# How do I decide the minimum number of link lengths and joints my manipulator must have from the configuration space?

I have a configuration space in 3D where my manipulator must reach certain points. Taking into account all the factors how do I decide the mininum number of links and joints my arm must have?

• Thanks, will take care now :) Commented Jul 6, 2020 at 4:53

If you should consider not just points, but poses (position and orientations). If you can write these as a table, where the columns are the coordinates

+-------+--------+--------+-------+---------+--------+--------+
| Point | X [mm] | Y [mm] | Z [mm]| A [deg] | B[deg] | C[deg] |
+-------+--------+--------+-------+---------+--------+--------+
| P1    |  100.0 |  200.0 | 400.0 |    90.0 |   20.0 |    0.0 |
| P2    | 1500.0 |  200.0 | 800.0 |    90.0 |   45.0 |    0.0 |
| P3    |   42.0 |  200.0 | 400.0 |     0.0 |   90.0 |    0.0 |
+-------+--------+--------+-------+---------+--------+--------+


In order to asses how many degree of freedom do you need, you have to see how many coordinates in the table are constant. e.g. the Y column is constant and the C column is constant so you might be able to use a four degrees of freedom robot. However, systems with limited amount of degrees of freedom might have coupled motions, meaning that one or more of the coordinates which are not controlled are not constant, but change in a way which can be calculated, but not controlled. These simply result form the specifics of the mechanism used.

In the example form the table, two transnational axes and two rotational axes and two rotational axes can be used, but four rotational axes would not be well suited.

As for dimensioning the linkages maximum distances between the points can be considered. The angular difference between the orientation can be used to dimension the angular range of the orientation axes. Dimensioning the linkages is heavily dependent on the chosen axis types and configuration, there is no generic answer if the axis configuration is not known.