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Mean horizontal angles, station O (P, Q, R)

Short answer

Adjusted angles: POQ=65°29′58.33″, QOR=156°35′48.33″, ROP=137°54′13.33″. Mean directions from face left and right give the angles; error of -40″ distributed equally.

Calculate the mean horizontal angles. If necessary, adjust them also.

InstrumentSighted toHCR Face LeftHCR Face right
OP00°00′00″179°59′40″
Q294°29′50″114°30′20″
R137°54′20″317°54′40″
P00°00′40″180°00′20″

Method. The mean direction of each target is found from the two faces: mean direction = [Face L + (Face R - 180°)]/2.

The readings decrease from P to Q to R and back to P, so the telescope was turned anticlockwise (the circle is graduated clockwise). Each angle is therefore (previous mean direction) - (next mean direction). The three angles at O must add up to 360°, and the second observation of P closes the horizon.

Inst.Sighted toFace LFace RMean direction
OP00°00′00″179°59′40″359°59′50″
Q294°29′50″114°30′20″294°30′05″
R137°54′20″317°54′40″137°54′30″
P (closing)00°00′40″180°00′20″0°00′30″

Mean angles

∠POQ=359°59′50″−294°30′05″=65°29′45″∠QOR=294°30′05″−137°54′30″=156°35′35″∠ROP=137°54′30″−0°00′30″=137°54′00″\begin{aligned} \angle POQ &= 359°59′50″ - 294°30′05″ = 65°29′45″\\ \angle QOR &= 294°30′05″ - 137°54′30″ = 156°35′35″\\ \angle ROP &= 137°54′30″ - 0°00′30″ = 137°54′00″ \end{aligned}

Sum = 359°59′20″; required = 360°00′00″; error = -0°00′40″ (the sum is 0°00′40″ less than 360°). The error is small, so it is distributed equally: correction = +0°00′13.33″ per angle.

AngleMean angleCorrectionAdjusted angle
POQ65°29′45″+0°00′13.33″65°29′58.33″
QOR156°35′35″+0°00′13.33″156°35′48.33″
ROP137°54′00″+0°00′13.33″137°54′13.33″
Sum359°59′20″+0°00′40″360°00′00″

Answer: adjusted angles POQ = 65°29′58.33″, QOR = 156°35′48.33″, ROP = 137°54′13.33″.

✓ Answer checked by Claude Opus 5.5 (high effort)

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