How to model a mobile manipulator?

[LIUHAITAO-CH]

i want to know how to modeling the mobile manipulator with float-base,as shown below

I can model the manipulator with traditional method,but i don't konw how to model the mobile . i would like to learn how pinocchio handles this kind of problem？

[jcarpent]

FOr the mobile platform, you can use PlanarJoint to work with X and Y translations and one rotation around the vertical axis.
Best,
Justin

[stephane-caron]

In the "type" attribute of the joint element there is a "planar" option, see urdf/XML/joint. Those will be loaded as planar joints in Pinocchio.

[LIUHAITAO-CH]

@stephane-caron thanks for your reply, i understand what you said. Is there any similar example（mobile manipulator） in pinocchio?

[LIUHAITAO-CH]

and，this considers the mobile platform as a whole. Have you directly considered the situation of four wheels?

[stephane-caron]

I don't know of examples in Pinocchio itself, but there are examples in libraries that rely on Pinocchio such as:

• PLACO: where there is a WheelTask suited to wheels, which can optionally be omnidirectional

omniwheel.mp4

• Pink: where there are examples for the Stretch RE1 mobile-base manipulator (disclaimer: I'm a maintainer of Pink) but no wheel task yet

mobile_stretch.mp4

[LIUHAITAO-CH]

Thanks, Let me examine the example you provided

[LIUHAITAO-CH]

i want to know how to modeling the mobile manipulator with float-base,as shown below

I can model the manipulator with traditional method,but i don't konw how to model the mobile . i would like to learn how pinocchio handles this kind of problem？

@stephane-caron hello，i have a question, this is my urdf,

<link name="world"/>

<link name="base_link">
    <inertial>
      <origin rpy="0 0 0" xyz="-25428000000000 -1005100 -2223600000000"/>
      <mass value="110"/>
      <inertia ixx="9.1864E+40" ixy="-4.7485E+35" ixz="-1.0505E+42" iyy="1.2105E+43" iyz="-4.1525E+34" izz="1.2013E+43"/>
    </inertial>

    <visual>
      <origin rpy="0 0 0" xyz="0 0 0"/>
      <geometry>
        <mesh filename="xxxxxxxxx/base_link.STL"/>
      </geometry>
      <material name="">
        <color rgba="0.75294 0.75294 0.75294 1"/>
      </material>
    </visual>

    <collision>
      <origin rpy="0 0 0" xyz="0 0 0"/>
      <geometry>
        <mesh filename="xxxxxxxxxxx/base_link_sc.STL"/>
      </geometry>
    </collision>

  </link>

  <joint name="base_link_planar_joint" type="planar">
    <origin xyz="0 0 0" rpy="0 0 0"/>
    <parent link="world"/>
    <child link="base_link"/>
  </joint>

  `other links, joints`

i can get q0 = by pin.neutral(model),

but i noticed that I set q0 = np.array([0, 0, 1, 0, xxxx]). The first two are to set the values ​​of x and y, but what does 1 mean? What do the third and fourth values ​​mean? I feel It seems to be the scaling factor, not the rotation angle of the z-axis？

thank you very much

[stephane-caron]

Hmm, the documentation page related to that would be Joint dynamics, but the information you are looking for seems to be missing.

The configuration vector for planar joints is $(x, y, \cos \theta, \sin \theta)$ where $\theta$ is the angle (to be precise: of the rotation from the child frame to the parent frame). I opened a proposal to add this comment to the docs in https://github.com/stack-of-tasks/pinocchio/pull/2211

[LIUHAITAO-CH]

@stephane-caron thanks，I never thought it was cos (theta), sin (theta), why not use one theta, but use two quantities?

[jorisv]

You can find some explanation here and here

[LIUHAITAO-CH]

Ok,thanks