Triangulation in three dimensions: Difference between revisions

Revision as of 03:38, 16 May 2011

Revision as of 12:37, 16 May 2011

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\left \Vert{ \mathbf{AB}-\cfrac{\mathbf{AB}\bullet\mathbf{BC}}{(BC)^2}\mathbf{BC}} \right \|

=\sqrt{(AB)^2-\cfrac{(\mathbf{AB}\bullet\mathbf{BC})^2}{(BC)^2}}</math><br />

<br />

:<math>

\mathbf{M_{AB}}=(M_{ABx},M_{ABy},M_{ABz})</math><br />

:<math>

\mathbf{M_{AC}}=(M_{ACx},M_{ACy},M_{ACz})</math><br />

:<math>

\mathbf{M_{BC}}=(M_{BCx},M_{BCy},M_{BCz})</math><br />

<br />

:<math>

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:<math>

{M_{BCz}} = {z_B}+ \left [\dfrac{(BD)^2+(BC)^2-(CD)^2}{2(BC)^2} \right ](z_C-z_B)

</math><br />

<br /><br />

:<math>

\mathbf{N_{AB}}=(N_{ABx},N_{ABy},N_{ABz})</math><br />

:<math>

\mathbf{N_{AC}}=(N_{ACx},N_{ACy},N_{ACz})</math><br />

:<math>

\mathbf{N_{BC}}=(N_{BCx},N_{BCy},N_{BCz})</math><br />

<br />

:<math>

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<br />

== Discussion ==

This is a robust solution incorporating all possibilities and accommodating non-intersecting cases with the closest answer using the least squares calculation of matrix H and '''b'''.

While the same result can be obtained using only a few of these many equations, all possibilities are treated here. If the three spheres do not intersect, the least squares solution finds the axis of symmetry of the three spheres, or the closest solution. If the three spheres do intersect, the solution requires only a few of these equations.<br />

<br />

== See also ==

* [[Sphere]]