Triangulation in three dimensions: Difference between revisions

Revision as of 23:50, 21 May 2011

Revision as of 00:09, 22 May 2011

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== Development ==

Three sticks of known lengths ''AD, BD, CD'' are anchored in the ground at known coordinates '''A, B, C'''. This development calculates the coordinates of the apex where the other ends of the three sticks will meet. These coordinates are given by the vector '''D'''. In the mirror case, '''D'''' is sub-apex where the three sticks would meet below the plane of '''A, B, C''' as well.

Calculating '''D''' and '''D'''', the projection<ref>Borisenko, A. I. and Tarapov, I. E., (1968) "Vector and Tensor Analysis", General Publishing Company, p. 6. ISBN 0-486-63833-2</ref> of '''AD''' onto '''AB''' and '''AC''', and the projection of '''BD''' onto '''BC''' results in,

<br />

[[image: triangulation_illust_02.gif]]

<br />

::By the [[law of cosines]].,

</math><br />

::<math>

(BD)^2 = (AB)^2+(AD)^2-2(AB)(AD)\cos(\angle{BAD})

</math><br />

::<math>

(CD)^2 = (AC)^2+(AD)^2-2(AC)(AD)\cos(\angle{CAD})

</math><br /><br />

::<math>

(CD)^2 = (BC)^2+(BD)^2-2(BC)(BD)\cos(\angle{CBD})

</math><br /><br />

<br />

Calculating '''D''' and '''D'''', theThe projection<ref>Borisenko, A. I. and Tarapov, I. E., (1968) "Vector and Tensor Analysis", General Publishing Company, p. 6. ISBN 0-486-63833-2</ref> of '''AD''' onto '''AB''' and '''AC''', and the projection of '''BD''' onto '''BC''' results in,

[[image: facesABD_ACD_BCD_2.gif]]

<br />

:<math>\mathbf

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{M_{BC}} = \mathbf{B}+ BD\cos(\angle{CBD})\dfrac{\mathbf{BC}}{\left \Vert \mathbf{BC} \right \|}=\mathbf{B}+ \left [\dfrac{(BD)^2+(BC)^2-(CD)^2}{2(BC)^2} \right ]\mathbf{BC}

</math><br />

<br />

[[image: facesABD_ACD_BCD_2.gif]]

::<math>

(BD)^2 = (AB)^2+(AD)^2-2(AB)(AD)\cos(\angle{BAD})

</math><br />

::<math>

(CD)^2 = (AC)^2+(AD)^2-2(AC)(AD)\cos(\angle{CAD})

</math><br />

::<math>

(CD)^2 = (BC)^2+(BD)^2-2(BC)(BD)\cos(\angle{CBD})

</math><br /><br />

::By the [[law of cosines]].

<br />

<br />

The three [[unit normal]]s to '''AB''', '''AC''' and '''BC''' in the [[Plane (geometry)|plane]] of ABC are:

<br /><br />

[[image: Intersect_normals_ABC_3.gif]]

<br />

:<math>

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

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\mathbf{M_{BC}}+m_{BC}\mathbf{N_{BC}}</math>

<br />

[[image: Intersect_normals_ABC_3.gif]]

<br />

Solving for ''m_AB'', ''m_AC'' and ''m_BC''