COMPASS
MAGNETIC DECLINATION
Magnetic declination at a place is the horizontal angle between the true meridian and the magnetic meridian shown by the needle at the time of observation. If the magnetic meridian is to the right side (or eastern side) of the true meridian, declination is said to be eastern or ; if eastern or positi positi ve it is to be the left side (or western side), the declination is said to be western or negative.
Mariners call declination by the name variation .
The declination at any particular location particular location can be obtained by establishing a true meridian from astronomical observations and then reading the co mpass while sighting along the true meridian.
I sogonic ogonic line is the line drawn through the points of same declination. The distribution of earth’s
magnetism is not regular and consequently, the isogonic lines do not form complete great circles, but radiating from the North and South magnetic regions they follow irregular paths. Agonic line is the line made up of points having a zero declination.
Variations in Declination
The value of declination at a place never remains constant but changes from time to time. There are four types of variations in. declination (a) Diurnal variation (b) Annual variation (c) Secular variation (d) Irregular.
(a) Diurnal variation The diurnal variation or daily variation is the systematic departure of the declination from its mean value during a period of 24 hours. It generally varies with the phase o f the sunspot period. The difference in declination between morning and afternoon is often as much as 10’ of arc. The extent of daily variations depends upon the following factors: (I) The Locality: More at magnetic poles and less at equator. (ii) Season of the year: Considerably more in sum mer than in Winter. (iii) Time : More in day and less in night. The rate of variation during 24 hours is variable. (iv) The amount of daily variation. Changes from year to year.
(b) Annual Variation
The variation which has a yearly period is known as annual variation. The declination has a yearly swing of about 1’ or 2’ in amplitude. It varies from place to place.
(c) Secular Variation Due to its magnitude, secular variation is most important in the work of surveyor. It appears to be of periodic character and follows a roller-coaster pattern. It swings like a pendulum. For a given place, the compass needle after moving continuo usually for a period of years in one direction with respect to the true North, gradually co mes to a standstill and then begins to move in opposite direction. Secular change from year to year is riot uniform for any given locality and is different for different places. Its period is approximately 250 years. in Pa ris, the records show a range from 11°E in 1680 to 22°W in 1820. This magnitude of secular variation , is very great, it is very important in the work of the surve yor, and, unless otherwise specified, it is the change commonly referred to.
(d) Irregular Variation
The irregular variations are due to what are known as ‘magnetic storms’, earthquakes and othe r solar influence. They may occur at any time and cannot be predicted. Changes of this kind amounting to more than a degree have been observed.
Determination of True Bearing
All important surveys are plotted with reference to true meridian since the direction of magnetic meridian at a place changes with time. If, however, the magnetic declination at a place, at the of observation is known, the true bearing can be calculated from the observed magnetic bearing by the following relation True bearing = Magnetic Bearing ± declination.
Use plus sign if the declination is to the East and minus sign if it is to the West.
The above rule is valid for whole circle bearings only. If however, a reduced bearing has been observed, it is always advisable to draw the dia gram and calculate the bearing.
Example 5.6. The magnetic bearing of a line is 48° 24’. Calculate the true bearing if
the magnetic declination is 5° 38’ East.
Solution. Declination= +5° 38’ True bearing = 48° 24’+5° 38’
= 54° 02’.
Example 5.6. The magnetic bearing of a line AB is S 28° 30’ E Calculate the true bearing if the declination is 7° 30’ West.
Solution. The position of true meridian, magnetic meridian and the line has been shown in Fig. Since the declination is to be West, the magnetic meridian w ill be to the West of true meridian.
Hence, true bearing = S 28° 30’ E+7° 30’ = S 36° 00’ E.
Example 5.7. In an old map, a line AB was drawn to a magnetic
Bearing 5° 30’, the magnetic declination at the time being 1° East. To What magnetic bearing should the line be set now if the present magnetic Declination is 8° 30’ East. Solution.
True bearing of the line
=5° 30’+1° = 6° 30’
Present declination
= +8° 30’ (East)
Now, True bearing
=Magnetic bearing+8° 30’
Magnetic bearing
=True bearing—8° 30’ =6° 30’—8° 30’=—2° (i.e. 2° in the anti-clockwise direction) = 358°.
Example 5.8. Find the magnetic declination at a place if the magnetic bearing of the sun at
noon is (a) 184° (b) 350° 20’. Solution.
(a) At noon, the sun is exactly on the geographical meridian. Hence, the true bearing of the sun at noon is zero or 180° depending Upon whether it is to the North of the place or to the South of the Place. Since the magnetic bearing of the sun is 184°, the true bearing Will be 180°. Now, True bearing
= Magnetic bearing + Declination
So,
= 184° + Declination
180°
Or Dec1ination
= -4° = 4° W.
(b) Since the magnetic bearing of the sun is 350° 20’, it is at
The North of the place and hence the true bearing of the sun, which is on the meridian, will be 360°. Now, true bearing Magnetic bearing +Declination 360°
= 350° 20’ + Declination
or Declination = 360° —350° 20’ = 9° 40’ = 9° 40’ E.
# LOCAL ATTRACTION A magnetic meridian at a place is established by a magnetic Needle which is uninfluenced by other attracting forces. However, Sometimes, the magnetic needle may be attracted and prevented from indicating the true magnetic meridian when it is in Proximity To certain magnetic substances. Local attraction is a term u sed
Denote any influence, such as the above, which prevents the needle from pointing to the Magnetic North in a given locality. Some of the sources of local attraction a re magnetite in the ground, wires carrying electric current, steel structures, railroad rails, underground iron pipes, keys, steel bowed spectacles, metal buttons etc., which may be lying on the ground nearby. Detection of Local Attraction
The local attraction at a particular place can be detected by observing the force and back bearings of each line and finding its difference. If the difference between the for and back bearing is 180°, it may be taken that both the stations are free from local Attraction, provided there are no observational and instrumental errors. Elimination of Local Attraction. -
If there is local attraction at a station, all the bearings measured at that place will be incorrect and the amount of error will he equal in all the bearings. There are two methods for eliminating the effects of local attraction. First Method.
In this method, the bearings of the lines are calculated on the basis of the bearing of that line which has a difference of 180° in its fore and back bearings. It is, however, assumed that there are no observational and other instrumental errors. The amount and direction of error due to local attraction at each of the affected station is found. If, however, there is n o such line in which the two bearings differ by 180°, the corrections should be made from the mean value of the bearing of that line in which there is least discrepancy between the back sight and fore sight readings. If the bearings are expressed in quadrantal system. The corrections must be applied in rd proper direction. In 1st and 3 quadrants, the numerical value of bearings increases in clockwise direction while they increase in anti-clockwise direction in “d and 4th quadrants. Positive corrections are applied clockwise and negative corrections counter-clockwise.
Examples 5.9 , 5.10 and S.11 completely illustrate the procedure for applying the corrections by the first method.
Second Method.
This is a more general method and is based on the fact that though the bearings measured at a station may be incorrect due to local attraction, the included angle calculated from the bearings will be correct since the amount of error is the same for all the bearings measured at that station. The included angles between the lines are calculated at all the stations. If the traverse is a closed one, the sum of the internal included angles must be (2s — 4) right angles. If there is any discrepancy in this, observational and instrumental errors also exist. Such error is distributed equally to all the angles. Proceeding now with the line the bearing of which differ by 180°, the bearings of all other lines are calculated, as illustrated in example 5.12.
Example 5.9. The fol1wing bearings were observed while traversing with a compass.
Mention which stations were affected by local attraction and determine the corrected bearings.
Solution On examining the observed bearings of the lines, it will be noticed that difference between b ack and fore bearings of the line DE is exactly 180°. Hence both stations D and E are free from local attraction and all other hearings measured at these stations are also correct Thus, the observed bearing of DC (i.e. 209 10) is correct. The correct bearing of CD will. therefore, be 209° 1O’— l80°’29’ 10’ while the observed bearing is 29° 45’. The error at C is therefore +35’ arid a correction —35’ must be applied to all the bearings measured at C. The correct beatings of CB thus becomes 277° 5’—35’216° 30’ and that of BC as 276° 30’—l80°=96° 30’. The observed bearing of BC is 96° 55’. Hence the error at B is +25’ and a correction of —25’ must be applied to all the bearings measured at B. The correct bearing of BA thus becomes 226° l0’-25’=225° 45’, and that of AB was 225° 45 —180’=45° 45’ which is the same as the observed one. Station A is therefore, free from local attraction.
The Results may be tabulated as under:
Line AB
Observed Bearing 45°45´
Correction 0 at A
Corrected Bearing 45°45´
BA
226°10´
-25´ at B
225°45´
BC
96°55´
-25´ at B
96°30´
CB
277°5´
-35´ at C
276°30´
CD
29°45´
-35´ at C
29°10´
DC
209°10´
0 at D
209°10´
DE
324°48´
0 at D
324°48´
ED
144°48´
0 at E
144°48´
Remark
Stations B and C are affected by Local Attraction
Example 5.10. Apply the corrections if the bearings of the previous examples are measured in quadrantal system as under: Line
F.B.
B.B.
Line
F.B.
B.B.
AB
N45°45´E
S46°10´W
CD
N29°45´E
S29°10´W
BC
S33°05´E
N82°55´W
DE
N35°12´W
S35°12´E
Solution
By inspection of observed bearings, stations D and E are free from local attraction and hence bearings of ED, DE and DC are correct. The correct bearing of CD will therefore be N29°10´E.Since the observed bearing of CD is N29°45´E, the magnetic needle at C is deflected by 35´ towards West. The corrected bearing of CB will therefore be N82°55´W+35°W. The corrected bearing of BC will be S83°30´E.Since the observed bearing of BC is S83°05´E,the needle a B is deflected by 25´ towards East. Hence the corrected bearing of BA will be S46°10´W-25´-S45°45´W.The bearing of line AB will be N45°45´E, which is the same as the observed one. Station A is therefore not affected by local attraction.
Example 5.11. The following bearings were observed in running a complete traverse: Line
F.B.
B.B.
AB
75°5´
254°20´
BC
115°20´
296°35´
CD
165°35´
345°35´
DE
224°50´
44°5´
EA
304°50´
125°5´
Solution
By inspection of the observed bearings, it will be noticed that stations C and D arc from free local attractions since the B.B. and F.B. of CD differ by 180°. All the bearings measured at C and D are, therefore, correct. Thus, the observed bearing of CD (i.e. 296° 35’) is correct. The correct bearing of BC will be 296° 35’ — 180° =116° 35´. Since the observed bearing of BC is 115° 20´, a correction of + 1° 15’ will have to be applied to the bearing of BA measured at B. Thus, the correct bear ing of BA becomes 254° 20’ + 1° 15’=255° 35’. The correct bearing of BA will, therefore, be 255° 35’—180°=75° 35’. Since the observed bearing of AB is 75° 5’ a correction of +30’ will have to be applied to the bearing of AR measured at A. Thus, the correct bearing of AE becomes 125° 5° +3´=125° 35’, the corrected bearing of LA will be 125° 35’ + 180°=305° 35’. Since. the observed bearing of EA is 304° 50’. a correction of +45’ will have to be applied to the hearing of ED measured at E. The correct bearing of ED will thus be 44° 5’+45’ =44° 50’. The correct bearing of DE will be 44°50’+180° =224° 50’, which is the same as observed one, since the station D is not affected by local attraction.
Thus, results may be tabulated as given below. Since the magnetic declination is + 5° 10’ E, the true bearings of the lines will be obtained by adding 5° 10’ to the corrected magnetic bearings.
Line
Observed Bearing
Correction
Corrected Bearing
True Bearing
AB
75°5´
+30´ at A
75°35´
80°45´
BA
254°20´
+1° 15´ at B
255°35´
260°45´
BC
115°20´
+1° 15´ at B
116°35´
121°45´
CB
296°35´
0 at C
296°35´
301°45´
CD
165°35´
0 at C
165°35´
170°45´
DC
345°35´
0 at D
345°35´
350°45´
DE
224°50´
0 at D
224°35´
230°0´
ED
44°5´
+45´ at E
44°50´
50°0´
EA
304°50´
+45´ at E
305°35´
310°45´
AE
125°5´
+30´ at A
125°35´
130°45´
Remarks
Stations A,B and E are affected by Local Attraction
Example 5.12. The Following are the bearings taken on a closed compass traverse: Line
F.B.
B.B.
AB
80°10´
259°0´
BC
120°20´
301°50´
CD
170°50´
350°50´
DE
230°10´
49°50´
EA
310°20´
130°15´
Compute the interior angles and correct them for observational errors. Assuring the observed bearing of the line CD to be correct adjust the bearing of remaining sides.
: Solution Angle A=Bearing of AE-Bearing of AB = 130°15´-80°10´=50°5´ Angle B=Bearing of BA-Bearing of BC = 259°-120°20´=138°40´ Angle C = Bearing of CB – Bearing Of CD = 301°50´-170°50´ = 131°0´ An gle D = Bearing of DC — Bearing of DE
= 350°50´ - 230°10´ = 120°40´ Angle E = Bearing of ED — Bearing of EA =49° 30’—310° 20’+360° = 99° 10’ Angle A +Angle B+ Angle C+ Angle D+ Angle E = 50° 5’+138° 40’+131° 0’+120° 40’+99° 10’ 35’ = 539° Theoretical sum =(2n — 4) 90° = 540° error= -25’.
Hence a correction of +5’ Is applied to all the angles. Th e corrected angles are Angle A = 50° 10’; Angle B= 138° 45’; Angle C=131° 5’; Angle D=120° 45’; and Angle E =99° 15’.
Starting with the corrected bearing of CD, all other bearings can be calculated as under Bearing of DE = Bearing of DC — Angle D =350° 50’— 120° 45’=230° 5’ Bearing of ED = 2 30° 5’— 180°=50° 5’ Bearing of EA =Bearing of ED — Angle E =50° 5’— -99° 15+360°=310° 50’ Bearing of AE 50’— 18O°=130° 50’ =310° Bearing of AB = Bearing of AE — Angle A = 130° 5O’— 50° 10’=80° 40’ = 80° 40’+ 180°=260° 40’ Bearing of BA Bearing of BC = Bearing of BA- Angle B = 260° 40´ - 138° 45´=121°55´ .‘. Bearing of CB =121° 55’ +18O°= 301°55´ Bearing of CD = Bearing Of CB – Angle C = 301°55´-131°5´=170° 50’ Bearing of DC = 170° 50´+180°=350°50´
# ERRORS IN COMPASS SURVEY
The errors may be classified as
(a) Instrumental errors (b) Personal errors (c) Errors due to natural causes.
(a) Instrumental errors. They are those which arise due to the faulty adjustments of the instruments. They may be due to the following reasons:
(1) The needle not being perfectly straight. (2) Pivot being bent. (3) Sluggish needle. (4) Blunt pivot point. (5) Improper balancing weight. (6) Plane of sight not being vertical, (7) Line of sight not passing through the centres of the ring.
(b) Personal errors. They may be due to the following reasons:
(1) Inaccurate leveling of the compass box.. (2) Inaccurate centring. (3) Inaccurate bisection of signals. (4) Carelessness in reading and recording.
(c) Natural errors. They may be due to the following reasons
(1) Variation in declination. (2) Local attraction due to proximity of local attraction forces. (3) Magnetic in the atmosphere due to clouds and Storms.. (4) Irregular variations due to magnetic storms etc.
