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Anti-clockwise traverse PQRSP missing data

Short answer

QR = 561.600 m; SP bearing = 313°43' (N46°17'W); deflection at P = 88°43' left, at S = 86°17' left. Coordinates relative to R: S (7171.409 mN, 2665.880 mE), P (7528.075 mN, 2292.875 mE), Q (7219.210 mN, 1984.011 mE).

In a four sided anti-clockwise traverse PQRSP the following information are given.

SideLength (m)Deflection angleBearingCo-ordinates
PQ436.80?S45°W?
QR?85°00' Left??
RS499.20100°00' Left?R = 2345 mE, 6789 mN
SP516.084???

a) Compute all the missing figures. b) Compute the co-ordinates of other points with respect to R.

The traverse P-Q-R-S-P is anticlockwise, so each deflection is to the left and bearing of a line = bearing of previous line − left deflection. Unknowns: length QR, bearing SP (length SP is given), deflections at P and S, and the coordinates.

Step 1: Bearings that can be fixed

  • PQ = S45°W = 180° + 45° = 225°00'
  • QR = 225°00' − 85°00' = 140°00'
  • RS = 140°00' − 100°00' = 40°00'

Latitude and departure of PQ and RS:

PQ: 436.80cos⁡225∘=−308.86,436.80sin⁡225∘=−308.86RS: 499.20cos⁡40∘=+382.41,499.20sin⁡40∘=+320.88\begin{aligned} \text{PQ}:&\ 436.80\cos225^\circ=-308.86,\quad 436.80\sin225^\circ=-308.86\\ \text{RS}:&\ 499.20\cos40^\circ=+382.41,\quad 499.20\sin40^\circ=+320.88 \end{aligned}

Their sums are a=73.545a=73.545 m and b=12.015b=12.015 m.

Step 2: Length of QR and bearing of SP

Closure requires PQ+QR+RS+SP=0\text{PQ}+\text{QR}+\text{RS}+\text{SP}=0. With QR of unknown length xx at 140∘140^\circ, the vector SP is −(a+xcos⁡140∘, b+xsin⁡140∘)-(a+x\cos140^\circ,\ b+x\sin140^\circ) and its length is 516.084 m:

(a+xcos⁡140∘)2+(b+xsin⁡140∘)2=516.0842(a+x\cos140^\circ)^2+(b+x\sin140^\circ)^2=516.084^2 x2+2(acos⁡140∘+bsin⁡140∘)x+(a2+b2−516.0842)=0x^2+2(a\cos140^\circ+b\sin140^\circ)x+(a^2+b^2-516.084^2)=0 x2+(−97.231)x+(−260789.4)=0 ⇒ x=561.600 m  (the negative root is rejected)x^2+(-97.231)x+(-260789.4)=0\ \Rightarrow\ x=561.600\ \text{m}\ \ (\text{the negative root is rejected})

So QR = 561.600 m.

Then SP: latitude =−(−356.665)=+356.665=-(-356.665)=+356.665, departure =−(+373.005)=−373.005=-(+373.005)=-373.005.

LSP=+356.6652+(−373.005)2=516.084 m (agrees with the given 516.084 m),θSP=tan⁡−1−373.005+356.665=313∘43′L_{SP}=\sqrt{+356.665^2+(-373.005)^2}=516.084\ \text{m (agrees with the given 516.084 m)},\qquad \theta_{SP}=\tan^{-1}\frac{-373.005}{+356.665}=313^\circ 43'

(North-west quadrant, N 46°17' W.)

LineLength (m)BearingLatitudeDeparture
PQ436.800225°00'-308.864-308.864
QR561.600140°00'-430.210+360.989
RS499.20040°00'+382.409+320.880
SP516.084313°43'+356.665-373.005
Sum+0.000+0.000

Step 3: Missing deflection angles

  • At P: bearing SP − bearing PQ = 313°43' − 225°00' = 88°43' left
  • At S: bearing RS − bearing SP = 40°00' − 313°43' + 360° = 86°17' left

Check: 88°43' + 85°00' + 100°00' + 86°17' = 360°00', as required for a closed anticlockwise traverse.

StationLeft deflection
P (SP to PQ)88°43' (computed)
Q (PQ to QR)85°00' (given)
R (QR to RS)100°00' (given)
S (RS to SP)86°17' (computed)

Step 4: Coordinates with respect to R (R = 2345 mE, 6789 mN)

Going round R to S to P to Q and back to R:

StationNorthing (m)Easting (m)
R6789.0002345.000
S7171.4092665.880
P7528.0752292.875
Q7219.2101984.011
R (check)6789.0002345.000

The traverse closes on R, so the work is checked.

Answer: QR = 561.600 m; SP bearing = 313°43'; deflection at P = 88°43' L, at S = 86°17' L; bearings PQ 225°00', QR 140°00', RS 40°00'. Coordinates: S (7171.409 N, 2665.880 E), P (7528.075 N, 2292.875 E), Q (7219.210 N, 1984.011 E).

✓ Answer checked by Claude Opus 5.5 (high effort)

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