Snellius–Pothenot problem
Given three known points A, B, C, an observer at an unknown point P observes that the line segment AC subtends an angle α and the segment CB subtends an angle β; the problem is to determine the position of the point P. (See figure; the point denoted C is between A and B as seen from P).
Historically it was first studied by Snellius, who found a solution around 1615.
by using the sum of the angles formula for the quadrilateral PACB.
The variable C represents the (known) internal angle in this quadrilateral at point C. (Note that in the case where the points C and P are on the same side of the line AB, the angle ∠C will be greater than π).
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 φ such that
(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 α and call the other (non-zero) angle β, reversing the roles of A and B as well.
An alternative approach to the zero angle problem is given in the algorithm below.)
Once x and y are known the various triangles can be solved straightforwardly to determine the position of P.[1] The detailed procedure is shown below.
Given are two lengths AC, BC, and three angles α, β, C, the solution proceeds as follows.
By the inscribed angle theorem the locus of points from which AC subtends an angle α is a circle having its center on the midline of AC; from the center O of this circle, AC subtends an angle 2α.
Similarly the locus of points from which CB subtends an angle β 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: This method of solution is sometimes called Cassini's method.
The only place trigonometry is used is in converting the angles to spreads.
Ventura et al. [3] 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 P happens to be located on the same circle as A, B, 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 α and β as from P (inscribed angle theorem).
Thus the solution in this case is not uniquely determined.
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.
If this condition is observed the computer/spreadsheet calculations should be stopped and an error message ("indeterminate case") returned.
(Adapted form Bowser,[4] exercise 140, page 203).
A, B, C are three objects such that AC = 435 (yards), CB = 320, and ∠C = 255.8 degrees.
Find the distances of P from A, B, C. (Note that in this case the points C and P are on the same side of the line AB, a different configuration from the one shown in the figure).
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 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.
[5] 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.
