TRACING CURVES AND TURNING OBSTACLES IN THE FIELD. ΧΧ. TO TRACE A CURVE ON THE GROUND WITH THE CHAIN ONLY. 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′. 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 × 0.09+ 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 P BC. The sine of half this angle multiplied by twice the given chord = 0.0327 × 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. B E arc CP. Prolong B C to D. tial distance due to the whole Then PC ED::BC: BD or BF; and, by the foregoing supposition, E D:EF::BC:BF. Combining these proportions, and cancelling ED, we have PC: EF:: BC2: B F2 .. PC EFX (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 dis tances. 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÷6625.95 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 P C = 3.14 feet: C will be a stake on the curve. 7. Next, run out the whole chain length from C to O in the range BC. To find OD, suppose the line N CT 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=TCB=P BC (XVI., 4); and (5) ON:ND::BC: CD..ON=NDX BC÷CD; i.e., O D =ND+ON=ND+[NDX (B C÷C D)]=5.95 × [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 CN 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 V E to be drawn tangent to the curve at E. Then the angle L EV is equal to the tangential angle of the curve; and the sub-tangential distance LV 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 × 40 ÷ 66 = 3.6 feet. equal to 5.95 X 402 662 : the distance LF the sub-chord E F established. 3.62.18 5.78 feet, thus obtained, and 40 feet, the point of tangent F may be = = 10. Next, set off UE FV 2.18 feet, and lay out FK in 'prolongation of the range U F; FK will be in the line of the terminal tangent. 11. This analysis has been somewhat minute and detailed. in order that the subject may be thoroughly understood. An instrument for measuring angles should always be used in railroad service: it greatly simplifies and abridges the labor of tracing field-curves, and gives more exact results. But occasions sometimes rise, in miscellaneous practice, when strict accuracy is not required, and the chain only can be had: the - young engineer should qualify against such occasions. XXI. TO TRACE A CURVE ON THE GROUND WITH 1. This, also, is best taught by an example. Let it be a general rule, in locating, to fix the intersection of tangents, and to set the tangent points, or the P. C. at least. from the P. I. There are exceptional conditions, as a steep hillside, timber or broken ground, a very long arc, unimportance of exact conformity to the project, and the like, which warrant its omission; but where these conditions do not obtain or are not prohibitory, and a snug fit is desirable, time will usually be saved by fixing the P. I. It often proves serviceable as a reference point during construction: on the location, it gives confidence in the work and an assurance of safe progress, which are well worth a little painstaking beforehand. 2. Having established the P. I., and found the intersection angle to measure, say, 66° 45', the first step is to find the apex distances so called, or tangent lengths IB, IF. These are each equal to R tan. I. If a 7° 30' curve be prescribed to close the angle, RX tan. I 764 X 0.659 12 Or, referring to Table XVI., the tangent corresponding to 66° 45' is found by interpolation to be 3774.6; dividing by 7.5, the rate of curvature in degrees and decimals, we have for the apex distance 503 feet, as above. 3. Before disturbing the instrument, which is presumed to stand in the range of the terminal tangent, measure I F, = 503 feet, and set the P. T. at F. Then direct the telescope to the last point fixed on the initial range AB, meas ure I B, = 503 feet, and set the P. C. at B. Move to B. 4. Suppose the P. C. to have fallen at a stake 2.50. In order to find the length of the curve, divide the intersection angle by the degree of curvature, having first reduced the minutes in each to hundreths of a degree by multiplying by 10 and dividing the product by 6. Thus the intersection angle becomes 66.75°, and the degree of curva |