enter image description hereThe intended purpose of the arm is to pick and place chess pieces.

I am planing to make a 3 DoFs system (only using 2 joints mounted on a revolving turntable), the link structure should be composed of rectangular pieces of acrylic sheet. My chessboard is 28cm x 28cm.

I don't know how to calculate the length of the 2 links so that the robotic arm end-effector can reach each and every square.

  • $\begingroup$ I think that a quick sketch of your design idea, precise the type of each joint (revolute or translational) would be necessary to answer your question properly $\endgroup$ – N. Staub Jun 29 '18 at 14:41

Assuming you put your manipulator along the middle of the chess board, your

(link1 + link2)^2 >= (28+6)^2+ (28/2)^2


(link1 + link2)^2 <= 6^2 + (28/2)^2

kinematically these two equations would suffice. you can choose any link1 and link2 values that satisfies the equation. But considering the dynamics it's preferable to choose l2 < l1


There are probably nice mathematical solutions, but they would require knowing the size of the chessboard and the placing of the arm. My personal opinion is that 3 DoF isn't enough.

What I'd do is to cut up and old box and make a fake 2d version of your arm and see if you can reach every square. Make the sections oversized so that you can cut them down as you test.

  • $\begingroup$ That is a nice idea, but I would prefer some good mathematical solutions; if 3 DoF is not enough what design would you suggest I use? $\endgroup$ – Roshan Jun 30 '18 at 5:58
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    $\begingroup$ The problem that I see is that the arm is 2 DoF on a turntable adding a 3rd DoF, but I'm ignoring the turntable for now. Think of the arm itself as a triangle. As the arm extends, the arm becomes straighter. Normal sized inexpensive arms with only 2 DoF will probably become so straight and low that other chess pieces will be knocked down. Another DoF on the arm will prevent this. Spending a few minutes with a cardboard prototype that you move by hand will quickly show you if I'm right or wrong. $\endgroup$ – NomadMaker Jun 30 '18 at 16:22
  • $\begingroup$ ok cool, I tried it out and it helped me come to a quick solution of what lengths to use! $\endgroup$ – Roshan Jun 30 '18 at 17:39
  • $\begingroup$ @Roshan - If this is the answer that answered your question, please mark it as accepted by clicking the check between the up and down arrows to the left of the answer (and also consider upvoting this and any other answers that helped!) $\endgroup$ – Chuck Sep 4 '18 at 14:06

If you place your mechanical arm between rows 4 and 5 or between columns e and d, the furthest point you have to reach would be sqrt(14^2 + (28+6)^2), which evaluates into 36.77cm, and I would say 2 pieces of 20cm would be almost enough, but just to be safe, I'd make 'em both 25cm and you may put it anywhere within 6cm from the board.

Material choice. Acrylic is very fragile and prone to cracking. The weight of the largest chess piece + the weight of all machinery (actuators, magnets, etc.) will easily crack it after a while. Use a more durable plastic instead.

Design choice. The arm you describe would be quite difficult to control, you have to move both joints at the same time with the different speeds, and if you miss and knock another piece off the board -- you lose the game. For speed-independent precise positioning I'd use X-Y table, something like this picture, where X and Y motions can be totally independent, precision is very high and you may easily put a few kilograms of equipment on the top and it will work. XY table

  • $\begingroup$ I don't understand how the length of the hypotenuse gives me the furthest point that it should reach. can you please explain that again . $\endgroup$ – Roshan Jun 30 '18 at 10:30
  • $\begingroup$ @Roshan I doubt that. What you're planning to use to control the angles in your robo-arm? Hardly it's cheaper than 2 stepper motors. $\endgroup$ – lenik Jun 30 '18 at 10:30
  • $\begingroup$ mg996R, 11kg/cm. Hmm, you're right it's not much of a difference actually. $\endgroup$ – Roshan Jun 30 '18 at 10:31
  • $\begingroup$ Can you explain the point regarding the furthest point, it just doesn't feel right. $\endgroup$ – Roshan Jun 30 '18 at 10:35
  • $\begingroup$ @Roshan 11kg/cm will become 220g at the end of 0.5m long arm. Are you sure the magnet mount + chess piece + the weight of the arm itself is less than that? $\endgroup$ – lenik Jun 30 '18 at 10:35

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