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Inverse and direct surveying problems online

Inverse problem: from the coordinates of two points find the grid bearing, quadrant bearing and horizontal distance. Direct problem: from a point, a bearing and a distance find the coordinates of the second point. The solution is shown step by step with the quadrant check.

Point 1: X₁, Y₁, m
Point 2: X₂, Y₂, m
Calculate

Direct and inverse problems

These are the two basic computations of plane surveying. The X axis points north and Y east; directions are given by the grid bearing α, measured clockwise from grid north, 0° to 360°.

  • Direct problem: given X₁, Y₁, bearing α and distance d, find X₂, Y₂.
  • Inverse problem: given two points, find α and d — e.g. the initial bearing of a traverse or setting-out data.

Direct problem

ΔX = d·cos α, ΔY = d·sin α, X₂ = X₁ + ΔX, Y₂ = Y₁ + ΔY.

Inverse problem

ΔX = X₂ − X₁, ΔY = Y₂ − Y₁; the quadrant bearing r = arctan (ΔY / ΔX) taken as an absolute value; the quadrant from the signs:

ΔXΔYQuadrantBearing
++NEα = r
−+SEα = 180° − r
−−SWα = 180° + r
+−NWα = 360° − r

d = √(ΔX² + ΔY²), check d = ΔX / cos α = ΔY / sin α; the back bearing is α ± 180°.

Example

X₁ = 6243.17, Y₁ = 4127.52, X₂ = 6098.44, Y₂ = 4305.91 m. ΔX = −144.73, ΔY = +178.39, r = 50°56′50″, quadrant SE, α = 180° − r = 129°03′10″, d = 229.717 m.

FAQ

How do I find the quadrant in the inverse problem?
From the signs of the increments: ΔX > 0, ΔY > 0 — NE (α = r); ΔX < 0, ΔY > 0 — SE (α = 180° − r); ΔX < 0, ΔY < 0 — SW (α = 180° + r); ΔX > 0, ΔY < 0 — NW (α = 360° − r).
What is the difference between a bearing and an azimuth?
An azimuth is measured from true north, a grid bearing from grid north (the X axis). They differ by the grid convergence γ: α = A − γ.
What is a quadrant bearing?
The acute angle (0°–90°) from the north or south end of the X axis to the line, with the quadrant name, e.g. S 50°56′50″ E (SE).
How do I find the distance between two points from coordinates?
d = √((X₂ − X₁)² + (Y₂ − Y₁)²). This is the horizontal distance.