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Accelerating reference frame

  1. May 5, 2013 #1
    I'm attempting to build a line follower robot and I'm currently in the process of building appropriate models.

    For the control system I need to define a coordinate system. The most convinient coordinate system from many point of views would be a coordinate system that moves along and changes direction with the robot, thus a rotating and accelerating reference frame.

    The question is if calculations regarding acceleration still would be valid if they are carried out in the same way as in a fixed reference frame.

    The calculations to be carried out are:

    ƩM = Jω' - Angular acceleration related to net torque applied
    ƩF = ma - Acceleration of center of gravity related to net force applied


    I've glanced some about information regarding the coriolis effect but I dont really understand it yet. No "loose" object are to be treated in the reference frame.
     
  2. jcsd
  3. May 5, 2013 #2

    A.T.

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  4. May 6, 2013 #3
    Aha. That was an interesting article.

    The force observed from an arbitrary accelerating and rotating coordinatesystem is

    Fb= Fa + F[itex]_{fic}[/itex]

    F[itex]_{fic}[/itex] = -(m[itex]_{ab}[/itex] + 2mƩv[itex]_{j}[/itex]u'[itex]_{j}[/itex] + mƩx[itex]_{j}[/itex]u´´[itex]_{j}[/itex])

    Fb is the appearent force that an observer in a rotating reference frame would think is acting on an object, while F is the "real" force an observer in an inertial reference frame would see and Ffic is the fictional force coming from the movements of the ref. system and m[itex]_{ab}[/itex] is the acceleration of the ref. system.

    I however want a coordinatsystem that is fixed both in position and angle to the robot at a point on the robot which defines position [0,0,0].
    The position and velocity in its "own" coordinatesystem would thus be 0.
    Will this zero all terms in the Ffic and leave Fb = F - m[itex]_{ab}[/itex] in this particular case?


    As the robot would see the acceleration and in combination the force "on itself" in this system as zero we would get back F = m[itex]_{ab}[/itex] if the world of math smiles to me this time?
     
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