Map Projections, Coordinate Conversion, and Meridian Convergence PDF - GE Board Exam Review 2024
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2024
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Summary
This document is a review of map projections for the GE Board Examination in 2024. It delves into various classifications of projections, their properties, and examples, including the differences between cylindrical, conical, and planar projections. It also provides explanations of distortions, conversions, and meridian convergence. The review is organized as a series of questions and answers, covering topics like preserved properties and examples like the Mercator or Winkel Tripel projection.
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04/07/2024 MAP PROJECTIONS, COORDINATE CONVERSION, AND MERIDIAN CONVERGENCE GE Board Examination Review 2024 Cartography 01 – 03 [Perez] 1...
04/07/2024 MAP PROJECTIONS, COORDINATE CONVERSION, AND MERIDIAN CONVERGENCE GE Board Examination Review 2024 Cartography 01 – 03 [Perez] 1 TOPIC CONTENTS Map Projections Geographic to Grid 01 Definitions, Classifications, Variants, Properties, Samples, History, PRS92 02 Process, Examples Grid to Geographic Meridian Convergence 03 Process, Examples 04 Process, Examples 2 1 04/07/2024 Map Projections PART 01 03 3 Earth and Globe “UNDEVELOPABLE SURFACE” 04 4 2 04/07/2024 Map Projection A systematic projection of all or part of the surface of a round body, especially Earth, on a plane (Snyder, 1987). Earth’s Surface Map Projection Map Series of mathematical transformations 05 5 Map Projection Defines how positions on the Earth’s curved surface are transformed into a flat map. 06 6 3 04/07/2024 Classifications of Map Projections Based on the Developable Surface Cylindrical Conical Planar/ Azimuthal 07 7 Classifications of Map Projections Mercator Projection [Cylindrical] 08 8 4 04/07/2024 Classifications of Map Projections Cylindrical Conical Planar 09 9 Variants of Map Projections: Case 10 10 5 04/07/2024 Variants of Map Projections: Case 11 11 Variants of Map Projections: Case 12 12 6 04/07/2024 Variants of Map Projections: Aspect Normal Transverse Oblique 13 13 Variants of Map Projections: Viewpoint 14 14 7 04/07/2024 Variants of Map Projections: Viewpoint 15 15 Properties of Map Projections Conformality Shapes of small features on the Earth are preserved; scale and direction on Earth and map are equal for small areas; useful for navigation and topographic mapping 16 16 8 04/07/2024 Properties of Map Projections Equal Earth (2018) Equivalence [Equal Area/Authalic] Areas on the map are always proportional to areas on the Earth’s surface; useful for area computation applications 17 17 Properties of Map Projections Equidistance Maintains scale along one or more lines, or from one or two points to all other points on the map Equidistant Cylindrical 18 18 9 04/07/2024 Properties of Map Projections Azimuthal Equidistant Azimuthality True directions are preserved; direction measurements on the map are the same as those made on the ground; useful for air and sea navigation 19 19 Understanding the Distortions Tissot’s Indicatrix A Tissot indicatrix contains circles at grid intersections and shows how they vary due to distortion from a map projection. 20 20 10 04/07/2024 Map Projections Aphylactic Compromise Combined Projections that Projections that Maps that are: attempt to preserve: preserve: enough of area, shape, neither distance, and direction so more than one property conformal that the earth looks right but preserves none of nor equal-area them 21 21 Question and Answer Question What property is preserved? Answer Equivalence [Hobo-Dyer cylindrical equal-area projection] 22 22 11 04/07/2024 Question and Answer Question What property is preserved? Answer Equidistance [Equirectangular projection] 23 23 Question and Answer Question What property is preserved? Answer Conformality [Mercator projection] 24 24 12 04/07/2024 Question and Answer Question What property is preserved? Answer Azimuthality + Equidistance and Equivalence [Azimuthal Equidistant and Azimuthal Equal-Area projections] 25 25 Question and Answer Question What property is preserved? Answer Conformality [Lambert Conformal Conic projection] 26 26 13 04/07/2024 Question and Answer Question What property is preserved? Answer Equivalence [Sinusoidal projection] 27 27 Question and Answer Question What kind of map is on the image? Answer Compromise Scale and Area [Winkel Tripel projection] 28 28 14 04/07/2024 Question and Answer Question What kind of map is on the image? Answer Aphylactic [Wagner V projection] 29 29 Question and Answer Question What kind of map is on the image? Answer Aphylactic/Compromise [Robinson projection] 30 https://map-projections.net/compare.php?p1=robinson&p2=wagner-5&w=1&sm=1 30 15 04/07/2024 Question and Answer Question What kind of map is on the image? Answer Aphylactic/Compromise [Dymaxion/Fuller projection] 31 31 Question and Answer Question What is the “BEST” map projection? Answer Authagraph [Compromise] 32 32 16 04/07/2024 Question and Answer Question What is the “BEST” map projection? Answer Authagraph [Compromise] 33 33 Question and Answer Question What is the “BEST” map projection? Answer Authagraph [Compromise] 34 34 17 04/07/2024 Philippine Reference System of 1992 35 35 History 01 02 03 04 USC&GS Local Datums Luzon Vigan Datum 13 different Vigan Astronomic Datum of 1901-1942 astronomic stations all 1901 (topographic stations used over the works) in isolated Integration country surveys of surveys 36 36 18 04/07/2024 History 05 06 07 08 Luzon Datum WP Grid UTM and PTM PTM 4 Zones Luzon Datum World Polyconic 1947-1962 of 1911 Grid (1952) National Most Extension to First Grid Civil Grid System popular central and (USC&GS, (Luzon southern part 1919) → UTM 1911) 37 37 History 09 10 11 12 PTM 5 Zones NAMRIA PRS92 Updating 1987 Philippine Densification PBC&GS → Reference System of 1992 of networks 1962 NAMRIA → → Adjustment new Use of new of Luzon observations Datum of 1911 technologies 38 38 19 04/07/2024 Luzon Datum of 1911 39 39 Luzon Datum of 1911 is defined by its origin near San Andres Point on ______________ ___________________ Marinduque Island in the Southern Tagalog Region, specifically at Station Balanacan (a port name). ____________ 40 40 20 04/07/2024 13-33-41 N LATITUDE 121-52-03 E LONGITUDE 009-12-37 AZIMUTH TO STATION BALTAZAR 41 41 0 m (original); 0.34 m (updated) GEOID/SPHEROID SEPARATION Clarke Spheroid of 1866 REFERENCE ELLIPSOID 98 measured baselines; 52 observed azimuths; 49 latitude and longitude stations CONTROLLED BY 42 42 21 04/07/2024 Universal Transverse Mercator 43 43 Universal Transverse Mercator Scale Factor at Origin False Easting 0.9996 500,000 Number of Zones Degrees Per Zone 60 6 44 44 22 04/07/2024 45 45 Philippine Transverse Mercator 46 46 23 04/07/2024 Philippine Transverse Mercator Reference Ellipsoid Map Projection Clarke 1866 T. Mercator Point of Origin Intersection of Equator and Central Meridian 47 47 Philippine Transverse Mercator Scale Factor at Origin False Easting 0.99995 500,000 Number of Zones Degrees Per Zone 5 2 48 48 24 04/07/2024 49 49 Geographic to Grid Coordinate Conversion PART 02 50 50 25 04/07/2024 Formulas to be used: N = I + (II)p2 + (III)p4 E = (IV)p + (V)p3 + (VI)p5 + 500,000.00 p = 0.0001 (λ - λCM) = 0.0001 (Δλ”) ` Δλ = λ - λCM should be expressed in seconds (example: 0⁰01’20” = 60”+20” = Δλ”= 80”) 51 51 Coordinate Transformation Process Step 1 Establish first the Zone and Central Meridian to be used in the coordinate conversion. 52 52 26 04/07/2024 Coordinate Transformation Process Step 2 Compute for the value of Δλ” using the equation Δλ = λ – λCM and convert the value to seconds (Δλ”). Step 3 The coefficient p can now be computed using the equation p = 0.0001Δλ”. 53 53 Coordinate Transformation Process Step 4 From the Geographic to Grid Tables found in Technical Bulletin No. 26, obtain the values of the Roman Numerals I, II, III, IV, V and VI corresponding to the given latitude, ϕ. This could be done by interpolation using the Diff. 1” column or by manual/calculator interpolation. 54 54 27 04/07/2024 Coordinate Transformation Process Step 5 Compute for the value of the Northing of the Station using the equation: N = I + (II)p2 + (III)p4 Step 6 Finally, solve for the Easting of the Station using the equation: E = (IV)p + (V)p3 + (VI)p5 + 500,000.00 55 55 Example 01 GEOGRAPHIC TO GRID COORDINATE CONVERSION 56 56 28 04/07/2024 Example 01 Determine the grid coordinates under Zone 3 of a point whose geographic coordinates are 14⁰00’00.00”N and 121⁰25’42.90” E. I = 1,699,469.335 II = 1,915.718 III = 1.738 IV = 298,224.054 V = 101.105 VI = 0.047 57 57 Example 01 Step 1: Establish the Zone and Central Meridian. 𝒁𝒐𝒏𝒆 𝑰𝑰𝑰 → 𝝀𝑪𝑴 = 𝟏𝟐𝟏° Step 2: Solve for p. 𝒑 = 𝟎. 𝟎𝟎𝟎𝟏(𝝀 − 𝝀𝑪𝑴)" 𝒑 = 𝟎. 𝟎𝟎𝟎𝟏(𝟏𝟐𝟏°𝟐𝟓′ 𝟒𝟐. 𝟗𝟎" − 𝟏𝟐𝟏°)(𝟑𝟔𝟎𝟎") = 𝟎. 𝟎𝟎𝟎𝟏(𝟏𝟓𝟒𝟐. 𝟗𝟎) 𝒑 = 𝟎. 𝟏𝟓𝟒𝟐𝟗 M 58 58 29 04/07/2024 Example 01 I = 1,699,469.335 II = 1,915.718 III = 1.738 IV = 298,224.054 V = 101.105 VI = 0.047 Step 3: Solve for N. 𝑵 = 𝑰 + 𝑰𝑰𝒑𝟐 + 𝑰𝑰𝑰𝒑𝟒 𝑵 = 𝟏, 𝟔𝟗𝟗, 𝟒𝟔𝟗. 𝟑𝟑𝟓 + (𝟏, 𝟗𝟏𝟓. 𝟕𝟏𝟖)𝑴𝟐 + (𝟏. 𝟕𝟑𝟖)𝑴𝟒 𝑵 = 𝟏, 𝟔𝟗𝟗, 𝟓𝟏𝟒. 𝟗𝟒𝟎 𝒎. 59 59 Example 01 I = 1,699,469.335 II = 1,915.718 III = 1.738 IV = 298,224.054 V = 101.105 VI = 0.047 Step 4: Solve for E. 𝑬 = 𝑰𝑽𝒑 + 𝑽𝒑𝟑 + 𝑽𝑰𝒑𝟓 + 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬 = 𝟐𝟗𝟖, 𝟐𝟐𝟒. 𝟎𝟓𝟒 𝑴 + 𝟏𝟎𝟏. 𝟏𝟎𝟓 𝑴𝟑 + (𝟎. 𝟎𝟒𝟕)𝑴𝟓 + 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬 = 𝟓𝟒𝟔, 𝟎𝟏𝟑. 𝟑𝟔𝟏 𝒎. 60 60 30 04/07/2024 Example 02 GEOGRAPHIC TO GRID COORDINATE CONVERSION 61 61 Example 02 Convert to grid coordinates. GIVEN: Latitude Longitude 12⁰ 11’ 09.307” N PPCS/TM PRS 92 Zone 122⁰ 41’ 43.162” E ZONE IV 62 62 31 04/07/2024 Example 02 Step 1: Establish the Zone and Central Meridian. 𝒁𝒐𝒏𝒆 𝑰𝑽 → 𝝀𝑪𝑴 = 𝟏𝟐𝟑° Step 2: Solve for Δλ”. 𝜟𝝀 = (𝝀 − 𝝀𝑪𝑴) ∆𝝀 = 𝟏𝟐𝟐°𝟒𝟏′ 𝟒𝟑. 𝟏𝟔𝟐" − 𝟏𝟐𝟑° = −𝟎°𝟏𝟖′ 𝟏𝟔. 𝟖𝟑𝟖“ ∆𝝀" = ∆𝝀 (𝟑𝟔𝟎𝟎") ∆𝝀" = −𝟏𝟎𝟗𝟔. 𝟖𝟑𝟖 63 63 Example 02 Step 3: Solve for p. 𝒑 = 𝟎. 𝟎𝟎𝟎𝟏(∆𝝀") 𝒑 = 𝟎. 𝟎𝟎𝟎𝟏(−𝟏𝟎𝟗𝟔. 𝟖𝟑𝟖) 𝒑 = −𝟎. 𝟏𝟎𝟗𝟔𝟖𝟑𝟖 M 64 64 32 04/07/2024 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 12⁰ 11’ 09.307” N 65 65 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” OPTION 1: Use Diff. 1” value. 𝑰𝝓 = 𝑰𝝓𝟏 + [𝑫𝒊𝒇𝒇. 𝟏"(𝝓 − 𝝓𝟏)"] 66 66 33 04/07/2024 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” OPTION 1: Use Diff. 1” value. 𝑰𝝓 = 𝟏, 𝟑𝟒𝟕, 𝟐𝟏𝟒. 𝟏𝟒𝟎+(𝟑𝟎. 𝟕𝟐𝟓𝟒𝟗)(𝟗. 𝟑𝟎𝟕) 𝑰𝝓 = 𝟏, 𝟑𝟒𝟕, 𝟓𝟎𝟎. 𝟏𝟎𝟐 A 67 67 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” OPTION 2: Manual interpolation. 𝝓 − 𝝓𝟏 𝑰𝝓 − 𝑰𝟏 = 𝝓𝟐 − 𝝓𝟏 𝑰𝟐 − 𝑰𝟏 68 68 34 04/07/2024 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” OPTION 2: Manual interpolation. 𝑰𝝓 − 𝑰𝟏 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” − 𝟏𝟐°𝟏𝟏′ 𝑿 = 𝟏𝟐°𝟏𝟐′ − 𝟏𝟐°𝟏𝟏′ 𝟏, 𝟑𝟒𝟗, 𝟎𝟓𝟕. 𝟔𝟔𝟗 − 𝟏, 𝟑𝟒𝟕, 𝟐𝟏𝟒. 𝟏𝟒𝟎 69 69 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” OPTION 2: Manual interpolation. 𝑰𝝓 − 𝑰𝟏 𝟗. 𝟑𝟎𝟕” 𝑿 𝟔𝟎" = 𝟏, 𝟖𝟒𝟑. 𝟓𝟐𝟗 → 𝑰𝝓 = 𝑰 𝟏 + 𝑿 70 70 35 04/07/2024 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” OPTION 2: Manual interpolation. 𝑰𝝓 = 𝟏, 𝟑𝟒𝟕, 𝟐𝟏𝟒. 𝟏𝟒𝟎 + 𝟐𝟖𝟓. 𝟗𝟔𝟐𝟎𝟕𝟑𝟒 𝑰𝝓 = 𝟏, 𝟑𝟒𝟕, 𝟓𝟎𝟎. 𝟏𝟎𝟐 A 71 71 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟗. 𝟑𝟎𝟕 OPTION 3: CalTech interpolation [Statistics Mode]. 𝑴𝒐𝒅𝒆 → 𝟑: 𝑺𝒕𝒂𝒕 → 𝟐: 𝑳𝒊𝒏 𝒐𝒓 𝑨 + 𝑩𝑿 72 72 36 04/07/2024 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝒚 𝟎 𝒙 𝟔𝟎 𝒙 = 𝟗. 𝟑𝟎𝟕 OPTION 3: CalTech interpolation [Statistics Mode]. 𝑰𝒏𝒑𝒖𝒕 𝒕𝒉𝒆 𝒙 𝒂𝒏𝒅 𝒚 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒏 𝒕𝒉𝒆 𝒕𝒂𝒃𝒍𝒆. 73 73 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝒚 𝟎 𝒙 𝟔𝟎 𝒙 = 𝟗. 𝟑𝟎𝟕 OPTION 3: CalTech interpolation [Statistics Mode]. 𝑪𝑨 𝒐𝒓 𝑨𝑪 → 𝟗. 𝟑𝟎𝟕 → 𝑨𝒑𝒑𝒔 𝒐𝒓 𝑺𝒉𝒊𝒇𝒕 + 𝟏 ෝ →= 𝟖 𝒐𝒓 𝟓: 𝑹𝒆𝒈 → 𝟓: 𝒚 A 𝑰𝝓 = 74 74 37 04/07/2024 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” OPTION 1: Use Diff. 1” value. 𝑰𝑰𝝓 = 𝑰𝑰𝝓𝟏 + [𝑫𝒊𝒇𝒇. 𝟏"(𝝓 − 𝝓𝟏)"] 𝑰𝑰𝝓 = 𝟏, 𝟓𝟒𝟔. 𝟒𝟒𝟓 + (𝟎. 𝟎𝟑𝟑𝟏𝟏)(𝟗. 𝟑𝟎𝟕)= 𝟏, 𝟓𝟒𝟔. 𝟕𝟓𝟑𝟏𝟓𝟓 B 75 75 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” 𝑰𝑰𝑰𝝓 = 𝑰𝑰𝑰𝒓𝒐𝒖𝒏𝒅𝒆𝒅 𝑰𝑰𝑰𝝓 = 𝟏. 𝟒𝟓𝟏 C 76 76 38 04/07/2024 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” OPTION 1: Use Diff. 1” value. 𝑰𝑽𝝓 = 𝑰𝑽𝝓𝟏 + [𝑫𝒊𝒇𝒇. 𝟏"(𝝓 − 𝝓𝟏)"] 𝑰𝑽𝝓 = 𝟑𝟎𝟐, 𝟐𝟗𝟎. 𝟎𝟕𝟐 + (−𝟎. 𝟑𝟏𝟒𝟓𝟖)(𝟗. 𝟑𝟎𝟕) = 𝟑𝟎𝟐, 𝟐𝟖𝟕. 𝟏𝟒𝟒𝟐 D 77 77 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” OPTION 1: Use Diff. 1” value. 𝑽𝝓 = 𝑽𝝓𝟏 + [𝑫𝒊𝒇𝒇. 𝟏"(𝝓 − 𝝓𝟏)"] 𝑽𝝓 = 𝟏𝟎𝟖. 𝟔𝟎𝟖 + (−𝟎. 𝟎𝟎𝟎𝟓𝟗)(𝟗. 𝟑𝟎𝟕) = 𝟏𝟎𝟖. 𝟔𝟎𝟐𝟓𝟎𝟖𝟗 X 78 78 39 04/07/2024 Example 02 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝝓𝟏 𝝓𝟐 𝝓 = 𝟏𝟐⁰ 𝟏𝟏’ 𝟗. 𝟑𝟎𝟕” 𝑽𝑰𝝓 = 𝑽𝑰𝒓𝒐𝒖𝒏𝒅𝒆𝒅 𝑽𝑰𝝓 = 𝟎. 𝟎𝟓𝟒 Y 79 79 Example 02 Step 5: Solve for N. 𝑵 = 𝑰 + 𝑰𝑰𝒑𝟐 + 𝑰𝑰𝑰𝒑𝟒 𝑵 = 𝑨 + 𝑩𝑴𝟐 + 𝑪𝑴𝟒 𝑵 = 𝟏, 𝟑𝟒𝟕, 𝟓𝟏𝟖. 𝟕𝟏𝟏 𝒎. 80 80 40 04/07/2024 Example 02 Step 6: Solve for E. 𝑬 = 𝑰𝑽𝒑 + 𝑽𝒑𝟑 + 𝑽𝑰𝒑𝟓 + 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬 = 𝑫𝑴 + 𝑿𝑴𝟑 + 𝒀𝑴𝟓 + 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬 = 𝟒𝟔𝟔, 𝟖𝟒𝟑. 𝟖𝟓𝟒 𝒎. 81 81 Example 03 GEOGRAPHIC TO GRID COORDINATE CONVERSION 82 82 41 04/07/2024 Example 03 Convert to grid coordinates. GIVEN: Latitude Longitude 006⁰ 27’ 17.436” N PPCS/TM PRS 92 Zone 121⁰ 35’ 22.548” E ZONE III 83 83 Example 03 Step 1: Establish the Zone and Central Meridian. 𝒁𝒐𝒏𝒆 𝑰𝑰𝑰 → 𝝀𝑪𝑴 = 𝟏𝟐𝟏° Step 2: Solve for Δλ”. 𝜟𝝀 = (𝝀 − 𝝀𝑪𝑴) ∆𝝀 = 𝟏𝟐𝟏°𝟑𝟓′ 𝟐𝟐. 𝟓𝟒𝟖" − 𝟏𝟐𝟏° = 𝟎°𝟑𝟓′ 𝟐𝟐. 𝟓𝟒𝟖" ∆𝝀" = ∆𝝀 (𝟑𝟔𝟎𝟎") ∆𝝀" = 𝟐𝟏𝟐𝟐. 𝟓𝟒𝟖" 84 84 42 04/07/2024 Example 03 Step 3: Solve for p. 𝒑 = 𝟎. 𝟎𝟎𝟎𝟏(∆𝝀") 𝒑 = 𝟎. 𝟎𝟎𝟎𝟏(𝟐𝟏𝟐𝟐. 𝟓𝟒𝟖) 𝒑 = 𝟎. 𝟐𝟏𝟐𝟐𝟓𝟒𝟖 M 85 85 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" Use CalTech Interpolation 86 86 43 04/07/2024 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" x y 0 713,153.359 60 714,996.290 87 87 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" 𝑰𝝓 = 𝟕𝟏𝟑, 𝟔𝟖𝟖. 𝟗𝟏𝟒𝟕 A 88 88 44 04/07/2024 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" x y 0 836.714 60 838.839 89 89 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" 𝑰𝑰𝝓 = 𝟖𝟑𝟕. 𝟑𝟑𝟏𝟓𝟐𝟓 B 90 90 45 04/07/2024 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" x y 0 0.817 60 0.819 91 91 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" 𝑰𝑰𝑰𝝓 = 𝟎. 𝟖𝟏𝟕𝟓𝟖𝟏𝟐 C 92 92 46 04/07/2024 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" x y 0 307,264.625 60 307,254.575 93 93 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" 𝑰𝑽𝝓 = 𝟑𝟎𝟕, 𝟐𝟔𝟏. 𝟕𝟎𝟒𝟓 D 94 94 47 04/07/2024 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" x y 0 118.130 60 118.110 95 95 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" 𝑽𝝓 = 𝟏𝟏𝟖. 𝟏𝟐𝟒𝟏𝟖𝟖 X 96 96 48 04/07/2024 Example 03 Step 4: Interpolate I, II, III, IV, V, & VI from the TB#26 Table. 𝟎 𝟔𝟎 𝒙 = 𝟏𝟕. 𝟒𝟑𝟔 𝝓 = 𝟔°𝟐𝟕′ 𝟏𝟕. 𝟒𝟑𝟔" 𝑽𝑰𝝓 = 𝟎. 𝟎𝟔𝟕 Y 97 97 Example 03 Step 5: Solve for N. 𝑵 = 𝑰 + 𝑰𝑰𝒑𝟐 + 𝑰𝑰𝑰𝒑𝟒 𝑵 = 𝑨 + 𝑩𝑴𝟐 + 𝑪𝑴𝟒 𝑵 = 𝟕𝟏𝟑, 𝟕𝟐𝟔. 𝟔𝟒𝟎 𝒎. 98 98 49 04/07/2024 Example 03 Step 6: Solve for E. 𝑬 = 𝑰𝑽𝒑 + 𝑽𝒑𝟑 + 𝑽𝑰𝒑𝟓 + 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬 = 𝑫𝑴 + 𝑿𝑴𝟑 + 𝒀𝑴𝟓 + 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬 = 𝟓𝟔𝟓, 𝟐𝟏𝟖. 𝟗𝟎𝟏 𝒎. 99 99 50 04/07/2024 MAP PROJECTIONS, COORDINATE CONVERSION, AND MERIDIAN CONVERGENCE GE Board Examination Review 2024 Cartography 01 – 03 [Perez] 100 Grid to Geographic Coordinate Conversion PART 03 101 101 1 04/07/2024 Formulas to be used: Φ = Φ’ – (VII)q2 + (VIII)q4 λ = (IX)q – (X)q3 + (XI)q5 + λCM q = 0.000001(E – 500,000.00) = 0.000001(E’) ` the result of (-VIIq2 + VIIIq4) and (IXq – Xq3 + XIq5) are in seconds 102 102 Coordinate Transformation Process Step 1 Determine the Central Meridian of the given Zone. 103 103 2 04/07/2024 Coordinate Transformation Process Step 2 Compute for the value of E’ using the equation: E’ = E – 500,000.00 Step 3 The coefficient q can now be computed using the equation: q = 0.000001(E’). 104 104 Coordinate Transformation Process Step 4 Using the Grid to Geographic Tables, interpolate the value of the footpoint (initial) latitude, Φ’. Do this by first finding the nearest I to the given Northing: I1 < N < I2, Φ1 < Φ’ < Φ2 105 105 3 04/07/2024 Coordinate Transformation Process Step 5 Interpolate the value of the corresponding footpoint latitude based of the range of the I values for the N: Φ’ - Φ1 = N - I1 Φ2 - Φ1 I2 - I1 106 106 Coordinate Transformation Process Step 6 From the Grid to Geographic Tables, obtain the values of the Roman Numerals VII, VIII, IX, X and XI corresponding to the computed footpoint latitude, Φ’. Again, this could be done by interpolation using the Diff. 1” column or by manual interpolation. 107 107 4 04/07/2024 Coordinate Transformation Process Step 7 Compute for the value of the latitude of Station using the equation: Φ = Φ’ – (VII)q2 + (VIII)q4 where the term (-VIIq2 + VIIIq4) is expressed in seconds. 108 108 Coordinate Transformation Process Step 8 Finally, solve for the longitude of the Station using the equation: λ = (IX)q – (X)q3 + (XI)q5 + λCM where the result of (IXq - Xq3 + XIq5) is in seconds 109 109 5 04/07/2024 Example 01 GRID TO GEOGRAPHIC COORDINATE CONVERSION 110 110 Example 01 Convert to geographic coordinates: N = 1,422,287.156 m; E = 524,937.241 m; Φ’ = 12⁰51’43”; Zone IV VII = 582.463 VIII = 6.154 IX = 33,167.397 X = 150.896 XI = 1.096 111 111 6 04/07/2024 Example 01 Step 1: Establish the Zone and Central Meridian. 𝒁𝒐𝒏𝒆 𝑰𝑽 → 𝝀𝑪𝑴 = 𝟏𝟐𝟑° Step 2: Solve for q. 𝒒 = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏(𝑬 − 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎) 𝒒 = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏(𝟓𝟐𝟒, 𝟗𝟑𝟕. 𝟐𝟒𝟏 − 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎) = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏(𝟐𝟒, 𝟗𝟑𝟕. 𝟐𝟒𝟏) M 𝒒 = 𝟎. 𝟎𝟐𝟒𝟗𝟑𝟕𝟐𝟒𝟏 112 112 Example 01 VII = 582.463 VIII = 6.154 IX = 33,167.397 X = 150.896 XI = 1.096 Step 3: Solve for 𝝓. 𝝓 = 𝝓′ − 𝑽𝑰𝑰𝒒𝟐 + 𝑽𝑰𝑰𝑰𝒒𝟒 𝝓 = 𝟏𝟐°𝟓𝟏′ 𝟒𝟑" + [(−𝟓𝟖𝟐. 𝟒𝟔𝟑𝑴𝟐 + 𝟔. 𝟏𝟓𝟒𝑴𝟒 )/𝟑𝟔𝟎𝟎"] 𝝓 = 𝟏𝟐°𝟓𝟏′ 𝟒𝟐. 𝟔𝟑𝟖" 𝑵 113 113 7 04/07/2024 Example 01 VII = 582.463 VIII = 6.154 IX = 33,167.397 X = 150.896 XI = 1.096 Step 4: Solve for 𝝀. 𝝀 = 𝑰𝑿𝒒 − 𝑿𝒒𝟑 + 𝑿𝑰𝒒𝟓 + 𝝀𝐂𝐌 𝝀 = [(𝟑𝟑, 𝟏𝟔𝟕. 𝟑𝟗𝟕𝑴 − 𝟏𝟓𝟎. 𝟖𝟗𝟔𝑴𝟑 + 𝟏. 𝟎𝟗𝟔𝑴𝟓 )/𝟑𝟔𝟎𝟎"] + 𝟏𝟐𝟑° 𝝀 = 𝟏𝟐𝟑°𝟏𝟑′𝟒𝟕. 𝟏𝟎𝟏" 𝑬 114 114 Example 02 GRID TO GEOGRAPHIC COORDINATE CONVERSION 115 115 8 04/07/2024 Example 02 Convert to geographic coordinates. N = 1,347,518.711 m GIVEN: Northing Easting E = 466,843.854 m PPCS/TM PRS 92 Zone ZONE IV 116 116 Example 02 Step 1: Establish the Zone and Central Meridian. 𝒁𝒐𝒏𝒆 𝑰𝑽 → 𝝀𝑪𝑴 = 𝟏𝟐𝟑° Step 2: Solve for E’. 𝑬′ = 𝑬 − 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬′ = 𝟒𝟔𝟔, 𝟖𝟒𝟑. 𝟖𝟓𝟒 − 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬′ = −𝟑𝟑, 𝟏𝟓𝟔. 𝟏𝟒𝟔 117 117 9 04/07/2024 Example 02 Step 3: Solve for q. 𝒒 = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏(𝑬′ ) 𝒒 = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏(−𝟑𝟑, 𝟏𝟓𝟔. 𝟏𝟒𝟔) 𝒒 = −𝟎. 𝟎𝟑𝟑𝟏𝟓𝟔𝟏𝟒𝟔 M 118 118 Example 02 Step 4: Find the nearest I to the given Northing. N = 1,347,518.711 m 𝝓𝟏 𝝓𝟐 𝝓′ = 𝐟𝐨𝐨𝐭𝐩𝐨𝐢𝐧𝐭 𝐥𝐚𝐭𝐢𝐭𝐮𝐝𝐞 119 119 10 04/07/2024 Example 02 Step 5: Interpolate the footpoint latitude, 𝝓′. Manual interpolation N = 1,347,518.711 m 𝝓′ − 𝝓𝟏 𝑵 − 𝑰𝟏 = 𝝓𝟐 − 𝝓𝟏 𝑰𝟐 − 𝑰𝟏 𝝓𝟏 𝝓𝟐 𝝓′ − 𝟏𝟐°𝟏𝟏′ 1,347,518.711 − 𝟏, 𝟑𝟒𝟕, 𝟐𝟏𝟒. 𝟏𝟒𝟎 = 𝟏𝟐°𝟏𝟐′ − 𝟏𝟐°𝟏𝟏′ 𝟏, 𝟑𝟒𝟗, 𝟎𝟓𝟕. 𝟔𝟔𝟗 − 𝟏, 𝟑𝟒𝟕, 𝟐𝟏𝟒. 𝟏𝟒𝟎 𝝓′ = 𝐟𝐨𝐨𝐭𝐩𝐨𝐢𝐧𝐭 𝐥𝐚𝐭𝐢𝐭𝐮𝐝𝐞 𝟔𝟎" 120 120 Example 02 Step 5: Interpolate the footpoint latitude, 𝝓′. Manual interpolation N = 1,347,518.711 m 𝟑𝟎𝟒. 𝟓𝟕𝟏 𝝓′ − 𝟏𝟐°𝟏𝟏′ = (𝟔𝟎") 𝟏, 𝟖𝟒𝟑. 𝟓𝟐𝟗 𝝓𝟏 = 𝟗. 𝟗𝟏𝟐𝟔𝟓𝟏𝟐𝟐𝟓" A 𝝓𝟐 𝝓′ = 𝟏𝟐°𝟏𝟏′ 𝟗. 𝟗𝟏𝟐𝟔𝟓𝟏𝟐𝟐𝟓" 𝝓′ = 𝐟𝐨𝐨𝐭𝐩𝐨𝐢𝐧𝐭 𝐥𝐚𝐭𝐢𝐭𝐮𝐝𝐞 121 121 11 04/07/2024 Example 02 Step 5: Interpolate the footpoint latitude, 𝝓′. CalTech interpolation N = 1,347,518.711 m x y 0 1,347,214.140 𝝓𝟏 60 1,349,057.669 𝝓𝟐 𝒙 = 𝟗. 𝟗𝟏𝟐𝟔𝟓𝟏𝟐𝟐𝟓 A 𝑵ෝ 𝝓′ = 𝐟𝐨𝐨𝐭𝐩𝐨𝐢𝐧𝐭 𝐥𝐚𝐭𝐢𝐭𝐮𝐝𝐞 𝝓′ = 𝟏𝟐°𝟏𝟏′ 𝟗. 𝟗𝟏𝟐𝟔𝟓𝟏𝟐𝟐𝟓" 122 122 Example 02 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use Diff. 1” value 𝑽𝑰𝑰𝝓′ = 𝑽𝑰𝑰𝝓𝟏 + [𝑫𝒊𝒇𝒇. 𝟏"(𝝓′ − 𝝓𝟏)"] 𝑽𝑰𝑰𝝓′ = 𝟓𝟓𝟎. 𝟕𝟗𝟑 + (𝟎. 𝟎𝟏𝟐𝟗𝟑)(𝑨) A 𝟗. 𝟗𝟏𝟐𝟔𝟓𝟏𝟐𝟐𝟓" 𝑽𝑰𝑰𝝓′ = 𝟓𝟓𝟎. 𝟕𝟗𝟑 + 𝟎. 𝟏𝟐𝟖𝟏𝟕𝟎𝟓𝟖𝟎𝟑 = 𝟓𝟓𝟎. 𝟗𝟐𝟏𝟏𝟕𝟎𝟔 B 123 123 12 04/07/2024 Example 02 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 𝑽𝑰𝑰𝑰𝝓′ = 𝑽𝑰𝑰𝑰𝒓𝒐𝒖𝒏𝒅𝒆𝒅 𝑽𝑰𝑰𝑰𝝓′ = 𝟓. 𝟖𝟎𝟐 C 124 124 Example 02 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 0 33,080.809 𝑰𝑿𝝓′ = 𝟗. 𝟗𝟏𝟐𝟔𝟓𝟏𝟐𝟐𝟓ෝ 𝒚 60 33,082.874 𝑰𝑿𝝓′ = 𝟑𝟑, 𝟎𝟖𝟏. 𝟏𝟓𝟎𝟏𝟔 D 125 125 13 04/07/2024 Example 02 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 0 149.015 𝑿𝝓′ = 𝟗. 𝟗𝟏𝟐𝟔𝟓𝟏𝟐𝟐𝟓ෝ 𝒚 60 149.060 𝑿𝝓′ = 𝟏𝟒𝟗. 𝟎𝟐𝟐𝟒𝟑𝟒𝟓 X 126 126 Example 02 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 𝑿𝑰𝝓′ = 𝑿𝑰𝒓𝒐𝒖𝒏𝒅𝒆𝒅 𝐗𝑰𝝓′ = 𝟏. 𝟎𝟔𝟓 Y 127 127 14 04/07/2024 Example 02 Step 7: Solve for 𝝓. 𝝓 = 𝝓′ − 𝑽𝑰𝑰𝒒𝟐 + 𝑽𝑰𝑰𝑰𝒒𝟒 𝝓 = 𝝓′ − 𝑩𝑴𝟐 + 𝑪𝑴𝟒 𝜙 ′ = 12°11′ 9.912651225" −0.6056371683" 𝝓 = 𝟏𝟐°𝟏𝟏′ 𝟗. 𝟑𝟎𝟕" 128 128 Example 02 Step 8: Solve for 𝝀. 𝝀 = 𝑰𝑿𝒒 − 𝑿𝒒𝟑 + 𝑿𝑰𝒒𝟓 + 𝝀𝐂𝐌 𝝀 = 𝑫𝑴 − 𝑿𝑴𝟑 + 𝒀𝑴𝟓 + 𝟏𝟐𝟑° ÷ 𝟑𝟔𝟎𝟎 −𝟎°𝟏𝟖′ 𝟏𝟔. 𝟖𝟑𝟖" 𝝀 = 𝟏𝟐𝟐°𝟒𝟏′ 𝟒𝟑. 𝟏𝟔𝟐" 129 129 15 04/07/2024 Example 03 GRID TO GEOGRAPHIC COORDINATE CONVERSION 130 130 Example 03 Convert to geographic coordinates. N = 743,980.762 m GIVEN: Northing Easting E = 531,249.025 m PPCS/TM PRS 92 Zone ZONE V 131 131 16 04/07/2024 Example 03 Step 1: Establish the Zone and Central Meridian. 𝒁𝒐𝒏𝒆 𝑽 → 𝝀𝑪𝑴 = 𝟏𝟐𝟓° Step 2: Solve for E’. 𝑬′ = 𝑬 − 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬′ = 𝟓𝟑𝟏, 𝟐𝟒𝟗. 𝟎𝟐𝟓 − 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬′ = 31,249.025 132 132 Example 03 Step 3: Solve for q. 𝒒 = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏(𝑬′ ) 𝒒 = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏(𝟑𝟏, 𝟐𝟒𝟗. 𝟎𝟐𝟓) 𝒒 = 𝟎. 𝟎𝟑𝟏𝟐𝟒𝟗𝟎𝟐𝟓 M 133 133 17 04/07/2024 Example 03 Step 4: Find the nearest I to the given Northing. N = 743,980.762 m 𝝓𝟏 𝝓𝟐 𝝓′ = 𝐟𝐨𝐨𝐭𝐩𝐨𝐢𝐧𝐭 𝐥𝐚𝐭𝐢𝐭𝐮𝐝𝐞 134 134 Example 03 Step 5: Interpolate the footpoint latitude, 𝝓′. CalTech interpolation N = 743,980.762 m x y 0 742,640.409 60 744,483.360 𝝓𝟏 𝝓𝟐 𝒙 = 𝟒𝟑. 𝟔𝟑𝟕𝟏𝟕𝟕𝟓𝟓" A 𝑵ෝ 𝝓′ = 𝐟𝐨𝐨𝐭𝐩𝐨𝐢𝐧𝐭 𝐥𝐚𝐭𝐢𝐭𝐮𝐝𝐞 𝝓′ = 𝟔°𝟒𝟑′ 𝟒𝟑. 𝟔𝟑𝟕𝟏𝟕𝟕𝟓𝟓" 135 135 18 04/07/2024 Example 03 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 𝑽𝑰𝑰𝝓′ = 𝟒𝟑. 𝟔𝟑𝟕𝟏𝟕𝟕𝟓𝟓ෝ 𝒚 0 300.564 60 301.317 𝑽𝑰𝑰𝝓′ = 𝟑𝟎𝟏. 𝟏𝟏𝟏𝟔𝟒𝟔𝟔 B 136 136 Example 03 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 𝑽𝑰𝑰𝑰𝝓′ = 𝟒𝟑. 𝟔𝟑𝟕𝟏𝟕𝟕𝟓𝟓ෝ 𝒚 0 3.108 60 3.115 𝑽𝑰𝑰𝑰𝝓′ = 𝟑. 𝟏𝟏𝟑𝟎𝟗𝟏𝟎𝟎𝟒 C 137 137 19 04/07/2024 Example 03 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 𝑰𝑿𝝓′ = 𝟒𝟑. 𝟔𝟑𝟕𝟏𝟕𝟕𝟓𝟓ෝ 𝒚 0 32,562.606 60 32,563.715 𝑰𝑿𝝓′ = 𝟑𝟐, 𝟓𝟔𝟑. 𝟒𝟏𝟐𝟓𝟔 D 138 138 Example 03 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 𝑿𝝓′ = 𝟒𝟑. 𝟔𝟑𝟕𝟏𝟕𝟕𝟓𝟓ෝ 𝒚 0 138.003 60 138.026 𝑿𝝓′ = 𝟏𝟑𝟖. 𝟎𝟏𝟗𝟕𝟐𝟕𝟔 X 139 139 20 04/07/2024 Example 03 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 𝑿𝑰𝝓′ = 𝟎. 𝟖𝟗𝟏 Y 140 140 Example 03 Step 7: Solve for 𝝓. 𝝓 = 𝝓′ − 𝑽𝑰𝑰𝒒𝟐 + 𝑽𝑰𝑰𝑰𝒒𝟒 𝝓 = 𝝓′ − 𝑩𝑴𝟐 + 𝑪𝑴𝟒 𝜙 ′ = 6°43′ 43.63717755" −0.2940330252" 𝝓 = 𝟔°𝟒𝟑′ 𝟒𝟑. 𝟑𝟒𝟑" 141 141 21 04/07/2024 Example 03 Step 8: Solve for 𝝀. 𝝀 = 𝑰𝑿𝒒 − 𝑿𝒒𝟑 + 𝑿𝑰𝒒𝟓 + 𝝀𝐂𝐌 𝝀 = 𝑫𝑴 − 𝑿𝑴𝟑 + 𝒀𝑴𝟓 + 𝟏𝟐𝟓° ÷ 𝟑𝟔𝟎𝟎 𝟎°𝟏𝟔′ 𝟓𝟕. 𝟓𝟕𝟏" 𝝀 = 𝟏𝟐𝟓°𝟏𝟔′ 𝟓𝟕. 𝟓𝟕𝟏" 142 142 Example 04 GRID TO GEOGRAPHIC COORDINATE CONVERSION 143 143 22 04/07/2024 Example 04 Solve for the Roman Numerals required to convert the grid coordinates (1,373,628.452 N, 593,895.237 E) of Station X in Zone 4 to its geographic coordinates. Use the following values from the TB No. 26. 144 144 Example 04 Step 1: Establish the Zone and Central Meridian. 𝒁𝒐𝒏𝒆 𝑰𝑽 → 𝝀𝑪𝑴 = 𝟏𝟐𝟑° Step 2: Solve for E’. 𝑬′ = 𝑬 − 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬′ = 𝟓𝟗𝟑, 𝟖𝟗𝟓. 𝟐𝟑𝟕 − 𝟓𝟎𝟎, 𝟎𝟎𝟎. 𝟎𝟎 𝑬′ = 𝟗𝟑,𝟖𝟗𝟓.𝟐𝟑𝟕 145 145 23 04/07/2024 Example 04 Step 3: Solve for q. 𝒒 = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏(𝑬′ ) 𝒒 = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏(𝟗𝟑, 𝟖𝟗𝟓. 𝟐𝟑𝟕) 𝒒 = 𝟎. 𝟎𝟗𝟑𝟖𝟗𝟓𝟐𝟑𝟕 M 146 146 Example 04 Step 4: Find the nearest I to the given Northing. N = 1,373,628.452 m 𝝓𝟏 𝝓𝟐 𝝓′ = 𝐟𝐨𝐨𝐭𝐩𝐨𝐢𝐧𝐭 𝐥𝐚𝐭𝐢𝐭𝐮𝐝𝐞 147 147 24 04/07/2024 Example 04 Step 5: Interpolate the footpoint latitude, 𝝓′. CalTech interpolation N = 1,373,628.452 m x y 𝝓𝟏 0 1,373,023.758 𝝓𝟐 60 1,374,867.319 𝒙 = 𝟏𝟗. 𝟔𝟖𝟎𝟏𝟗𝟓𝟎𝟏" A 𝑵ෝ 𝝓′ = 𝐟𝐨𝐨𝐭𝐩𝐨𝐢𝐧𝐭 𝐥𝐚𝐭𝐢𝐭𝐮𝐝𝐞 𝝓′ = 𝟏𝟐°𝟐𝟓′ 𝟏𝟗. 𝟔𝟖𝟎𝟏𝟗𝟓𝟎𝟏" 148 148 Example 04 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 𝑽𝑰𝑰𝝓′ = 𝟏𝟗. 𝟔𝟖𝟎𝟏𝟗𝟓𝟎𝟏ෝ 𝒚 0 561.663 60 562.440 𝑽𝑰𝑰𝝓′ = 𝟓𝟔𝟏. 𝟗𝟏𝟕𝟖𝟓𝟖𝟓 149 149 25 04/07/2024 Example 04 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 𝑽𝑰𝑰𝑰𝝓′ = 𝟏𝟗. 𝟔𝟖𝟎𝟏𝟗𝟓𝟎𝟏ෝ 𝒚 0 5.923 60 5.931 𝑽𝑰𝑰𝑰𝝓′ = 𝟓. 𝟗𝟐𝟓𝟔𝟐𝟒𝟎𝟐𝟔 150 150 Example 04 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 𝑰𝑿𝝓′ = 𝟏𝟗. 𝟔𝟖𝟎𝟏𝟗𝟓𝟎𝟏ෝ 𝒚 0 33,110.005 60 33,112.114 𝑰𝑿𝝓′ = 𝟑𝟑, 𝟏𝟏𝟎. 𝟔𝟗𝟔𝟕𝟔 151 151 26 04/07/2024 Example 04 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 Use CalTech Interpolation x y 𝑿𝝓′ = 𝟏𝟗. 𝟔𝟖𝟎𝟏𝟗𝟓𝟎𝟏ෝ 𝒚 0 149.648 60 149.694 𝑿𝝓′ = 𝟏𝟒𝟗. 𝟔𝟔𝟑𝟎𝟖𝟖𝟏 152 152 Example 04 Step 6: Interpolate VII, VIII, IX, X, & XI for 𝝓′. 𝝓𝟏 𝝓′ 𝝓𝟐 𝑿𝑰𝝓′ = 𝟏. 𝟎𝟕𝟔 153 153 27 08/07/2024 MAP PROJECTIONS, COORDINATE CONVERSION, AND MERIDIAN CONVERGENCE GE Board Examination Review 2024 Cartography 01 – 03 [Perez] 154 Meridian Convergence PART 04 155 155 1 08/07/2024 Meridian Convergence Correction The Grid North has its direction of the central meridian, but elsewhere, a meridian does not align with Grid North. Thus, in general, the grid (projected) azimuth of a line will not equal the true (geodetic/ astronomic) azimuth of that line. To convert from grid to true azimuth, a convergence factor must be applied. 156 156 Meridian Convergence Correction Δα = (λ-λCM) sin Φ (if only one point is given) where: Φ and λ are the latitude and longitude of the point; λCM is the longitude of the central meridian; (λ-λCM) is the difference of longitudes; and Δα is the convergence correction. The correction Δα is added to the computed grid azimuth algebraically: True Azimuth = Grid Azimuth + Δα 157 157 2 08/07/2024 Meridian Convergence Correction Δα = (Δλ) sin Φave (if two points forming a line are given) where: Φ and λ are the latitude and longitude of the endpoints of the line; Δλ is the difference between the longitudes of the endpoints of the line; Φave is the average of the latitudes; and Δα is the convergence correction. The correction Δα is added to the computed grid azimuth algebraically: True Azimuth = Grid Azimuth + Δα 158 158 True Azimuth = Grid Azimuth + Δα 159 159 3 08/07/2024 True Azimuth = Grid Azimuth + Δα 160 Meridian Convergence Correction 161 161 4 08/07/2024 Example 01 MERIDIAN CONVERGENCE 162 162 Example 01 Compute for the meridian convergence (Δα) of line AB located in Zone III given the geographic coordinates of point A: 14⁰00’00.00”N, 121⁰25’42.90” E. GIVEN: Latitude Longitude PPCS/TM PRS 92 Zone ∆𝜶 = (𝝀 − 𝝀𝑪𝑴)𝒔𝒊𝒏𝝓 𝝀𝐂𝐌 = 𝟏𝟐𝟏° = (𝟏𝟐𝟏°𝟐𝟓′ 𝟒𝟐. 𝟗𝟎" − 𝟏𝟐𝟏°)𝒔𝒊𝒏(𝟏𝟒°) = (𝟎°𝟐𝟓′ 𝟒𝟐. 𝟗𝟎")𝒔𝒊𝒏(𝟏𝟒°) ∆𝜶 = +𝟎°𝟔′𝟏𝟑. 𝟐𝟔" 163 163 5 08/07/2024 Example 02 MERIDIAN CONVERGENCE 164 164 Example 02 The following PRS92 geographic and grid coordinates of stations MMA-1 and MMA-6 are provided as reference stations: Determine: A. the Grid Azimuth from Station MMA-1 to station MMA-6. B. the Meridian Convergence in seconds. C. the Geodetic Azimuth from station MMA-1 to station MMA-6 if the arc-to-chord correction is negligible. 165 165 6 08/07/2024 Example 02 A. Grid Azimuth from Station MMA-1 to Station MMA-6. |Δ𝐸| |496,954.655 − 504,138.600| 𝛽 = tan−1 = tan−1 |Δ𝑁| |1,614,525.125 − 1,607,761.469| 𝛽 = 𝑁 460 43′ 33.58" 𝑊 𝑷𝒐𝒍( 𝜟𝑵 , 𝜟𝑬 ) 166 166 Example 02 A. Grid Azimuth from Station MMA-1 to Station MMA-6. 𝐺𝑟𝑖𝑑 𝐴𝑧𝑖𝑚𝑢𝑡ℎ = 1800 − 460 43′ 33.58" 𝐺𝑟𝑖𝑑 𝐴𝑧𝑖𝑚𝑢𝑡ℎ = 𝟏𝟑𝟑°𝟏𝟔′ 𝟐𝟔. 𝟒𝟐" 167 167 7 08/07/2024 Example 02 B. Meridian Convergence in seconds. 𝛥𝛼 = (𝛥𝜆) 𝑠𝑖𝑛 𝛷𝑎𝑣𝑒 = (1210 02′ 23.14210" − 1200 58′ 18.24728") ∗ 140 32′ 13.66238" + 140 35′ 59.15786" sin 2 𝛥𝛼 = 0°1′1.6"(3600) = 𝟔𝟏. 𝟔" 168 168 Example 02 C. Geodetic Azimuth from station MMA-1 to station MMA-6 if the arc-to-chord correction is negligible. 𝐺𝑒𝑜𝑑𝑒𝑡𝑖𝑐 𝐴𝑧𝑖𝑚𝑢𝑡ℎ = 𝑇𝑟𝑢𝑒 𝐴𝑧𝑖𝑚𝑢𝑡ℎ = 𝐺𝑟𝑖𝑑 𝐴𝑧𝑖𝑚𝑢𝑡ℎ + 𝛥𝛼 = 133°16′ 26.42" + 61.6" = 𝟏𝟑𝟑𝟎 𝟏𝟕′ 𝟐𝟖. 𝟎𝟐" 169 169 8 08/07/2024 Example 03 MERIDIAN CONVERGENCE 170 170 Example 03 Solve for the grid azimuth of a line on the west side of the CM, given that the geodetic azimuth is 245⁰31’16.37” and the convergence correction is 34.12”. Assume that the arc-to-chord correction is negligible. True Azimuth = Grid Azimuth + Δα 245⁰31’16.37” = Grid Azimuth + Δα 245⁰31’16.37” = Grid Azimuth + (-34.12”) 245⁰31’16.37” + 34.12” = Grid Azimuth = 245⁰31’50.49” 171 171 9 08/07/2024 Example 04 MERIDIAN CONVERGENCE 172 172 Example 04 Solve for the grid azimuth of a line on the east side of the CM, given that the geodetic azimuth is 245⁰31’16.37” and the convergence correction is 34.12”. Assume that the arc-to-chord correction is negligible. True Azimuth = Grid Azimuth + Δα 245⁰31’16.37” = Grid Azimuth + Δα 245⁰31’16.37” = Grid Azimuth + 34.12” 245⁰31’16.37” - 34.12” = Grid Azimuth = 245⁰30’42.25” 173 173 10 08/07/2024 Example 05 MERIDIAN CONVERGENCE 174 174 Example 05 The following PRS92 geographic and grid coordinates of stations MMA-1 and MMA-6 are provided as reference stations: Determine: A. the Grid Azimuth from Station MMA-1 to station MMA-6. B. the Meridian Convergence in seconds. C. the Geodetic Azimuth from station MMA-1 to station MMA-6 if the arc-to-chord correction is negligible. 175 175 11 08/07/2024 Example 05 The following PRS92 geographic and grid coordinates of stations BHL-1 and CBU-1 are provided as reference stations: Determine: A. the Grid Azimuth from Station BHL-1 to station CBU-1. B. the Meridian Convergence in arc seconds. C. the Geodetic Azimuth from station BHL-1 to station CBU-1 assuming that the arc-to-chord correction is negligible. 176 176 Example 05 A. Grid Azimuth from Station BHL-1 to Station CBU-1. |Δ𝐸| |603,303.482 − 593,626.159| 𝛽 = tan−1 = tan−1 |Δ𝑁| |1,215,141.963 − 1,062,419.629| 𝛽 = 𝑁 030 37′ 32.62" 𝐸 𝑷𝒐𝒍( 𝜟𝑵 , 𝜟𝑬 ) 177 177 12 08/07/2024 Example 05 A. Grid Azimuth from Station BHL-1 to Station CBU-1. 𝐺𝑟𝑖𝑑 𝐴𝑧𝑖𝑚𝑢𝑡ℎ = 1800 + 030 37′ 32.62" 𝐺𝑟𝑖𝑑 𝐴𝑧𝑖𝑚𝑢𝑡ℎ = 𝟏𝟖𝟑°𝟑𝟕′𝟑𝟐. 𝟔" 178 178 Example 05 The following PRS92 geographic and grid coordinates of stations MMA-1 and MMA-6 are provided as reference stations: Determine: A. the Grid Azimuth from Station MMA-1 to station MMA-6. B. the Meridian Convergence in seconds. C. the Geodetic Azimuth from station MMA-1 to station MMA-6 if the arc-to-chord correction is negligible. 179 179 13 08/07/2024 Example 05 B. Meridian Convergence in arc seconds. 𝛥𝛼 = (𝛥𝜆) 𝑠𝑖𝑛 𝛷𝑎𝑣𝑒 = (1230 56′ 42.72537" − 1230 51′ 10.60199") ∗ 090 36′ 26.39543" + 100 59′ 16.07254" sin 2 𝛥𝛼 = 0°0′ 59.37051" = 𝟓𝟗. 𝟒" 180 180 Example 05 The following PRS92 geographic and grid coordinates of stations MMA-1 and MMA-6 are provided as reference stations: Determine: A. the Grid Azimuth from Station MMA-1 to station MMA-6. B. the Meridian Convergence in seconds. C. the Geodetic Azimuth from station MMA-1 to station MMA-6 if the arc-to-chord correction is negligible. 181 181 14 08/07/2024 Example 05 C. Geodetic Azimuth from station BHL-1 to station CBU-1 assuming that the arc-to-chord correction is negligible. 𝐺𝑒𝑜𝑑𝑒𝑡𝑖𝑐 𝐴𝑧𝑖𝑚𝑢𝑡ℎ = 𝑇𝑟𝑢𝑒 𝐴𝑧𝑖𝑚𝑢𝑡ℎ = 𝐺𝑟𝑖𝑑 𝐴𝑧𝑖𝑚𝑢𝑡ℎ + 𝛥𝛼 = 183°37′32.6" + 59.4" = 𝟏𝟖𝟑𝟎 𝟑𝟖′ 𝟑𝟐. 𝟎" 182 182 Example 05 The following PRS92 geographic and grid coordinates of stations MMA-1 and MMA-6 are provided as reference stations: Determine: A. the Grid Azimuth from Station MMA-1 to station MMA-6. B. the Meridian Convergence in seconds. C. the Geodetic Azimuth from station MMA-1 to station MMA-6 if the arc-to-chord correction is negligible. 183 183 15 08/07/2024 END OF TOPIC Thank you for listening. 184 Extra Questions ? RANDOM TOPICS IN CARTOGRAPHY & FRIENDS 185 185 16 08/07/2024 QUESTION AND ANSWER During the 1960-1980 period, the basic framework of the triangulation network of the Philippines was, maintained by the _____ with _____ accuracy level. a. Bureau of Lands, 1st order b. Bureau of Coast and Geodetic Survey, 1st order c. Bureau of Lands, 2nd order d. Bureau of Coast and Geodetic Survey, 2nd order 186 QUESTION AND ANSWER Equivalent to a planimetric map with a uniform scale except that instead of lines and symbols, images convey the information. a. rectified photograph b. orthophotograph c. registered photograph d. pictograph 187 17 08/07/2024 QUESTION AND ANSWER In real-time kinematic (RTK) GNSS surveys, a base station at the known point transmits corrections to the ______. a. Observer b. Rover c. Office d. Satellite 188 QUESTION AND ANSWER When the ______ direction of the earth in elongated, ______ ellipsoid of revolution is generated. a. North-south, a problate b. North-south, an oblate c. East-west, a prolate d. East-west, an oblate 189 18 08/07/2024 QUESTION AND ANSWER The angle of the earth’s rotational axis with respect to the orbital plane of the earth around the sun is known as: a. deflection of the vertical b. centripetal angle c. obliquity of ecliptic d. gravity angle 190 QUESTION AND ANSWER Triangulation baseline were measured accurately and carefully using _____, an alloy which is highly resistant to change in length caused by changes in _____. a. Fiberglass, temperature b. Invar, humidity c. Invar, temperature d. Fiberglass, humidity 191 19 08/07/2024 QUESTION AND ANSWER _______ errors in GNSS surveying can be ______ through careful procedures and proper site reconnaissance. a. Site-dependent, maximized b. Site-dependent, minimized c. Receiver timing, maximized d. Receiver timing, minimized 192 QUESTION AND ANSWER A ______ in the value of a ______ angle between a back-sight and any other survey point. a. Bearing, counterclockwise b. Bearing, clockwise c. Direction, counterclockwise d. Direction, clockwise 193 20 08/07/2024 QUESTION AND ANSWER Astronomic azimuth is based on the true shape and rotation of the earth; whereas, ____ azimuth is based on the ____ approximation of the earth’s shape. a. Geodetic, graphical b. Grid, graphical c. Grid, mathematical d. Geodetic, mathematical 194 QUESTION AND ANSWER A transit without a vertical circle and telescope level. a. Plain Transit b. City Transit c. Geodimeter d. Mining Transit 195 21 08/07/2024 QUESTION AND ANSWER The vertical angle which the magnetic needle makes with the horizontal due to uneven magnetic attraction from the magnetic poles. a. magnetic declination b. agonic angle c. magnetic dip d. isogonic angle 196 QUESTION AND ANSWER The angle that a magnetic meridian makes with the true meridian. a. magnetic declination b. agonic angle c. magnetic dip d. isogonic angle 197 22 08/07/2024 QUESTION AND ANSWER Imaginary line passing through points having the same magnetic dip. a. contour line b. isogonic line c. agonic line d. isoclinic line 198 QUESTION AND ANSWER Imaginary line passing through points having the same magnetic declination. a. contour line b. isogonic line c. agonic line d. isoclinic line 199 23 08/07/2024 QUESTION AND ANSWER Imaginary line passing through points with zero declination. a. contour line b. isogonic line c. agonic line d. isoclinic line 200 24 08/07/2024 Remote Sensing GE BOARD EXAMINATION REVIEW 2024 CARTOGRAPHY 04-05 1 TOPIC CONTENTS Foundations of Remote Sensing Energy Interactions 01 Processes, EMR, Physical Laws in RS 02 With Atmosphere and Earth Surface Materials, Spectral Signature Remote Sensing Systems Digital Image Processing 03 Active and Passive RS, Sensors and Platforms, Satellite Imaging and RS 04 General Steps, Georeferencing, Classification, Accuracy Assessment Systems, Digital Images, Resolutions 2 2 1 08/07/2024 Foundations of Remote Sensing PART 01 3 What is “Remote Sensing”? “… is the measurement or acquisition of some property of an object or phenomenon, by a recording device that is not in physical contact with the object or phenomenon under study” 4 2 08/07/2024 Processes Involved D A. Energy Source or Illumination B. Radiation and the Atmosphere C. Interaction with the Target A D. Recording of Energy by the Sensor E. Transmission, Reception, and Processing F. Interpretation and Analysis B B G. Application C E F G 5 EMR and Remote Sensing In remote sensing, a detector measures the EM radiation/energy that is reflected/ emitted from the Earth’s surface materials. Radiation Source travel time from the Sun to Earth is about 8 minutes 6 6 3 08/07/2024 Electromagnetic Wave Wave Model: Waves propagate through space in the form of sine waves. These waves are characterized by two fields, electrical (E) and magnetic (M), which are perpendicular to each other. Energy travels through atmosphere at the speed of light (c) at a certain wavelength ()ג ◦ c = 299,790,000 m/s 7 Electromagnetic Wave Velocity is the speed of light, c=3 x 108 m/s Wavelength ) )גis the length of one wave cycle, is measured in meters (m) or some factor of meters such as ◦ centimeters (cm) 10-2 m ◦ micrometers (μm) 10-6 m ◦ nanometers (nm) 10-9 m Frequency (v) refers to the number of cycles of a wave passing a fixed point per unit of time. Normally measured in Hertz (Hz). 8 4 08/07/2024 Electromagnetic Wave c = vλ (frequency x wavelength) ◦ The shorter the wavelength, the higher the frequency ◦ The longer the wavelength, the lower the frequency 9 Electromagnetic Spectrum The continuum of energy that ranges from m to nm in wavelength, travels at the speed of light, and propagates through a vacuum such as outer space. 10 5 08/07/2024 Electromagnetic Spectrum Cosmic Rays Visible Thermal IR TV & Radio UV Near IR X Rays Microwave 10-7 10-6 10-5 10-4 10-3 10-2 10-1 1 101 102 103 104 105 106 107 108 wavelength, m Blue 0.4 - 0.5 um Green 0.5 - 0.6 um Comprises 2% Red 0.6 - 0.7 um of EM Spectrum 11 Region Name Wavelength Range Details Very narrow zone of EMR. Ultraviolet (UV) 0.30-0.38 μm It has short wave lengths. Largely scattered by atmospheric particles. These regions can be perceived by human eyes
