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8. It is true, that, on a uniform declivity, he might aid conjecture by taking a rod distant half the width of road-bed, or 10 feet, from the centre stake, ascertain thus the slope of the ground surface as well as the cutting at that point; and with these data, knowing also the formation slope, approximate the point sought by solving the terminal triangle of the proposed section, indicated by dotted lines in the figure. But, in practice, he will find it the quicker and better way to approximate the point by vividly imagining the underground forma. tion lines; or by vividly imagining a level section, the upper surface of which shall coincide with his instrument height, 15.5 feet above grade. This gives him a point in the air, 10 + 15.5 = 25.5 feet out from the centre stake, level with the instrument, as the limit of the imaginary section; and from that point he can pretty well judge where a line corresponding to the formation slope must meet the ground.

9. Suppose him, by either method, or even by random guess, to think that 10 feet for half the road-bed, and 10 more for the slope, looks about right. The formation slope being 1 to 1, this implies a cutting of 10 feet at the side stake, and a rod, therefore, of 15.5 – 10.0 = 5.5 feet. Taking a rod accordingly, 20 feet out, measured horizontally from the centre stake, he finds it to be 11.0 instead of 5.5, indicating that he has gone too far down hill. Let him now reason that the rod of 11.0 corresponds to a cutting of 15.5 – 11.0 = 4.5 feet, and that a cutting of 4.5 feet corresponds to a distance out of 10 + 4.5 = 14.5 feet. Try, then, a rod 14.5 feet out. It proves to be 9.0, corresponding to a cutting of 15.5 – 9.0 = 6.5, instead of 4.5 feet, and a distance out of 16.5 instead of 14.5 feet. Try, next, 16.5 feet out; the rod there, of 10.0 instead of 9.0, shows him again to be in error on the down-hill side of his object; but the lessening error shows also that he is approaching it, and that a few more like trials will reach it.

10. Recurring to his first error with the 11.0 feet rod, he cannot fail to observe after a little practice, since the ground ascends thence toward the centre line, that the side stake must fall farther out than the point where his second trial was made; that its true position, in fact, divides the distance between those points of observation into two parts which are to one another directly as the inclinations of the formation slope and the ground surface. By deg:'ers he will grow skilful in

observation, will place a slope stake on the second or third trial, without conscious effort of mind.

11. Next, suppose the level at B, 25.5 feet above grade, commanding the upper slope.

Note the change of ground 11 feet out, and take a rod there, recording the observation. The cutting at that point is 25.5 - 9.5 = 16 feet, corresponding to a distance out for the side stake of 10 + 16 26 feet, if the ground were level. A trial rod 26 feet out reads 7.8, corresponding to a cutting of 25.5 – 7.8= 17.7 feet, and a distance out for the side stake of 10 + 17.7 = 27.7 feet, showing that the point sought is still beyond. A repetition of such trials will finally fix it; but, as in the case of the lower slope, practice will speedily lessen the number and abridge the labor of them.

12. The foregoing section would be noted in the field book as follows:

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Example No. 2. 13. In the annexed figure, representing an embankment 14 feet wide on top, with side slopes of 14 to 1, the first thing to attract attention is that the instrument is 1 foot below grade,

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and that, therefore, 1.0 is to be added to all rods, in order to find the height of embankment above the points at which rods are taken.

14. Consider the down-hill side. The engineer, with the centre stake to aid him in forming an airy image of the pro posed 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 ļ, the horizontal reach of 23 feet corresponds to a vertical height of f 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 + 2 +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 = 23.5 feet. Approximating thus, by shorter and shorter steps, ne finally reaches the point sought.

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:

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Example No. 3. 17. Here is a case, partly in rock excavation, slope ; to 1; partly in embankment, slope 14 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

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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 (22.5 = 4) for slope; total, 14.6 feet. The trial rod, however, al that distance, instead of reading 0, reads 6 feet, indicating a cut 22.5 — 6.0 16.5 feet deep, and a distance out corresponding thereto of 9.0 + (16.5 ; +) 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 (7.6 – 0) = 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:


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2.5 - 3.5


+ 15.0






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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 arc consuming so small an angle may be treated as a parabola, in which the external secant B F 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



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

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