chain 100 feet long being the unit generally adopted by American engineers for field ineasurements, any circular arc having that radius, of 5,730 feet, is called a one-degree curve, for the reason that one chain is equivalent to an arc of one degree at the circumference. 2. The circumferences of circles vary directly as their radii: hence, in any circular arc struck with half that radius, or 2,865 feet, one hundred feet at the circumference would sul)lend an angle of two degrees at the centre. Such an arc is called a two-degree curve. If one-third of the primary radius of 5,730 feet, or 1,910 feet, be used, the arc is called a threedegree curve; and so on. 3. It should be borne in mind, however, that these measurements are supposed to be made around the arc itself, and not on lines of chords. Since field measurements with the chain are always made on the lines of the chords, which are shorter between given points at the circumference than the lines of the arcs, as a bowstring is shorter than the bow, it is plain that, in advancing towards the centre of the one-degree curve by a series of concentric circles having radii equal to one-half, one-third, &c., of the primary radius, the chord 100 feet long differs more and more in length from the arc subtended by it, the bow being more and more arched in relation to the string. Thus, in the circle having a radius equal to one-twentieth of the primary radius, the chord 100 feet long subtends an angle of 20° 06', at the centre, instead of 20°, and the arc is 100.5 feet in length, instead of 100 feet. In order, therefore, that the chord of 100 feet may subtend arcs of 1o, 2o, 3o, &c., in regular succession, the radii of these successive arcs must be somewhat greater than the above method by subdivision of the primary radius would inake them; though, as might be inferred from the extreme case given by way of illustration, the difference is not appreciable in ordinary field practice, and radii, together with all the functions dependent on them, may usually be held to vary as the degree of curvature, or central angle per 100 feet chord, varies. XVIII. TO FIND THE RADIUS, THE APEX DISTANCE, THE LENGTH, THE DEGREE, ETC., OF A CURVE. M K K F H B В N 1. Let E B, A O be two straight lines intersecting at E. Lay off equal distances, E A, E B; erect perpendiculars at A and B, meeting at G, and connect A B, EG. From the centre G, with radius GA, draw the curve A HB. The point E will be the P. I.; A and B, tangent points; EA, E B, the tangents, or apex distances, which denote by T; EH, the external secant, or S; HN, the versed sine, or V. Let the long chord A B, connecting the tan gent points, be called C, and G A or G B, the radius, R. Call the deflection angle to a chord of 100 feet D, as before. 2. By XVI. 3 and 4, angle E AB=EBA=AGE=EGB = { 1. 3. GIVEN THE INTERSECTION ANGLE I AND RADIUS R, TO FIND THE TANGENT T. T=R x tan. & I. Example. R=1,910.1, I= 35° 24'. Then T=R tan. I=1,910.1 X 0.3191 = 609.5. 4. Measure from the P. I. equal distances, EM, EF, along the tangents. Measure, also, MF and EK, the distance from E to the middle point of MF. Then, by reason of similarity in the triangles MEK, EAG, MK:EK:: AG:AE::R:T Example. Then R=1910.1 EK= 60.8 T= 609.6 3.281056 2.785065 5. If 100-feet chords be used, find the tangent in Table XVI. corresponding to the given angle I. Divide that tabular tangent by the degree of curvature corresponding to the given radius: the quotient will be the required tangent. Thus, Tab. Tan. corresponding to 35° 24' = 1,828.7, which, divided by 3, the degree of curvature, gives 609.6, the tangent sought. 6. GIVEN THE INTERSECTION ANGLE I AND TANGENT T, TO FIND RADIUS R. Transposing the equation in (3), R=T= tan. 41=TX cot. f I. Example. T= 609.6, I=35° 24' R=T cot. } I=609.6 X 3.1334=1910.1. By a like transposition of the equation in (4), R=TX MK; E K. 7. If 100-feet chords be used, find in Table XVI. the tangent corresponding to the given angle 1. Divide that tabular tangent by the given tangent; the quotient will be the degree of curvature in degrees and decimals. The radius corresponding to this degree of curvature may be found by (12), by Table X., or, with sufficient accuracy for ordinary practice, by dividing 5,730, the radius of a 1o curve, by it. Thus, in the foregoing example, the tabular tangent corresponding to 35° 24' is 1,828.7. Dividing by 609.6, we have 3 for the degree of curvature; and 5,730 divided by 3 gives 8. GIVEN THE INTERSECTION ANGLE I AND CHORD AB=C, CONNECTING THE TANGENT POINTS, TO FIND RADIUS R. AN=} AB=1C; AGN=11. Example. I= 35° 24', C=1161.4. Then R=1C ; sin. } ], 5$0.7 = 0.304 1910.2. 9. If 100-feet chords be used, find in Table XVI, the chord corresponding to the given angle 1. Divide that chord by the given chord, for the degree of curvature in degrees and decimals. Determine the corresponding radius by (17), by Table X., or, for ordinary practice, by dividing 5,730 by it. Thus, in the foregoing example, the tabular chord corresponding to angle 35° 24' would be 3,484.2, which, divided by the given chord, 1,161.4, gives 3 for the degree of curvature, and 5,730 divided by 3 makes the radius R=1,910 feet. 10. GIVEN THE INTERSECTION ANGLE I AND THE DEGREE OF CURVATURE OR DEFLECTION ANGLE D, WITH 100-FEET CHORDS, TO DETERMINE THE LENGTH OF THE LONG CHORD C, THE VERSED SINE V, THE EXTERNAL SECANT S, OR THE TANGENT T. Take from the proper column in Table XVI., number corresponding to the intersection angle, and divide it by the degree of curvature: the quotient will be the length required. Example. A 4o curve, I= 50° 16'; to find the several functions above named. Table XVI. gives the designated functions of a 1o curve as follows: C 4,867.3, V 542.4, S 599.3, T 2,688.2. Dividing by 4 the degree of curvature, we have for the corresponding functions of a 4o curve as follows: C 1,216.8, V 135.6, S 149.8, T 11. GIVEN C, V, S, OR T, OF ANY CURVE, AND D, THE DEGREE OF CURVATURE, TO FIND THE INTERSECTION ANGLE, I. Multiply the given function C, V, S, or T, by the degree of curvature, D: the product will be found in the proper column of Table XVI., corresponding to the required angle. Example 1. Given T=515, D=5°; to find I. Then T XD= 2,575, which corresponds in Table XVI. to 48° 24' =I. Example 2. Given C=1,656, D=3°; to find I. Then C XD= 4,968, which corresponds in Table XVI. to 51° 23' = I. 12. GIVEN C, V, S, OR T, OF ANY CURVE, AND THE INTERSECTION ANGLE I, TO FIND THE DEGREE OF CURVATURE D. Take from the proper column of Table XVI. the number corresponding to the given angle I, and divide that tabular number by the length of the given part; the quotient will be D, in degrees and decimals. Example 1. Then 1,136.3 : 587 1.935 = 1° 56' = D. Example 2. Then 208 • 64 = 3.25 = 3° 15' = D. 13. GIVEN THE INTERSECTION ANGLE I, AND DEFLECTION ANGLE D, TO FIND THE LENGTH OF THE CURVE. Divide I by D: the quotient will be the number of chord lengths in the curve. If the degree of curvature is a whole number, the inore con |