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CE 111 Circular Curve Lab

Spring 2005

Objective: Develop skills in determining positions of circular curves for field layout using standard

taping techniques and off-curve EDM measurements. Apply knowledge of spreadsheet

development learned in traverse laboratories to compute circular curve coordinate and layout data

for field survey.

Data

PI Station = 32+ 30.30 with coordinates of 170.34E, 284.51N (your coordinate values and

distances only need to be shown to the nearest 0.01 foot).Azimuth of BC to PI tangent = 30O 00 00

Delta = 36O 00 00 and the curve is to the right

D = 8O 00 00

Computations required

1) Complete sufficient manual computations of the circular curve to identify the following:

BC and EC stations around the curve, as well as the EC station coming along the tangent.

Deflection angle from BC to station 34+00

Coordinates of Station 34+00The answers are included in the following table, although the individual calculations to developthem are not included here

In the hand computations they were only asked to complete a small portion of the table below.

The BC and EC stations, the deflection angles and the Coordinates of Station 34 are shown in thetable. The only element not shown there is the EC station coming along the tangent from the PI

Station EC from PI = 32+30.30 + T = 32+30.30 + 2+32.71 = 34+63.01

In the table EC around the curve was found to be 34+ 47.49

.

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Sheet3

Simple Circular curves revised 1/2/03 (Spiral length is 0 for simple circular curves)

Given R, Delta, PI sta and coordinates, azimuth from PC to PIPlace needed data in shaded cells Distances may be m or ft.

deg 8 716.1975

Givens: spirallen= 0

PI Sta = 3230.30m X coord Y coord

Delta = 36deg PI = 170.348 284.507

PC or TS 53.99 82.98

Azi PC to PI = 30deg PT or ST 382.94 379.16

Radius = 716.197m Radius pt. 164.47 -275.12of CC

Curve to : mid pt cc 195.01 317.38

Right =+1 1 Back AZ= 210.00deg

Left = -1

L of Cir C 450.00 Partial Chords:

Tangent = 232.71 Nominal length = 50.00

L.C. of cc 442.63 true chord = 49.99

mid ord = 35.05 mid ord = 0.44

Ext. = 36.86

Increment = 50

arc dist defl angleDEFLECTION ANGLE chord

STA Sta (m) from PC from PC degrees min from PC X Coord Y Coord

BC or SC 2997.59 0 0 0.00 0.00 0 0.00 53.99 82.98

30 3000 2.41 0.096264 0.00 5.00 47 2.41 55.20 85.06

30.5 3050 52.41 2.096264 2.00 5.00 47 52.39 81.83 127.3631 3100 102.41 4.096264 4.00 5.00 47 102.32 111.35 167.71

31.5 3150 152.41 6.096264 6.00 5.00 47 152.12 143.61 205.89

32 3200 202.41 8.096264 8.00 5.00 47 201.73 178.46 241.74

32.5 3250 252.41 10.09626 10.00 5.00 47 251.10 215.72 275.06

33 3300 302.41 12.09626 12.00 5.00 47 300.17 255.22 305.71

33.5 3350 352.41 14.09626 14.00 5.00 47 348.86 296.76 333.52

34 3400 402.41 16.09626 16.00 5.00 47 397.13 340.13 358.37

CS or EC 3447.59 450.00 18 18.00 0.00 0 442.63 382.94 379.16

2) General spreadsheetA basic spreadsheet layout with initial computations will be provided by the instructor

The results of the spreadsheet are included on the previous page

3) Angle and distance measurements for EDM

The surveyor found a convenient location in the field to use to set out the curve from a

single point that was not on the curve. The following measurements were made from this positionto two control points A and B, with the instrument at I.

Page 1

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Instrument at I. Distance IA = 477.86 feet, Distance IB = 499.26 feet. Points A and B were

both west of the instrument. The angle AIB was 75O 59 00.

a) In the office the coordinates of the control points were found to be:

A: 398.44 E, 215.65 N B: 432.66 E, 877.13 N.

Calculate the position the surveyor plans to occupy by using two strategies:1. Use the measured distance data in conjunction with the calculated AB distance

2. Use distance AI and the measured angle.

1) Find length and azimuth of A B using basic inversing equations learned in the traverseprogramming:

Length = 662.36 Azimuth = 20 57 41

2) Strategy 1: Find coordinate of I using only the distancesCosine law can be applied to determine Angle I and the Sine law or Cosine Law can be used

to determine the other two angles: The values are shown in Table 1 along with the solutions using

the measured angle, the control line and one of the measured field distances.

A

B

I

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Strategy 2: Find coordinate of I using Angle AIB, and distances AB and AISine law can be used for all remaining unknowns.

The results are shown in Table 1 and Table 2.

Strategy 3: Although not required, the calculated coordinates using line IB, AB and angle AIB

are also shown.

(Note: the process of computing the coordinates of an occupied point using field measurements isreferred to as resection. Many modern total stations have the software to do this for the user. Not

all total stations in our laboratory have this capability).

Strategy 1 requires application of the cosine law to find the first angle, then cosine law or sine law

can be used The three angles should be:

A = 48 d 44 11 B = 45d 58 48 I = 85d 19 01

The coordinates of I can be found by using the coordinates of A and the knowledge that theAzimuth to I must be 20 57 41 + 480 44 11

Strategy 2 can begin with the sine law to find Angle B (using angle I plus the distances IA and AB.All results are shown in Tables 1 and 2 below.

Only Strategy 1 and 2 were asked for in the lab instructions.

Table 1 Computed angles and sides for Instrument location

Distance or Angle Strategy 1 Strategy 2 Strategy 3

IA 477.86 (measured) 477.86 (measured) 477.94

IB 499.26 (measured) 499.33 499.26 (measured)

Angle A 480 41 56 480 41 28 480 41 53

Angle B 45

0

58 34 45

0

58 37 45

0

59 07Angle I 850 19 30 850 8900(measrd) 850 1900 (measrd)

Table 2 Coordinates of Total Station (I)

Strategy Easting Northing

Strategy 1 773.25 512.08

Strategy 2 773.29 512.02

Strategy 3 773.31 512.13

Average 773.28 512.08

Relying on distance measurements produced a calculated angle 30 seconds less than the measuredangle. This difference would be considered very large for an instrument that apparently can

measure to the nearest second. Of course, the angle may have been correct and the difference is

due to errors in the distance measurements. Fortunately, because the triangle was a reasonably

strong triangle, the coordinates were reasonably consistent. (Small angles, especially those lessthan 20 degrees when the sine of the angle is needed in computations, are undesirable). Even in

our situation, the computed Northing differed by over 0.10 feet, dependent on the data used to

compute the coordinate.

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b) Using the coordinates from part a, establish a table to show the azimuth and distance fromthe instrument to points A and B and all the curve points including the PI, BC, EC, and every half

station along the curve.

If coordinates of I are taken to be 773.28E, 512.08N the Azimuth to point A (398.44 E, 215.65 N )

is Azi = 2700 Atan( (215.65-512.08)/(398.44- 773.28) = 2310 39 45Dist = sqrt(deltax^2 + deltaY^2) = 477.89

All other distances and azimuths from I could be calculated in a similar manner. Since the students

had already written a program to inverse coordinates to distances and azimuths, I would expectmost of them to copy segments of that program to make a spreadsheet that would work here.

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Table 3 Distance and Azimuth data for Circular curve from Total Station

Coordinates delta E delta Ndist tonex sta Azimuth Azimuths

STATION E Nto nextsta point

(radians) degrees minutes sec

INSTRUMENT 773.28 512.08

Contrl point 398.44 215.65-

374.840-

296.430 477.89 4.043272 231 39 45

BC Sta. 53.99 82.98-

719.293-

429.100 837.56 4.174505 239 10 53

30.00 55.19 85.06-

718.087-

427.018 835.46 4.175905 239 15 42

30.50 81.83 127.37-

691.454-

384.713 791.27 4.204658 240 54 33

31.00 111.35 167.71-

661.935-

344.369 746.16 4.232675 242 30 52

31.50 143.61 205.90-

629.673-

306.183 700.17 4.259796 244 4 6

32.00 178.45 241.74-

594.827-

270.340 653.38 4.285811 245 33 32

32.50 215.72 275.06-

557.565-

237.016 605.85 4.310441 246 58 12

33 255.21 305.71-

518.069-

206.371 557.66 4.333308 248 16 49

33.5 296.75 333.52-

476.532-

178.557 508.89 4.353881 249 27 32

34 340.12 358.37-

433.156-

153.708 459.62 4.371395 250 27 45

EC 382.93 379.16

-

390.352

-

132.920 412.36 4.384191 251 11 44

PI 170.34 284.51-

602.940-

227.570 644.46 4.351486 249 19 18

SummaryThe spreadsheet programs took time to develop, but the principles used were those learned

from previous activity. Similar problems encountered in the future (class project and other) will be

easily addressed. In reviewing the final layout problem, the surveyor may need to re-consider thesetup location as several distances are long; sighting the point may be difficult unless the optics of

the instrument are good. In any event, to verify the correctness of the layout the points will need to

be checked from another location or by a different method.