Triangulation in three dimensions: Difference between revisions

Revision as of 03:54, 17 May 2011

Revision as of 05:32, 17 May 2011

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Suppose three sticks of known length are anchored in the ground at known coordinates. 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'''. '''D'''' is the vector to the coordinates of the apex where the three sticks would meet below the plane of '''A, B, C''' as well. This problem also calculates the axis of symmetry of three intersecting spheres. See the article [[Trilateration|Trilateration]] for the spherical case. Even if the three spheres don't intersect, this problem still calculates their axis of symmetry. This axis is the line drawn between '''D''' and '''D''''. To proceed, three [[sphere]]s of known centers '''A, B, C''' and known [[Radius|radii]] ''AD, BD, CD'' intersect at two points '''D''' and '''D''''. Similarly, three sticks of known lengths ''AD, BD, CD'' are planted in the ground at known [[coordinates]] '''A, B, C'''. The other ends meet at a calculated [[apex]]. 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,

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[[image: Triangulation illusttriangulation_illust_02.gif]]

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:<math>\mathbf

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</math><br />

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[[image: FacesABD_ACD_BCDfacesABD_ACD_BCD_2.gif]]

::<math>

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

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

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[[image: Intersect_normals_ABC_2Intersect_normals_ABC_3.gif]]

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

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== Example ==

[[image: Example_00_triangExample_00_triang_02.gif]]

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== See also ==