Hip joint torque in walking mechanism

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Discussion Overview

The discussion revolves around calculating the torque required at the hip joint of a hexapod walking mechanism, specifically focusing on the tripod gait. Participants explore the dynamics and static equilibrium of the robot's leg movements, addressing both theoretical and practical aspects of torque calculations in relation to the robot's design and motion.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant proposes using the formula Torque = I x alpha for calculating the torque at the hip joint, suggesting that the inertia should be divided by three since three joints act together.
  • Another participant emphasizes the need to consider static moments due to body mass and the position of the legs, arguing that torque calculations must account for the forces acting on each leg and the dynamics involved.
  • A participant describes their approach to static equilibrium, detailing calculations for restoring moments and forces acting on various components of the robot, questioning the contribution of the hip joint torque to static equilibrium.
  • Concerns are raised about the limitations of the robot's hip joint design, noting that it only allows for one degree of motion, which may restrict the robot's ability to move horizontally without slipping.

Areas of Agreement / Disagreement

Participants express differing views on the adequacy of the hip joint's design and the calculations for torque. There is no consensus on the best approach to determine the torque required for the hip joint, and the discussion remains unresolved regarding the dynamics of the robot's movement.

Contextual Notes

Limitations include assumptions about the robot's design and the degrees of freedom in the hip joint, as well as the complexity of including the mass of leg components in static calculations. The discussion also highlights the need for further exploration of the dynamics involved in the robot's movement.

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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
 
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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.

https://en.wikipedia.org/wiki/Moment_(physics)
 
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?
 
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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