Solution: Meridian Conversion

Problem (1)

In 1875 the magnetic bearing of a line was recorded as N 32°45'E and the declination was 8°12'W. The current declination is 2°30'E.

Sketch of provided information

(a) What is the current magnetic bearing?

Sketch

MBrng_Now = N 22°03' E

(b) What was the true bearing in 1875?

Sketch

TBrng = N 24°33' E

(c) What is the current true bearing?

True meridian doesn't change so current true bearing same as it was in 1875: TBrng = N 24°33' E.

Problem (2)

In 1890, the magnetic bearings of lines AB and BC were S30º50’E and N85º30’E, respectively. The delcination then was 3º00’E.

In 1960, the magnetic bearing of line AB was  S21º10’E.

(a) What was the magnetic bearing BC in 1960?

The angle between lines AB and BC doesn't change over time. Compute the angle from the 1890 directions and use it with line AB's 1960 magnetic direction to obtain line BC's 1960 magnetic direction.

Sketch referenced to 1890 magnetic north:

Sketch referenced to 1960 magnetic north

Subtract 1960 magnetic bearing of line AB from 116°20'

Since γ exceeds 90°, line BC crosses from the NE quadrant to the SE quadrant. 

1960 MBrng_BC = S84°50'E 

(b) What was the declination in 1960?

Because line AB is referenced to both magnetic meridians, the change from 1890 to1960 is the amount the declination changed. Knowing the 1890 declination, the 1960 decllination can be computed.

Compute how much line AB's magnetic bearing changed from 1890 to 1960

Sketch:

The magnetic meridian rotated 14°40' to the west, going past the true meridian. 

Sketch

1960 declination was 11°40'W 

Problem (3)

When performing a survey of Lot 6, Surveyor Black started at a point on the center line of Oak Street, then turned 90° left to go to the first lot corner, point A. She then proceeded to survey the Lot exterior. She assumed a N 20°00'00" E bearing for the the street center line. Her final adjusted directions are shown below.

After checking recoreds at the Highway Department, she discovered the center line bearing is actually N 14°28'35" E.  Based on this new information, what are the corrected lot line bearings?

Sketch of provided information

Each lot line bearing must be corrected by 5°31'25"

Line AB

Line BC

Line CD

Line DA

Answers

Problem (4)

The true azimuth of line DE is 287°15'35". The angle to the right at E from D to F is 121°38'55". Grid convergence at D is +2°15'45" and at E is +1°59'10". What is the true azimuth of line EF?

Sketch of provided data

Compute grid azimuth of line DE

Compute grid azimuth of line EF, then its true azimuth.