The Snellius–Pothenot problem is a problem in planar surveying. Given three known points A, B and C, an observer at an unknown point P observes that the segment AC subtends an angle and the segment CB subtends an angle ; the problem is to determine the position of the point P.. Since it involves the observation of known points from an unknown point, the problem is an example of resection. Historically it was first studied by Snellius, who found a solution around 1615.
Formulating the equations
First equation Denoting the angles CAP as x and CBP as y we get: by using the sum of the angles formula for the quadrilateralPACB. The variable C represents the internal angle in this quadrilateral at point C. . Second equation Applying the law of sines in triangles PAC and PBC we can express PC in two different ways: A useful trick at this point is to define an auxiliary angle such that With this substitution the equation becomes We can use two known trigonometric identities, namely to put this in the form of the second equation we need We now need to solve these two equations in two unknowns. Once x and y are known the various triangles can be solved straightforwardly to determine the position of P. The detailed procedure is shown below.
Solution algorithm
Given are two lengths AC and BC, and three angles, and C, the solution proceeds as follows.
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If the coordinates of A: xA,yA and C: xC,yC are known in some appropriate Cartesian coordinate system then the coordinates of P can be found as well.
Geometric (graphical) solution
By the inscribed angle theorem the locus of points from which AC subtends an angle is a circle having its center on the midline of AC; from the center O of this circle AC subtends an angle. Similarly the locus of points from which CB subtends an angle is another circle. The desired point P is at the intersection of these two loci. Therefore, on a map or nautical chart showing the points A, B, C, the following graphical construction can be used:
Draw the segment AC, the midpoint M and the midline, which crosses AC perpendicularly at M. On this line find the point O such that. Draw the circle with center at O passing through A and C.
Repeat the same construction with points B, C and the angle.
Mark P at the intersection of the two circles
This method of solution is sometimes called Cassini's method.
The following solution is based upon a paper by N. 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:
now let:
the following equation gives two possible values for :
choosing the larger of these values, let:
finally we get:
The indeterminate case
When the point P happens to be located on the same circle as A, B and C, the problem has an infinite number of solutions; the reason is that from any other point P' located on the arc APB of this circle the observer sees the same angles alpha and beta as from P. Thus the solution in this case is not uniquely determined. The circle through ABC is known as the "danger circle", and observations made on 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 APBC is cyclic iff a pair of opposite angles are supplementary i.e. iff. If this condition is observed the computer/spreadsheet calculations should be stopped and an error message returned.
Solved examples
. A, B and C are three objects such that AC = 435, CB = 320, and C = 255.8 degrees. From a station P it is observed that APC = 30 degrees and CPB = 15 degrees. Find the distances of P from A, B and C.. Answer: PA = 790, PB = 777, PC = 502. A slightly more challenging test case for a computer program uses the same data but this time with CPB = 0. The program should return the answers 843, 1157 and 837.
Naming controversy
The British authority on geodesy, George Tyrrell McCaw wrote that the proper term in English was Snellius problem, while Snellius-Pothenot was the continental European usage. McCaw thought the name of Laurent Pothenot did not deserve to be included as he had made no original contribution, but merely restated Snellius 75 years later.
