Snellius–Pothenot problem

Denoting the (unknown) angles {{math|∠CAP}} as {{mvar|x}} and {{math|∠CBP}} as {{mvar|y}} gives:

$$x+y\; =\; 2\; \backslash pi\; -\; \backslash alpha\; -\; \backslash beta\; -\; C$$

by using the sum of the angles formula for the quadrilateral {{mvar|PACB}}. The variable {{mvar|C}} represents the (known) internal angle in this quadrilateral at point {{mvar|C}}. (Note that in the case where the points {{mvar|C}} and {{mvar|P}} are on the same side of the line {{mvar|AB}}, the angle {{math|∠C}} will be greater than {{pi}}).

Applying the law of sines in triangles {{math|△PAC}} and {{math|△PBC}}, we can express {{overline|{{mvar|PC}}}} in two different ways:

$$\backslash overline\{PC\}\; =\; \backslash frac\{\backslash overline\{AC\}\; \backslash sin\; x\; \}\{\backslash sin\; \backslash alpha\}\; =\; \backslash frac\{\backslash overline\{BC\}\; \backslash sin\; y\}\{\backslash sin\; \backslash beta\}.$$

A useful trick at this point is to define an auxiliary angle {{mvar|φ}} such that

$$\backslash tan\; \backslash phi\; =\; \backslash frac\{\{\backslash rm\; BC\}\; \backslash sin\; \backslash alpha\}\{\{\backslash rm\; AC\}\; \backslash sin\; \backslash beta\}.$$

(A minor note: one should be concerned about division by zero, but consider that the problem is symmetric, so if one of the two given angles is zero one can, if needed, rename that angle {{mvar|α}} and call the other (non-zero) angle {{mvar|β}}, reversing the roles of {{mvar|A}} and {{mvar|B}} as well. This will suffice to guarantee that the ratio above is well defined. An alternative approach to the zero angle problem is given in the algorithm below.)

With this substitution the equation becomes

$$\backslash frac\{\backslash sin\; x\}\{\backslash sin\; y\}=\backslash tan\; \backslash phi.$$

Now two known trigonometric identities can be used, namely

\tan \left(\tfrac{\pi}{4} - \phi\right) = \frac{1 - \tan\phi}{\tan\phi+1}\ , \qquad

\frac{\tan\tfrac12(x-y)}{\tan\tfrac12(x+y)} = \frac{\sin x - \sin y}{\sin x + \sin y}\ ,

to put this in the form of the second equation;

$$\backslash tan\backslash tfrac12(x-y)\; =\; \backslash tan\backslash tfrac12(\backslash alpha\; +\; \backslash beta\; +\; C)\; \backslash tan\; \backslash left(\backslash tfrac\{\backslash pi\}\{4\}-\backslash phi\backslash right).$$

Now these two equations in two unknowns must be solved. Once {{mvar|x}} and {{mvar|y}} are known the various triangles can be solved straightforwardly to determine the position of {{mvar|P}}.Bowser: A treatise The detailed procedure is shown below.

Given are two lengths {{mvar|{{overline|AC}}, {{overline|BC}}}}, and three angles {{mvar|α, β, C}}, the solution proceeds as follows.

• calculate $\backslash phi=\; \backslash mathsf\{atan2\}\backslash left(\; \backslash overline\{BC\}\; \backslash sin\; \backslash alpha,\backslash \; \backslash overline\{AC\}\; \backslash sin\backslash beta\; \backslash right),$ where atan2 is a computer function, also called the arctangent of two arguments, that returns the arctangent of the ratio of the two values given. Note that in Microsoft Excel the two arguments are reversed, so the proper syntax would be = atan2(AC*\sin(beta), BC*\sin(alpha)). The atan2 function correctly handles the case where one of the two arguments is zero.
• calculate $K\; =\; 2\; \backslash pi\; -\backslash alpha-\backslash beta-C.$
• calculate $W\; =\; 2\; \backslash cdot\; \backslash arctan\backslash left[\; \backslash tan(\backslash tfrac\{\backslash pi\}\{4\}\; -\; \backslash phi)\; \backslash ,\; \backslash tan\backslash tfrac12(\backslash alpha+\backslash beta+C)\backslash right].$
• find $x\; =\; \backslash frac\{K+W\}\{2\},\; \backslash \; y\; =\; \backslash frac\{K-W\}\{2\}.$
• find $\backslash overline\{PC\}\; =$

\begin{cases}

\dfrac{\overline{BC} \sin y}{\sin \beta} & \text{if } |\sin \beta| > |\sin \alpha|, \\[4pt]

\dfrac{\overline{AC} \sin x}{\sin \alpha} & \text{otherwise.}

\end{cases}

• find $\backslash overline\{PA\}\; =\; \backslash sqrt\{\backslash overline\{AC\}^2\; +\; \backslash overline\{PC\}^2\; -\; 2\; \backslash cdot\; \backslash overline\{AC\}\; \backslash cdot\; \backslash overline\{PC\}\; \backslash cdot\; \backslash cos(\backslash pi-\backslash alpha-x)\}.$ (This comes from the law of cosines.)
• find $\backslash overline\{PB\}\; =\; \backslash sqrt\{\backslash overline\{BC\}^2\; +\; \backslash overline\{PC\}^2\; -\; 2\backslash cdot\; \backslash overline\{BC\}\; \backslash cdot\; \backslash overline\{PC\}\; \backslash cdot\; \backslash cos(\backslash pi-\backslash beta-y)\}.$

If the coordinates of $A:\; x\_A,y\_A$ and $C:\; x\_C,y\_C$ are known in some appropriate Cartesian coordinate system then the coordinates of {{mvar|P}} can be found as well.

By the inscribed angle theorem the locus of points from which {{overline|{{mvar|AC}}}} subtends an angle {{mvar|α}} is a circle having its center on the midline of {{overline|{{mvar|AC}}}}; from the center {{mvar|O}} of this circle, {{overline|{{mvar|AC}}}} subtends an angle {{math|2α}}. Similarly the locus of points from which {{overline|{{mvar|CB}}}} subtends an angle {{mvar|β}} is another circle. The desired point {{mvar|P}} is at the intersection of these two loci.

Therefore, on a map or nautical chart showing the points {{mvar|A, B, C}}, the following graphical construction can be used:

• Draw the segment {{overline|{{mvar|AC}}}}, the midpoint {{mvar|M}} and the midline, which crosses {{overline|{{mvar|AC}}}} perpendicularly at {{mvar|M}}. On this line find the point {{mvar|O}} such that $\backslash overline\{MO\}\; =\; \backslash tfrac\{\backslash overline\{AC\}\}\{2\; \backslash tan\; \backslash alpha\}.$ Draw the circle with center at {{mvar|O}} passing through {{mvar|A}} and {{mvar|C}}.
• Repeat the same construction with points {{mvar|B, C}} and the angle {{mvar|β}}.
• Mark {{mvar|P}} at the intersection of the two circles (the two circles intersect at two points; one intersection point is {{mvar|C}} and the other is the desired point {{mvar|P}}.)

This method of solution is sometimes called Cassini's method.

The following solution is based upon a paper by N. J. Wildberger.{{cite journal |title=Greek Geometry, Rational Trigonometry, and the Snellius – Pothenot Surveying Problem |journal=Chamchuri Journal of Mathematics |volume=2 |number=2 |pages=1–14 |year=2010

|url=http://www.math.sc.chula.ac.th/cjm/sites/www.math.sc.chula.ac.th.cjm/files/01-CJM-Wildberger-2011-Mar-03_0.pdf |author=Norman J. Wildberger}} It has the advantage that it is almost purely algebraic. The only place trigonometry is used is in converting the angles to spreads. There is only one square root required.

• define the following:$$\backslash begin\{align\}$$

& s(x) = \sin^2x \\[4pt]

& A(x,y,z) = (x+y+z)^2 - 2(x^2 + y^2 + z^2) \\[2pt]

&\begin{alignat}{5}

r_1 &= s(\beta) &\qquad Q_1 &= \overline{BC}^2 \\

r_2 &= s(\alpha) & Q_2 &= \overline{AC}^2 \\

r_3 &= s(\alpha + \beta) & Q_3 &= \overline{AB}^2

\end{alignat}\end{align}

• now let:$$\backslash begin\{align\}$$

R_1 &= \frac{r_2 Q_3}{r_3} \qquad R_2 = \frac{r_1 Q_3}{r_3} \\[4pt]

C_0 &= \frac{(Q_1 + Q_2 + Q_3) (r_1 + r_2 + r_3) - 2 (Q_1 r_1 + Q_2 r_2 + Q_3 r_3)}{2 r_3} \\[4pt]

D_0 &= \frac{r_1 r_2 A(Q_1,Q_2,Q_3)}{r_3}

\end{align}

• the following equation gives two possible values for {{math|R{{sub|3}}}}:$$(R\_3\; -\; C\_0)^2\; =\; D\_0$$
• choosing the larger of these values, let:$$\backslash begin\{align\}$$

v_1 &= 1 - \frac{(R_1 + R_3 - Q_2)^2}{4 R_1 R_3} \\[4pt]

v_2 &= 1 - \frac{(R_2 + R_3 - Q_1)^2}{4 R_2 R_3}

\end{align}

finally:$$\backslash begin\{align\}$$

\overline{AP}^2 &= \frac{v_1 R_1}{r_2} = \frac{v_1 Q_3}{r_3} \\[4pt]

\overline{BP}^2 &= \frac{v_2 R_2}{r_1} = \frac{v_2 Q_3}{r_3}

\end{align}

Ventura et al. {{Cite journal |last=Ventura |first=Jorge |last2=Martinez |first2=Fernando |last3=Manzano-Agugliaro |first3=Francisco |last4=Návrat |first4=Aleš |last5=Hrdina |first5=Jaroslav |last6=Eid |first6=Ahmad H. |last7=Montoya |first7=Francisco G. |date=2024-05-27 |title=A novel geometric method based on conformal geometric algebra applied to the resection problem in two and three dimensions |url=https://doi.org/10.1007/s00190-024-01854-1 |journal=Journal of Geodesy |language=en |volume=98 |issue=6 |pages=47 |doi=10.1007/s00190-024-01854-1 |issn=1432-1394|doi-access=free |hdl=11012/249518 |hdl-access=free }} solve the planar and three-dimensional Snellius-Pothenot problem via Vector Geometric Algebra and Conformal Geometric Algebra. The authors also characterize the solutions' sensitivity to measurement errors.

When the point {{mvar|P}} happens to be located on the same circle as {{mvar|A, B, C}}, the problem has an infinite number of solutions; the reason is that from any other point {{mvar|P'}} located on the arc {{overarc|{{mvar|APB}}}} of this circle the observer sees the same angles {{mvar|α}} and {{mvar|β}} as from {{mvar|P}} (inscribed angle theorem). Thus the solution in this case is not uniquely determined.

The circle through {{mvar|ABC}} is known as the "danger circle", and observations made on (or very close to) this circle should be avoided. It is helpful to plot this circle on a map before making the observations.

A theorem on cyclic quadrilaterals is helpful in detecting the indeterminate situation. The quadrilateral {{mvar|APBC}} is cyclic iff a pair of opposite angles (such as the angle at P and the angle at {{mvar|C}}) are supplementary i.e. iff $\backslash alpha+\backslash beta+C\; =\; k\; \backslash pi,\; (k=1,2,\backslash cdots)$. If this condition is observed the computer/spreadsheet calculations should be stopped and an error message ("indeterminate case") returned.

(Adapted form Bowser,Bowser: A treatise exercise 140, page 203). {{mvar|A, B, C}} are three objects such that {{overline|{{mvar|AC}}}} = 435 (yards), {{overline|{{mvar|CB}}}} = 320, and {{math|∠C}} = 255.8 degrees. From a station {{mvar|P}} it is observed that {{math|∠APC}} = 30 degrees and {{math|∠CPB}} = 15 degrees. Find the distances of {{mvar|P}} from {{mvar|A, B, C}}. (Note that in this case the points {{mvar|C}} and {{mvar|P}} are on the same side of the line {{mvar|AB}}, a different configuration from the one shown in the figure).

Answer: {{overline|{{mvar|PA}}}} = 790, {{overline|{{mvar|PB}}}} = 777, {{overline|{{mvar|PC}}}} = 502.

A slightly more challenging test case for a computer program uses the same data but this time with {{math|∠CPB}} = 0. The program should return the answers 843, 1157 and 837.

File:Plaquete huis Willebrord Snellius.jpg

The British authority on geodesy, George Tyrrell McCaw (1870–1942) wrote that the proper term in English was Snellius problem, while Snellius-Pothenot was the continental European usage.{{cite journal |first=G. T. |last=McCaw |title=Resection in Survey |journal=The Geographical Journal |volume=52 |issue=2 |year=1918 |pages=105–126 |doi= 10.2307/1779558|jstor=1779558 |url=https://zenodo.org/record/1449312 }}

McCaw thought the name of Laurent Pothenot (1650–1732) did not deserve to be included as he had made no original contribution, but merely restated Snellius 75 years later.

• Solution of triangles
• Triangulation (surveying)

{{reflist}}

• Gerhard Heindl: Analysing Willerding’s formula for solving the planar three point resection problem, Journal of Applied Geodesy, Band 13, Heft 1, Seiten 27–31, ISSN (Online) 1862-9024, ISSN (Print) 1862-9016, DOI: [https://doi.org/10.1515/jag-2018-0037]

• Edward A. Bowser: A treatise on plane and spherical trigonometry, Washington D.C., Heath & Co., 1892, page 188 [https://books.google.com/books?id=9MlHAAAAIAAJ Google books]

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Category:Trigonometry

Category:Surveying

Category:Mathematical problems