Two vectors, Two coordinate systems

I have a question that I would like to ask the group given I am not an engineer. This is not a homework problem, but I am trying to figure out how to determine the angle between two cables attached to two rigid structures. Please let me explain what I am talking about. I have two vectors (2 cables attached to a fixed beam - they will always have tension and will never deform out of shape of a line) and the other end they both are attached to a non-fixed box (I guess you can think of it as rubber cords on the wall attached to a box). I will always know the orientation of one cable (CABLE A) that is, the angle relative to the box (coordinate system of the box). I know the INITIAL angle of the second cable (CABLE B), however, the angle is the angle relative to the beam. I would like to know if there is a way to solve the angle of CABLE B in the boxes coordinate system at 1) the initial position and 2) if the box is moved to a new location. This seems like it would be simple, but I don't know how to solve this. So, I started it by trying to transform the vector of CABLE B into the coordinate system of the box (going from the fixed beam to the box). I think that would give me the initial position/angle of CABLE B in the boxes coordinate system. I really do not know how to write the transformation matrix to do this, but think it would be something

CABLE B X Y * [Rotation Matrix] = CABLE B's new XYZ coordinates Z

Then how would I need to get the angels with respect to the boxes coordinate system.

Then if the box moves, how would this all work as well? Please help!

Reply to
animalover1
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Dear animalov...:

On May 29, 9:23=A0am, snipped-for-privacy@live.com wrote: =2E..

=2E..

Treat the two cables as rigid members, the points where they connect as pin (swivel) joints, the two tie points on the box as a single rigid member, and the two tie points "in the world" or "on the crane" as a rigid member.

Then your problem simplifies to a classical "four bar mechanism" or "four bar linkage".

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David A. Smith

Reply to
dlzc

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