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equal to 5.95 x 402 • 662 - 5.95 x 0.367 : 2.18 feet. Wia the distance LF= 3.6 + 2.18 = 5.78 feet, thus obtained, and the sub-chord E F = 40 feet, the point of tangent F may be established.

10. Next, set off UE=FV 2.18 feet, and lay out FK in prolongation of the range UF; FK wil be in the line of the terminal tangent.

11. This analysis has been somewhat minute and detailed, in order that the bject 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

TRANSIT AND 100-FEET CHAIN.

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 X tun. £ I. If a 70 30' curve be prescribed to

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Or, referring to Table XVI., the ap. dist. 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, measure 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

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former by the latter, we have 890 feet for the length of the

curve.

Or, the intersection angle 66° 45' is equivalent to 4003', and the degree of curvature 7° 30' is equivalent to 450': dividing the former by the latter, we have 890 feet for the length of the curve, as before.

5. Adding 8.90 to 2,50, the number of the P. C., the P. T. is found to fall at stake 11.40. Let the rear chainman make a note of this, that there may be no mistake in the terminal plus.

6. Next, to determine the proper deflections from the line of tangent at B, bear in mind that the deflection for a whole chain is half the degree of curvature; and that, in field-curves of more than 300 feet radius, the deflections for sub-chords, or plusses, may, without material error, be held to vary directly as the sub-chords themselves; that is to say, the sub-deflec-. tions due to 30, 60, and 80 feet, for instance, will be, to the deflection due to 100 feet, as 30, 60, and 80 are to 100.

7. Thus, in the example, 7° 30' being the degree of curvature, one-half of this, or 3° 45', will be the deflection due to a chord of 100 feet; and 5 of this, or a deflection of 1° 52 from the line of tangent at B, will fix stake 3, 50 feet distant on the curve.

8. The following is a simple rule for finding sub-deflections:

Multiply the sub-chord in feet by the rate of curvature in degrees and decimals: three-tenths of the product will be the sub-deflection in minutes.

Thus, in the example, 50 X 7.5 = 375, and 375 X 0.3 = 112.5! - 1° 521', as before.

9. Having set stake 3, stakes 4 and 5 will be fixed by successive deflections of 3° 45'. In establishing stake 5, the index will read, 1° 52' + 3° 45' + 3° 45' = 9° 224' = angle C B 5.

10. Suppose the instrument moved to 5. See that the verpier has not been disturbed, backsight to B, and deflect go 22/' right; i.e., double the index angle. The index will now read, 180 45/ the angle ICD; and the telescope will be directed along the line CD, tangent to the curve at 5, for the reason that the angle B 5 C has been made equal to the angle C B5 (XVI. 4).

Proceed with successive deflections of 3° 45' from this tangent, and set stakes 6, 7, 8, and 9, at intervals of 100 feet.

stake, the index will read, 18° 45' + 4 times the constant angle 3° 45', 18° 45' + 150 angle ICD + angle D 59, · 330 45'. In order to place the telescope in the line DE, tangent to the curve at 9, it is now necessary to turn an angle to the right, from backsight to 5, equal to D95 D59 15°; i.e., the vernier should be moved from 33° 45' to 33° 45' + 15o = 48° 45'. The telescope will then be in tangent at 9.

12. A simple rule for finding the index angle which shall place the telescope in tangent at any point on the curve is as follows:

From double the index angle which fixed the given point, subtract the index reading in tangent at the last turning-point : the remainder will be the required index angle.

Thus the index angle which established stake 9 was 330 45'. Double this angle will be 67° 30'; subtracting 18° 45', the reading in tangent at the last turning-point, we have 48° 45', the required index angle, as before.

The reasons for the rule will be obvious" from an examination of the figure.

13. Being in tangent, then, at 9, and the index reading 480 45', a deflection of 3° 45' will fix 10: a further deflection of 30 45' will fix 11, and the index will stand at 48° 45' + 7° 30' 56° 15'. 14. To find the deflection corresponding to the sub-chord 11

40 feet: by the foregoing rule (8), the degree of curvature, 7.5, multiplied by 40, the length of the sub-chord in feet, gives a product of 300, three-tenths of which amount to 90 minutes = 1° 30'. Adding 1° 30' to 56° 15', makes the index angle 57° 45' to fix the P. T. at 11.40.

15. Move to the P. T. at 11.40, see that the vernier has not been disturbed, and backsight to 9. By the foregoing rule (12), double the index angle, 57° 45', less the angle in tangent at 9, the last turning-point, 48° 45',

115° 30' — 48° 45', the index angle in tangent at the P. T., the total angle consumed by the curve. The work thus proves itself.

16. The preceding example would appear in the field-book as follows:

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66° 45',

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