PROBLEMS IN FIELD LOCATION. XXVI. HOW TO PROCEED WHEN THE P. C. IS INACCESSIBLE. 1. Suppose, for example, a projected 5° 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 B D, around the curve, is 280 feet, corresponding to an angle EBD of 7° at the circumference, or an angle of 14° at the centre. The chord of a 1° curve consuming this angle, by Table XVI., is 1,396.6 feet; that of a 5° 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, — As the sum of the sides given=359.3 A C 7.444543 = 199.3. 2.299507 Adding half the difference to half the sum, the larger angle, A, is found to be 5° 26'; subtracting half the difference from half the sum, the smaller angle, D, is found to be 1° 33'. The 93 length of the side A D may be found in like manner by trigonometrical proportion; or, perhaps more simply, thus: We are now prepared, from our point A, to deflect the angle 5° 26' 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° 33 R places the telescope on line DB; a further deflection of 7° 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 10 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 50 curve, which may be measured on the ground. Moving the instrument to C, turning 14° R, and laying off the line C D 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. B C E Suppose a 2° 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 A C, = 320 feet, includes an angle of 6° 24'. The tangent of a 1° curve corresponding to this angle in Table XVI. is 320.34 feet; that of a 2° curve is therefore 160.2A B. 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 The tri E. Measure the angle DE B, and the distance DE. angle BED may then be solved. If BE is found to be 670 feet, CE 670 160.2 509.8, and stake E32.20 +509.8, XXVII. HOW TO PROCEED WHEN THE P. C. C. IS INACCESSIBLE. 1. Suppose a 4° curve, A B, compounding at B into a 60 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 wish to reach C, on the second curve, by means of the straight line ADC. The arc A B, covering 225 feet of a 4° curve, subtends an A D angle of 9o. A D is half the chord of twice this angle. = By Table XVI., the chord of 18° on a 1° curve is 1,792.7 feet. That of a 4° curve is therefore 448.2 feet, half of which 224.1, = A D. The versin. of 18° on a 1° 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 4° curve. In order to find what angle on the 6° 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 6° 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 B C. Thus angle at C = 11° 02', and 406.85 feet. The angle = are found the angle at A = 9°, the the distance AC 224.1+182.75, = 11° 02' corresponds to a length of 1.84 feet on the 6° 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 A D = 9°, as above, find in Table XVI. the |
