Imágenes de páginas
PDF
EPUB

PROBLEMS IN FIELD LOCATION.

XXVI.-XXXVII.

PROBLEMS IN FIELD LOCATION.

XXVI.

HOW TO PROCEED WHEN THE P. C. IS INACCES

SIBLE.

E

с

B

1. Suppose, for example, a projected 5o curve, beginning at stake 24.20, or B in the diagram.

FIRST METHOD. - At any point A, which we will assume to be stake 23.40, set up the transit. Let it be judged that stake 27, marked D in the diagram, must fall on accessible ground. Then the distance BD, around the curve, is 280 feet, corresponding to an angle E B D of 7° at the circumference, or an angle of 14° at the centre. The chord of a 1o curve consuming this angle, by Table XVI., is 1,396.6 feet; that of a 5o curve, B D in the figure, is one-fifth of this, or 279.3 feet. In the triangle A B D are thus known the sides A B, B D, and the sum of the angles at A and D, which sum is equal to the angle EBD.

Hence, by trigonometry,

А

As the sum of the sides given=359.3 AC
Is to their difference

= 199.3.
So is tan. sum of angles at base = 3° 30' .
To tan. | their difference

=1° 561

7.444543 2.299507 8.786486

8.530536

Adding half the difference to half the sum, the larger A, is found to be 5° 26}'; subtracting half the difference from half the sum, the smaller angle, D, is found to be 1° 333'. The

93

length of the side A D may be found in like manner by trigonometrical proportion; or, perhaps more simply, thus:

BD x nat. cos. D=DF= 279.2.
BA X nat. cos. A = AF = 79.6.

AF+FD= AD= 358.8.

We are now prepared, from our point A, to deflect the angle 50 264' R, and lay out the line A D to the point D on the curve. Moving the instrument to that point, and backsighting to A, a deflection of 1° 334' R places the telescope on line D B; a further deflection of 70 places it in tangent at D, and the curve may thence be traced in both directions.

2. SECOND METHOD. — Having, as in the first method, judged that stake 27, marked D, must fall on accessible ground, and thus determined the central angle subtended by the arc BD, refer to Table XVI. for the tangent of a 1° curve, corresponding to 14°, the given angle. It proves to be 703.5 feet. One-fifth of this, 140.7 feet, is the tangent or apex distance, BC, of a 5o curve, which may be measured on the ground. Moving the instrument to C, turning 14° R, and laying off the line CD=BC, the point D on the curve is ascertained.

3. The preceding methods are manifestly applicable to the ends also of curves, as well as the beginnings. A case not

unfrequent in practice may be added in conclusion of the subject.

Suppose a 2o curve terminating at C, in marsh or stream not measurable directly. Let C fall at stake 32.20. At any convenient point A, say stake 29, place the transit with telescope in tangent. The arc AC, 320 feet, includes an angle of 6° 24'. The tangent of a 1o curve corresponding to this angle in Table XVI. is 320.34 feet; that of a 2o curve is therefore 160.2=AB. Move to B, deflect 6° 24' R into the range of the terminal tangent, and fix E on the opposite shore. Fix also

D, and note the angle EBD. Move to E. Measure the angle DE B, and the distance D E. The triangle BED may then be solved. If B E is found to be 670 feet, C E= 670 — 160.2 = 509.8, and stake E = 32,20 + 509.8,

3

с

E

XXVII.

HOW TO PROCEED WHEN THE P. C. C. IS INAC

CESSIBLE.

G

B

E

F

А

[ocr errors]

1. Suppose a 4o curve, A B, compounding at B into a 6o curve BC.

FIRST METHOD. - Place the transit at any point A, say stake 34. Let the proposed P. C. C. fall at stake 36.25. Assume that we wishi to reach C, on the second curve, by means of the straight line ADC. The arc A B, covering 225 feet of a 4o curve, subtends an angle of 90. A D is half the chord of twice this angle.

By Table XVI., the chord of 18° on a 1o curve is 1,792.7 feet. That of a 4o curve is therefore 448.2 feet, half of which 224.1, = AD. The versin. of 18° on a 1o curve, by the same table, is 70.54 feet; one-fourth of which, or 17.635, is the versin. BD, corresponding to the same angle on a 4o curve. In order to find what angle on the 6o curve this versin. B D, - 17.635 feet, corresponds to, multiply it by 6, and seek the product, 105.81, in Table XVI., where it is found, nearly enough for field-practice, opposite the angle 22° 04. The chord of that angle, on a 1o curve, is seen at the same time in the adjoining column to be 2,193.2 feet; on a 6o curve it is therefore 365.5 feet, one-half of which, 182.75 feet, = DC, and one-half of 22° 04' = 11° 02', the angle covered by the arc BC. Thus are found the angle at A = 9°, the angle at C 11° 02', and the distance AC= 224.1 + 182.75, 406.85 feet. The angle 11° 02' corresponds to a length of 1.84 feet on the 6o curve; C, therefore, falls at stake 36.25 + 1.84 38.09. With these data the field-work is obvious.

2. SECOND METHOD. — Having reached the point A, and determined the arc AP=9°, as above, find in Table XVI. the

« AnteriorContinuar »