Resection by three point problem

Three-point problem: given three well-defined points A, B and C, plotted on the sheet as a, b, c, locate on the sheet the station P where the table is set up and from which A, B and C are all visible. It is a resection method of plane tabling. Three solutions are used:

1. Mechanical method (tracing paper method)
2. Graphical method (Bessel's method, or the method of perpendiculars)
3. Trial and error method (Lehmann's method)

Mechanical (tracing paper) method

1. Fix a sheet of tracing paper over the drawing sheet. Mark a point p on it to represent the station P.
2. Set up the table at P, level it, and orient it roughly with the compass.
3. With the alidade centred on p, sight A, B and C and draw rays on the tracing paper.
4. Loosen the tracing paper and move it until the three rays pass through a, b and c at the same time.
5. Prick the point p through onto the drawing sheet. This gives the position of P.
6. Orient the table by placing the alidade along p-a and turning the board until A is bisected. Check by sighting B and C.

      a         b         c      (on sheet)
       \        |        /
        \       |       /
         \      |      /
            rays from p (tracing paper)

Trial and error (Lehmann's) method

1. Set up the table at P, level it and orient it by eye or compass as accurately as possible.
2. Pivoting on a, b and c, sight A, B and C and draw the three rays back. If the orientation is wrong, they form a small triangle of error.
3. Estimate the position of P (the point p) inside or near the triangle using the rules below. Re-orient the board by placing the alidade along p-a and turning the table until A is sighted.
4. Repeat until the triangle of error reduces to a point. That point is the position of P.

Lehmann's rules for locating p:

1. If P lies inside the great triangle ABC, p lies inside the triangle of error; otherwise it lies outside it.
2. When p is outside, it lies on the same side of all three rays, that is, to the right of all of them or to the left of all of them, when looking along each ray towards its station.
3. The distance of p from each ray is proportional to the distance of that station from the table. So p is nearest to the ray of the nearest station and farthest from the ray of the farthest station.

     ray a        ray b
         \      /
          \____/
          /\  /\   <- small triangle
         /  \/  \     of error
      ray c

Strong and weak positions

• The position is well fixed when P is inside the great triangle ABC, or outside it but well away from the circle through A, B and C.
• If P lies on or very near the circle through A, B and C (the "danger circle"), the three rays meet in a single point for any orientation of the table, so the problem has no definite solution. Another set of points should then be chosen.