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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

B

с

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 tho intermediate points

Example 1. Elevation at A= : + 94.0; AB=+1 in 100; BC=-1 in 100; AD, DC, each = 150 feet or 1.5 stations of 100 feet each.

Hence BD= 1.5; and FD=0.75 feet.

Ordinate 8= 0.75 X 0.55 0.41.

Ordinate 4 0.75 X 0.89 = 0.67.
Elevation of grade at 8 —8=94.0 + 0.41 = 94.41
Elevation of grade at 4 - 4= 94.0 + 0.67 = 94.67.
Elevation of grade at D 94.0 + 0.75 = 94.75.

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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, ilivided into 50 feet spaces, tlie 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 • S= 0.15 per 50 feet.

BE=BD - *CH=2 - 0.6 = 1.4.

.:. FE=1.4 ; 2=0.7.
Ordinates at 9 — 9=0.7 X 0.44 = 0.31.
Ordinates at 6 – 6= 0.7 X 0.75 0.52.
Ordinates at 3 – 3= 0.7 X 0.94 = 0.66.
Mid-ordinate

- 0.70.

The elevations then along the chord AC, ascending at the rate of 0.15 per 50 feet, will be:

A 9 6 3 0 3 6 9 с

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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

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of division corresponding therefore to ordinates 3, 6, and 9 of he ordinate diagram CH=1 X 2= 2 feet.

Ascent from A to C along chord AC=CH: 8 = 0.25 per 60 feet.

BE - BD – CH=1 foot.

.:. FE=1 ; 2=0.5.
Ordinates 9 - 9= 0.5 X 0.44 = 0.22.
Ordinates 6 — 6= 0.5 X 0.75 = 0.37.
Ordinates 3 —3= 0.5 x 0.94 = 0.47.
Mid.ordinate

- 0.50.

The elevations then along the chord A C, ascending at the rute of 0.25 per 50 feet, will be:

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And the grade elevations on the curve will be:

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Example 4. Elevation at A = + 94.0; AB 1 in 100; BC, AH, xvel; AD, BC, each 150 feet, divided into 50-feet spaces, the oints of division corresponding therefore to ordinates 8 and 4 or the initial diagram CH=1 X 1.5=1.5.

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Descent from A to C along chord AC=CH : 6 = 0.25.

EB=DB - DE=1.5 – 0.75 = 0.75

.:. FE= 0.75 ; 2= 0.375
Ordinates 8-8= 0.375 X 55 - 0.21
Ordinates 4 – 4= 0.375 X 89 = 0.33
Mid ordinate

= 0.37

The elevations then along the chord A C, descending at the rate of 0.25 per 50 feet, will be:

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And the grade elevations on the curve will be:

94.0 93.54 93.17 92.88 92.67 92.54 92.5

The figures are drawn much distorted, in order to make the illustration clear.

2. With profile paper at hand, a convenient and quite suf. ficient determination of the grade elevations on a vertical curve may be made by drawing the gradients to a scale of 2 . feet to an inch vertical, and 50 feet to an inch horizontal. By means of the curve protractor (Art. XXV. 1) a suitable arc may

XIV.

THE TRANSIT.

1. Should the vernier and circle plates be out of parallel, -should one or the other be sprung, a defect shown by a slight rocking motion when the rims are pinched alternately on om posite sides, — the instrument must be sent to the shop fo repair. This is a common disease of transits in their old age instrument-makers need to study its cause and cure.

2. TO ADJUST THE LEVEL TUBES.

Bring the bubbles to the middle of them by means of the levelling screws. Turn the top of the instrument horizontally half way round. If the bubbles then err, reduce the error onehalf with the small adjusting screws attached to the tubes, and one-half with the levelling screws. Repeat urtil the adjustment is perfect.

3. TO ADJUST THE VERTICAL HAIR SO THAT IT SHALL RE

VOLVE IN A PLANE, AND MARK BACKSIGHT AND FORESIGHT POINTS IN THE SAME STRAIGHT LINE.

Try, first, by reference to the corner of a well-built house, or to a plumb-line, whether the hair be truly vertical. If it is not, loosen the four small capstan head screws on the outside of the barrel slightly, and gently tap the topmost one right or left, until the adjustment is effected.

4. Then, after bringing the four screws to a snug bearing again, direct the cross-hair to the edge of some well-defined object, as a chain pin, or stake, placed 400 or 600 feet distani. Upset the telescope, and place a like mark at about the same distance, and level in the opposite direction. Unclamp. Revolve the instrument horizontally on its spindle half way round, and direct the cross-hair to the point first observed. Again upset the telescope. If the cross-hair now strikes aside from the second mark, correct one-quarter of the error by

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