Whatever.
I don't do pissing contests, so I'll be moving on.....
Here are some things the reader will need to know to follow the discussion to come:
A general vector is of the form xi +yj + zk with 3 degrees of freedom.
A class of normalized vectors (vectors of unit length 1) representing all possible directions in 3-D can be written with only 2 degrees of freedom, using angles theta, phi as defined in conventional spherical coordinates as shown
For my purposes, I will change the definiition of theta: Theta is the angle the unit vector makes with the plane z=0. Theta=0 means a horizontal viewing vector
p in all my equations.
For all unit vectors r=1. The reader can easily see that the class all possible unit vectors in the above graphic would form a spherical shell of radius 1. Any point on the shell represents a unique, different direction and can be represented by a specific phi, theta pair of numbers
for all -90<=theta<=90
and all -180=phi<180
Rewritten in terms of 2 angular coordinates any directional vector becomes
A*cos(theta)*cos(phi)
i + cos(theta)*sin(phi)
j + sin(theta)
k
The advantage to using coords phi, theta is that both angles have a simple, practical meaning in the case of WTC1 antenna orientation and any derived displacement vectors.
>>>>>>>>>>>>>>>>>>>>>.
Vector projected towards a viewer: We are interested in how any vector in 3-D is projected onto any viewpoint.
Vector equations for basic perpendicular and parallel vector projections of any vector onto any projection are given in the first section at the following link:
http://en.wikipedia.org/wiki/Vector_projection
We want the thing called "vector rejection" in the very first section. And here she is:
Displacement vector definition:
a is what I will call D(from frame #, to frame #)
The displacement vector D between point in frame 120 and 220, for example, can be written..
D(120, 220)
This defines a unique vector in 3-D.
Coordinate system definition: Center of antenna before leaning is the positive z axis, basic cartesian system with positive x through the south wall outward and positive y through the east wall outward. Origin undertermined until necessary.
Viewing vectors definition: In the vector rejection equation b(with a "hat") is a unit vector which points from the eyes of an observer from any particular viewpoint to the object being looked at, in this case various points on the WTC1 upper perimeter and antenna. But instead of "b", I will call my viewing vectors p. We will have 3 projection vectors, all of unit length and pointing from the cameras of the Sauret, NBC NW and NBC NE viewpoints to the objects being looked at.
Projection vectors can be written as
p(phi, theta)
or
p = A*[cos(theta)*cos(phi)i + cos(theta)*sin(phi)j + sin(theta)k]
where A is just a constant to normalize the vector.
>>>>>>>>>>>
So, the projection of vector D into any viewpoint is
D(projection) = D - (D*p)p
>>>>>>>>>>>>>..
3 working equations become:
D(Sauret projection) = D - (D*p)p
D(NW projection) = D - (D*p)p
D(NE projection) = D - (D*p)p
Each equation has a different p.
Two of these equations should be enough to determine D completely, but three is more fun. It's a party.
We can define the unit vector d as being parallel to D but of unit length,
so D =( magnitude of D)*d and we can do the same for all plane projections of D
>>>>>>>>>>>>>>>>>>>>>>
For any viewpoint, the "vector rejection" can be derived for any vector D as
D(projection) = D - (D*p)p. we can write p in terms of it's components as
p = pxi + pyj + pzk, where px^2 + py^2 + pz^2 = 1
and D = Dxi +Dyj + Dzk
The vector equation becomes 3 scalar equations by equating coefficients. These equations can be written in matrix form rather nicely because they are all linear in Dx, Dy and Dz.
This gives us a 3x3 matrix where each coeficient is written in terms of px, py and pz only.
It can be thought of as a "transformation matrix", transforming any 3 coefficients Dx, Dy, Dz to their "vector rejection" coefficients. It would be more correct to call it a vector operator.
Matrix entries:
a(1,1) =1-px^2 ....... a(1,2)= -px*py....... a(1,3)= -px*pz
a(2,1)= -px*py....... a(2,2)=1-py^2 ....... a(2,3)= -py*pz
a(3,1)= -px*pz....... a(3,2)= -py*pz....... a(3,3)= 1-pz^2
if the coefficients are rewritten as px=p1, py=p2 and pz=p3, the coefficients can be written
a(m,m)=1-pm^2 for m = 1, 2, 3
and for the cross terms,
a(m,n) = -pm*pn for all m, n from 1 to 3 where m, n are not equal.
Basically the 3x3 matrix operates on a 1x3 matrix with coefficients Dx, Dy, Dz, changing the 1x3 matrix into the components of the corresponding "vector rejection".
If you plug in any D vector in terms of it's components, after the matrix operation you get the 3 components of it's "vector rejection".
It is important to remember that many different vectors
D can map onto the exact same "vector rejection", so there is no 1:1 correspondence between Dx, Dy, Dz and the components of it's vector rejection. We would never expect to be able to reconstruct an exact vector D from a single rejection. Common sense tells us that we need at least 2 viewpoints to reconstruct a specific vector D from knowledge of it's vector rejections since whole classes of vectors can look identical from a single viewpoint.
A vector rejection has lost information of it's original vector that it cannot get back. This means the 3x3 vector operator cannot be inverted to yield unique Dx, Dy, Dz values from knowledge of it's vector rejection. There is no inverse operation that allows us to revert the vector back to it's original state.
By choosing the viewer coordinate system carefully, the vector components labelled "projection" and "rejection" can take on a very simple form, as well the viewing vector
p
