Content deleted Content added
m replacing File:Triangulation illust 02.gif with commons equivalent File:Triangulation illust.gif (report errors here) |
m WP:CHECKWIKI error fixes + general fixes using AWB (7896) |
||
Line 3:
| orphan=April 2011
| confusing=April 2011
▲{{expert|date=April 2011}}
'''Triangulation in three dimensions''' is a method of finding the ___location of a point in [[three dimensional|three dimensions]] based on other known coordinates and distances, it is commonly used in [[surveying]] and [[astronomy]].
Line 13 ⟶ 12:
== Development ==
[[File:Triangulation illust.gif|600px|thumb|Figure 1.
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.
<br />
Line 23 ⟶ 22:
:<math>
(BD)^2 = (AB)^2+(AD)^2-2(AB)(AD)\cos(\angle{BAD})
</math
:<math>
(CD)^2 = (AC)^2+(AD)^2-2(AC)(AD)\cos(\angle{CAD})
</math
:<math>
(CD)^2 = (BC)^2+(BD)^2-2(BC)(BD)\cos(\angle{CBD})
</math
<br />
<br />
Line 37 ⟶ 35:
<br />
[[image:
<br />
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,
Line 43 ⟶ 41:
:<math>\mathbf
{M_{AB}} = \mathbf{A}+ AD\cos(\angle{BAD})\dfrac{\mathbf{AB}}{\left \Vert \mathbf{AB} \right \|}=\mathbf{A}+ \left [\dfrac{(AD)^2+(AB)^2-(BD)^2}{2(AB)^2} \right ]\mathbf{AB}
</math
:<math>\mathbf
{M_{AC}} = \mathbf{A}+ AD\cos(\angle{CAD})\dfrac{\mathbf{AC}}{\left \Vert \mathbf{AC} \right \|}=\mathbf{A}+ \left [\dfrac{(AD)^2+(AC)^2-(CD)^2}{2(AC)^2} \right ]\mathbf{AC}
</math
:<math>\mathbf
{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:
The three [[unit normal]]s to '''AB''', '''AC''' and '''BC''' in the [[Plane (geometry)|plane]] of ABC are:
<br />
Line 70 ⟶ 67:
\mathbf{N_{BC}}=\cfrac{\mathbf{AB}-\cfrac{\mathbf{AB}\bullet\mathbf{BC}}{(BC)^2}\mathbf{BC}}{\left \Vert{ \mathbf{AB}-\cfrac{\mathbf{AB}\bullet\mathbf{BC}}{(BC)^2}\mathbf{BC}} \right \|}
</math>
Then the three [[vector]]s intersect at a common point:
:<math>\mathbf{M_{AB}}+m_{AB}\mathbf{N_{AB}}=
\mathbf{M_{AC}}+m_{AC}\mathbf{N_{AC}}=
Line 80 ⟶ 76:
<br />
Solving for ''m<sub>AB</sub>'', ''m<sub>AC</sub>'' and ''m<sub>BC</sub>''
:<math>\begin{vmatrix}
m_{AB} \\
Line 87 ⟶ 83:
\end{vmatrix}
=(H^{T}H)^{-1}H^{T}\mathbf{g}</math>
== Spreadsheet formula ==
A [[spreadsheet]] command for calculating this is,
Line 93 ⟶ 89:
:PRODUCT(PRODUCT(MINVERSE(PRODUCT(TRANSPOSE H, H)), TRANSPOSE H), '''g''')
An example of a spreadsheet that does complete calculations of this entire problem is given at the '''External links''' section at the end of this article.
The matrix H and the matrix '''g''' in this ''[[least squares]] solution''<ref>Leon, Steven J. (1980) "Linear Algebra", Macmillan Publishing Co., Inc., p. 152. ISBN 0-02-369870-5</ref> are,
:<math>
H=
Line 124 ⟶ 120:
M_{BCz}-M_{ABz} \\
\end{vmatrix}
</math>
<br />
Alternatively, solve the system of equations for ''m<sub>AB</sub>'', ''m<sub>AC</sub>'' and ''m<sub>BC</sub>'':
<math>
\begin{align}
Line 135 ⟶ 131:
\end{align}
</math><br />
The unit normal to the plane of ABC is,
<br />
:<math>\mathbf{N_D}=\dfrac{\mathbf{AC}\times\mathbf{AB}}{\left \Vert{\mathbf{AC}\times\mathbf{AB}} \right \|}</math>
== Solution ==
Line 152 ⟶ 147:
\end{cases}
</math>
<math>
Line 164 ⟶ 157:
</math>
<br />
Line 170 ⟶ 162:
<br />
:<math>
M_{AB}D=AD\sqrt{1- \left [\dfrac{(AD)^2+(AB)^2-(BD)^2}{(AB)(AD)} \right ]^2}</math
:<math>
M_{AC}D=AD\sqrt{1- \left [\dfrac{(AD)^2+(AC)^2-(BD)^2}{(AC)(AD)} \right ]^2}</math
:<math>
M_{BC}D=BD\sqrt{1- \left [\dfrac{(BD)^2+(BC)^2-(CD)^2}{(BC)(BD)} \right ]^2}</math
== Condition for intersection ==
Line 182 ⟶ 172:
If ''AD, BD, CD'' are assigned according to the arrangement,
<br />
:<math>AD \le BD \le CD</math
Then ''AD, BD, CD'' intersect if and only if,<br />
:<math>AD+BD \ge AB \ge BD-AD</math
:<math>AD+CD \ge AC \ge CD-AD</math
:<math>BD+CD \ge BC \ge CD-BD</math
<br />
Viz, if <br />
Line 193 ⟶ 183:
''CD''=r<sub>C</sub>=radius of sphere centered at '''C''',<br />
such that, <br />
:<math>r_A \le r_B \le r_C</math
then the three spheres intersect if and only if,<br />
:<math>r_A+r_B \ge AB \ge r_B-r_A</math
:<math>r_A+r_C \ge AC \ge r_C-r_A</math
:<math>r_B+r_C \ge BC \ge r_C-r_B</math
== Decoding vector formulas ==
:<math>
\mathbf{A}=(x_A, y_A, z_A)</math
:<math>
\mathbf{B}=(x_B, y_B, z_B)</math
:<math>
\mathbf{C}=(x_C, y_C, z_C)</math
<br />
:<math>
\mathbf{AB}=(x_B-x_A, y_B-y_A, z_B-z_A)</math
:<math>
\mathbf{AC}=(x_C-x_A, y_C-y_A, z_C-z_A)</math
:<math>
\mathbf{BC}=(x_C-x_B, y_C-y_B, z_C-z_B)</math
<br />
:<math>\mathbf{AC}\bullet\mathbf{AB}=(x_C-x_A)(x_B-x_A)+(y_C-y_A)(y_B-y_A)+(z_C-z_A)(z_B-z_A)</math
:<math>\mathbf{AB}\bullet\mathbf{AC}=(x_C-x_A)(x_B-x_A)+(y_C-y_A)(y_B-y_A)+(z_C-z_A)(z_B-z_A)</math
:<math>\mathbf{AB}\bullet\mathbf{BC}=(x_B-x_A)(x_C-x_B)+(y_B-y_A)(y_C-y_B)+(z_B-z_A)(z_C-z_B)</math
<br />
:<math>
AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2}</math
:<math>
AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2+(z_C-z_A)^2}</math
:<math>
BC=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2+(z_C-z_B)^2}</math
<br />
:<math>
\left \Vert{ \mathbf{AC}-\cfrac{\mathbf{AC}\bullet\mathbf{AB}}{(AB)^2}\mathbf{AB}} \right \|
=\sqrt{(AC)^2-\cfrac{(\mathbf{AC}\bullet\mathbf{AB})^2}{(AB)^2}}</math
<br />
:<math>
\left \Vert{ \mathbf{AB}-\cfrac{\mathbf{AB}\bullet\mathbf{AC}}{(AC)^2}\mathbf{AC}} \right \|
=\sqrt{(AB)^2-\cfrac{(\mathbf{AB}\bullet\mathbf{AC})^2}{(AC)^2}}</math
<br />
:<math>
\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 />
:<math>
\mathbf{M_{AB}}=(M_{ABx},M_{ABy},M_{ABz})</math
:<math>
\mathbf{M_{AC}}=(M_{ACx},M_{ACy},M_{ACz})</math
:<math>
\mathbf{M_{BC}}=(M_{BCx},M_{BCy},M_{BCz})</math
<br />
:<math>
Line 254 ⟶ 243:
:<math>
{M_{ABz}} = {z_A}+ \left [\dfrac{(AD)^2+(AB)^2-(BD)^2}{2(AB)^2} \right ](z_B-z_A)
</math
:<math>
Line 265 ⟶ 253:
:<math>
{M_{ACz}} = {z_A}+ \left [\dfrac{(AD)^2+(AC)^2-(CD)^2}{2(AC)^2} \right ](z_C-z_A)
</math
:<math>
Line 276 ⟶ 263:
:<math>
{M_{BCz}} = {z_B}+ \left [\dfrac{(BD)^2+(BC)^2-(CD)^2}{2(BC)^2} \right ](z_C-z_B)
</math
:<math>
\mathbf{N_{AB}}=(N_{ABx},N_{ABy},N_{ABz})</math
:<math>
\mathbf{N_{AC}}=(N_{ACx},N_{ACy},N_{ACz})</math
:<math>
\mathbf{N_{BC}}=(N_{BCx},N_{BCy},N_{BCz})</math
:<math>
{N_{ABx}}=\cfrac{{(x_C-x_A)}-\cfrac{\mathbf{AC}\bullet\mathbf{AB}}{(AB)^2}(x_B-x_A)}{\sqrt{(AC)^2-\cfrac{(\mathbf{AC}\bullet\mathbf{AB})^2}{(AB)^2}}}
</math
:<math>
{N_{ABy}}=\cfrac{{(y_C-y_A)}-\cfrac{\mathbf{AC}\bullet\mathbf{AB}}{(AB)^2}(y_B-y_A)}{\sqrt{(AC)^2-\cfrac{(\mathbf{AC}\bullet\mathbf{AB})^2}{(AB)^2}}}
</math
:<math>
{N_{ABz}}=\cfrac{{(z_C-z_A)}-\cfrac{\mathbf{AC}\bullet\mathbf{AB}}{(AB)^2}(z_B-z_A)}{\sqrt{(AC)^2-\cfrac{(\mathbf{AC}\bullet\mathbf{AB})^2}{(AB)^2}}}
</math
:<math>
{N_{ACx}}=\cfrac{{(x_B-x_A)}-\cfrac{\mathbf{AB}\bullet\mathbf{AC}}{(AC)^2}(x_C-x_A)}{\sqrt{(AB)^2-\cfrac{(\mathbf{AB}\bullet\mathbf{AC})^2}{(AC)^2}}}
</math
:<math>
{N_{ACy}}=\cfrac{{(y_B-y_A)}-\cfrac{\mathbf{AB}\bullet\mathbf{AC}}{(AC)^2}(y_C-y_A)}{\sqrt{(AB)^2-\cfrac{(\mathbf{AB}\bullet\mathbf{AC})^2}{(AC)^2}}}
</math
:<math>
{N_{ACz}}=\cfrac{{(z_B-z_A)}-\cfrac{\mathbf{AB}\bullet\mathbf{AC}}{(AC)^2}(z_C-z_A)}{\sqrt{(AB)^2-\cfrac{(\mathbf{AB}\bullet\mathbf{AC})^2}{(AC)^2}}}
</math
:<math>
{N_{BCx}}=\cfrac{{(x_B-x_A)}-\cfrac{\mathbf{AB}\bullet\mathbf{BC}}{(BC)^2}(x_C-x_B)}{\sqrt{(AB)^2-\cfrac{(\mathbf{AB}\bullet\mathbf{BC})^2}{(BC)^2}}}
</math
:<math>
{N_{BCy}}=\cfrac{{(y_B-y_A)}-\cfrac{\mathbf{AB}\bullet\mathbf{BC}}{(BC)^2}(y_C-y_B)}{\sqrt{(AB)^2-\cfrac{(\mathbf{AB}\bullet\mathbf{BC})^2}{(BC)^2}}}
</math
:<math>
{N_{BCz}}=\cfrac{{(z_B-z_A)}-\cfrac{\mathbf{AB}\bullet\mathbf{BC}}{(BC)^2}(z_C-z_B)}{\sqrt{(AB)^2-\cfrac{(\mathbf{AB}\bullet\mathbf{BC})^2}{(BC)^2}}}
</math
<br />
The equation of the line of the axis of symmetery of 3 spheres is,
:<math>
\cfrac{x-(M_{ABx}+m_{AB}N_{ABx})}{(y_C-y_A)(z_B-z_A)-(y_B-y_A)(z_C-z_A)}=
\cfrac{y-(M_{ABy}+m_{AB}N_{ABy})}{(x_B-x_A)(z_C-z_A)-(x_C-x_A)(z_B-z_A)}=
\cfrac{z-(M_{ABz}+m_{AB}N_{ABz})}{(x_C-x_A)(y_B-y_A)-(x_B-x_A)(y_C-y_A)}
</math
<br />
:<math>
Line 334 ⟶ 317:
x_B-x_A & y_B-y_A & z_B-z_A \\
\end{vmatrix}
</math
<br />
:<math>
=\Big((y_C-y_A)(z_B-z_A)-(y_B-y_A)(z_C-z_A),\ (x_B-x_A)(z_C-z_A)-(x_C-x_A)(z_B-z_A),\ (x_C-x_A)(y_B-y_A)-(x_B-x_A)(y_C-y_A)\Big)
</math
<br />
:<math>
Line 347 ⟶ 330:
\Big((x_C-x_A)(y_B-y_A)-(x_B-x_A)(y_C-y_A)\Big)^2
\end{Bmatrix}^\frac{1}{2}
</math
== Example ==
Line 375 ⟶ 358:
* [[Linear algebra]]
* [[Matrix algebra]]
== References ==
<!--- See [[Wikipedia:Footnotes]] on how to create references using <ref></ref> tags which will then appear here automatically -->
Line 380 ⟶ 364:
<!--- Categories --->
== External links ==▼
* [http://jons-math.bravehost.com/triangulation/triangulation03.xls Example of a spreadsheet that calculates solutions to this problem]▼
[[Category:Mathematics articles related to geometry]]
[[Category:Vector calculus]]
[[Category:Linear algebra]]
[[Category:Trigonometry]]
▲== External links ==
▲* [http://jons-math.bravehost.com/triangulation/triangulation03.xls Example of a spreadsheet that calculates solutions to this problem]
| |||
