Fieldwork No 15 Tad - Fundamentals of Surveying Laboratory PDF

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Summary

Fieldwork No. 15 LAYING-OUT A COMPOUND CURVEName : Medina, Treb Asle Doen M. Weather : Cloudy Group No. : Place : Usc-Tc Field Designation : Recorder, Instrument Man Time, Start : 8:00 A Instructor : Engr. Imadyl Damuag End : 10:00 AI. ObjectivesTo lay out a compound curve consisting of two simple c...

Description

Fieldwork No. 15 LAYING-OUT A COMPOUND CURVE Name Group No. Designation Instructor

: Medina, Treb Asle Doen M. : : Recorder, Instrument Man : Engr. Imadyl Damuag

Weather Place Time, Start End

: Cloudy : Usc-Tc Field : 8:00 A.M : 10:00 A.M

I. Objectives To lay out a compound curve consisting of two simple curves with different radii by deflection angles method

II.

Instruments 1 unit - Engineers transit 1 roll - string 10 pcs - marking pins

III.

1 pc - Plumb bob 1 pc - 50-meter tape

Theory

A compound curve consists of two or more simple curves connected one after the other. The point where the curves are connected is called PCC which stands for point of compound curve. At that point the curves have the same tangent thus the term common tangent of the compound curve. The length of the common tangent is equal to the sum of the length of the tangent of the first curve, T1, and the length of the 2nd tangent, T2. The point of intersection of the tangents of the first curve is called PI1 and the point of intersection of the tangents of the 2nd curve is called PI2. Obviously the distance from PI1 to PI2 is the common tangent of the compound curve. The deflection angle of the common tangent with respect to the main backward tangent of the compound curve is equal to the central angle of the 1st curve and the deflection angle of the common tangent with respect to the main forward tangent of the compound curve is the same as the central angle of the 2nd curve. The compound curve can be laid out using deflection angles. For the first curve the deflection angles would be reckoned from to the main backward tangent and those of the second curve from the common tangent.

L

IV.

Procedure

A. Preliminary Calculations

1. The following data that was given by the instructor were used to layout the desired compound curve-centerline in the field. Sta PI1 = 100 + 00 1040 00’ ( central angle of curve)

Deflection angle at PI, I

=

Deflection angle I1

=

500

Deflection angle, I2

=

54o

Length of the curve, Lc1

=

35 m

Length of the curve, Lc2

=

37 m

(the central angle of the first curve)

2. The surveyors were asked to compute the value of the following curve parameters: Radius of the curve, R1

=

40.11 m

Radius of the curve, R2

=

39.26 m

Length of Tangent, T1

=

18.70 m

Length of Tangent, T2

=

20.00 m

Degree of curve, D1

=

40o 06’ (subtending a 20- meter arc length)

Degree of curve, D2

=

39o 15’ (subtending a 20- meter arc length)

3. The suveyors tabulated the computed deflection angles corresponding to every point on the curve and the corresponding lengths of subchord which will be shown in part V. B. Laying out the curve 1. The surveyors set up and leveled the instrument at a point of intersection of the tangents of the first curve (PI 1) and, using the calculated tangent distances and angle of intersections, they established the beginning of the curve (PC), the point of compound curve (PCC), the point of intersection of the tangents of the 2nd curve (PI2), and the end of the curve PT. 2. Then the surveyors transferred the instrument to the PC and laid out the first curve starting from the PC to the PCC using the deflection angles and subchord lengths. 3. The surveyors transferred the instrument at the PCC and laid out the 2nd curve from the PCC to the PT using the deflection angles from the common tangent and subchord distances. 4. The surveyors measured the length of the long chord, C1 and C2, subtended by the first and second curves that is from P.C.C to BC and P.C.C to E.C. and the distance from the PC to the PT for measuring purposes.

V.

Findings

Station Station

Deflection angles Deflection angles

Length of subchord (m) Length of subchord (m)

Remarks Remarks

10+006

00° 00’ 00”

0.00

PCC

10+016

07° 19’ 01”

10.00

4th even station

10+026

14° 45’ 24”

20.00

5th even station

10+036

22° 29’ 43”

30.00

6th even station

10+050

27o 00’00’’

34.67

PT

Station 1st Even station 2nd Even station 3rd Even station 4th Even station 5th Even station 6th Even station From PC Chord(m ) 10 20 30

Experimental deflection angle

Measured deflection angle

07° 09’ 40” 14° 26’ 13” 21° 57’ 32” 07° 19’ 01” 14° 45’ 24” 22° 29’ 43”

07° 12’ 20’’ 14° 24’ 00” 20° 30’ 40” 07° 04’ 40’’ 16° 08’ 40” 24° 20’ 40”

Theoretical Value Experimental Value

Percent Error

7’12’20’’ 14’24’00’’ 20’30’40’’

0.61% 0.26% 7.06%

From PT Chord(m) 10 20 30

7’9’39.45’’ 14’26’13’’ 21’57’38.75’’

Theoretical Value 7’4’40’’ 16’8’40’’ 24’20’40’’

Experimental Value 7’19’0.61’’ 14’45’23.51’’ 22’27’42.68’’

Percent Error 3.37% 8.60% 7.73%

 Sample Computations: Deflection angle at I2 ; Assumed value = 54

R=

Difference 00° 02’ 40” 00° 02’ 13” 01° 26’ 52” 00° 14’ 21” 01° 23’ 16” 01° 50’ 57”

L I π 180

( )

Radius R1 (1st curve) R1 = 40.11m 35 R1 = π 50 180

( )

Radius R2 (2nd curve)

R2 =

37 π 54 180

( )

R2 = 39.26

* Values for I1 and I2 were assumed and Values for L1 and L2 were assigned by the instructor.

Stationing of PC Sta. PC = Sta. PI1 – T I1 T 1 =R 1 tan 2

( )

T = 18.70m Sta. PC = 100+00 – 18.70 = 99+81.3

( )

50 T 1 =40.11 tan 2

Deflection Angles of Even Stations from PC α=sin−1

( 2l R)

α=sin−1

( 2 ×1040.11) =7 °9 39.45' '

α=sin−1

( 2 ×2040.11) =14 °26 13 ' '

α=sin

α=sin

−1

ch

'

'

−1

( 2 ×3040.11) =21° 57 38.75 ' ' '

( 2 ×4040.11) =29° 54 33.56 ' ' '

Deflection Angles of Even Stations from PCC α=sin−1

( 2l R)

α=sin−1

10 ( 2 ×39.26 )=7 ° 19 0.61 ' '

α=sin−1

30 ( 2 ×39.26 )=22 °27 42.68 ' '

α=sin−1

20 ( 2 ×39.26 )=14 ° 45 23.51' '

α=sin−1

40 ( 2 ×39.26 )=30 °37 31.78 ' '

ch

'

'

'

'

Length of Chord from PCC to PC

( I2 )

Let B = distance from PC to PCC Sine Law

1

T 1 =R 1 tan

( )

50 T 1 =40.11 tan 2

18.70 B = sin 25 sin 130 x = 33.90m

T1 = 18.70m

Length of Chord from PCC to PT T 2 =R2 tan

( )

Let A = distance from PT to PCC Sine Law

I2 2

T 2 =39.26 tan

( ) 54 2

y = 36.25m

T2 = 20.00m

Distance from PC to PT Cosine Law 76 ° cos ¿ z 2=51.342 +50.252−2 (51.34 )(50.25 )¿ z = 62.55m Let t1 = distance from PC to PI t1 =40.11 tan

20.00 A = sin 130 sin 25

( 1042 )

t1 = 51.34m Let t2 = distance from PT to PI

( 1042 )

t2 =39.26 tan

¿

Theoretical value−Experimental Value ∨x 10 0 Theoretical Value

At 20m Chord length (PC) 14 ’ 24 ’ 00 ’ ’−14 ’ 26 ’ 13 ’ ’ ∨x 100 = ¿ 14 ’ 24 ’ 00 ’ ’ 0.256 or 0.26% At 10m chord length (PT) 7 ’ 4 ’ 40 ’ ’−7 ’ 19 ’ 0.61 ’ ’ ∨x 100 = 0.0337 or 3.37% ¿ 7 ’ 4 ’ 40 ’ ’ The procedure of the fieldwork was well explained, even if we had taken some time visualizing the experiment, it was just really easy to do with the help of the drawing that was provided and our instructor. For improvements, I think we may need a little time to discuss the procedure as a group so that the experiment would be done faster and smoothly.

V.

Conclusion A compound curve is appropriate in a gradual change of direction and elevation for roadways, railroads and highways and it includes intersection curves for turning vehicles. They are mostly used because of the terrain’s curvature which are too steep that prevent transportation vehicles to move easily. To lessen sudden transition between two points in a steep setting, surveyors and engineers use compound curves which contains a common tangent. Compound curves perfectly fits the topography comparatively better than the simple curves. This method is used in laying out a curve to provide a smooth transition when there is a gradual increase in the height of two points, BC and BT. It is also used when you are laying out a curve that is connected to a straight line. In the fieldwork, there are several factors that caused the errors like the instrument that we are using has a horizontal movement that is faulty, measuring errors due to sag, temperature and pull and personal errors caused by the surveyors.

VII.

Sketch

L c1 =35

PI

Common tangent

I

m

Lc2=35m PI1

I1=50°PCC

I2=54° PI2

PC

PT

I2 =54°

R2 = 39.26m

R1=40.11m

I1 = 50°

Documentation

Photo of our Data as proof in checking the angles.

View from point PC

View from point PT...