GRB 603 MAP PROJECTIONS EXERCISE 6 Question given

GRB 603: MAP PROJECTIONS EXERCISE 6

Question given: Draw a graticule for the International Projection (reduced ¼) on a scale 1: 4, 000 (original 1: 1, 000) for the area with latitudinal extent from 24◦N to 28◦N. The distance between 24◦N - 28◦N is 442. 90 mm. The coordinates of the intersections of parallels and meridians are given below: Latitudes 24◦ 28◦ Coordinates Longitudes from Central Meridian 1◦ 2◦ 3◦ x 101. 76 203. 62 305. 31 y 0. 36 1. 45 3. 25 x 98. 37 196. 75 295. 75 y 0. 40 1. 61 3. 63 #Please note some changes, which I think could possibly be printing errors: 1. Original scale is 1: 1, 000 and not 1: 100 ◦, 000 2. Instead of longitudinal extent, I think it should be latitudinal extent as that is what is mentioned immediately next to it. 3. There should have been a longitudinal extent in the question, though International Projection uses a 6 ◦ extent for meridians in one sheet for the given latitudinal extent. However, all longitudes intersect the given latitudes. We will therefore take it more generically without mentioning any specific longitudinal extent. 4. The value for the y coordinate of 24 ◦ latitude is 0. 36 instead of 0. 46, as that is the value mentioned in most of books I could consult. 5. All table values are in millimetres.

Let us know about the projection theoretically: ‘This is a modified polyconic projection. Following the decision of International Map Committee held in 1909, the projection was introduced for the topographical maps of the whole world on a scale of 1 : 1, 000’ (R. L. Singh, 1979, p. 281). This projection was developed so that parts of the entire world can be prepared in sheets that exactly fit in with each other along all sides to create a consolidated map of the world. Each sheet however was to be prepared independently with different Central meridians. There a total of 2222 sheets that make up the world. (Misra, 1989, p. 115). This is because this projection covers: EXTENT ZONE NUMBER OF SHEETS Latitudinal extent of 4◦ and longitudinal extent of 6 ◦ Between 60 ◦ N and S Therefore, there it requires 15 sheets to cover the Northern Hemisphere (60 ÷ 4 = 15) + 15 sheets to cover the Southern Hemisphere Total = 30 sheets. Longitudinally, 180÷ 6 = 30 sheets for the Eastern Hemisphere and 30 sheets for the Western Hemisphere Total = 60 sheets Therefore, there a total of 30 x 60 = 1800 sheets to cover the entire world between 60 ◦ N and S Latitudinal extent of 4◦ and longitudinal extent of 12◦ (meaning two sheets clubbed into one). Between 60 ◦ - 88◦ N and S Here, it requires 7 sheets to cover the Northern Hemisphere (28 ÷ 4 = 7) + 7 sheets to cover the Southern Hemisphere Total = 14 sheets. Longitudinally, 180 ÷ 12 = 15 sheets for the Eastern Hemisphere and 15 sheets for the Western Hemisphere Total = 30 sheets Therefore, there a total of 14 x 30 = 420 sheets to cover the entire world between 60 ◦ - 88◦ N and S Between 88 ◦ - 90◦ N and S, the polar caps are covered using 2 circular sheets. TOTAL SHEETS NEEDED TO COVER THE ENTIRE WORLD = 1800 + 420 +2 = 2222 SHEETS

The differences, which are also the improvisations of International Projection with respect to Polyconic Projection are as follows: POLYCONIC INTERNATIONAL ‘In the polyconic, sheets east and west of each other do not fit together properly because the marginal meridians are curved’. (Singh, 2007, p. 281) as shown below: ‘In the International Projection they have been made to fit on all sides’ (Singh, 2007, p. 281) as shown below: (Singh, 2007, p. 281 -82) ‘all the parallels are equally and correctly divided into equal parts and the meridians are drawn as regular curves by joining the points of division thus obtained’ (Singh, 2007, p. 281) ‘only the top and bottom parallels are divided into equal parts at true distances and the meridians are drawn as straight lines, passing through the corresponding points thus marked along them. As the marginal meridians become straight lines, the adjoining sheet on all sides may be fitted together with little distortion. But in doing so the intermediate parallels become very slightly short. ’ (Singh, 2007, p. 281) ‘the scale is correct along the central meridian as it is divided truly, and along other meridians it becomes gradually enlarged towards the margins. ’ (Singh, 2007, p. 281 -82) ‘this defect has been minimised by making the scale true along two meridians on each sheet, one lying 2° west and the other, 2° east of the central meridian. Thus the scale along the central meridian becomes - slightly too small and that along the marginal meridians, slightly too long. The parallels are true distances apart only along the two true meridians. ’ (Singh, 2007, p. 282) ‘the parallels are not concentric but their centres lie on the central meridian produced. ’ (Singh, 2007, p. 283) ‘The central meridian and the top and bottom parallels are constructed first with the help of the two computed tables. ’ (Singh, 2007, p. 283). The tables are given in the next slide.

(Misra, 1989, p. 115). Properties: 1. All meridian are straight lines. 2. All parallels are curved. 3. Each parallel is treated as a standard parallel. 4. They are plotted according to corrected length. 5. Parallels intersect the central meridian at right angles and other meridians at oblique angles. 6. The two meridians at 2 ◦ East and West from the central meridian are true to scale. 7. Distances around extreme meridians are slightly longer, others slightly shorter. (Misra, 1989, p. 116 -17). (Misra, 1989, p. 116).

Let us now start construction: Step I: The question says “The distance between 24◦N to 28◦N is 442. 90 mm”. Let us consult Table 8. 2, which gives us “Length of Central Meridian in International Projection in millimetres”. If we look into the 24◦ - 28 ◦ row, we find the corrected length = 442. 91. These two statements and values are basically the same, which is (as shown in red line): 28◦N 442. 91 mm A 24◦N B Therefore, the first step is to draw this straight vertical line of length 444. 91 mm ÷ 10 = 44. 491 cms. , which is very long to draw on normal sized paper. Therefore, the reduced ¼ scale has been given in the question. Thus, the length to be drawn becomes: 44. 491 ÷ 4 = 11. 123 cms. Let us call this line AB. This is our Central Meridian. So, let us first draw a straight vertical line AB (Central Meridian) of length 11. 12 cms in the centre of our paper.

Step II: Next, we have to draw 2 straight lines (of any length) at right angles to AB at points A and B. Let us call these lines LM and UV respectively. These are our reference lines for the two extreme parallels: LM for 28 ◦ N and UV for 24 ◦ N. These are NOT the parallels we need. L A M ◦ 11. 12 cm 90 90 U B ◦ V

Step III: From A, towards east and west, on the line LM and from B, east and west on the line UV, we now have to mark the x coordinate values, which we will get from table given in the question. (Also refer to Table 8. 3). Concerned Line on Point from x coordinate values (from given table) Parallels which to be in millimetres marked 1◦ 2◦ 3◦ 24 ◦ N UV B 101. 76 203. 62 305. 31 28 ◦ N LM A 98. 37 196. 75 295. 75 Concerned Parallel Therefore, x coordinate values in centimetres reduced to given scale are: 1◦ 2◦ 3◦ 24 ◦ N 10. 176 ÷ 4 = 2. 544 20. 363 ÷ 4 = 5. 091 30. 531 ÷ 4 = 7. 632 28 ◦ N 9. 837 ÷ 4 = 2. 459 19. 675 ÷ 4 = 4. 918 29. 575 ÷ 4 = 7. 394 These points will give us the positions of our meridians on the extreme parallels only at 1 ◦, 2 ◦ and 3 ◦ distance on either side of the Central Meridian.

Let us mark the points thus obtained on LM as c, d, e, f, g, h and on UV as o, p, q, r, s, t L e d c A f h ◦ 90 M 2. 45 cms Therefore, the lengths will be: 4. 91 cms Line Segment values (as calculated in the previous table) each being 1 ◦ apart 1◦ 2◦ 3◦ 24 ◦ N Bo = Br = 2. 544 cms Bp = Bs = 5. 091 cms Bq = Bt = 7. 632 cms 28 ◦ N Ac = Af = 2. 459 cms Ad = Ag = 4. 918 cms Ae = Ah = 7. 394 cms 7. 39 cms 11. 12 cm Concerned Parallel g 7. 63 cms 5. 09 cms 2. 54 cms q U p 90 o B ◦ r s t V

L 0. 091 cms e 0. 01 cms 0. 040 cms d A c 0. 091 cms 0. 01 cms f 0. 040 cms g h M Step IV: From these point, now we have to erect (NOT DROP) perpendiculars with lengths = y coordinates of the concerned parallels using values from the table given in the question. Also refer to Table 8. 3. Therefore, the lengths will be: Concerne d Parallel 24 ◦ N 28 ◦ N Length of the perpendiculars as calculated from the given y coordinate values Calculation of value as per given scale 1◦ 2◦ 3◦ Given value (in mm) 0. 36 1. 45 3. 25 Calculated value in cms reduced to scale (in cms) 0. 036 ÷ 4 = 0. 009 0. 145 ÷ 4 = 0. 036 0. 325 ÷ 4 = 0. 081 Given value (in mm) 0. 40 1. 61 3. 63 Calculated value in cms reduced to scale (in cms) 0. 04 ÷ 4 = 0. 01 0. 161 ÷ 4 = 0. 040 0. 363 ÷ 4 = 0. 091 Therefore, as we can see the values are very small, this is due to ¼ reduction. If we can use a diagonal scale, we will be able to plot the minute variations. Uq 0. 081 cms p 0. 036 cms 0. 009 cms o 0. 081 cms B 0. 009 cms r s 0. 036 cms t V

L 28 ◦ N A e d c f g h M Step V: Join the tops of these perpendiculars with smooth curves (arcs of circles). The tops of these perpendiculars are the points of intersection between the concerned parallels and meridians. The curves should touch the Central Meridian at the points A and B. The centre of the curves should lie on the Central Meridian (AB). These curves are our needed extreme parallels of : 28 ◦ N and 24 ◦ N Uq p o B r s t V

Step VI: Join the points e with q to get the 3 ◦ meridian on the west and h with t to get the 3 ◦ meridian on the east. Join the points d with p to get the 2 ◦ meridian on the west and g with s to get the 2 ◦ meridian on the east. Join the points c with o to get the 1 ◦ meridian on the west and f with r to get the 1 ◦ meridian on the east. Join with straight lines. Join from top of the curves, ignore the straight horizontal lines now. Thus, we have got all our meridians. Let us name them generally since no longitudinal extent is given in the question. 28 ◦ N A e d Meridian at 3 ◦ distance on the West Uq c f Meridian at 1 ◦ distance on the West Meridian at 2 ◦ distance on the West p o Central Meridian L g r M Meridian at 3 ◦ distance on the East Meridian at 2 ◦ distance on the East Meridian at 1 ◦ distance on the East B h 24 ◦ N s t V

Step VII: Now, we have to get the other parallels for which we have to divide each of the meridians into 4 equal parts. This can simply be done by measurement. Like: Meridians Length of the meridian (in cms) Meridian at 3 ◦ distance on the East and West Length of the central meridian was = 11. 12 cms. To which we added 0. 091 cms at 28◦ N and subtracted 0. 081 cms at 24◦ N in the form of the perpendiculars we erected. Therefore, in total we added 0. 01 cms (0. 091 – 0. 081) to get the meridian at 3◦ distance from the central meridian, which is thus = 11. 12 + 0. 01 = 11. 13 cms. Meridian at 2 ◦ distance on the East and West Meridian at 1 ◦ distance on the East and West Similarly, length of the meridian at 2◦ distance from the central meridian = 11. 12 + (0. 040 - 0. 036) = 11. 124 cms Similarly, length of the meridian at 1 ◦ distance from the central meridian = 11. 12 + (0. 01 - 0. 009 = 11. 121 cms Length of each segment (in cms) 11. 13 ÷ 4 = 2. 783 cms Therefore, use this length to divide the Meridian at 3 ◦ distance on the East and West into 4 equal parts. 11. 124 ÷ 4 = 2. 781 cms Therefore, use this length to divide the Meridian at 2 ◦ distance on the East and West into 4 equal parts. 11. 121 ÷ 4 = 2. 780 cms Therefore, use this length to divide the Meridian at 1 ◦ distance on the East and West into 4 equal parts. Once, we make the divisions, we can join the points using smooth curves. This will give us our intermediate parallels. Now, we need to label, write the title, scale etc. and complete our projection.

INTERNATIONAL PROJECTION 28 ◦ N 27 ◦ N Central Meridian 27 ◦ N 26 ◦ N 25 ◦ N 24 ◦ N Meridian at 3 ◦ distance on the West Meridian at 2 ◦ distance on the West Meridian at 1 ◦ distance on the East Original scale 1: 1, 000 ; reduced scale 1: 4, 000 Meridian at 2 ◦ distance on the East Meridian at 3 ◦ distance on the East 24 ◦ N

References: Singh, R. L. , 1979, Elements of Practical Geography, Kalyani Publishers. Misra, R. P. and Ramesh, A. , 1989, Fundamentals of Cartography, Revised and Enlarged edition, Concept Publishing Company, New Delhi. THANK YOU