Geodetic Mapping Individual Assignments

Assignment 04 1. Given the geodetic position of point Q referring to the WGS84 reference ellipsoid as follows: ϕQ = -6° 34’ 47,84875’’ λQ = +107° 52’ 32,06722’’ hQ = +734,881 m Calculate the geodetic position of point Q, but expressed on the Bessel 1841 reference ellipsoid! Use iterative formulas and closed formulas when converting from 3D Cartesian coordinate system to geodetic coordinate system! Use computational software!! Answer : Known data • ϕQ = -6° 34’ 47,84875’’ • λQ = +107° 52’ 32,06722’’ • hQ = +734,881 m • Parameter ellipsoid WGS 84 Bessel a = 6.378,137 km a = 6.377,397 km b = 6.356,752 km b = 6.356,078 km How to work A. Conversion from Geodetic WGS 84 to Cartesian 3D a. Convert ϕQ and λQ to degrees ϕQ = - ( 6 + 34/60 + 47,84875/3600 ) ° = - 6,579957° λQ = ( 107 + 52/60 + 32,06722/3600 ) ° = 107,875574° b. Looking for eccentricity (e) eWGS84 = e’ 2−𝑏2 √𝑎 = = 0,006694 𝑎 2−𝑏2 √𝑎 WGS84 = 0,006739 𝑏 c. Find the magnitude of the main vertical curve (N) on the WGS 84 ellipsoid. aWGS 84 NWGS 84 = ( 1 – e2 ( sin ϕQ2))1/2 = 6.378,417,345586 m d. Finding 3D Cartesian Coordinates Coordinates X X = ( N + h ) cos ϕQ cos λQ = ((6.378.417,345586 + 734,881)cos(-6,579957°)cos(107,875574°)) m = - 1.945.188,522990 m Coordinates Y Y = ( N + h ) cos ϕQ sin λQ = (( 6.378.417,345586 +734,881 ) cos (- 6,579957° ) sin (107,875574°)) m = 6.031.209,316099 m Coordinates Z Z = [ N ( 1 – e2 ) + h ] sin ϕQ = -726.091,973950 m So the coordinates of point Q on the 3D Cartesian plane are (X, Y, Z ) = (1.945.188,522990 ; 6.031.209,316099 ; -726.091,973950) m. B. Conversion from Cartesian to Geodetic Bessel 1841dari Kartesia ke Geodetik Bessel 1841 a. Finding longitude 𝑋 λ = tan-1 ( ) 𝑌 = -72,124425° Since λ is in quadrant 2, then,λ = 72,124425° + 180° = 107,875574° = 107°52’32,064’’ Iteration formula a. Finding the eccentricity value of Bessel 1841 eBessel = e’ Bessel = 2−𝑏2 √𝑎 = 0,006674 𝑎 2−𝑏2 √𝑎 = 0,006719 𝑏 b. Find the p value p = √𝑋2 + 𝑌2 = 6.337.132,175091 m c. Looking for Geodetic coordinates (Latitude) on the Bessel 1841 ellipsoid 𝑍 ϕ = tan-1 ( ( 1 – e2 ( 𝑁 ))) Bessel 𝑝 𝑁+ℎ Continue iterating until the values of ϕ and h are constant or have a difference < 0.0001. *First Iteration ( Set h = 0 ) 𝑍 ϕ = tan-1 ( ( 1 – e2 ( Bessel (1) 𝑝 𝑁 ))) 𝑁+ℎ = - 6.579831° NBessel (1) = ( 1 – e 2 aBessel ( sin2 (ϕ Bessel (1) Bessel ) )1/2 = 6.377.676,619543 m p hBessel (1) = - NBessel (1) cos ϕBessel (1) = 1.473,984907 m *Second Iteration ( Set h = hBessel (1) ) NBessel (1) 𝑍 ϕ Bessel (2) = tan-1 ( ( 1 – e2 ( 𝑝 )) NBessel (1) + hBessel (1) = -6,579821° NBessel (2) = ( 1 – e aBessel 2 ( sin2 (ϕ Bessel (2) Bessel ) )1/2 = 6.377.676,618686 m p hBessel (2) = - NBessel (2) cos ϕBessel (2) = 1.473,855721 m The difference ϕ between the first and second iterations is 1.0126° and the difference h is 0.129186 m, so the third iteration is carried out again. *Third Iteration ( Set h = hBessel (2) ) ϕ Bessel (3) 𝑍 NBessel (2) 𝑝 NBessel (2) + hBessel (2) = tan-1 ( ( 1 – e2 ( )) = -6,579821° NBessel (3) = ( 1 – e aBessel 2 ( sin2 (ϕ Bessel Bessel (3) ) )1/2 = 6.377.676,618686 m p hBessel (3) = cos ϕBessel (3) - NBessel (3) = 1.473,855732 m The difference ϕ between the third and second iterations is -8.872761° and the difference h is 1.131929 m, so the fourth iteration is carried out again. *Fourth Iteration ( Set h = hBessel (3) ) 𝑍 NBessel (3) 𝑝 NBessel (3) + hBessel (3) = tan-1 ( ( 1 – e2 ( ϕ Bessel (4) )) = -6,579821° NBessel (4) = ( 1 – e 2 aBessel ( sin2 (ϕ Bessel (4) Bessel ) )1/2 = 6.377.676,618686 m p hBessel (4) = - NBessel (4) cos ϕBessel (3) = 1.473,855732 m The difference ϕ between the third and fourth iterations is 1.346145° and the difference h is 9.313225 m, so the fifth iteration is carried out again. *Fifth Iteration ( Set h = hBessel (4)) NBessel (4) 𝑍 ϕ Bessel (5) = tan-1 ( ( 1 – e2 ( 𝑝 )) NBessel (4) + hBessel (4) = -6,579821° NBessel (5) = ( 1 – e aBessel 2 ( sin2 (ϕ Bessel (5) Bessel ) )1/2 = 6.377.676,618686 m p hBessel (5) = - NBessel (5) cos ϕBessel (5) = 1.473,855732 m The difference ϕ between the third and fourth iterations is 0° and the difference h is 0 m, so the iteration stops at the fifth iteration. Then the latitude and height of point Q are, ϕ = -6,579821° = - 6°34’47,3556’’ h = 1.473,855732 m Closed Formula a. Finding Bessel eccentricity eBessel = e’ = 2−𝑏2 √𝑎 = 0,006674 𝑎 2−𝑏2 √𝑎 Bessel = 0,006719 𝑏 b. Find the p value p = √𝑋2 + 𝑌2 = 6.337.132,175091 m c. Finding Ɵ Ɵ = tan-1 ( 𝑍𝑎 𝑝𝑏 ) = -6,558028° d. Looking for Geodetic coordinates (Latitude) on the Bessel 1841 ellipsoid Z + e’2 b sin3Ɵ ϕBessel = tan-1 ( p – e2 a cos3Ɵ ) = -6,579821° = - 6°34’42,3556’’ e. Find the magnitude of the main vertical curve (N) on the Bessel 1841 ellipsoid aBessel NBessel = ( 1 – e Bessel2 ( sin ϕ Bessel 2))1/2 = 6.378.417,345586 m f. Finding the height on the Bessel ellipsoid 1841 hBessel = p - NBessel cos ϕBessel = 1.473,855732 m From the calculation of the closed formula and iteration, the geodetic coordinates of point Q with the Bessel 1841 reference are λ Q = 107°52’32.064’’, ϕ = - 6°34’47.3556’’, and h = 1,473.855732 m. 2. Given the geodetic position of point Q which refers to an ellipsoid with dimension GRS80 and refers to a particular 3-D Cartesian coordinate system, say system 1. The coordinates of point Q are: 𝜑𝑄 = −60 34′ 47,84875 " 𝜆𝑄 = +1070 52′ 32,06722 " ℎ𝑄 = +734,881 𝑚 Calculate the geodetic position of point Q which is also stated with respect to the ellipsoid GRS80, but the ellipsoid is oriented to another 3-D Cartesian coordinate system, say system 2, and its three axes are kept parallel to each other. The displacement vectors of the starting points of the two Cartesian coordinate systems are: 𝑋2 − 𝑋 1 = +1,000 𝑚 𝑌2 − 𝑌1 = −2,000 𝑚 𝑍 2 − 𝑍 1 = +2,000 𝑚 Use iterative formulas and closed formulas when converting from a 3D Cartesian coordinate system to a geodetic coordinate system! Use computing software! Answer : Known Data • ϕQ = -6° 34’ 47,84875’’ • λQ = +107° 52’ 32,06722’’ • hQ = +734,881 m • Parameter ellipsoid GRS80 a = 6.378.137 m b = 6.356.752,314 m • Displacemen Vector 𝑋2 − 𝑋 1 = ΔX = 1 𝑚 𝑌2 − 𝑌1 = ΔY = − 2 𝑚 𝑍 2 − 𝑍 1 = ΔZ = 2 𝑚 How To Work A. Conversion from GRS80 Geodetic to 3D Cartesian a. Convert ϕQ and λQ to degrees ϕQ = - ( 6 + 34/60 + 47,84875/3600 ) ° = - 6,579957° λQ = ( 107 + 52/60 + 32,06722/3600 ) ° = 107,875574° b. Looking for eccentricity (e) 2−𝑏2 e= √𝑎 e’ = √𝑎 𝑎 2−𝑏2 𝑏 = 0.006694 = 0,006739 c. Finding the magnitude of the main vertical curve (N) a N= ( 1 – e2 ( sin ϕQ2))1/2 = 6.378,417,345588 m d. Finding 3D Cartesian Coordinates X Coordinates X = ( N + h ) cos ϕQ cos λQ = ((6.378.417,345586 + 734,881)cos(-6,579957°)cos(107,875574°)) m = - 1.945.188,522990 m Y Coordinates Y = ( N + h ) cos ϕQ sin λQ = (( 6.378.417,345586 +734,881 ) cos (- 6,579957° ) sin (107,875574°)) m = 6.031.209,316102 m Z Coordinates Z = [ N ( 1 – e2 ) + h ] sin ϕQ = -726.091,973904 m So the coordinates of point Q on the 3D Cartesian are (X, Y, Z ) = (1,945,188.522990;6,031,209.316102 ; -726,091.973904) m.. B. Convert from 3D Cartesian to another 3D Cartesian XQ = X + ΔX = -1.945.187,522990 m YQ = Y + ΔY = 6.031.207,316102 m ZQ = Z + ΔZ = -726.089.973904 m Then the coordinates of point Q on another 3D Cartesian are (XQ, YQ, ZQ ) = (1,945,187.522990 ; 6,031,207,316102 ; -726,089.973904) m. C. Conversion from other Cartesian to Geodetic GRS80 b. Finding longitude 𝑋 λ = tan-1 ( ) 𝑌 = -72,124425° Since λ is in quadrant 2, then, λ = -72,124425° + 180° = 107.875571° = 107°52’32,556’’ Iteration Formula a. Find P value p = √𝑋2 + 𝑌2 = 6.337.129,964692 m b. Looking for Geodetic coordinates (Latitude) 𝑍 ϕ = tan-1 ( ( 1 – e2 ( 𝑝 𝑁 𝑁+ℎ ))) Continue the iteration until the values of ϕ and h are constant or have a difference < 0,0001. *First Iteration ( Set h = 0 ) 𝑍 ϕ = tan-1 ( ( 1 – e2 ( 𝑁 ))) (1) 𝑝 𝑁+ℎ = - 6.579947° a N (1) = ( 1 – e2 ( sin2 (ϕ (1) ) )1/2 = 6.378.417,344687 m p h (1) = - N (1) cos ϕ = 732,520369 m *Second Iteration ( Set h = h (1) ) 𝑍 -1 2 ϕ(2) = tan ( ( 1 – e ( 𝑝 N (1) N(1) + h (1) )) = -6,579942° a N(2) = ( 1 – e2 ( sin2 (ϕ(2) ) )1/2 = 6.378.417,344258 m p h(2) = - N(2) cos ϕ (2) = 732,455973 m The difference ϕ between the first and second iterations is 5.047556° and the difference h is 0.064395 m, so the third iteration is carried out again.. *Third Iteration ( Set h = h (2) ) 𝑍 N (2) ϕ(3) = tan-1 ( ( 1 – e2 ( 𝑝 )) N(2) + h (2) = -6,579942° a N(3) = ( 1 – e2 ( sin2 (ϕ(3) ) )1/2 = 6.378.417,344258 m p h(3) = - N(3) cos ϕ (3) = 732,455979 m The difference ϕ between the third and second iterations is -4.436794° and the difference h is 5.660578 m, so the fourth iteration is carried out again.. *Fourth Iteration ( Set h = h (3) ) N (3) ϕ(4) = tan-1 ( 𝑍 ( 1 – e2 ( 𝑝 )) N(3) + h (3) = -6,579942° a N(4) = ( 1 – e2 ( sin2 (ϕ(4) ) )1/2 = 6.378.417,344258 m p h(4) = - N(4) cos ϕ (4) = 732,455979 m The difference ϕ between the third and fourth iterations is 3.816665° and the difference h is 9.313225 m, so the fifth iteration is carried out again. *Fifth Iteration ( Set h = h (4) ) 𝑍 N (4) ϕ(5) = tan-1 ( ( 1 – e2 ( 𝑝 )) N(4) + h (4) = -6,579942° a N(5) = ( 1 – e2 ( sin2 (ϕ(5) ) )1/2 = 6.378.417,344258 m p h(5) = - N(5) cos ϕ (5) = 732,455979 m The difference ϕ between the third and fourth iterations is 0° and the difference h is 0 m, so the iteration stops at the fifth iteration. . Then the latitude and altitude of point Q are,, ϕ = -6,579942° = - 6°34’47,7912’’ h = 732,455979 m Closed formula a. Find P Result p = √𝑋2 + 𝑌2 = 6.337.129,964692 m b. Find Ɵ Ɵ = tan-1 ( 𝑍𝑎 𝑝𝑏 ) = -6,558074° c. Search for Geodetic coordinates (Latitude) Z + e’2 b sin3Ɵ tan-1 ϕl = ( p – e2 a cos3Ɵ ) = -6,579942° = - 6°34’47,7912’’ d. Find the magnitude of the main vertical curve (N) a N= ( 1 – el2 ( sin ϕ2))1/2 = 6.378.417,344258 m e. Seeking height h = p cos ϕ = 732,455979 m -N From closed formula calculations and iterative formulas, the geodetic coordinates of point Q with the GRS 80 reference are λQ = 107°52'32.556'', ϕQ = - 6°34'47.7912'', and hQ = 732.455979 m. 3. Given the geodetic position of point P with reference to the ellipsoid with WGS84 dimensions. The coordinates of point P are: 𝜑𝑃 = −6034′ 47,84875 " 𝜆𝑃 = +1070 52′ 32,06722 " ℎ𝑃 = +734,881 𝑚 To determine the geodetic position of point Q which is also stated referring to the WGS84 ellipsoid, then the baseline vector is determined from P to Q through relative GPS satellite observations. Calculate the geodetic position of point Q! ∆𝑋 = −1577,302 𝑚 ∆𝑌 = +32,484 𝑚 ∆𝑍 = +53,056 𝑚 Use iterative formulas and closed formulas when converting from a 3D Cartesian coordinate system to a geodetic coordinate system! Use computing software! Answer Known Data • ϕP = -6° 34’ 47,84875’’ • λP = +107° 52’ 32,06722’’ • hP = +734,881 m • Parameter ellipsoid WGS 84 a = 6.378.137 m b = 6.356.752,3142 m • Vector Baseline ∆𝑋 = −1577,302 𝑚 ∆𝑌 = 32,484 𝑚 ∆𝑍 = 53,056 𝑚 How To Work A. Conversion from Geodetic WGS 84 to Cartesian 3D a. Convert ϕQ and λQ to degrees ϕQ = - ( 6 + 34/60 + 47,84875/3600 ) ° = - 6,579957° λQ = ( 107 + 52/60 + 32,06722/3600 ) ° = 107,875574° b. Looking for eccentricity (e) e= 2−𝑏2 √𝑎 𝑎 = 0.006694 e’ = 2−𝑏2 √𝑎 𝑏 = 0,006739 c. Find the magnitude of the main vertical curve (N) on the WGS 84 ellipsoid N= a ( 1 – e2 ( sin ϕQ2))1/2 = 6.378,417,345588 m d. Finding 3D Cartesian Coordinates X Coordinates X = ( N + h ) cos ϕQ cos λQ = ((6.378.417,345586 + 734,881)cos(-6,579957°)cos(107,875574°)) m = - 1.945.188,522990 m Y Coordinates Y = ( N + h ) cos ϕQ sin λQ = (( 6.378.417,345586 +734,881 ) cos (- 6,579957° ) sin (107,875574°)) m = 6.031.209,316109 m Z Coordinates Z = [ N ( 1 – e2 ) + h ] sin ϕQ = -726.091,973950 m So the coordinates of point P on the 3D Cartesian are (XP, YP, ZP) = (- 1,945,188.522990; 6,031,209.316109; -726,091.973950) m. B. Convert from 3D Cartesian to another 3D Cartesian XQ = X + ΔX = -1.946.765,824990 m YQ = Y + ΔY = 6.031.241,800099 m ZQ = Z + ΔZ = -726.038,917950 m Then the coordinates of point Q on another 3D Cartesian are (XQ, YQ, ZQ ) = (1,946,765,824990 ; 6,031,241,800099 ; -726,038,917950) m.. C. Conversion from other Cartesian to Geodetic WGS 84 a. Finding Longitude 𝑋 λP = tan-1 ( ) 𝑌 = -72,110944° Since λ is in quadrant 2, then, λP = -72,124425° + 180° = 107,889055° = 107°53’20,598’’ Iteration Formula a. Find P Result p = √𝑋2 + 𝑌2 = 6.337.647,420661 m b. Search for Geodetic coordinates (Latitude) 𝑍 ϕ = tan-1 ( ( 1 – e2 ( 𝑁 ))) P 𝑁+ℎ 𝑝 Continue the iteration until the values of ϕP and h are constant or have a difference < 0,0001. *First Iteration ( Set h = 0 ) 𝑍 ϕ = tan-1 ( ( 1 – e2 ( 𝑁 ))) (1) 𝑝 𝑁+ℎ = -6,578956° a N (1) = ( 1 – e2 ( sin2 (ϕ (1) ) )1/2 = 6.378.417,344687 m p h (1) = - N (1) cos ϕ = 1.240,762950 m *Second Iteration ( Set h = h (1) ) N (1) 𝑍 -1 2 ϕ(2) = tan ( ( 1 – e ( 𝑝 )) N(1) + h (1) = -6,578947° a N(2) = ( 1 – e2 ( sin2 (ϕ(2) ) )1/2 = 6.378.417,344258 m p h(2) = - N(2) cos ϕ (2) = 1.240,653907 m The difference ϕ between the first and second iterations is 8.547738° and the difference h is 0.109042 m, so the third iteration is carried out again. . *Third Iteration ( Set h = h (2) ) N (2) ϕ(3) = tan-1 ( 𝑍 ( 1 – e2 ( 𝑝 N(2) + h (2) = -6,578947° a N(3) = ( 1 – e2 ( sin2 (ϕ(3) ) )1/2 = 6.378.417,259877 m p h(3) = cos ϕ (3) - N(3) )) = 1.240,653917 m The difference ϕ between the third and second iterations is -7.510569° and the difference h is 9.582377 m, so the fourth iteration is carried out again. . *Fourth Iteration ( Set h = h (3) ) N (3) ϕ(4) = tan-1 ( 𝑍 ( 1 – e2 ( 𝑝 )) N(3) + h (3) = -6,578947° a N(4) = ( 1 – e2 ( sin2 (ϕ(4) ) )1/2 = 6.378.417,259877 m p h(4) = - N(4) cos ϕ (4) = 1.240,653917 m The difference ϕ between the third and fourth iterations is 6.679164° and the difference h is 1.862645 m, so the fifth iteration is carried out again.. *Fifth Iteration ( Set h = h (4) ) 𝑍 N (4) ϕ(5) = tan-1 ( ( 1 – e2 ( 𝑝 )) N(4) + h (4) = -6.578947° a N(5) = ( 1 – e2 ( sin2 (ϕ(5) ) )1/2 = 6.378.417,259877 m p h(5) = - N(5) cos ϕ (5) = 1.240,653917 m The difference ϕ between the third and fourth iterations is 0° and the difference h is 0 m, so the iteration stops at the fifth iteration.. Then the latitude and altitude of point Q are,, ϕP = -6.578947° = - 6°34’44,2092’’ hP = 1.240,653917 m Closed Formula a. Find P Result p = √𝑋2 + 𝑌2 = 6.337.647,420661 m b. Find Ɵ Ɵ = tan-1 ( 𝑍𝑎 𝑝𝑏 ) = -6,557090° c. Search for Geodetic coordinates (Latitude) Z + e’2 b sin3Ɵ tan-1 ϕl = ( p – e2 a cos3Ɵ ) = -6.578947° = - 6°34’44,2092’’ d. Finding the magnitude of the main vertical curve (N) a N= ( 1 – el2 ( sin ϕ2))1/2 = 6.378.417,259877 m e. Find the Height p h = -N cos ϕ = 1.240,653917 m From the calculation of the closed formula and iteration formula, the geodetic coordinates of point P with WGS 84 reference are λP = 107°53’20.598’’, ϕP = 6°34’44.2092’’, and hP = 1,240.653917 m..