264. Calculation of an Extensive Survey (fig. 17c), where the First has been made. Calculated 77,6502 77,6512|68,5125|68,5466|| 77,651 | 77,651 Here we find that line K A, which theoretically should close on A, wants but 1,3 links. To find the Most Westerly Station. By looking to fig. 17d, it will appear that either the point S or P is the most westerly. Point S 25,919 west of the assumed point L. Therefore the point S is the most westerly station, through which, if the first meridian be made to pass the area, can be found by the second method. To Find the Meridian Distances. When the first mer. passes through the most westerly station, we add the eastings and subtract the westings. When the first mer. is through the most easterly station, we add the westings and subtract the eastings. When the first mer. passes through the polygon, we add the eastings in that part east of the first mer., and subtract them in that part west of that mer. We also subtract the westings in that part east of that mer., and add them west of it. Meridian is made the Base Line A B, at each of which a Station 8666 910000 Required area == 1935, chains, or 193,5867 acres. VARIATION OF THE COMPASS. 264a. base line through it, such as A M. variation from the true meridian. take a fly-sheet and remeasure the third time if the two surveys differ. possible to make two surveys of a mile in length to agree within one foot. With a fifteen feet pole they agree very closely. In surveying an estate such as that shown in fig. 17c, we run a We refer the base line A M to permanent objects as follows: 0° 00/ On the S. W. corner of St. Paul's tower, 15° 11/ On the S.E. corner of the Court House (main building), 27° 10/ On the S.W. corner of John Cancannon's Mill, 44° 16/ 276° 15/ On the N.E. corner of John Doe's stone house, us 311° 02/ Any two or three of these, if remaining at a future date, would enable determine the base A M, to which all the other lines may be referred. The variation of the compass is to be taken on the line at a station where there is no local attraction, the station ought to be at same distance from buildings. We find the magnetic bearing of A M N. 64° 10' E., as observed at the hour of 8 A. M., 8th December, 1860, at a point 671 links north of station A, on the base line A M. Thermometer 40°, and Barometer 29 inches. Let the latitude of station Polar distance of Pole Star (Polaris) (Declination of Polaris being 53° 45′ 00 1° 25/30/ 88° 34′ 30, ... its polar distance is found by taking the declination from 90.) To Find at what time Polaris will be at its Greatest Azimuth or Elongation. 2646. Rule. To the tan. of the polar dist. add the tan. of the lat.; from the sum take 10. The remainder will be the cosine of the hour The time here means sidereal. angle in space, which change into time. To Find the Greatest Azimuth or Bearing of Polaris. 264c. Rule. To radius 10 add sine of the polar distance; from the sum take the cosine of the latitude. The remainder will be the sine of the greatest azimuth. To Find the Altitude of Polaris when at its Greatest Azimuth. 264d. Rule. To the sine of the latitude add 10; from the sum take the cosine of the polar distance. The difference will be the log. sine of the altitude. In the above example we have lat. 53° 45', and its tan. = 10,1357596 Polar distance 88° 3/05// 1° 25′ 30′′, and its tangent hour angle in space, whose cosine 8,3957818 8,5315414 This changed into time gives 5 h., 52 m., 12,3 s. This gives the time from the upper meridian passage to the greatest elongation. t To Find when Polaris will Culminate or Pass the Meridian of the Station on Line A M, being on the Meridian of Greenwich on the 8th Dec., 1860. 264e. From Naut. Almanac, star's right ascension Sun's right ascension of mean sun (sidereal time) Sidereal time, from noon to upper transit 1h. 08m. 43,58. 17 09 59,9 7 58 52,6 5 00 01 2 58 52 Sidereal time, from upper transit to greatest azimuth Sidereal time from noon to greatest eastern azimuth Now, as this is in day time, we cannot take the star at its greatest eastern elongation, but by adding 5h. 52m. 12,3s. to 7h. 58m. 52,6s., we find the time of its greatest western azimuth 13h. 51m. 4,9s. from the noon of the 8th December, and by reducing this into mean time, by table xii, we have the time by watch or chronometer. To Find the Altitude and Azimuth in the above. 264f. Lat. 53° 45′ N., sine+10+19,906575 N. polar dist. 1° 25′ 30, COS. 9,999866 sine + 10+18,395648 9,771815 8,623833 2° 24′ 37′′/. Cos. sine True altitude 53° 46′ 27′′. Alpha and Beta are termed the pointers, or guards, because they point out the Pole star, which is of the same (second) magnitude, and nearly on the same line. The distance from Alpha Ursamajor to the Pole star Benetnach * URSAMAJOR, or DIPPER, or THE PLOUGH, at its under transit. is about five times the distance between the two pointers. When Alioth and Polaris are on the same vertical line, the Pole star is supposed to be on the meridian. Although this is not correct, it would not differ were we to run all the lines by assuming it on the meridian; but as we sometimes take Polaris at its greatest azimuth, both methods would give contradictory results. 264g. Alioth and Polaris are always on opposite sides of the true pole. This simple fact enables us to know which way to make the correction for the greatest azimuth. (For more on this subject, see Sequel Canada Surveying, where the construction and use of our polar tables will be fully explained.) Variation of the Compass. 264h. Variation of the compass is the deviation shown by the north end of the needle when pointing on the north end of the mariner's compass and the true north point of the heavens; or, it is the angle which is made by the true and magnetic meridians. Rule 1. Count the compass and true bearings from the same point north or south towards the right. Take the difference of the given bearings when measured towards the east or towards the west; but their sum when one bearing is east and the other west. When the true bearing is to the right of the magnetic, the variation is east. When the true bearing is to the left of the magnetic, the variation is west. Here we have the true bearing at 300°, counting from N. to right, and the magnetic bearing at bearing is to the east of the magnetic. 290°, counting from N. to right. Rule 2. From the true bearing subtract the magnetic bearing. If the remainder is +, the variation is east; but if the remainder or difference is, the variation is west. Example 6. True bearing N. 5° E. 5 from N. to right, 365 from N. to right. N. 60° 40′ E. Here we call the east +, and the west negative —; and by the method of subtracting algebraic quantities, we change the sign of the lower line, and add them. Example 8. Let true bearing N. 16° W. and magnetic bearing = N. 6° W. 10° = variation 10° west. |