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Hip joint torque in walking mechanism

  1. Jun 19, 2015 #1
    Hi everyone,
    I am making a hexapod ( six legged walking mechanism ) which walking by keeping 3 alternate legs in air and other 3 on ground, pushing the body ahead, also known as tripod gait. The basic structure of model and one leg are shown in the pictures. I want to calculate the torque requires by servo whose axis is alligned to axis1, i.e. hip joint. Hip joint of 3 alternating legs will act together to push the body forward. I have calculated the angles of rotation and mass moment of inertia of the whole structure.
    Torque= I x alpha
    Can I simply use the above formula? I think i will have to divide the Inertia by 3 as 3 hip joints will act together.

    If you imagine the motion, alternate 3 legs on the ground forming a triangle rotate about the tip touching the ground due to rotation of hip joint about axis 1.

    I am a confused how to approach this problem.

    Looking forward to the solution!

    Thanks :)
    n one leg 3 axes.png model stage 1.jpg
  2. jcsd
  3. Jun 20, 2015 #2


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    Welcome to PF.
    The legs you have drawn appear to have planar hip and knee joints so I do not see how it can walk anywhere in 3+3 mode. I guess you will sort that out later.

    There are two things at play here. The moment due to the static body mass and the position of the leg must be countered by torque from the servo motor, and the dynamics of the leg components and body movement. The static moment is not a "moment of inertia" which is only needed for dynamic computations. Here I consider only the static force.

    Consider the body mass, m, typically one third, m/3, will be carried by each leg as a force, f = m/3 * g. The horizontal distance, d, along the surface from below the hip to the foot contact is effectively a torque arm. The moment at the hip will be m/3 * g * d. That assumes the foot does not slide on the surface.The worst case value for d will be the total length of the leg.

    It gets more complex if you include the mass of the leg components in the static calculation because the position of the centroid of each leg will decide how the leg mass is distributed between the foot and the hip. Then there is also the mass of the servo actuators etc.

  4. Jun 21, 2015 #3
    Thanks for the reply Baluncore!

    For static equilibrium I calculated restoring moments about axis 2 and 3, which will make sure the robot stands.
    Here are its calculations,
    torque diagram.png
    This diagram shows one leg and half chassis.
    W1= force acting at the centre of chassis

    W2= force due to weight of vertical joint servo

    W3= force due to weight of 1st revolute joint servo

    W3= force due to weight of link 2

    W5= force due to weight of 2nd revolute joint servo

    W6= force due to weight of link 3

    N= normal reaction from ground

    L= distance between chassis centre and axis of vertical joint ( hip joint)

    l1, l2, l3 lengths of link 1, 2 and 3

    a1, a2 : angles made by link 2 and link 3 with the horizontal

    Simply by using following conditions for static equilibrium I calculated torque at the two revolute joints at point B and C
    Σ Fy = 0
    Σ MB = TB
    Σ MC = TC

    Now what remains is the hip joint at point A. Does the torque of hip joint contribute to static equilibrium? assuming the the robot is on a flat surface with no inclination.

    In dynamics, the hip joint performs 2 motions:
    1. moving the leg forward in air ( lifting and placing on the ground is done by other servo)
    2. Pushing the body ahead when 3 legs are on the ground (three hip joints will act together)

    for first case I think i can simply use ,
    torque= I x alpha, knowing mass moment of inertia of one leg and angular acceleration from speed.

    What I cannot figure out is calculation of 2nd case.
    What concepts and equations do i need to apply there?
  5. Jun 21, 2015 #4


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    A human hip joint is a ball and socket joint. You show only one degree of motion possible in the robot hip joint.
    Once the three feet are placed on the surface there is no possibility of horizontal body movement without foot slip.
    The centre of the body must remain above a fixed point on the surface. Rotation of the body would not be possible.
    If movement is prohibited by insufficient degrees of freedom in the structure then there can be no dynamics analysis.
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