CogoKit

Traverse adjustment

Locating a new point by intersection from two traverse stations

A point on the far side of a river is fixed by directions observed from two established traverse stations, then verified independently by the sine rule and by a third-angle check.

Apply· about 24 minutes by hand· 5 steps

Given

  • Two established traverse stations: STA 12 at N 8240.65, E 6115.30 and STA 13 at N 8905.44, E 6742.88, US survey feet.
  • A new point P, a rock outcrop across a river, cannot be occupied or taped. It is visible from both stations.
  • At STA 12, with a backsight on STA 13, the angle turned clockwise to P is 31°54′20″.
  • At STA 13, with a backsight on STA 12, the angle turned counter-clockwise to P is 84°45′10″.
  • Both angles were turned in two sets with a 5 second instrument.
  • Coordinates to 0.01 ft, angles to 1 second.

Required

  • The azimuth and length of the base line STA 12 to STA 13.
  • Azimuths from each station to P.
  • The coordinates of P.
  • The distances from each station to P.

Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.

Worked solution

Base line by inverse

The base line of an intersection is the line between the two occupied stations, and it comes from the published coordinates rather than from any field measurement. Its length and direction carry directly into the strength of the fix.

Use the surveying inverse: azimuth = atan2(ΔE, ΔN) with northing first.

ΔN = 8905.44 − 8240.65 = 664.79 ft ΔE = 6742.88 − 6115.30 = 627.58 ft distance = √(664.79² + 627.58²) = √(441,945.7 + 393,856.7) = √835,802.4 = 914.22 ft azimuth STA 12 to STA 13 = atan2(627.58, 664.79) = 43°21′03″ back azimuth STA 13 to STA 12 = 43°21′03″ + 180° = 223°21′03″

Azimuths to the new point

At STA 12 the angle was turned clockwise from the backsight, so it adds to the base azimuth. At STA 13 the angle was turned counter-clockwise, so it subtracts from the back azimuth.

Booking the hand of each angle is the whole of the field discipline here. Reverse one of them and the two rays intersect on the wrong side of the base line, giving a plausible-looking point in entirely the wrong place. A rough sketch to scale, made before the crew leaves the site, is the defence.

azimuth STA 12 to P = 43°21′03″ + 31°54′20″ = 75°15′23″ azimuth STA 13 to P = 223°21′03″ − 84°45′10″ = 138°35′53″

Intersect the two rays

Two lines, each given by a point and a direction, meet where the two parametric expressions for position agree. Solve for the distance along the first ray, then run forward from STA 12.

Writing u for the unit vector of a direction, with components cos of the azimuth in northing and sin of the azimuth in easting, the distance along the first ray is the cross product of the base vector with the second direction, divided by the cross product of the two directions.

u₁ = (cos 75°15′23″, sin 75°15′23″) = (0.254494, 0.967074) u₂ = (cos 138°35′53″, sin 138°35′53″) = (−0.750089, 0.661337) denominator = u₁ₑ u₂ₙ − u₁ₙ u₂ₑ = (0.967074)(−0.750089) − (0.254494)(0.661337) = −0.893698 t₁ = 1018.68 ft along azimuth 75°15′23″ from STA 12 N(P) = 8240.65 + 1018.68 cos 75°15′23″ = 8240.65 + 259.25 = 8499.90 ft E(P) = 6115.30 + 1018.68 sin 75°15′23″ = 6115.30 + 985.14 = 7100.44 ft

Distances to P

Inverse from each station to the computed point. These are the numbers a later crew will use to set the point out again, or to check it with a reflectorless observation.

The angle at P, which no one observed, is what remains of the triangle: 180 degrees less the two observed angles.

STA 12 to P: ΔN = 259.25, ΔE = 985.14, distance = 1018.68 ft, azimuth 75°15′23″ STA 13 to P: ΔN = −405.54, ΔE = 357.56, distance = 540.66 ft, azimuth 138°35′53″ angle at P = 180°00′00″ − 31°54′20″ − 84°45′10″ = 63°20′30″

Strength of the fix

An intersection is strongest when the two rays cross near a right angle and weakest when they are nearly parallel. Here they cross at 63°20′30″, which is comfortable: an error of 5 seconds in either observed angle moves P by about 0.03 ft.

Estimate that directly. Five seconds is 0.0000242 radians; times the 1018.68 ft ray from STA 12 gives 0.025 ft of lateral movement, and the geometry divides that by the sine of the intersection angle, 0.8937, to give about 0.028 ft along the ground.

If the intersection angle had been 15 degrees instead, the same 5 second error would produce 0.095 ft, and at 5 degrees it would produce 0.28 ft. That sensitivity, not the instrument, is what governs whether an intersection is worth doing.

5″ = 0.0000242 radians lateral shift on the longer ray = 0.0000242 × 1018.68 = 0.025 ft divided by sin(63°20′30″) = 0.8937: about 0.028 ft on the ground

Answer

  • Base line STA 12 to STA 13: 914.22 ft on azimuth 43°21′03″.
  • Azimuth STA 12 to P = 75°15′23″; azimuth STA 13 to P = 138°35′53″.
  • P is at N 8499.90, E 7100.44.
  • STA 12 to P = 1018.68 ft; STA 13 to P = 540.66 ft.
  • The rays intersect at 63°20′30″, a strong fix; a 5 second error in either direction moves P about 0.03 ft.

Check

Solve the same triangle by the sine rule, which uses no coordinates at all. STA 12 to P = 914.22 × sin(84°45′10″) / sin(63°20′30″) = 914.22 × 0.995809 / 0.893698 = 1018.68 ft, and STA 13 to P = 914.22 × sin(31°54′20″) / sin(63°20′30″) = 914.22 × 0.528521 / 0.893698 = 540.66 ft. Both match the coordinate solution to the hundredth.

The three angles of the triangle must sum to 180 degrees: 31°54′20″ + 84°45′10″ + 63°20′30″ = 180°00′00″.

Inverse back from P to each station and confirm the reverse azimuths: P to STA 12 is 75°15′23″ + 180° = 255°15′23″, and P to STA 13 is 138°35′53″ + 180° = 318°35′53″. Running 1018.68 ft from P on 255°15′23″ returns to N 8240.65, E 6115.30.

A field check worth having: if any single distance to P can be measured, even roughly, compare it with the computed one. A reflectorless shot from STA 13 reading within a tenth of 540.66 ft confirms that neither angle was booked with the wrong hand, which is the error this method is most exposed to.

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