Re: Two vectors, two coordinate systems
- From: "Androcles" <Headmaster@xxxxxxxxxxxxxxxx>
- Date: Thu, 29 May 2008 19:17:33 +0100
<animalover1@xxxxxxxx> wrote in message
news:65480d7f-8b9b-4582-9166-5561cfb3a5c8@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
|I have a question that I would like to ask the group given. 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!
What you need is a translation matrix combined with a rotation matrix.
http://www.kwon3d.com/theory/transform/transform.html
--
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
1/2[tau(A)+tau(A')]= tau(B)
where
A = (0,0,0,t)
A' =(0,0,0,t+x'/(c-v) +x'/(c+v))
B = (x',0,0,t+x'/(c-v))
x' = x-vt
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
"Easy. He did not say that." - cretin van lintel.
Androcles
.
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