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17. This mode of running curves secures a record of each step in the proceeding; so that, if any error occurs, it can readily be detected. At each turning-point, the number in the “ tangentcolumn' must correspond with the central angle due to the length of curve to that point; and at the P. T. that number must correspond with the total central angle. The work can thus be checked with facility during its progress, and checks itself at the end.

18. The young transitman is recommended to rule blanks after the pattern given, and exercise himself thoroughly in computing the parts, and recording the field-notes' of various curves assumed at will: drawings are not necessary.

19. Another method, and in some respects a better one, is, before starting on a curve, or during its progress, to record for all its stations the deflections which would locate them if the instrument remained at the P. C. Obviously the final deflection thus recorded would be half the central angle of the whole curve; and, if the instrument were placed anywhere on the curve, a B. S. to P. C. and deflection of half that central angle would locate the P.T. The same reasoning will apply to any subdivisions of the curve which may be found convenient in field work, the deflection angles for plusses being reckoned from the P. C. and recorded as in the case of whole stations. I am indebted to Mr. Robert Burgess, C.E., for recommending this method. Following is his illustration of it as applied to the preceding example. This illustration serves also to exemplify another form for field-notes.

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20. The transitman, at P. C., Sta. 2 + 50, sets his vernier at zero, takes his B. S. and locates the stations in order to 5,

zero and backsights to P. C. A deflection of go 221' will now place bim in tangent; a deflection of 13° 07f' will locate Sta. 0, and 'so on, using the recorded angles. Suppose a tree to mask the site of Sta. Placing a stake as near thereto as possible, to hold distance; go on deflecting to 9, the vernier reading 24° 221', where another point is given. Moving now to 9, the transitman observes the general rule, applicable to this method, namely, to set his vernier at each new instrument point, 10 the angle recorded opposite the point of his proposed backsight. In this case that point is Sta. 5. He therefore sets the vernier at 9° 221', backsights to 5, and turns to 20° 37' to locate 8. The vernier at 24° 22' would put him in tangent, and the suc. cessive recorded deflections complete the curve. The transit is then moved to P. T., the vernier set at 24° 221', the recorded angle opposite last point at 9, according to the above rule, backsight taken on that point, and a fipal deflection to 33° 227 turns off the tangent ahead.

21. The chief advantages of this method are the completeness of the record for subsequent use; its adaptation to a retracing of the line backward as well as forward, which please observe particularly; the little labor it imposes in mental arith. metic; and its simplicity, permitting any one who can read figures and turn an angle to relieve the transit on occasion.

The writer was raised on the first method; is still partial to it, out of a certain loyal feeling to the elder generation from whom it descended to him; but it must be owned the children, in some things, have surpassed the fathers, -as needs they should, else were there no progress, -and this seems to be one of those things.



1. Draw CF parallel to A B. Let lines BC, CE, FG, cut these parallels at equal inclinations. Call this angle I. Then BC = CE FG. BE BD + DE = 2 BD. But BD BC cos. I, .. BE = 2 BC cos. I. EG=CF. BG=EG + BE=CF + 2 BC cos. I.

Example Suppose B to be a stake 24.50 on the tangent A B, and that a deflection left of 10° be made there for 200 feet to a point C. Set transit at C, vernier reading 10° left. B S to B, and deflect 20° right. Vernier will now read 10° right, and telescope will be in line C E. Make CE= 200 feet. Move to E. See that vernier still reads 10° right. BS to C, and turn 10° left. Vernier will now read zero, and telescope will be in line EG, or tangent A B prolonged.

Distance BE=2B C cos. I 2 (200 cos. 10°) = 400 x .985 394 feet. Then E= 24.50 + 394, stake 28.44 on tangent A B prolonged.

If a parallel line C F were run, a deflection of 10° right would be made at each of the points C and F. If C F were 250 feet, then B G would be = 250 + 394 = 644 feet, and the point G would fall at stake 30.94 on tangent A B prolonged.

2. If angle I 60°, the other conditions of above method being observed, triangle BH E will be equilateral, and BE

BH= HE. If the parallel DC or DF be run, BE

BD + DC, and BG BD + DF. For field work see last example.

3. In turning obstacles by Ā

either of these methods, the angles should be measured with extreme niceness. Handle the instrument lightly, to avoid jarring the vernier; and, if possible, observe well-defined distant objects in the several short ranges, that the lines of foresight and backsight may







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In locating, the following method is preferable to those given above, and should always be used on long tangents.

4. Having established points A and B on the centre line, the farther apart the better within limits of distinct vision, set off the equal rectangular distances AE, BF, ranging clear of the obstacle. Place the transit at E or F, fix points G and H on the forward range, and, rectangularly to these points, establish others on the forward range of the centre line at C and D. The offset distances should be measured very carefully with the rod, or with a steel tape if they exceed in length the pocket rule which every engineer should have about him.






1. Fix a point on tangent A B prolonged at E. Lay off at B a perpendicular of any convenient length. Move the instrument to D, make the angle B D A =BD E, and mark the point of intersection A. By reason of symmetry in the triangles A D B, B.DE, A B = B E, and may be measured on the ground.

2. Or, fix the point E, and lay off the perpendicular B D as before. Move to D, direct the telescope to E, turn a right angle E DC, mark the point of intersection C, and measure C B. Then, by reason of similarity in the triangles C BD, DBE, CB : BD :: BD : BE, .. BE=BD2 = BC.

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Example. Suppose BD to be 60 feet, and B C 40 feet. Then BD 2 = BC= 3600 = 40 = 90 feet

BE. 3. Or, with the instrument at D, measure the angle B D E.



Example. 120 feet, angle BDE= 54° 40'. BD tan. BDE= 120 X 1.41 - 169.2 feet =BE. 4. Or, without an instrument, lay off any convenient lines BF.

or CH. Mark the middle point D. Line out H G, parallel to A B. Mark on it the point G in range with D and E. Then GF=BE, or GH=C E.

5. Should the use of a right angle be inconvenient, turn any angle EBD

X, measure BD about equal by estimation to B E, if the ground permits, and set a point D. Move to D, and measure angle B D E = y..


Then the angle B E D, or 2,=180 — (x+y), and, by trigonometry, sin. z: sin. y :: BD: BE, .. BE=BD sin. y · sin. z.

Example Let x =

44° 02', y=71° 48', BD= 300 feet.

Then 2 = 1900 (a + y) = 1800 (44° 02 + 71° 48') 1800 115° 50' = 64° 10'. BE= B D sin. y sin. z = 300 sin. 71° 48' – sin. 64° 10' : - 300 X .95 • .95 = 316.6 feet.

The calculation by logarithms would be as follows:


Log. 300
Log. sin. 71° 48':

Log. sin. 64° 10'
Log. 316.6. Diff.



If E is invisible from B, extend the line D B towards C, until a line CE clears the obstacle. The point E must then be established by intersection of the sides C E, D Е, in triangle CDE. Supposing the extension BC to have been 120 feet, the side CD will be 420 feet, the angle y 71° 48'; and, by a calculation similar to the above, the side D E, opposite angle x in the lesser triangle, identical with D E in the larger one, will be found to be 231.7 feet. The sum of the angles at the base C E of the triangle C D E = 1800 Y 1800 - 71° 48' 1080

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