3
$\begingroup$

A robotic arm should pick a cuboid up of a table, rotate it around its vertical axis and put it down on all possible positions. How many degrees of freedom are at least necessary? (All coordinates, that should be reached by the robotic arm, are in its workspace. It is not allowed to put the cuboid down and pick it up, once the robot has it )

  • The answer is 4 (3 translatory and 1 rotatory).

But I don’t understand why. I thouhgt that it should be 3. 2 prismatic joints: 1 to pick the cuboid up, and another one to move it anywhere on the table. 1 revolute joint to rotate the cuboid around its vertical axis. => 2 translatory and 1 rotatory.

$\endgroup$
4
$\begingroup$

The answer is four. The arm needs to be able to move up and down, left and right, forward and backward, and rotate. The robot will need 3 translatory degrees of freedom and 1 rotatory.

$\endgroup$
4
$\begingroup$

You specified that you need to be able to pick a cuboid up [off] a table(1 dof), rotate it around its vertical axis (1 dof), and put it down on all possible positions - assuming this means, "put it down at any (x,y) coordinate", then you need two more degrees of freedom - one for $x$, one for $y$. Note the implementation may vary - you could use Cartesian (x,y) actuators or you could use Polar (r,$\theta$), but whatever the combination you still wind up with two degrees of freedom.

$\endgroup$
  • $\begingroup$ This is close but not entirely correct logic. Lifting the object does not necessarily require its own DOF. Consider a human-like arm: this is capable of foreward/reverse motion as well as up/down depending on the angles. @holmeski's answer, below, is correct. $\endgroup$ – Ryan Loggerythm Jul 17 '15 at 1:10
  • $\begingroup$ But a human arm also uses three DOFs for this task: The should is has 2dof and the ellbow adds another one. Although the arm (without a hand or wrist) only has to joints, it has three dof. $\endgroup$ – FooTheBar Jul 17 '15 at 6:10
  • $\begingroup$ @Ryan - I'm not sure if you realize you're arguing against yourself. How does the shoulder manage to be able to go up and down AND forward and reverse? It's not "depending on the angles", it's that it is constrained to a sphere of radius armLength and the up/down motion, like forward/backward motion, are projections of the spherical motion. $\endgroup$ – Chuck Jul 17 '15 at 10:58
  • $\begingroup$ For the lifting task this is arguably acceptable, but the shoulder does not allow placement of the cube anywhere along the forward/backward path because that path is not along the plane of the table - you would be dropping the object because the shoulder lacks the degree of freedom to lower the object. Could you get by with one shoulder and one elbow? Yes. How many degrees of freedom is that? 3 for the shoulder + 1 for the elbow = 4. $\endgroup$ – Chuck Jul 17 '15 at 10:59
  • 2
    $\begingroup$ This is an exercise to get new engineering students to stop think about one particular solution and start thinking about the problem. The problem is how do you move the cube. Define the degrees of freedom, then look at what methods exist that provide those degrees of freedom. Spherical/rotary, spherical/linear, 4 x rotary, etc. Some of the things to consider are one dof joints you wouldn't ordinarily consider right away, like screws, rack and pinion, etc. $\endgroup$ – Chuck Jul 17 '15 at 11:00
1
$\begingroup$

Good answers already, so I just thought I'd address this bit:

But I don’t understand why. I thought that it should be 3. 2 prismatic joints: 1 to pick the cuboid up, and another one to move it anywhere on the table. 1 revolute joint to rotate the cuboid around its vertical axis. => 2 translatory and 1 rotatory.

Not quite correct: though it seems from your question that only three would be required, in reality you need a fourth to rotate the "gripper" to pick up the item. UNLESS you have some sort of magnetic pickup, or a specially shaped item like a cylinder - then three might work as a special case, but that would no longer be a general purpose manipulator.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.