Accelerating reference frame

by MechatronO
Tags: accelerating, frame, reference
MechatronO is offline
May5-13, 09:14 AM
P: 25
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.
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A.T. is offline
May5-13, 12:40 PM
P: 3,543
You need to include inertial forces:
MechatronO is offline
May6-13, 02:22 PM
P: 25
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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