Difficulties Faced during Ranging Simple Curves | Surveying

After reading this article you will learn about the difficulties faced during ranging simple curves.

Case I:

When the point of intersection of the tangents is inaccessible Fig. (11.17):

It so commonly happens that the intersection point (B) becomes inaccessessible due to obstacle like lake, river or wood etc.

Let AB and BC be the two tangents intersection at the point B, and T₁and T₂ the tangent points. Since the intersections point (B) is used for measuring the deflection angle (ɸ) and also for locating the tangent points T₁ and T₂, the field- work be arranged to measure ɸ and to locate T₁and T₂without going at B.

Producers:

(i) Select two indivisible points M and N suitably on the tangents AB and BC respectively and measure MN accurately.

(ii) Measure the angles AMN (α) and CNM (β) by setting the theodolite at M and N respectively.

(iii) Solve the ∆ BMN by the sine rule for the distance BM and BN.

(iv) Calculate the tangent lengths BT₁ = BT₂  = R tan φ/2

(v) Now the distance MT₁ = BT₁-BM  and NT₂ = BT₂-BN

(vi) Measure the distance MT₁ from M along BA, thus locating the first tangent point T₁. Similarly locate T₂ by measuring NT₂ from N along BC.

When a clear line MN is not available, a traverse is run between M and N to find the length and bearing of the line MN. Then from the known bearings of the tangents and the calculated bearing of the line MN, calculate the angles α and β and proceed as above.

Case II:

When the complete curve cannot be set out from the starting point due to the obstructed vision (Fig. 11.18):  

When the curve is short and the ground is free from obstructions like buildings, trees etc., the whole curve can be set out from the first tangent point (T₁). However it is not always possible to have short curves and ground free from obstacles. In such cases the instrument has to be set up at one or more points along the curve. The following procedure may be adopted in such cases.

Procedure:

Let D, E, F, G, H etc. be the points on the curve, and G is the last point located from the instrument at T₁ when it is found necessary to shift the instrument. Let deflection angle for G be ∆₄, then deflection angle for the next point H should be Δ₅.

(i) Shift the instrument and set it up at G.

(ii) Set the vernier A to zero, take a back sight on T₁ and transit the telescope. The line of sight is thus directed along GG’ i.e. along T₁ G produced.

(iii) Loose the upper clamp and set the vernier to read the deflection angle ∆₅, thus directing the line of sight along GH. [By Geometry, ∠G’GH = ∠BT₁H. (Please see proof at the end)].

(iv) Locate H by measuring a distance equal to chord length along GH.

(v) Then using the same tabulated deflection angles, continue the setting out of the curve from G as usual.

For using the above method, the instrument should be in perfect adjustment.

Alternative Method:

Suppose G is last point located from the instrument at T₁.

(i) When the instrument is at T₁, fix a point G’ in line T, G produced.

(ii) Shift the instrument to G, set the vernier to zero and bisect G

(iii) Loosen the vernier plate and set the vernier to ∆₅ for locating the next point H.

Proof:

Draw the tangent G₁ GG₂ at G meeting the first tangent at G₁ and produce TG of G.  

Case III:

When an obstacle intervenes on the curve (fig.11.19): 

Suppose the first three points have been located from T₁ in the usual manner, F being the last point located on the curve and the chaining is obstructed between F & G. Let H be the next point on the curve which is visible from T₁, and ∆₅ is its total deflection angle.

Procedure:

(i) Calculate length of the chord T₁H by the relation T₁H = 2R sin ∆₅.

(ii) Set the vernier to the deflection angle ∆₅, the line of sight is thus directed along T₁H. Measure the distance T₁H along this direction, thus fixing the point H on the curve.

(iii) Locate the remaining points as usual. The point G which was left, will be located after removing the obstruction.

Alternatively, set out the curve T₂G in the reverse direction from T₂.

Example 3:

A railway curve is to tangential to each of the following line:

Find the radius of the curve and the tangents lengths.

Solution:

Refer to fig 11.22: 

∠ABC= Bearing of BA-Bering of BC

=180°- 90° =90°

∠BCD= Bearing of CB-Bearing of CD

=270°-140° = 130°

Let the bisectors of the angle ABC and BCD meet at O which is then the center of the curve, From O draw perpendicular OT₁ ,OT₂ and OT ₃ to AB,BC and CD respectively.T₁,T₂and T₃are the tangent points.