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3.1 Force and Moment Transformation
Description of the angular velocity of link ί+1 with respect from {ί+1},
.
ί+1 ω ί+1 = ί+1 R ί ω ί + θ ί+1 ί+1 Ž ί+1 (3.1)
ί
The linear velocity of the origin of frame {ί+1} is the same as that of the origin of from {ί} plus a new component caused by rotational velocity of link ί. Therefore it will become:
ί ν ί+1 = ί ν ί + ί ω ί X ί P ί +1 (3.2)
By multiplying both side with ί+1 R this can be compute:
ί
ί ν ί+1 = ί+1 R ( ί ν ί + ί ω ί X ί P ί +1 ) (3.3)
ί
Write (3.1) and (3.2) in matrix form to transform general velocity vectors in frame {A} to their description in frame {B}. Since the two frames involved here are rigidly connected θ ί+1 appearing in (3.1) is set to zero:
B ν B B R B R PBORG X A ν A A A
=
(3.4)
B ω B 0 BR A ω A
A
Then the cross product of the matrix operator:
0 Px Py
PX = Px 0 Px (3.5)
Px Px 0
(3.4) is relates velocity in one frame to those in another, this is called a velocity transformation and symbol use as Tν . The velocity transformation which maps velocities in {A} in to velocities in {B}, (3.4) can be express in compactly as:
B ν B = A Tν . A ν A (3.6)
B
By invert (3.4) compute the description of velocity in terms of {A} when given the values in {B} :
A ν A = B Tν . B ν B (3.7)
A
This mapping of velocities form frame to frame depend on A T must be interpreted as instantaneous results unless the relationship between the two frames is B
static. From (3.8) and (3.9) transfer general force vectors written in terms of {B} into frame {A}:
ί ƒ ί = ί R ί+1 ƒ ί+1 (3.8)
ί+1
ί n ί = ίR ί+1 n ί+1 + ί P ί+1 x ί ƒ ί (3.9)
ί+1
A F A A R 0 B F B
= B
A N A A R PBORG X AR A R B N B (3.10)
B B B
This can be express in compactly as:
A F A = A Tƒ . B F B (3.11)
B
Tƒ is force and moment transformation. Velocity and force transformation are similar they relate and forces in different coordinate system. So this can be expressed as:
B Tƒ = A Tν T
A B