# Drive Sizing 101

So, the emergence of Dr. Joe back to these forums inspired me to look into all of his and my old posts in the technical forum. I have noticed that I have not done my part on the instruction and guidance in these forums as I used to. With that in mind for the entire design season I will be posting these little educational lessons based on the actual design and analysis that 217 is doing this year.

Now, I will not be sharing all of the information and assumptions that we are using as I think that is part of the lesson. I promise that after several others post with what they think are the proper assumptions and analysis steps, I will share more detail.

First lesson: Drive sizing for an arm that lifts a 150ish pound robot

Objective: Lift our robot at the end of the match in less than 5 seconds.

Assumptions:

1. Robot Weight (W) is 150 lbs
2. Center of Gravity (CG) is approximately in the middle of the robot
3. Robot is a long body (37" from front to rear)

Final Decisions:

1. Use 1 Fisher-Price (F-P) motor as that is the highest power motor available to use after the motors were allocated of other tasks (Using all 5 CIMs and one other F-P).

2. Final gear ratio from the F-P to the arm joint is about 1500:1

The question has two parts:

1. Why did we select 1500:1?

2. What type of lift are we using (scissor, rotational arm, winch, etc)?

Again, I will share all of our calculations and thought process after some discussion in this thread.

Have fun,

Paul

First off, I like how you posted this themed as more of a “guess what I am doing” lesson/game than a “Hey what is everyone else doing?” to provoke thought. It is a nice departure from the standard fare.

Can you please clarify if the FP motor is attached to the KOP FP gearbox, and then adding 1500:1, or is it 1500:1 directly from the FP. I am assuming you mean the latter since you were cocise on the rest of your comments.

Also, is your 5 second requirement only for the “lift” or also the deployment of your attachment mechanism

To throw out some FP Specs:
Stall torque: 0.45 Nm
Free speed: 15,500 RPM
Peak Power: 184 Watts @ 10.4VDC

I will throw my guess out that this is an arm based on the following:

1. You state “Drive sizing for an arm that lifts a 150ish pound robot”, so I am guessing that might be literal
2. The FP motor has the highest speed to power ratio after the CIM. Since this works well into any of your three types, nothing definitive there.
3. Your requirement of five seconds would seem difficult to achieve with an FP 1500:1 with a winch from a speed perspective. A scissor lift seems possible, but only if it is already deployed, or if it is not required to hange from the horizontal bar, but instead from a side pole. From my experience controlling scissor lifts from JLG, the weight to lift speed ratio is surprisingly slow if you want it to be done in a controlled (ie. safe) way.

Neat stuff, Paul!

Some major gear reduction going on there… my best guess would be that you’re simply rotating the robot up 90 degrees (or perhaps less) with your transmission directly driving whatever grabby-type arm you have planned. It takes about 3 seconds to raise a robot to the desired height at max power (with an FP), which fits somewhat closely with the reduction you chose.

Of course, you guys won’t be bothered trying to grab the top of the bar… I bet one vertical bar is sufficient.

I’m wondering why you guys are using such a wimpy motor though… don’t you have 4 CIMs sitting in your drivetrain that won’t be doing anything during the lifting period?

Can you please clarify if the FP motor is attached to the KOP

FP gearbox, and then adding 1500:1, or is it 1500:1 directly from the FP.

Directly to the F-P.

I’m wondering why you guys are using such a wimpy motor though… don’t you have 4 CIMs sitting in your drivetrain that won’t be doing anything during the lifting period?

We have our reasons, but the “wimpy” motor has plenty of power to do the job and powering it with more powerful motors will cause us to upsize all the components due to the stall condition. The total weight to just add a F-P was less.

I started doing a little research on getting an FP 1500:1 and I may be missing something, but it looks like a gigantic task of gearing this without adding a ton of weight. I would love to see how you plan on accomplishing this, but I have a feeling you may be reluctant to give up your secret recipe .

I encourage everyone to try and solve this instead of just guessing. (its more fun/rewarding that way)

I tried a few quick equations (by no means correct) and came up with my guesses. Highlight line below to see.
Either a scissor jack (more likely) that grabs the top bar, or a 0.5 metre rotational arm that attaches to a vertical bar on the tower.

The robot is lifted from Center of Gravity
5 seconds is just lift time (does not include time to raise mechanism)
Wouldn’t use a winch with a 0.4 metre radius :rolleyes:

Judging by the reduction, winch can be discounted from the possible ideas because you would get a ridiculously slow lift speed geared that low. That brings us to arm. At 1500:1 a FP motor can spin at about 60 deg/sec.

At 1/3 stall torque, a 14" arm supporting the weight of the robot would require approximately the reduction you described. Added to a lightweight apparatus raising the robot height to 60", this comes within grabbing distance of the bar. This would not bring the robot clear of the platform when raised, though. If, however, the robot is starting from the top of the bump, it could easily clear the platform when raised with only a 50" or so robot. This brings into question the raised center of gravity when navigating to the top of the bumps.

With a scissor jack, the linear speed of the motor would be multiplied by the number of links in the jack. Added to the fact that a lead screw would make a very convenient way to achieve the large reduction you want. Since the scissor lift also can achieve a large gain in height with a small retracted space, as well as the large reduction and issues with the other two designs, this is my guess for #2. As for #1, I am unsure of the math behind lead screw reductions, so I will leave that for someone else.

1.) FP has a Torque of 0.45 Nm, which is equal to 3.982 in*lbs.

``````3.982 in*lbs x 1,500(gear ratio) = 5,973 in*lbs
``````
``````5,973 in*lbs / 150 lbs (weight of Robot) = 39.82 inches
``````
``````15,600 RPM / 1,500 (gear ratio) = 10.4 RPM
``````
``````10.4 RPM / 60 seconds = 0.173 RPS
``````
``````0.173 RPS * 5 Seconds = 0.867 of a Revolution = 312 degrees
``````

2.) 30-36" Four-Bar-Linkage Arm (assuming you have a ratcheting mechanism to prevent you from slipping)

That is my guess.

-Clinton-

Here’s my guess -

Assuming long, you’ve got 35 inches or so in your bot to stow an arm. I’m also assuming you need to stay under 17.5 to go under the platform, which means you can’t really “slant” you arm from front to back to get increased length while stowed.

This gives you a 3 foot horizontal bar. If you use that bar and hook it on the vertical tube, you can pull down on the end. With a tiny bit of clever engineering, you can “lock” the bar in place just using that torsional force, then spin your robot up and over it (inverting your robot) and placing it high enough to score - while only using the one motor to pivot the arm.

So, if you put another 1/2 length arm, hinged from the center of the bot, and put the motor at the end of that connected to the longer 35 inch arm, you could release the shorter arm, say 16 inches long. It will pop up on the gas shock. This plus your 16 inch height robot height is more than enough to get it above the scoring height limit. Then you simply click your horizontal arm into place against the vertical bar and invert your robot.

Tell me you’re going to invert your robot. I’m drooling to see it

PS: This fits reasonably well. At 10 RPM roughly, you rotate 1 time every 6 seconds. 16 inches, the length of the shorter arm that rises from the center of your robot is about right for the torsional arm you’ll be working with. It packages within the long driver base with an inch or two to spare too.

I agree with Tom’s guess…

When do we hear the answer?

We looked at doing that exact thing, but the torsional force on the tower scared us away - over half a ton that your end effector would have to hold. I’m sure it’s doable, but we didn’t want to deal with the high forces.

O.K. here are some of the answers:

1. We are NOT going in the tunnel. 1/2" of clearance on either side is just not enough for us and making our robot skinnier is just not something we are willing to do.

2. We are using a single rotational joint to do our lift.

3. We are inverting ourselves

Calculations:
We make sure our design point is at Max Power. As the torque goes down the speed picks up and we have plenty of torque at start-up.

F-P speed at max power = 7800 RPM
F-P torque at max power = .1648 ft-lb
CG location from pivot = 18"
Robot Weight = 150 lbs
Required Torque = 225 ft-lbs (150 lbs * 1.5 ft)

F-P Torque at 1500:1 at max power = 223.45 ft-lbs
F-P Speed at 1500:1 at max power = 31 deg / sec

Rotation angle = 90 degrees so total time is ~3 seconds.

Now at various points along the rotation the speed will change based on the CG moving closer to the pivot. We can also always add Latex tubing if our CG is further out than we think.

For those looking for the extra credit, what Efficiency did 217 use in their calculations?

Nice. I would challenge someone to now tell us how much torque will be on the vertical tower post in worst case, and how thick the end effector will need to be to handle that torque…

I am going to say somewhere in the 60%-70% range, but it could also be as low as 50%. The reason I would guess 50% is it is half of the full power, so if it could do it in the set constraints at 50%, then at 100% or anything greater than 50%, it would be guaranteed every time.

Tom,

The force on the latch is much lower than you think. The design we are using is made such that the contact forces on the vertical pole are greatly reduced.

Paul

I’m very anxious to see your comp bot then. With our initial design, when the bot was out at the end of the 36 inch arm plus the 18 inch riser, we had 54 inches (call it 4.5 feet). With the 150 pound robot at the end, that’s 675 foot-lbs. However, our initial end effector was only 6 inches tall, so the torsion on the top point was 675 ft-lbs / .5 ft = 1350 lbs. Of course, we were trying to package within a 16 inch tall robot, which is why we limited it to 6 inches tall.

Since you’re not worried about the tunnel, I suspect you didn’t have that constraint regarding packaging.