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Derivation of the equations of APF

  1. Apr 14, 2015 #1
    I am working to use the artificial potential field method for path planning of mobile robot; actually I found in one of references the following description about this method:

    the artificial potential field method uses a scalar function called the potential function. This function has two values, a minimum value, when the mobile robot is at the goal point and a high value on obstacles. The function slopes down towards the target point, so that the mobile robot can reach the target by following the negative gradient of the potential field. The potential force has two components: the first one is attractive force and second one is repulsive force. The goal position generates an attractive force which makes the mobile robot move towards it while obstacles produce a repulsive force, the combination of the attractive force to the destination and the repulsive forces away from the obstacles drive the mobile robot in a safe path to the target point

    The attractive potential takes the form:

    Uatt (q)=1/2 * ζ * d2 (q,qgoal) … (1)

    Where ζ is proportional coefficient , d(q,qgoal) is the Euclidean distance between the mobile robot q and the position of the goal point qgoal. The attractive force on robot is determined as the negative gradient of attractive potential field and takes the following form

    Fatt (q)=-∇Uatt (q) =- ζ (q - qgoal) …(2)

    Fatt(q) is a vector directed toward qgoal with magnitude linearly related to the distance from q to qgoal.

    The repulsive function is defined as :

    Urep (q) = 1/2 * ƞ * [1/d(q,qobs) - 1/d0 ]2 ........... if d(q,qobs )≤ d0
    ... (3)

    Where q is the robot position and qobs is the obstacle position. d0 is the positive constant denoting the distance of influence of the obstacle. d(q,qobs) The distance between the mobile robot and obstacle. ƞ is the proportional coefficient. The repulsive force is the negative gradient of this repulsive potential fields function.

    Frep (q)=-∇Urep (q) = ƞ * [(1/d(q,qobs ) - 1/d0] * [(q-qobs)/ d3(q,qobs)] ......... if d(q,qobs )≤ d0

    ... (4)

    My question is about equations 2 and 4 which they represent the negative gradient of equations 1 and 3 respectively, as you know that negative gradient of function is the derivative of the function, but when I am trying to derivative equations 1 and 3 that didn't give the same result in the equations 2 and 4 , so could anyone help me in the problem?
  2. jcsd
  3. Apr 16, 2015 #2
    Why there is no any answer for my question? Is my question ambiguous and unclear ???

    I really need help , Please
  4. Apr 16, 2015 #3
    You could perhaps try and show us what you have done. What do you get instead of eq 2 and 4?
  5. Apr 16, 2015 #4
    Thanks a lot for reply

    Actually, according to derivation rules and from my poor knowledge in derivation of equations when I derived equation (1) the result was:

    Fatt (q)=-∇Uatt (q) = - ζ * d(q,qgoal) and this result doesn't match equation (2)

    when I derived equation (3) the result was :

    Frep (q)=-∇Urep (q)= - [1/d(q,qobs) - 1/d0] * [ (d(q,qgoal))` / d2(q,qgoal) ]

    and also this result doesn't match equation (4)
  6. Apr 17, 2015 #5
    Do you know what the function ##d(q,q_{goal})## is?
  7. Apr 17, 2015 #6
    d(q,qgoal) is the Euclidean distance between the mobile robot q and the position of the goal point qgoal

    I think the problem is in this type of function , how to derive this type ?
  8. Apr 17, 2015 #7
    How have you differentiated (1) if you do not know how ## d(q,q_{goal} ## looks like?
  9. Apr 17, 2015 #8
    I have not differentiated (1), I found it in reference book with its derivation (eq. 2) and when I try to derive it the result was not match with eq. 2 . and also with eq. 3
  10. Apr 18, 2015 #9
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