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Classical Physics
Rigid body mechanics and coordinate frames
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[QUOTE="AbsoluteUnit, post: 6626246, member: 698957"] Hello all, I have some issues understanding the inertial-frame (or global-frame, G-frame) versus the body-frame (B-frame) when it comes to simulating the motion of a rigid body in 2 dimensions (planar body mechanics) in a system of ODEs. I have been self-learning from textbooks on simulating rigid body motion, so I sometimes misunderstand concepts, since some books are a bit vague when they assume the reader already have the technical knowledge to make the necessary connections. I hope someone can clear up any misunderstandings that I have here. --------------- Given a state space, Q and its gradient Q' describing the planar mechanics of a rigid body: Q = [ V[SIZE=3]Bx[/SIZE](t), V[SIZE=3]By[/SIZE](t), ψ(t) ][sup]T[/sup] ... (1) Q' = [ a[SIZE=3]Bx[/SIZE](t), a[SIZE=3]Bx[/SIZE](t), ψ'(t) ][sup]T[/sup] ... (2) [SIZE=4]Where[/SIZE] [INDENT]Bx := Quantity expressed in body-frame (B-frame) coordinates[/INDENT] [INDENT]V := Velocity[/INDENT] [INDENT]a(t) := Acceleration[/INDENT] [INDENT]ψ, ψ'(t) := Body angular velocity and acceleration (respectively)[/INDENT] The values of a(t) and ψ'(t) are being generated by some process in the B-frame at each time point (t) as forces and moments, e.g. vehicle accelerating/braking and steering. Essentially, [LIST] [*]Some books (seem to) suggest that it is possible to perform integration directly[B] in the B-frame [/B]on Q'(t) (ref. (2)) to solve for Q(t) (ref. (2)) at every time point, t. Then, the G-frame positions of the rigid body, (X,Y,ψ) can be obtained by rotating Q to the G-frame coordinates via the transformation: [/LIST] [CENTER]V[SIZE=3]Gx[/SIZE] = V[SIZE=3]Bx[/SIZE]*cos(ψ) - V[SIZE=3]By[/SIZE]*sin(ψ) V[SIZE=3]Gy[/SIZE] = V[SIZE=3]Bx[/SIZE]*sin(ψ) + V[SIZE=3]By[/SIZE]*cos(ψ)[/CENTER] [INDENT=2]Then integrate V[SIZE=3]Gx[/SIZE], V[SIZE=3]Gy[/SIZE] to get (X,Y) in G-frame.[/INDENT] [INDENT=2][/INDENT] In short: [CENTER]Fully solve the ODE in the B-frame first, then rotate the velocities to the G-frame and integrate to obtain the G-frame positions: (X,Y,ψ)[/CENTER] --------------- Part of me thinks that this interpretation of mine doesn't really make sense, because Newton's First Law only applies directly in the G-frame, but not in the B-frame. I suspect that for every time point, t, one should instead: [LIST=1] [*]Compute the forces and moments that give rise to a[SIZE=3]Bx[/SIZE](t), a[SIZE=3]Bx[/SIZE](t), and ψ'(t) [*]Transform (a[SUB]Bx[/SUB](t), a[SUB]Bx[/SUB](t), ψ'(t)) to G-frame coordinates [*]Solve for V[sub]Gx[/sub], V[sub]Gy[/sub], ψ'(t) [/LIST] So that at every time point, t, the forces are applied in the Inertial frame (instead of the body frame). Is that the correct way to think about this? Or is there a technicality that I am not getting here and both methods are actually equivalent? Thank you for your insights, AU [/QUOTE]
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