= centre stake to aid him in forming an airy image of the proposed section, judges that the natural surface and the formation slopes will meet 30 feet out. Of this distance, 7 feet are due to half the road-bed, and 23 feet to horizontal reach of the embankment slope. The slope being 14 to 1, or å, the horizontal reach of 23 feet corresponds to a vertical height of of 23 15.3 feet; and, since the instrument is 1 foot below grade, to a rod at the supposed embankment base of 153 1.0 14.3 feet. But the rod at that point is only 11 feet, to which, if 1 foot, the distance of instrument below grade, be added, the height of embankment would be 12 feet. He may then, as in the case of the upper slope in Example No, 1, try a rod at the distance out corresponding to the 11 feet rod, or 12 feet embankment. This distance would be 7+ 12 +6 25 feet, where, on trial, the rod proves to be 10 feet, instead of 11 feet, corresponding to an embankment height of 10+1= 11 feet, and to a distance out of 7+11+5.5 feet. Approximating thus, by shorter and shorter steps, ne finally reaches the point sought. = 23.5 15. The process in fixing the upper slope stake is similar to that used in fixing the lower one in Example No. 1. The several steps are designated by small letters in the figure, and a detail of them is not thought necessary. 16. This section would be noted in the field book as fol lows: 17. Here is a case, partly in rock excavation, slope to 1; partly in embankment, slope 1 to 1; road-bed 17 feet wide, 9 feet of which are on the right of the centre line and 8 feet on the left. 18. For the lower slope suppose the instrument height at A to be 6.5 feet above grade; centre cutting 2.5 feet. Find first, with a 6.5 feet rod, the grade point, to left of centre line, which proves to be 2.5 feet out. Note it, and set a stake feet out and 10.0 6.5 = 3.5 feet below grade. Then set the lower slope stake as in Example No. 2, observing that in this case the instrument is above grade, and that its height above grade is to be deducted from the rod at any point in order to obtain the height of grade above such point. = 19. Move the instrument to B, say 22.5 feet above grade. This elevation, if the cutting on that side be deemed to equal it, corresponds to a distance out of 9 feet for road-bed added to (22.54) for slope; total, 14.6 feet. The trial rod, however, at that distance, instead of reading 0, reads 6 feet, indicating a cut 22.56.0 16.5 feet deep, and a distance out corresponding thereto of 9.0+(16.5÷4) 13.1 feet. Trying again at this distance out, the rod reads 7.6 instead of 6 feet, requiring a further movement towards the centre line of (7.66) ÷ 4 0.4 feet. Thus by approximations much more rapid than in the case of a flatter formation slope, the point is soon fixed. 20. The field record of the above is as follows: = = XIII. VERTICAL CURVES. DIAGRAM. GIVING THE ORDINATES OF A PARABOLA AT INTERVALS OF TO THE SPAN, THE MIDDLE ORDINATE BEING UNITY. 1. Suppose gradients descending right and left at an equal rate from the summit B, and that it is required to truncate the summit with a vertical curve extending 150 feet each way. A circular are consuming so small an angle may be treated as a parabola, in which the external secant BF is equal to the versed sine FD. Referring to the above diagram, ordinates 4 and 8 will be seen to correspond to the ordinates between B chord AC and the curve in this instance, which ordinates therefore will be equal to the middle ordinate FD multiplied by 0.89 and 0.55 respectively. Adding these multiples to the grade elevation at A, the elevations of the intermediate points Example 1. = +1 in 100; BC= == - 1 in = 150 feet or 1.5 stations of 100 feet each. +94.0; AB: = Elevation of grade at 88 94.0+ 0.41 = 94.41. Elevation of grade at 4 — 4 = 94.0+ 0.67 = 94.67. Elevation at A +94.0. AB=+1 in 100; BC=— 0.4 in 100; A H, level; AD, DH, each = 200 feet, or 2 stations, divided into 50 feet spaces, the points of division corresponding therefore to ordinates 3, 6, and 9 of the preceding diagram. CH 1 X 2 0.4 X 2 = 2.0 -0.8 = 1.2 feet. Ascent from A to C along chord AC = CH÷8=1.2 ÷ 0.15 per 50 feet. Ꭶ = The elevations then along the chord AC, ascending at the rate of 0.15 per 50 feet, will be:: to which add the ordinates just found: 0.0 0.31 0.52 0.66 0.70 0.66 0.52 0.31 0.0 and the grade elevations on the curve will be: 94.0 94.46 94.82 95.11 95.30 95.41 95.42 95.36 95.2 Example 3. Elevation at A = +94.0; A B =+ 1 in 100; BC, AH, level. AD, BC, each 200 feet divided into 50-feet spaces, the points of division corresponding therefore to ordinates 3, 6, and 9 of he ordinate diagram C H = 1 X 2 = 2 feet. Ascent from A to C along chord AC CH÷8 = 0.25 per 50 feet. The elevations then along the chord A C, ascending at the rate of 0.25 per 50 feet, will be: And the grade elevations on the curve will be: |