TRACING CURVES AND TURNING OBSTACLES IN THE FIELD. XX. TO TRACE A CURVE ON TIIE GROUND WITH THE CHAIN ONLY. Example. From a point B, 18 feet in advance of A, on tangent A B, to trace a curve of 367 feet radius to the right, with chords 66 feet long, and consuming an angle of 34° 27'. 63 2. First, dividing half the unit chord, or 33 feet, by the radius, 367 feet (XVIII., 18), we have 0.09+ for the sine of the tangential angle, corresponding to an angle of 5° 10': the deflection angle, therefore, is 10° 20'. The tangential distance corresponding to the angle 5° 10', and chord 66 feet, is equal (XVIII., 22) to twice the chord multiplied by the sine of half the tangential angle, 132 X 0.04507 = 5.95 feet. The deflection distance (XVIII., 19) is equal to twice the chord multiplied by the sine of half the deflection angle, - 132 X 0.09+ 11.88, say 11.9 feet. 3. To find the length of the curve (XVIII., 13): Divide the total central angle by the degree of curvature. The central angle, 34° 27', is equivalent to 2067 minutes; dividing by 620, the number of minutes in the deflection angle, we have 3.33, the number of chord lengths in the curve, = 3} chains 220 feet. If A be a stake numbered 2, then the point of curvature, B, will be 2.18, and the point of tangent, F, will fall at 2.18+ 3.22 stake 5.40. 4. To determine the tangential distance CP, to the first stake on the curve, either of two methods may be used: First, The sine of any tangential angle is equal to half the chord which limits the angle on one side divided by radius. The limiting chord BC in this instance is equal to 66 — 18 = 48 feet; half of 48, therefore, or 24 feet, divided by radius, 367 feet, gives 0.0654, the sine of 3° 45' = tangential angle PBC. The sine of half this angle multiplied by twice the given chord - 0.0327 X 96 = 3.14 feet, the tangential distance CP. 5. Secondly, CP may be found as follows, assuming that the functions of small angles vary directly as the angles themselves, and vice versa. Let BF be a portion of the curve. Make the tangent B E equal to the chord B F, and strike the arc EF. Draw the sub chord BC, and strike the arc CP. Prolong B C to D. E F may be taken as the tangential distance due to the whole chord BF, and PC the tangen F Then PC:ED::BC:BD or BF; and, by the foregoing supposition, ED:EF::BC:BF. Combining these proportions, and cancelling E D, we have PC:EF:: BC? : BF2.. PC=EF X (BC ; BF)2. In words, the tangential distance for a sub-chord is to that for a whole chord as the square of the sub-chord is to the square of the whole chord. The same is true of deflection distances. -6. In the example we are treating, the tangential distance for the whole chord of 66 feet has been found to be 5.95 feet; that for 48 feet, therefore, is 5.95 x 482 = 662 = 5.95 x 0.528 = 3.14, as before. Stretch the 48 feet of chain from B to P, in prolongation of tangent A B, and mark the point P; then step aside, and stretch from B to C, making the distance PC= 3.14 feet: C will be a stake on the curve. 7. Next, run out the whole chain length from C to ( in the range BC. To find OD, suppose the line NCT to be drawn tangent to the curve at C. Then N D may be considered the tangential distance due to the whole chord, 5.95, as above determined. The angle O CN=TC B=PBC (XVI., 4); and (5) ON:ND::BC:CD.: 0 N=ND XBC:C D; i.e., OD =ND+ON=ND+[NDX (BC=CD)]=5.95 X (1+(48 ;- 66)]=5.95 X 1.727=10.27. 8. The point N may be fixed otherwise by laying off BT = CP, and running out the chain length C N in the range C T. The point D on the curve may then be fixed by making ND equal to 5.95 feet, the tangential distance. Next run out the chain to M, in the range CD; make ME equal to the deflection distance, 11.9 feet, and fix the point E. The points C, D, and E will be stakes 3, 4, and 5 on the curve. 9. To set the point of tangent, F, at stake 5.40, prolong the chord line D E for 40 feet to L, and suppose VE to be drawn tangent to the curve at E. Then the angle L E V is equal to the tangential angle of the curve; and the sub-tangential distance L V is to the whole tangential distance due to the 66feet chord, as the sub-chord is to the whole chord (5); i.e., LV=5.95 x 40 = 66 = 3.6 feet. |