Mechanics and Construction > Mechanics and Construction

Servos

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Yea, the shoulder would be the base joint. I am just saying do not be limited to just one servo per joint.

The reason I suggested a worm gear is because to hold an arm up requires constant power, but with a worm gear you only need power to move. It is very common to use worm gears in robot base (shoulder) joints.   :)

YAN-1:
Hi. Well I am facing some trouble here. I found a servo motor that will give up to 24.7 Kg.cm of torque and it was rated under 'monster servo, the servo that will move just about anything'! I saw a couple of robotic arms on some sites that use servos that are weaker than this and that worked just fine. But when I attempted to do a simple calculation for the robotic arm I intend to build, I found that if I used this monster servo for the shoulder joint, it won't be enough to lift the arm at all!  :(

To make sure that I am not doing the math wrong, I sketched the arm and presented the equation I used for calculating the torque carried by the shoulder joint in the worst case. Here, L1, L2, and L3 are the links, S1, S2, and S3 are the servos, G is the gripper, and the load is a glass of water. The numbers I assumed are as follows:

Mass of each link = 300 g
Mass of each servo = 150 g
Mass of gripper = 300 g
Mass of load = 200 g

With these numbers, I get a torque of 5.5 Nm that will have to be provided by the shoulder servo. This corresponds to a torque of 56 Kg.cm, which obviously cannot be supplied by the monster servo! Am I doing it wrong? I don't think I can make the links any shorter in order to reduce the torques because then the arm won't be practical. Please help me out. Thanks a lot.

Nichola V. Abdo

So your equation is correct. You forgot to give me lengths, so I just assumed 10cm each. I assume they are shorter?

Plugging in your numbers, I get:

(.3*10/2+.3*(10/2+10)+.3*(10/2+10+10)+.15*10+.15*20+.3*30+.2*30)*9.81 = 323.73 kg*cm required

And this is what you get if you dont have the gripper and water weight:

(.3*10/2+.3*(10/2+10)+.3*(10/2+10+10)+.15*10+.15*20)*9.81 =
176.58 kg*cm required

So you calculations all look correct to me. Perhaps why your calculation looks so large is that you have two really heavy loads at the very end of your arm, making the torque nearly twice as high.

Possible solutions:
- Make your shoulder joint the translating one (or at least not a vertical motion). That way you can make it rigid, and all your torque will then end up at S2, so it will be much less. Can you grab the glass from the top instead of the side?

- Use two servos at your base (your calculations say 2 servos will be more than enough)

- Go to using a larger DC motor

In all cases I still highly recommend using a worm gear . . .

YAN-1:
Hi. Well thanks a lot. Aren't the numbers you got supposed to be in N.cm (you multiplied by 9.81)? I'm sorry I forgot to give you the lengths of the links. L1 and L2 are 20 cm and L3 is 5 cm. What I don't understand is how does an arm like this one http://www.crustcrawler.com/products/arm6.php?prod=11 operate with these servos. They are smaller than the one I was talking about. Can't I use a gear set at the shoulder with the same servo? And how do I mount 2 servos together if I am to use more than one at the same joint? Thanks a lot.

Nichola Victor Abdo

oops sorry you are right, they should be in N*cm

so I recalculated with your numbers and got what you got:
(.3*20/2+.3*(20/2+20)+.3*(5/2+20+20)+.15*20+.15*40+.3*45+.2*45) =
56.25 kg*cm or 5.52N*m

56.25 (kg cm) = 781.17 inch ounces

But that robot arm uses a 343 inch ounces capable arm and claims to lift 403.41g plus its on body weight . . .

Looking at the arm, based on given weight and length, and an estimated center of mass at 1/2 distance (probably more), it would require at least
2.34 lbs * 16 oz/lbs * (17.2"/2) = 321 oz inches
just to lift itself . . .

Based on the servo specs, it can only lift another 22 (343-321 = 22) inch ounces (or 1.58 kg cm) at full extension. Your glass would require 200g*17.2" = 8.74kg cm if their arm were to lift it. So their robot arm would not be able to lift your load.

So the reason why their robot arm claims to lift so much is that the rating is NOT at full extension ("max lift capability" just means how much it can lift, not at what distance). If the robot arm was retracted half way, the torque required would also be halved (since the distance is half). Theoritically, the arm can lift its maximum amount when it is completely vertical because the distance is near zero.

Attached are two pics. The first shows two servos moving one arm. The other shows an idea I got of basically a servo using a string to move the arm up and down. The reason it would work is because of mechanical advantage, but would involve a list of other problems so its just food for thought. You may just have to redesign your arm a different way? Perhaps try to reduce arm weight by using different materials or putting holes in it?