CogoKit

Traverse adjustment

Traverse with two missing lengths solved from the closure conditions

Two sides of a five-sided traverse have known directions but no measured lengths. The two closure conditions give a pair of linear equations that solve both lengths at once.

Analyze· about 26 minutes by hand· 5 steps

Given

  • A closed five-sided traverse A-B-C-D-E-A. All five directions were observed; two of the five lengths could not be measured because a river crosses both lines.
  • A-B azimuth 62°15′00″ 425.60 ft
  • B-C azimuth 152°20′00″ 316.85 ft
  • C-D azimuth 235°10′00″ length unknown
  • D-E azimuth 281°30′00″ 254.75 ft
  • E-A azimuth 340°45′00″ length unknown
  • The angles have already been adjusted, so the azimuths are treated as exact.
  • Distances to 0.01 ft.

Required

  • The two missing lengths, C-D and E-A.
  • The complete latitude and departure table.
  • A statement of when this method fails.

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

Set up the two conditions

The traverse closes, so the latitudes sum to zero and the departures sum to zero. Each unknown length enters both conditions, multiplied by the cosine of its azimuth in the latitude equation and by the sine in the departure equation.

Writing x for the length of C-D and y for the length of E-A, the two conditions are a pair of simultaneous linear equations. Two conditions, two unknowns: the problem is determinate, which is why exactly two lengths and no more can be recovered this way.

latitude: Σlat(known) + x cos(235°10′00″) + y cos(340°45′00″) = 0 departure: Σdep(known) + x sin(235°10′00″) + y sin(340°45′00″) = 0 cos(235°10′00″) = −0.571191, sin(235°10′00″) = −0.820817 cos(340°45′00″) = +0.944089, sin(340°45′00″) = −0.329691

Latitudes and departures of the measured courses

Resolve the three courses whose lengths are known. Their sums are the constants in the two equations.

Carry three decimals. The determinant of the system is close to unity here, so the solution is well conditioned and three decimals is ample, but that is a property of these two azimuths rather than a general rule.

The three measured courses
CourseAzimuthDistance (ft)Latitude (ft)Departure (ft)
A-B62°15′00″425.60198.166376.651
B-C152°20′00″316.85−280.623147.122
D-E281°30′00″254.7550.789−249.636
Sums997.20−31.668274.137

Solve the pair of equations

Substituting the sums gives two equations in x and y. Solve them by determinants or by elimination; determinants are less error prone by hand because each quantity is written down once.

Both lengths come out positive, which they must: a negative length would mean the direction of that course had been taken backwards.

−0.571191 x + 0.944089 y = 31.668 −0.820817 x − 0.329691 y = −274.137 determinant = (−0.571191)(−0.329691) − (0.944089)(−0.820817) = 0.188316 + 0.774925 = 0.963241 x = [31.668 × (−0.329691) − 0.944089 × (−274.137)] / 0.963241 = 248.369 / 0.963241 = 257.85 ft y = [(−0.571191)(−274.137) − (−0.820817)(31.668)] / 0.963241 = 182.578 / 0.963241 = 189.55 ft

Complete the table

Put the two solved lengths back into the traverse, rounded to 0.01 ft as they will be recorded, and resolve their latitudes and departures. Both columns must now sum to zero, and they do to within four thousandths of a foot.

That near-zero is not an independent check. It is the equation that was solved. The few thousandths that remain are only the rounding of the two computed lengths from 257.847 and 189.546 ft to the hundredth.

Completed traverse
CourseAzimuthDistance (ft)Latitude (ft)Departure (ft)
A-B62°15′00″425.60198.166376.651
B-C152°20′00″316.85−280.623147.122
C-D235°10′00″257.85−147.282−211.648
D-E281°30′00″254.7550.789−249.636
E-A340°45′00″189.55178.952−62.493
Sums1444.600.002−0.004

When the method fails

The determinant is the sine of the angle between the two unknown directions. Here that angle is 340°45′00″ − 235°10′00″ = 105°35′00″, whose sine is 0.9632, close to the best possible value of 1 at ninety degrees.

As the two unknown courses approach parallel, the determinant approaches zero and the solution blows up. Two parallel unknown courses cannot be separated at all: any length added to one and subtracted from the other closes equally well. If the two missing courses are within about twenty degrees of parallel, the computed lengths are worthless no matter how carefully the arithmetic is done.

The other failure mode is silent. As with any omitted measurement, all the error in the three measured courses is absorbed by the two computed ones, and no closure remains to reveal it. Measure a diagonal if there is any way to do so.

angle between the unknown directions = 340°45′00″ − 235°10′00″ = 105°35′00″ sin(105°35′00″) = 0.963241, which is the determinant computed above

Answer

  • C-D = 257.85 ft (257.847 ft before rounding).
  • E-A = 189.55 ft (189.546 ft before rounding).
  • Completed latitudes: 198.166, −280.623, −147.282, 50.789, 178.952 ft, summing to 0.002.
  • Completed departures: 376.651, 147.122, −211.648, −249.636, −62.493 ft, summing to −0.004.
  • Total perimeter 1444.60 ft.

Check

Substitute both rounded lengths back into the original equations, not into the table. Latitude: −31.668 + 257.85(−0.571191) + 189.55(0.944089) = −31.668 − 147.282 + 178.952 = 0.002. Departure: 274.137 + 257.85(−0.820817) + 189.55(−0.329691) = 274.137 − 211.648 − 62.493 = −0.004. Both residuals are the rounding of the lengths and nothing else.

Verify the determinant a second way. It must equal the sine of the angle between the two unknown directions: sin(105°35′00″) = 0.963241, which is the value computed from the four trigonometric functions.

Both answers must be positive lengths of plausible size for the figure. They are, at 257.85 ft and 189.55 ft against a perimeter of 1444.60 ft.

Coordinate check: accumulating all five courses from A at 1000.00 / 1000.00 returns to A within 0.005 ft, which confirms the arithmetic of the table but, as always with omitted measurements, says nothing about the field work.

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