume the arm is extended horizontally. If you apply any counterweights, factor their torque in here as well.

Now you have (motor torque) + (moment due to gravitation force) + (optional counterbalance applied to the arm) = (moment of inertia) * (desired angular acceleration). There is only one unknown here, the motor torque, which you can solve for directly.

The motor torque is transmitted by the chain and sprockets. Consider the larger sprocket. (Chain tensile force) * (sprocket radius) = (motor torque). Now you can solve for the tensile force in the chain, and determine whether one loop of #35 chain will get the job done.

Caveats:

1. I havenât done dynamics in a year, so if someone sees something wrong with my solution please correct me.
2. There are other factors at play. For example, if you put a lot of force through a loose chain, youâre liable to skip teeth (and eventually destroy teeth) on the sprockets rather than lift the arm. Chain also stretches over time, so a mechanism that is tight when first assembled can loosen up over a period of time with applied force. Another example, if your arm needs a ton of torque, it is conceivable that you could shear the sprocket off your shaft or twist the shaft itself. This is why it is important to consider how you plan on transmitting torque from the sprocket to the arm (you could always bolt it on or something like that).
3. The motor torque in these calculations is the torque applied to the arm about the point of rotation (ie. the combination of the motor, gearbox, and any other chain/belt reductions). You can use this number to determine the configuration of your powertrain if so desired.
4. I believe that if it is worth doing, it is worth overdoing when it comes to powertrain design. As other people have stated, there isnât really a down side to over powering your mechanism so long as the extra torque doesnât destroy the gearbox, snap the chain, or consume your weight or monetary budgets.

If any of this is confusing (it is not easy stuff), you should definitely consider roping in a physics teacher at your high school. Itâs a good way to get them involved with the team, and also a good resource for yâall to have to help solve other mechanical design problems.

I hope this helps, and good luck!

Let me jump in on this and ask about our scenario. We are a second year team and this is our first time building and rotary arm mechanism. Our arm is about 24" long and weighs 8-10 pounds.

We plan to have a #35 15T sprocket on the VP gearbox driving a 60T sprocket on the arm.

It looks like the max recommended gear ratio for a 2 stage VP on a CIM is 50:1. Would that be sufficient to move the arm?

Take a look at JVNâs design calculator. With a ratio of 50:1 and a cim rotating at 6000 rpm free speed, your arm is going to try to turn itself into propellor. For comparison, we are looking at between 700:1 and 1000:1 as a final gearing for a 775 driven rotary mechanism.

After the reduction on the chain stage the final ratio would be 200:1, right? How would we achieve a greater reduction without exceeding the recommended limits on the VP?

CIM free speed is 5300, so geared at 200:1 that would be around 27 rpm. Still too much?

Itâs more about what your programmers can control with motion magic and PID.

Change 27 RPM into a language that you can estimate. 27 Rev/min * 1 min/60 seconds * 360 degrees / 1 rev = 162 degrees a second. At free speed. However, you wonât be traveling at free speed because youâll be loaded.

Again, take a look at JVNâs design calculator. It will give you insight into the speed of your mechanism when it is loaded. If you donât look at the loaded calculations, you canât be sure youâll even be able to move the arm.

After putting your details into JVN calculator I get:
to turn 90 degrees
Motor drawing 9.92A

If you use 12T to 60T sprocket reduction it gets to

In my opinion both are suitable reductions.

You can not use motor free speed to do these calculations as the speed changes when load is applied.

Thank you.
I think I understand.
Overkill will work.

I fondly remember the day I have left high school a bit over 29 years ago:

Problem with being an âoldâ engineer is that you donât get to do this kind of calculations at work. Mentoring FRC team helps to stay sharp and learn from younger people.

Ok. Thatâs what I was thinking as I played around with JVNâS calculator and AriMBâs linked above. I just wanted some reassurance that I was looking at it correctly.

Iâm guess we could also slow it down some in programming by decreasing the voltage to the motor or whatever witch magic programmers do (PID/motion magic)?

If you are not using some form of closed loop control, that arm speed will be much too fast. Are your programmers comfortable with PID?

I dont know what closed loop control meansâŚbut I do know that the programmers have used PID in the past.

I fondly remember the day I graduated engineering school 36 years agoâŚthe stuff I learned, I mostly forgot, unfortunately. Explaining this stuff to students now gives me a good excuse to try to remember some of it.

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PID is a form of closed loop control. Closed loop just means there is some type of feedback, such as a position encoder

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Ah, gotcha! Yeah, we are planning to use a SRX Mag encoder with a Talon SRX controller. I know the programmers have talked about using motion magic to control the arm.

These images can hopefully help explain the differences between open loop and closed loop control

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Yeah, now I feel dumb. Iâve been spoiled with a good programming mentor and good programming students, so I just leave all that up to them.

Weâve also considered using a potentiometer and limit switches along with the mag encoder as added insurance.

So the potentiometer/encoder along with the limit switches are what provide the feedback portion of the close loop control.

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In addition to what Lil Lavery stated, using two sprockets can allow the shaft to be a dead axle i.e. it only supports weight and does not transmit torque.

When a single sprocket is used on a â2 beam armâ, with the sprocket in between the two arms, the shaft is a live axle and transmits torque from the sprocket to the two arms. if the torque is high enough, the shaft will twist and deform permanently.

A team in our area used such an arrangement in 2015 with a long arm to lift a stack of totes. Their hex shaft ended up looking like decorative wrought iron railings.

Earlier this week, I asked a Tech Support person at AndyMark for the maximum torque spec of their solid aluminum hex shaft material and he said they did not have one. You may want to test your configuration before committing to it.