- SVY3107 GEODETIC SURVEYING B - 2008 TUTORIAL/PRACTICAL DISTANCE REDUCTIONS
Obtain copy of Distred.XLS In Windows Explorer click right button on this file - Properties - General - Attributes Make sure that Distred.XLS is not Read Only. Go to `Distance Reduction' sheet and expand the spreadsheet using Tools - Options tick on Outline Symbols. Cell E27 should contain the following formula: =Ellipsoids!B7 Change this if necessary. Save this change. Change the file attributes to Read Only. Go to Ellipsoids tab. Ellipsoids - in cell B5 enter number of ellipsoid - enter 3 - arrow down Distance reduction - note we are now on ANS We will now check the example from the study book. Example A distance has been measured between A and B with microwave EDME, corrected for atmospheric conditions (assume first and second velocity), scale factor, and first arc-tochord, as 43517.3346m. The RL of A is 658.37m and of B 725.25m AHD. The heights of the instruments are 1.47m at A and 1.83m at B. Determine the spheroidal distance between the 2 stations. Approximate azimuth = 45 N(ANS) is 14.563 for A and 14.682 for B N(WGS84) is 42.047 for A and 42.166 for B Assume = -30 Answer (ANS) = 6351399.358, =6383504.066 and R = 6367411.244 d2 = 43517.3346 l = 43517.2827 (Slope correction) s" = 43512.5438 (AHD correction) s' = 43512.6285 (Chord-to-arc correction) s (d4) = 43512.528 (Sea level to spheroid correction)
Now use the spreadsheet to solve the problem and (hopefully) get the same answer. For first station (no need to change the name) H = 658.37 + 1.47 = 659.84 For second station H = 725.25 + 1.83 = 727.08 Input d2 = 43517.3346 The answer you get for should be 43512.528 which agrees with our manual calculations in the example. Read the notes at the bottom of `Distance Reduction'. Now we will check our chord-to-arc correction: Assume H = 0 and N = 0 (ie. the measurement is at mean sea level and the geoid coincides with the ellipsoid at that point). Therefore d2 observed will equal d3. Input d2 (now our chord distance) = 43512.5438 (distance at MSL - AHD = 0) What is the arc distance? The answer should be 43512.6285 giving us = 0.0847 Which is the same as our manual chord-to-arc correction!
Now change ellipsoids to GRS80 Is there any change in the distance? Why not? What does this tell us? (That R is not critical?) Now reduce our original distance to GRS80 The spheroidal distance should be 43512.341 This distance is significantly smaller than the ANS spheroidal - distance Why? Now change to WGS84 ellipsoid. Is there any change? What do we deduce form this? Tools - Options - Tick `outline symbols' - OK Click the + buttons at the top and LHS and top to reveal intermediate steps Note line 8 - Ellipsoidal height (h) = H + N Note line 14 = d3 Therefore we now have a new way of calculating chord-to-arc correction Change to ANS Change N values back to ANS values d2 = 43517.3346 s - d3 = chord-to-arc correction = 0.0847 (the same result as before) Set H = 0, N = 0, and d2 = 43512.5456 as before Note that d2 does indeed equal d3 Go down to lines 26 to 30 On RHS an inspection of the formula tells us that the three values represent , , R in that order Try the following inputs = 0 We find that (radius of curvature in meridian) = R = 90 We find that (radius of curvature in prime vertical) = R Change ellipsoids to GRS80 note that now = 6351377.104 = 6383480.918 Change ellipsoids to WGS84 note that and have not changed What does this tell us? Make the following changes and note what happens to and values = 0 (a = and and are minimum at equator) = 1 = 89 = 90 ( = = maximum at the poles) Do you understand the reasons for this? Try different values for when = 90 Set = -2730' N = 12 (ANS-AHD maximum) H = 0 (assume at MSL) What error is caused in a 10km traverse due to assuming that N=0 instead of 12 or -12? Your answer should be 18 or 20 mm What if we assumed N = 6 (average difference) Your answer should be 8 or 10 mm Assume that we are doing a 10km traverse at the Gold Coast that we want to reduce to the spheroid (ANS). How accurately do we have to know the RLs of our stations to be sure the error due to this source will be no greater that the measuring accuracy of our EDM (5mm+5ppm)? Note we will ignore N values for this exercise - assume the geoid is coincident with AHD at this point so N=0 For a 10km line 5mm+5ppm represents 55mm Your answer should be approximately 30m for RLs Does this change if we are using the GRS80 ellipsoid (assume N=42.5)?
Does this answer change if we are measuring the traverse in Toowoomba on the ANS at an RL of 680m AHD? Using RL = 680, d2 = 10000, s = 9998.933 Using RL = 715, d2 = 10000, s = 9998.878, therefore = 0.055 Your answer should be approximately 35m for RLs Does this change if we are using the GRS80 ellipsoid (assume N=42.5)? Does this answer change if we are measuring the traverse at an RL of 2000m AHD? Close Distred.XLS and do not save changes