Traverse

A traverse is a series of consecutive lines whose ends have beenmarked in the field, and whose lengths and directions (angle,bearing or azimuth) have been determined from measurements.

Closed Traverses

Geometrically open,mathematically closedtraverse

Geometrically andmathematically closedtraverse

TraverseOpen Traverse

Open traverses should be avoided because they offer nomeans of checking for errors or mistakes

Geometrically andmathematically Open

Closed TraverseStep 1: Angle Misclosure

For a closed loop traverse with ninternal angles, the check that isused is:

Check the error with thePermissible misclosure (c)!

= n " 2( ) *1800#

!

c = K n

n is the number of anglesK is a constant depends on the levelof accuracy.

Closed Traverse

Check Interior Angle Closure

Observed Adjusted

A = 1000 45’ 37” 1000 45’ 35” B = 2310 23’ 43” 2310 23’ 40” C = 170 12’ 59” 170 12’ 56” D = 890 03’ 28” 890 03’ 27” E = 1010 34’ 24” 1010 34’ 22”

Total = 540000’ 11” = 5400 00’ 00”

Should = 5400 00’ 00” = (n-2)*180Misclosure = 00’ 11” = 11”

Adjustment = 11/5 = -2.2” per angle

Example 10.1 (textbook)

Closed TraverseStep 2: Compute Azimuths/BearingsExample 10-2 (textbook)

Azimuths are horizontal anglesmeasured clockwise from anyreference meridian (North).

The bearing is the actual horizontalangles between a reference meridian(North or South) and the line.

Closed Traverse

Adjusted Angles

A = 1000 45’ 35” B = 2310 23’ 40” C = 170 12’ 56” D = 890 03’ 27” E = 1010 34’ 22”

Step 2: Compute Azimuths/BearingsExample 10-2 (textbook)

Azimuths of AW = 2340 17’ 18”Measured angle WAE = 1510 52’ 24”

AB = 2340 17’ 18”+ 1510 52’ 24”+ 1000 45’35”- 3600 = 126055’17”

BA = 126055’17” + 1800 = 306055’17”BC = 306055’17”+ 2310 23’ 40”

= 538018’58” -3600 = 1780 18’58”CB = 1780 18’58” + 180 = 358018’58”etc …

Closed TraverseStep 3: Compute Latitudes and Departures

!

"X = L sin#

"Y = L cos#

Once all the azimuths are calculated, traverse closure is checkedby computing the departure, or easting (ΔX) and latitude, ornorthing (ΔY) of each line.

Example 10-3 (textbook)

where L is the horizontal length

Closed TraverseStep 4: Linear Misclosure and Relative Precision

Example 10-3 (textbook).

If all angles and distances weremeasured perfectly, the algebraic sumof the departures of all lines in thetraverse should equal zero. Likewise,the algebraic sum of all latitudes shouldequal zero.

!

Cx = "#X

Cy = "#Y

where: Cx = total closure distance of XCy = total closure distance of Y

Closed TraverseStep 4: Linear Misclosure and Relative Precision

Example 10-3 (textbook).

The error of linear closure (E) isdetermined using Pythagorean’sTheorem as:

!

E = Cx

2+Cy

2

!

Relative Pr ecision =E

P

P is the traverse perimeter ortotal length

Closed TraverseStep 5: Traverse Adjustment (Compass Rule)

Example 10-4 (textbook)

For any closed traverse the linearmisclosure must be adjusted (ordistributed) throughout the traverse to“close” or “balance” the figure.

ABx_Corr is amount of adjustmentfor length AB in the X direction,ABy_Corr is amount of adjustmentfor length AB in the Y direction,

!

ABx _Corr = Cx

AB

P

ABy _Corr = Cy

AB

P

Closed TraverseStep 6: Calculate Final Coordinates

Example 10-4 (textbook)

Using adjusted lats and deps and thecoordinates of your starting point (A),compute coordinates of all traversepoints (B, C, D, E)

X = X coordinate of the previous point + ΔXY = Y coordinate of the previous point + ΔY

Field ProcedureObjective♦ Learn the principles of running a closed

field traverse.♦ Learn how to compute a traverse (by hand

or using Excel) and properly adjust themeasured values of a closed traverse toachieve mathematical closure.

♦ Determine the error of closure and computethe accuracy of the work.

♦ Establish horizontal control points for theArea A.

A

B

C

D

Area A

Field ProcedureArea A♦ Azimuth of AB is given (232o42’40”).♦ Easting and Northing of A (X,Y)

♦ (324540.698 m, 5270414.258 m)♦ Newfoundland-MTM Zone 1

♦ Start your traverse on one corner of thearea you staked out.

♦ Measure the horizontal angle and distancebetween the two adjacent points.

♦ Each horizontal angle should be measuredusing the telescope in direct and reverseposition. Record the average of the angles.

♦ Compute the misclosure for the geometryand check that the internal angles of yourparcel sum to (n-2) *180 (should be within ±30’ of 360°).

A

B

C

D

Area A

N

CP1

CP2