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+00
The answers are included in the following table, although the individual calculations to develop them 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 the table. 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
.
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 PI
Place needed data in shaded cells / Distances may be m or ft.
deg / 8 / 716.1975
Givens: / spirallen / = 0
PI Sta = / 3230.30 / m / X coord / Y coord
Delta = / 36 / deg / PI = / 170.348 / 284.507
PC or TS / 53.99 / 82.98
Azi PC to PI / = 30 / deg / PT or ST / 382.94 / 379.16
Radius = / 716.197 / m / Radius pt. / 164.47 / -275.12 / of CC
Curve to : / mid pt cc / 195.01 / 317.38
Right =+1 / 1 / Back AZ= / 210.00 / deg
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 angle / DEFLECTION 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 / 127.36
31 / 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 spreadsheet
A 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 position to two control points A and B, with the instrument at I.
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 traverse programming:
Length = 662.36 Azimuth = 20 57’ 41”
2) Strategy 1: Find coordinate of I using only the distances
Cosine 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.
Strategy 2: Find coordinate of I using Angle AIB, and distances AB and AI
Sine 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 is referred 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 the Azimuth 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 3IA / 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 / 450 58’ 34” / 450 58’ 37” / 450 59’ 07”
Angle I / 850 19’ 30” / 850 89’00”(meas’rd) / 850 19’00” (meas’rd)
Table 2 Coordinates of Total Station (I)
Strategy / Easting / NorthingStrategy 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 measured angle. 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 less than 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.
b) Using the coordinates from part a, establish a table to show the azimuth and distance from the 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’ 45”
Dist = 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 expect most of them to copy segments of that program to make a spreadsheet that would work here.
Table 3 Distance and Azimuth data for Circular curve from Total Station
Coordinates / delta E / delta N / dist to nex sta / Azimuth / AzimuthsSTATION / E / N / to next sta / 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
Summary
The 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 the setup 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.