A mathematical investigation on the distance-preserving property of an equidistant cylindrical projection
MMap pro jection: A mathematical investigationon the distance-preserving property of anequidistant cylindrical pro jection
Bingheng Yang [email protected] a r X i v : . [ m a t h . G M ] J a n ontents A.1 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31A.2 General model for equidistant cylindrical projection . . . . . 32A.2 The Haversine formula . . . . . . . . . . . . . . . . . . . . . . 32A.3 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 bstract
A normal world map can be defined as a representation on a flat surfacethat shows the features of the earth, such as the area of Australia or thelatitude of the arctic circle, in their respective forms, sizes, and relationshipsaccording to some convention of representation.All world maps have inevitable errors : Some maps fail to represent cor-rect directions, some give us wrong displays of areas. In other words, thereexists no such maps that could show all the right characteristics of the Earthas the globe would. This is due to the distortion caused by different projec-tion methods. This research work aims to explore the distortions in distance in equidis-tant cylindrical projection. Although the projection is described as ”equidis-tant”, i.e. distance-preserving, it is far from error-free. By constructing arealistic and appropriate model of the projection, this work will demonstrate to what extend are distortions in distance present in equidistantcylindrical projection using mathematical methods such as linear model-ing, differentiation and trigonometric relationships.The investigation is conducted by examining the projection performinggeometrical and trigonometrical analyses. The horizontal bending that oc-curs in the projection process can be assessed by performing a geometricanalysis using Tissot’s indicatrices. In addition, the concept of the sphericalcoordinates, alongside with trigonometrical identities, can be used to illus-trate the route from a point to another as a curve. With a combination ofthe knowledge extracted from the examination of the projection using thosetwo theories, this research aims to fully unravel the degree of distortion indistance in equidistant cylindrical projections. Vox,
Why all world maps are wrong? (2016, December 2) Numberphile,
A strange map projection (Euler spiral) (2018, November 13) ntroduction A history of cartography
Cartography , or the study of making maps, is one of the oldest researchconcepts that human practiced. For thousands of years, mathematicians andastronomers have been successful to accumulate world maps that accuratelydisplay properties of Earth to scale. The idea of maps already existed duringthe Babylonian era around 2300 B.C, and the Greek developed world mapdesigns to an advanced level.Later, during the end of the Middle Ages, the first concept of a worldmap, the so called T-O map, was invented. For the next few hundred years,exploration trips, improvement in technology and the increasing market needlead to a rapid development phase of world maps. Different designs werecreated by talented cartographers for a number of purposes. Some of theworld map projections are for example, the Mercator projection and theEckert projections.After Soviet engineers launched the first satellite Sputnik 1, the designprocess of world map projections became much easier. The rise of GIS -Geographic Information Systems - enabled the creation of more accuratemaps and newer concepts such as digitalized maps, e.g. Google maps. James S. Aber,
Brief history of maps and cartography , retrieved March 2019. http://academic.emporia.edu/aberjame/map/h_map/h_map.htm ur map In our projection, we first assume the Earth to be spherical and the radius ofthe Earth to be . We aim to represent a projection using Cartesianxy-coordinates, with the x-axis describing the East-West direction and they-axis demonstrating the North-South direction. The scale factors, whichindicate how does the real distance between two points on the surface ofthe Earth correspond to the distance displayed by the projection, are alsorepresented by mathematical formulas.We must be aware of the properties and features of the projections. Inthe following sections, the formation method of the map projection is inves-tigated, from which the formulas of an equidistant cylindrical projection arefound. All numerical values for angles will be in radians . All numericalvalues for distances will be in kilometers . Equidistant cylindrical
Equidistant cylindrical projection is one of the first world map projectionscreated. It was first established by Claudius Ptolemy, and improved by latergenerations. The purpose of the projection is to give an accurate presen-tation of the distance between two points on the surface of the Earth: Bymultiplying the distance between two points on the map with a scale factor,the result should correspond to the real distance.A cylindrical projection is created by first inscribing a sphere intoa right cylinder without the top and bottom surfaces, then transferringthe points on the spherical grid to the surface of the cylinder. Finally,the side surface of the cylinder is unfolded, resulting a rectangular projec-tion. We can see the construction process in figure 2. This will give usa rectangular-shaped map, such as in figure 1: In equidistant cylindricalprojection, the shape of the world is rectangular. By name, the projectionshould be distance-preserving, meaning that the distance on the map corre-sponds to the real distance on the globe. From figure 2, we can observe anumber of properties that are unique for equidistant cylindrical projection:All vertical meridians are parallel to each other and all horizontal parallelsare also parallel to each other. The gap between individual meridians and Wikipedia,
Equidistant cylindrical projection, retrieved December 2018. Eotvos Lorand University,
Cylindrical projections , retrieved July 2019. http://lazarus.elte.hu/cet/modules/guszlev/cylin.htm
M P
Figure 2: The sphere is placed in a topless and bottomless right circularcylinder. Individual points on the sphere are transferred onto the side of thecylinder; An image of a rectangle is generated after the cylinder is unfolded,because the side surface of a right cylinder demonstrates a rectangle. Merid-ians (marked with M) and standard parallels (marked with P) intersect eachother at right angles and they are equally spaced. | M | = | P | holds for allmeridians and parallels in both the globe and the projection. Projection method
In this section, the way how individual points are transferred onto the cylin-der is explained and justified.Figure 2 implies that the sides of the cylinder are tangents to the sphere.To demonstrate this observation, figure 3 shows the front view of figure 2.The circle with diameter RS - which indicates the sphere - has a tangent BB (cid:48) , which denotes the side of the cylinder.In figure 3, O denotes the middle point of the sphere, and let points S and P be on the sphere. AA (cid:48) is a line crossing the diameter of the circle. Itis said in earlier sections that points S and P have to be transformed ontothe side of the cylinder, i.e. tangent BB (cid:48) . Since the circle and BB (cid:48) touch atpoint S , S is on the side of the cylinder. However, in the case of point P , itis projected to P (cid:48) .Another point Q on the sphere (circle) is selected such that Q has thesame latitude as P and Q is always located on AA (cid:48) . Lines OP and OS forman angle φ . When point P travels anti-clockwise to approach line AA (cid:48) , Q translate upwards and meets P at Q (cid:48) , during when the position of P (cid:48) on BB (cid:48) also shifts upwards. Additionally, as | P P (cid:48) | increases, angle φ increases when6 A BSφQ P P (cid:48)
R A (cid:48) B (cid:48) φQ (cid:48) Figure 3: A front view of figure 2. The angle φ is preserved in both 2D and3D illustrations. P approaches Q (cid:48) .From figure 3, the scale factors can be deduced. As figure 4 shows, thesphere can be rotated in a way so that point Q is transformed to the positionof point P in figure 3, and point Q will be projected similarly to the pro-jection of point P (page 7). Thus, distortion in distance between two pointson the projection is independent of their horizontal coordinates. Therefore,only vertical translation can affect the distance-preserving ability of the pro-jection.Figure 2 also shows that a rectangular map projection is formed. Let thehorizontal side of the projection be the length and the vertical side of theprojection be the width. Since the side surface of the cylinder is a tangent tothe sphere, the length of the projection should correspond to the perimeterof the circle in figure 3: if we consider | RS | to be on the sphere (figure 3), itcorresponds to an half-circle that has the radius of the original sphere; thehalf-circle would always be on the cylinder, hence the length of the cylinderequals the perimeter of the half-circle in figure 3. Similarly, the width of theprojection should be set to include all projected points such as P (cid:48) : whenthe radius of the circle (figure 3) increases, the position of P (cid:48) moves up inrelative terms. Hence, both the length and the width of the projection aredependent on the radius of the sphere.7igure 4: The sphere and cylinder in figure 2 from different perspectives. Asthe third cylinder illustrates, the sphere can be rotated so that point Q canbe on a position equivalent to the position of point P in the second cylinder.Hence, the position of point P in figure 3 can be recreated for all points alongthe same longitude, such as point Q . Hence, the points P and Q undergo thesame extent of disposition, causing the Euclidean distance | P Q | (figure 3) todistort. Since P and Q share the same longitude, it seems that distortion indistance is independent from the longitudinal coordinates.8 RQQ (cid:48)
P P (cid:48) φ Figure 5: Above view of figure 4. The greater circle stands for the circularbase of the right cylinder, and the smaller circle illustrates the above imageof the inscribed sphere. Points P and Q are located on a circle with radius r ,and the circular base of the cylinder has a radius R . φ is the common latitudeof all points P and Q on the sphere and P (cid:48) and Q (cid:48) on the projection. Forming the projection formulas
Using the information acquired in the last section, we strive to formulate ap-propriate formulas for an equidistant cylindrical projection using the Carte-sian grid. The points of longitude are denoted by λ and the points of latitudeare denoted by φ . Let the relative scale factor of the meridian be h and therelative scale factor of the standard parallel be k .As mentioned earlier, horizontal translations will not cause distortion indistance. Hence, h is a constant. To make further calculations simpler, wecan let h = 1. On the other hand, to calculate k , the relationship between | P Q | and | P (cid:48) Q (cid:48) | must be known; this can be deduced from figure 5. Becausethe scale factor is calculated asEuclidean distance between two points on the projectionEuclidean distance between two points on the surface of the Earth , we deduce that k = | P (cid:48) Q (cid:48) || P Q | from figure 5. Also, | P Q | = rφ and | P (cid:48) Q (cid:48) | = Rφ .Thus, See glossary for definition. = | P (cid:48) Q (cid:48) || P Q | = Rφrφ = Rr .
Notice that the relationship between R and r can be determined using fig-ure 3. Since R is the radius of the base circle of the cylinder, it corresponds to | OS | , which equals | OP | . Similarly, r stands for P Q . Using trigonometricalidentities, we deduce r = R cos φ . Therefore, k = Rr = RR cos φ = 1cos φ . Hence, the scale factors of the projection are h = 1 (1) k = 1cos φ (2)Next, a rectangular projection is built. Since the size of the projectiondepends on the radius of the sphere (figure 2), let the radius of the spherebe R .One way to create the model is to define the intersection of the centralmeridian and standard parallel to be at the origin. By this action, the centralmeridian is chosen to be the Greenwich line ( λ = 0 ) and the standard parallelis chosen to be the equator ( φ = 0 ) . Consequently, the x-coordinate is depen-dent on λ and the y-coordinate on φ . Recall from figure 2 that all meridiansare equal in length and all parallels are equal in length; Furthermore, themeridians are a half of the sphere’s outer perimeter and the parallels areequal to the sphere’s outer perimeter. Hence, we deduce that the length ofeach individual meridian must be πR and the length of each individual par-allel must be 2 πR , so the length of the projection is going to be 2 Rπ andthe width of the projection is going to be Rπ . The latitude coordinates ofthe Earth varies between 90 ◦ N and 90 ◦ S and the longitude coordinates shiftsfrom 180 ◦ E to 180 ◦ W; To make modeling easier, we let latitude points southof the standard parallel and longitude points west of the prime meridian totake negative values in radians. Therefore, the coordinates of a randompoint located on the projection satisfy y = Rφ, − π ≤ φ ≤ π = Rλ, − π ≤ λ ≤ π. (4)In this way, φ max = π and λ max = 2 π , so the length of the meridians andparallels will be πR and 2 πR . Hence, by combining equations 1, 2, 3 and 4,the equations for equidistant cylindrical projection are: y = Rφx = Rλh = 1 k = 1cos φ − π ≤ φ ≤ π , − π ≤ λ ≤ π. Where • x is the coordinates of East-West axis of the map • y is the coordinates of North-South axis of the map • h is the relative scale factor of the meridian • k is the relative scale factor of the standard parallel • R is the radius of the Earth • φ is latitude in radians • λ is longitude in radians.As noted earlier, in this particular model, the prime meridian is set to be theGreenwich line, which has the longitude of 0 ◦ , corresponding to the y-axis.Contrarily, the standard parallel is selected to be the equator. All pointsalong the Equator have 0 ◦ of latitude, therefore aligning to the x-axis. Amodel of the projection is shown by figure 6.From the formulas we can already see one possible cause of distortion: k ,the relative scale factor of the horizontal parallels, is a variable. This means11y( − Rπ, − Rπ ) ( Rπ, − Rπ )( − Rπ, Rπ ) ( Rπ, Rπ )Figure 6: Equidistant cylindrical projection in xy-coordinates.that the distance between two points with different longitude will not alwayscorrespond to the real distance. When k = 1, cos φ = 0 ↔ φ = 0, meaningthat the error in distance is minimized at the equator. The distance distortionis at the maximum in the poles, since lim φ → π − φ = ∞ and lim φ →− π + φ = ∞ . Distortion in distance
About distortion Distortion is and will always be present in map projections. There existsno maps that can perfectly preserve and represent the most important geo-metrical identities of the Earth to scale: the latitude and longitude angles,the surface area of regions, and the distance between two points on the earth.The distortion will be or will not be significant depending on the purpose ofthe map. For example, if we want to use a map for sailing from Gibraltar toCape Town, we can ignore the irrelevant area distortion as long as the mapis conformal so the direction angle is always correct. On the other hand, thedistortion in distance in a ”distance-preserving” projection must be taken Rice University,
Mapping the sphere , retrieved April 2019. https://math.rice.edu/~polking/cartography/cart.pdf y P P d Figure 7: P and P on a modeled equidistant cylindrical projection. d demonstrates the distance between P and P .into account. In the following section, the distance distortion in equidistantcylindrical projection is demonstrated. How can the distortion be seen?
Suppose we want to determine the distance between two points, P and P ,using the formulas of equidistant cylindrical projection. According to theformulas (page 7), by recognizing the longitude and latitude coordinates ofany two points, we can deduce their x- and y-coordinates. Therefore, wecan arbitrarily select the coordinates of P and P : For example, let P =(24 . ◦ E, 23 . ◦ N) and P = (39 . ◦ W, 3 . ◦ S). Their relative distance on themap is illustrated in figure 7. To calculate the distance between P and P ,we need to first change the longitude and latitude in the coordinates intoradians.P = (24 . ◦ E, 23 . ◦ N) = (0 . ..., . ... )P = (39 . ◦ W, 3 . ◦ S) = ( − . ..., − . ... )By using the formulas of equidistant cylindrical projection (page 7), wecan calculate the values of x and y-coordinates. x = Rλ = 6371 × . ... ≈ . x = Rλ = 6371 × ( − . ... ) ≈ − . y = Rφ = 6371 × . ... ≈ . = Rφ = 6371 × ( − . ≈ − . d = (cid:113) ( x − x ) + ( y − y ) . Hence, the distance betweenP and P is d = (cid:112) (2702 ... − ( − ... )) + (2601 ... − ( − ... )) = 7675 . ≈ . However, by using
Movable type script , an engine that is able to generatereal distances between two points on the surface of Earth, we observe that thereal distance between our randomly selected P and P is 7502 km. Thus,our numerically calculated result has an error of 7676 − × ≈ . Movable type scripts,
Calculate distance, bearing and more between latitude/longitudepoints
Investigation methods
In the previous section, we have demonstrated the existence of distance dis-tortion in a equidistant cylindrical projection. We will investigate this issueby performing a geometrical analysis and a trigonometrical analysis to scru-tinize the distance distortions of equidistant cylindrical projection. In thisway, we aim to construct general methods to determine the extent of errorin distance.
Geometrical analysis
In previous sections, we have shown that k , the relative scale factor of thehorizontal parallels, is a variable; the change in distortion is dependenton the value of latitude. Recall from page 9 that lim φ → π − φ = ∞ andlim φ →− π + φ = ∞ : the horizontal distortion will increase out of bounds inpolar areas.But why is this the case? Going back to figure 2, we notice that the pro-jection is constructed such that all parallels are assumed to have the samelength; The mapmaker chooses a standard parallel, and other parallels willbe presumed to have the same length. However, from figure 8, it can beobserved that it is not the case. The length of parallels (except the stan-dard parallel) are artificially changed, and distortion in distance is already15igure 9: Tissot’s indicatrices in a sphere.Figure 10: Tissot’s indicatrices in equidistance cylindrical projection. Noticethe distortion at polar areas illustrated by the bending of indicatrices; theindicatrices become ellipses due to horizontal flexion.inevitable. Therefore, we can deduce that artificial bending, also called as flexion , occurs in equidistant cylindrical projection, which is due to thenature of its construction process. Tissot’s indicatrix
There exists a various of methods to investigate the flexion which occurs inequidistant cylindrical projections. In this research,
Tissot’s indicatrices David M. Goldberg, J. Richard Gott III (2007),
Flexion and Skewness in Map Pro-jections of the Earth John P. Snyder (1987),
Map projections - A working manual y MO ABr rθ xy M (cid:48) A (cid:48) B (cid:48) Or r θ (cid:48) Figure 11: An illustration of the extend of distortion of indicatrices causedby flexion due to the faulty nature of equidistant cylindrical projection. Asmall circle with radius r experiences flexion and becomes an ellipse with awidth r and a height r .are used to illustrate the distortion in different regions. Tissot’s indicatricesare circles with a tiny radius. They are constructed at given locations on asphere, (see figure 9) and then the sphere undergoes cylindrical projection.As figure 10 illustrates, the spheres closer to polar areas experience the mosthorizontal flexion and become elliptical. This observation corresponds to ourformula for horizontal scale factor (page 7). Next, we examine the indicatrixin more detail and figure out to what extend does flexion cause distortionsto appear.To clarify the transformation of the spheres in figure 8 to ellipses in figure9, the process is modeled, as shown in figure 10. In figure 10, we consider asmall unit circle to be drawn on an arbitrary part of the surface of the Earth.By considering the circle to be small, we wish to generate an overall picturefor distortions in distance between any two points on the projection, such asin figure 10: Greater bending shows more significant distortion in distance.The sphere is projected according to the methods of an equidistant cylindricalprojection (see page 5 and figure 2). As shown in figure 3, in addition with Wenping Jiang, Jin Li (2014),
The Effects of Spatial Reference Systems on the Pre-dictive Accuracy of Spatial Interpolation Methods y MPO AB A (cid:48) B (cid:48) M (cid:48) θ θ (cid:48) Figure 12: A combined version of the bending of an indicatrix in figure 11.the relationship k = φ , it can be deduced that the indicatrix is bent in thehorizontal direction, which can be seen as a stretch from OA to OA (cid:48) in figure11. Consequently, a vertical stretch also occurs, resulting OB to decreasesto OB (cid:48) . As a result, the original circle is deformed and becomes an ellipse( r (cid:54) = r (cid:54) = r ).Consider point M to be on the Earth’s surface. Due to horizontal flexion, M translates to M (cid:48) that is the point of intersection of lines [ M P ] and theellipse. In addition, angle θ decreases as ∆ OM A is deformed and becomes∆ OM (cid:48) A . This phenomenon, know as an angular distortion, will affect thedistance between two points on the projection. As a result, | OM | (cid:54) = | OM (cid:48) | ,therefore the distance between points O and M is not preserved. Therefore,the distortion in distance is || OM | − | OM (cid:48) || . The overall combination of thetwo diagrams of figure 11 can be summarized using figure 12.18 aximum angular and distance distortion In this section, the maximum angular distortion will be calculated and evalu-ated. Moreover, the horizontal distortion in distance due to a not-up-to-scalescale factor is analyzed.Consider ω = θ − θ (cid:48) where ω is the angular distortion. As can be seenin figure 11, the distortion is dependent on the coordinates of the point M (cid:48) ,which is projected according to the scale factors of the projection (see page11). Recall from page 8 that y = Rφx = Rλh = 1 k = 1cos φ Thus, the latitude φ must be taken into account when computing themaximum angular distortion. Figure 13 represents the triangle OM P infigure 11. In figure 13, it can be seen that (cid:126)OM = (cid:126)OP + (cid:126)P M ; the point OM is transformed horizontally by 1 and vertically by 1. However, thosetransformations do not necessarily correspond to a distance of 1 unit on themap projection; the scale factors must be taken into account. Since thescale factors are h = 1 and k = φ respectively, the length of the verticalcomponent P M is 1 and the length of the horizontal component OP is cosφ on the projection, because the horizontal scale factor is φ instead of 1.Next, consider | P M (cid:48) | = a where point M (cid:48) is the position of point M on theprojection and 0 ≤ a ≤
1. It follows that | M (cid:48) M | = 1 − a .Using figure 12, we aim to utilize all of its valuable information to cal-culate the maximum value of ω . First, we notice that tan ω = tan( θ − θ (cid:48) ).Using the compound angle identity for tangent, we gettan( θ − θ (cid:48) ) = tan θ − tan θ (cid:48) θ tan θ (cid:48) .
19s can be seen from figure 12, tan θ = 1 φ = cos φ and tan θ (cid:48) = a φ = a cos φ. Therefore,tan θ − tan θ (cid:48) θ tan θ (cid:48) = cos φ − a cos φ φ · a cos φ = cos φ (1 − a )1 + a cos φ . Keep in mind that we are most interested in the maximum angle distor-tion. Thus, it would be plausible to differentiate the expression above withrespect to φ ; the maximum may appear when the zeros of the derivative areretrieved. ddφ cos φ (1 − a )1 + a cos φ = (1 − a ) − sin φ · (1 + a cos φ ) − ( − sin φ cos φ − sin φ cos φ ) · cos φ · a (1 + a cos φ ) = (1 − a ) − sin φ · (1 + a cos φ ) + sin 2 φ · cos φ · a (1 + a cos φ ) = (1 − a ) − sin φ − a sin φ cos φ + a sin 2 φ cos φ (1 + a cos φ ) = (1 − a ) − sin φ − a sin φ cos φ + 2 a sin φ cos φ · cos φ (1 + a cos φ ) = (1 − a ) 2 a sin φ cos φ − sin φ − a cos φ sin φ (1 + a cos φ ) = (1 − a ) (2 a − a ) sin φ cos φ − sin φ (1 + a cos φ ) = (1 − a ) a sin φ cos φ − sin φ (1 + a cos φ ) The zeros of the derivative are the zeros of the numerator. Notice that1 − a is a constant since the expression cos φ (1 − a )1 + a cos φ is differentiated in respectto φ . a sin φ cos φ − sin φ = 0 ⇔ sin φ = a sin φ cos φ Two solutions, sin φ = 0 and a cos φ = 1, are found. Solving sin φ = 0yields φ = 0 or φ = π , but comparing the solutions to figure 13, it seems20 M (cid:48) MPθ θ (cid:48) ω φ a − a Figure 13: The relationship between θ , θ (cid:48) and ω in relation to the changes ofthe value a .both are invalid: Since | M P | (cid:54) = 0, φ (cid:54) = 0. Additionally, (cid:54) M OP (cid:54) = π since∆ M OP is a right traingle. Therefore, sin φ (cid:54) = 0 and its solutions are invalid.On the other hand, the expression a cos φ = 1 illustrates a more interest-ing result. Solving the equation gives cos φ = a . Then, we can also substitutethis result back to the original formula:cos φ (1 − a )1 + a cos φ = a (1 − a )1 + a ( a ) = a − a + 1 = 1 − a a = tan( θ − θ (cid:48) ) = tan ω. Solving the expression for ω results ω = arctan 1 − a a . which shows the identical as substituting φ = 0 equation. Unlike theresults found from substituting φ = 0, the result retrieved using cos φ = a isvalid, since cos φ (cid:54) = 0 when 0 ≤ a ≤ − a a in the range of0 ≤ a ≤
1, we notice that a ∈ [0 , ⇒ arctan 1 − a a ∈ [0 , . ... ] . The range tells that the maximum angular distortion is ω ≈ . a = 0. This result implies that the distortion is at itsminimum along the standard parallel of the projection, which is selected tobe the equator. 21igure 14: The graph of arctan − a a in the range of 0 ≤ a ≤
1. It can be seenthat arctan − a a is monotonously decreasing in the interval in question.Furthermore, we want to calculate dda arctan 1 − a a , since the maximumof arctan 1 − a a may occur at the zero of its derivative within a ∈ [0 , − a a is in the form f ( g ( a )) where f ( a ) = arctan a and g ( a ) = 1 − a a . Additionally, recall that dda f ( g ( a )) = f (cid:48) ( g ( a )) + g (cid:48) ( a ). Hence, dda arctan 1 − a a = − − a − a ( a +1) (1 − a ) (1+ a ) + 1 = − a +1) (1 − a ) +(1+ a ) (1+ a ) = − a +1) − a + a +1+2 a + a ( a +1) = −
22 + 2 a = −
11 + a . It seems that −
11 + a has no real zeroes in its defined domain ( a ∈ R ).Hence, the expression has no zeroes within the range a ∈ [0 , −
11 + a < a ∈ [0 , − a ) is decreasing whenthe value of a increases, therefore the value a = 0 will give the maximumangular distortion. In conclusion, ω = 0 .
785 is indeed, the maximum angulardistortion possible.From page 12, we recall that the rate of horizontal distortion will increasewithout bound at polar areas; this phenomenon can also be shown using22he concept of Tissot’s indicatrix. By generalizing the lengths of OP and P M in figure 12 for all possible longitudinal and latitudinal coordinates,we can easily calculate the length of OM , which is the hypotenuse of thetriangle. Let OP = λ φ and M P = φ , where λ and φ are the differencesin longitudinal/latitudinal coordinates and − π ≤ λ ≤ π and − π ≤ φ ≤ π .According to Pythagoras’ rule, | OM | = (cid:112) φ + λ sec φ .However, the distance OM is distorted on the map projection: The scalefactors are not taken into account in our model (page 7, figure 3). Therefore,the projection of OP - which is OP (cid:48) - is equal to λ . Subsequently, the projectedline OM (cid:48) will be equal to (cid:112) λ + φ in the 2D plane.As mentioned in page 18, the maximum distortion in distance is equal to || OM | − | OM (cid:48) || = (cid:112) φ + λ sec φ − (cid:112) λ + φ . Examining the expressionwithin the range − π ≤ φ ≤ π , we notice that when φ = 0, (cid:112) φ + λ sec φ − (cid:112) λ + φ = 0; the distance on the equator - our standard parallel - is per-fectly preserved. Additionally, when φ → ± π , (cid:112) φ + sec φ − (cid:112) φ → ∞ ,implying that the distortion in distance is infinitely large in polar areas.In conclusion, this section proves that flexion arises due to a doubtfulassumption: The projection assumes all parallels to be homogeneous straightlines equal in length. Unlike meridians, not all parallels are equal in length,as figure 8 demonstrates. To make all parallels equal in length, points on theEarth are projected onto the surface of the cylinder as figure 3 illustrates.Thus, as proven on page 10, horizontal distances are distorted by a factorof 1cos φ . This operation causes artificial bending, due to which the shapeof the indicatrices change. As shown in pages 16-18, the bending of Tissot’sindicatrices will determine the extent of distortions in distance (values of (cid:112) φ + sec φ − (cid:112) φ when − π ≤ φ ≤ π ). Moreover, the errors in distanceare dependent only on one variable - φ , and cannot be alleviated by changingother variables. Therefore, as the distance distortion in the horizontal axisexists and cannot be canceled out by other bending, distance cannot becompletely preserved. 23 MPλ φ φ (cid:112) φ + λ sec φ O P (cid:48) M (cid:48) λ φ (cid:112) λ + φ Figure 15: A generalized illustration of the lengths of the sides OP , M P and OM in figure 12. On the projection, the projected side OM (cid:48) forms anotherrelationship with sides OP (cid:48) and M (cid:48) P (cid:48) , from which the distortion in distancecan be demonstrated. Trigonometrical analysis
In this section, we will further examine the relationship between the ”straightline” distance between two points on an equidistant cylindrical projection andthe real distance on the Earth’s surface predominately using trigonometryand sinusoidal identities.First, let sphere with a radius of 1 be in a xyz-coordinate system, asshown in figure 16. Arbitrarily selected points P and P lie on the surfaceof the sphere. If an equidistant cylindrical projection is generated from thesphere, the distance between P and P , would be (cid:112) ( λ − λ ) + ( φ − φ ) ,according to the Pythagoras’ formula. (See figure 7)However, we can observe that the de facto shortest path connecting P and P is not a line but a curve along the surface of the sphere, since it is im-possible to travel through the Earth. Therefore, we cannot purely examinethe straight-line Euclidean distance which we calculate using the formulasof equidistant cylindrical projection. We can convert the Cartesian coordi-nates into spherical coordinates , which can be used to treat the routetaken from P to P as a curve. Using the spherical coordinate system, wewant to compare the calculated ”straight-line” distance between two pointson equidistant cylindrical projection to the real distance between those twopoints on the surface of the Earth. Paul Dawkins,
Calculus III: Spherical coordinates , retrieved June 2019. http://tutorial.math.lamar.edu/Classes/CalcIII/SphericalCoords.aspx yz r = 1P P Figure 16: a unit sphere in xyz-coordinate system with arbitrary points P and P on its surface. It can be understood that the shortest distancefromP to P is a curvature along the surface of the sphere. Spherical coordinates
We can convert the Cartesian coordinates x , y and z into spherical coor-dinate notation.In figure 17, let (cid:126)OP be a vector in 3D plane and the Euclidean distance | (cid:126)OP | = R. It can be interpreted that the x , y and z components of (cid:126)OP are (cid:126)Ox , (cid:126)Oy and (cid:126)QP respectively. Let the x , y and z coordinates of point P and Q be ( x , y , z ) and ( x , y ,
0) respectively. The spherical coordinate systemis able to represent those Cartesian coordinates in terms of angles λ and φ .As figure 15 states, ∆ OP Q , ∆ Ox Q and ∆ Oy Q are all right triangles.Hence, trigonometrical identities can be used to solve for the magnitudes ofthe components of vector (cid:126)OP , which are (cid:126)Ox , (cid:126)Oy and (cid:126)P Q .To solve for | (cid:126)Ox | , we notice that | (cid:126)Ox | = cos λ | (cid:126)OQ | , where | (cid:126)OQ | =cos φ | (cid:126)OP | . By substituting | (cid:126)OP | = R and simplifying the expression, | (cid:126)Ox | = cos λ | (cid:126)OQ | = cos λ cos φ | (cid:126)OP | = R cos λ cos φ. University of Utah,
Cylindrical and spherical coordinates , retrieved June 2019. yz Pλ φROx y z Q Figure 17: An illustration of spherical coordinates:Since both ∆ Ox Q and ∆ Oy Q are right triangles, share a common side OQ , and sides Oy (cid:107) Qx and Ox (cid:107) y Q , it seems that ∆ Ox Q and Oy Q aresimilar triangles. Hence, because of alternating angles, (cid:54) OQy = λ . Thus,the other components, | (cid:126)Oy | and | (cid:126)P Q | , can be calculated: | (cid:126)Oy | = sin λ | (cid:126)OQ | = R sin λ cos φ | (cid:126)P Q | = R sin φ In conclusion, it seems that the spherical coordinates are for all pointswith Cartesian coordinates ( x, y, z ) can be written as x = R cos λ cos φy = R sin λ cos φz = R sin φ Using the concept of spherical coordinates, the distance between pointsP and P can be calculated using the longitude λ and latitude φ . First, we26ant to calculate the straight-line distance between P and P (Cf. figure7) by treating the two points as projections onto a unit sphere. Using thePythagorean theorem, the straight-line distance between P and P in theunit sphere can be deduced: d = ( x − x ) + ( y − y ) + ( z − z ) . By converging the variables x , y and z into spherical coordinates we get d = ( x − x ) + ( y − y ) + ( z − z ) = R [(cos λ cos φ − cos λ cos φ ) +(sin λ cos φ − sin λ cos φ ) +(sin φ − sin φ ) ]= R [cos λ cos φ − λ cos λ cos φ cos φ + cos λ cos φ + sin λ cos φ − λ sin λ cos φ cos φ + sin λ cos φ + sin φ − φ sin φ + sin φ . ]Recall that sin a + cos a = 1 and cos( a − b ) = cos a cos b + sin a sin b .Thus, rearranging and simplifying the terms results us d = R [cos λ cos φ + sin λ cos φ + cos λ cos φ +sin λ cos φ + sin φ + sin φ − λ cos λ cos φ cos φ − λ sin λ cos φ cos φ − φ sin φ ]= R [cos φ (cos λ + sin λ ) + cos φ (cos λ + sin λ ) + sin φ + sin φ − φ φ cos ( λ − λ ) − φ sin φ ]= R [cos φ +sin φ +cos φ +sin φ − φ φ cos( λ − λ ) − φ sin φ ]= R (2 − φ cos φ cos( λ − λ ) − φ sin φ ) . Therefore, by examining the sphere in spherical coordinates, it is deter-mined that the straight-line distance between two points on the surface of asphere is d = (cid:112) R (2 − φ cos φ cos( λ − λ ) − φ sin φ )= R (cid:112) − φ cos φ cos( λ − λ ) − φ sin φ , where R is the radius of the Earth. Kansas State University,
Distance between points on the Earth’s surface , retrievedJune 2019. elationship between straight-line distance and real distance Next, we want to examine the relationship between the real surface dis-tance between two points on a sphere and the straight-line distance that weobtained (see spherical coordinates ). In figure 18, we form a model con-taining both the straight-line distance denoted as d , and the real distance,denoted as D , on the surface of the sphere between points P and P . Theradius of the sphere is assumed to be R . The angle bisector of α is perpen-dicular to line d and bisects d , because in the triangle ∆P P the angle (cid:54) α is isosceles.From the figure, we observe that the arc length - which is the real distancebetween P and P is D , is equal to Rα . We may link d and D together if weare able to calculate α . Using trigonometric identities, we are able to write α with respect to d , the ”straight-line” distance between P and P . First,we see that sin α = d R . By using double-angle identities,sin α = sin(2 · α α α α = 2 · d R = dR and cos α = (cid:113) − sin α = (cid:113) − ( d R ) , wecan simplify sin α assin α = dR · (cid:114) − ( d R ) = dR (cid:114) − d R = dR (cid:114) R − d R = d R √ R − d . Thus, as D = Rα, we conclude that D = Rα = R arcsin( d R √ R − d ) . Hence, there exists a relationship between the ”straight-line” Euclideandistance between two arbitrary points and the real distance between thosetwo points along the surface of the Earth. We can now use this relationshipto calculate the real distance between the points P = (24.3 ◦ E, 23.4 ◦ N) ≈ (0.424,0.408) and P = (39.2 ◦ W, 3.67 ◦ S) ≈ ( − . , . and P is calculated using theexpression d = R (2 − φ cos φ cos( λ − λ ) − φ sin φ ). Thus,28 P d Dα α RR d Figure 18: A sketch of points P and P , their ”straight-line” distance d andtheir real distance D . d P ,P = R (cid:112) − φ cos φ cos( λ − λ ) − φ sin φ = 6371 · (cid:112) (2 − .
408 cos( − . . − ( − . − .
408 sin( − . ≈ · √ − . − . ≈ . D = R arcsin( d R √ R − d )= 6371 · arcsin( 7091 . · √ · − . ) ≈ · . ≈ in page 15 ( d = 7502)gives us a percentage error of − · . . The percentage error in distance calculated using equidistant cylindricalprojection formulas is 2 .
28% (see page 10), which is significantly greaterthan using the formula D = R arcsin( d R √ R − d ) to calculate the samedistance. Looking back to the assumptions made when creating the model, itseems that a key feature of the projection is based on a fallacy: the straight-line distance between two arbitrary points on the projection (figure 7) isassumed to always correspond to the real distance between those two pointson Earth. There are many examples that disprove the assumption, which asshown below.As demonstrated by the calculations on page 13, the real distance be-tween P and P is shorter than what the projection claims. Now, we are Movable type scripts,
Calculate distance, bearing and more between latitude/longitudepoints
Example 1.
Consider arbitrarily selected points A = (41 . ◦ W, 9 . ◦ S) ≈ ( − . , − . . ◦ E, 21 . ◦ N) ≈ (0 . , . d A,B = (cid:112) ( Rλ − Rλ ) + ( Rφ − Rφ ) = 10495 . However, by using the formula calculated using spherical coordinates, the”straight-line distance” between points A and B are found to be d = R (cid:112) − φ cos φ cos( λ − λ ) − φ sin φ = 9214and the real distance separating A and B along the Earth is D = R arcsin( d R √ R − d ) = 9715 < . Example 2.
Consider another set of randomly selected points C =(170 . ◦ E, 65 . ◦ N) = (2.98,1.13) and D = (152 . ◦ W, 64 . ◦ N) = (-2.67,1.13).According to equidistant cylindrical projection, d C,D = (cid:112) ( Rλ − Rλ ) + ( Rφ − Rφ ) = 35760Once again, by examining the distance using the formula deduced usingspherical coordinates, we get d = R (cid:112) − φ cos φ cos( λ − λ ) − φ sin φ = 1689 . and the real distance between points C and D is D = R arcsin( d R √ R − d ) = 1694 << . From the two examples above, we can deduce that equidistant cylindricalprojection does not have the ability to perfectly represent the actual distancebetween any two points on the surface of the Earth. The projection is able torepresent a straight-line route connecting two points as seen in figures 5 and6, but the length of the route may deviate from what the actual shortest routeis. Therefore, it can be said that the formula D = R arcsin( d R √ R − d )30s able to generate a more accurate result for the real distance between P and P than the formulas of the equidistant cylindrical projections. Hence, itcan be concluded that equidistant cylindrical projections fails to demonstratethe actual route between two arbitrary points on the surface of the Earthas curves, thus distances become less accurately preserved in equidistantcylindrical projections.In conclusion, the straight-line distance which equidistant cylindrical pro-jection illustrates differs from the real distance on the surface of the sphere;as a result, we can deduce that equidistant cylindrical projection is unable toperfectly represent the geodesic distance between two points: the projectionmust contain distortion in distance. 31 onclusion From this research, we can conclude that the equidistant cylindrical pro-jection cannot fully preserve distance between two arbitrary points on thesurface of the Earth. We have successfully found out how can distortions indistance be detected, and to what extent does the projected distance varyfrom the real distance. Finally, this research demonstrates that the distor-tion in distance is minimum at the equator and the maximum at polar areasin equidistant cylindrical projection. Thus, the aim of this work is achieved.It is undeniable that the projection has some distance-preserving quali-ties. However, the design of the projection has major flows that cause distor-tion in distance. By examining the distance between two arbitrary points onthe modeled projection using different methods, such as the Tissot’s indica-trix and the spherical coordinates, it is deduced that the projection containsa relevant amount of distortion in distance.In sections
Geometrical analysis and
Trigonometrical analysis , the degreeof distance distortion in the equidistant cylindrical projection is shown fromdifferent aspects. Based on the result of this research, it can be said thatthere are two main questionable assumptions made during the map-makingphase:1. All parallels are assumed to have the same length.2. The ”straight-line” route between two points is assumed to be the short-est possible route between those two points on the surface of the Earth.As explored in this research, those two assumptions are directly respon-sible for the inaccurate illustration of distance between two points on theprojection. Additionally, there are other assumptions made in the makingof equidistant cylindrical projections, such that the Earth is assumed to beperfectly spherical and the radius of the Earth is exactly 6371 kilometers.Further investigations can be conducted to determine the effect of differ-ent assumptions on the distance-preserving ability of equidistant cylindricalprojections. 32 ppendix
Glossary • Map projection: A representation of the properties of the sphericalEarth on a flat surface. • Cylindrical projection: A projection that is formed by centering asphere into a right cylinder and individual points on the sphere aretransferred onto the side surface of the cylinder. The end product is arectangular sheet, parallels and meridians intersect each other in rightangles. • Equidistant cylindrical projection: A type of cylindrical projection thatis meant to describe distances between any two arbitrary points, whichshould correspond to their actual distance on the Earth. • Meridian: Any great circle of the earth that vertically passes throughthe poles and any given point on the earth’s surface. • Standard parallel: Any great circle of the earth that is horizontallyparallel to the equator. • Scale factor: A value that corresponds to the distance of a specific routetaken to connect two points on a map projection in respect of the realdistance of those two points on the Earth. • Distortion: Any angular, areal or distant error that is present in mapprojections. • Geodesic: Denotation of the shortest possible distance between twopoints on a sphere or other curved surface. • Tissot’s indicatrix: An infinitely small circle drawn on the surface ofEarth. The circle’s shape is distorted once the Earth is projected ontoa flat sheet. • Flexion: Artificial bending of structures.33 eneral model for equidistant cylindrical projection
Note: the formulas listed in page 10 are those that are used in this research,but they are not the general formulas for equidistant cylindrical projections;the central meridian and standard parallel are selected to be specific values.According to Snyder (1987), the generalized formulas for any equidistantcylindrical projections are y = Rφx = R ( λ − λ )cos φ h = 1 k = cos φ cos φ − π ≤ φ ≤ π , − π ≤ λ ≤ π Where • λ is the selected central meridian • φ is the selected standard parallel.The other variables are explained in page 11.In Snyder’s model, the values of k - which causes distortion in distancein the first place - and x - which, due to k , suffers from artificial bending -are written to be dependent on the central meridian λ and standard parallel φ , which are arbitrarily selected. John P. Snyder (1987),
Map projection - A working manual he Haversine formula In an earlier subsection, the shortest route between two points on the surfaceof the sphere is examined as a curve along the surface of the Earth. This dis-tance can also be denoted using the Haversine formula , which is definedas haver sin α = sin α − cos α , from which we determine cos α = 1 − α, where α is the angle in figure 18.In this section, the real distance between two arbitrary points on thesurface of the Earth will be further examined using the Haversine formula,which is already briefly discussed in an earlier section. Using the expressioncos α = 1 − α , we can convert the relationship of the ”straight-line”distance between two points and the actual distance between them on theEarth.From page 24, we recall that d = R (2 − φ cos φ cos( λ − λ ) − φ sin φ ) . Substituting cos λ − λ = 1 − λ − λ , the expres-sion can be rewritten as d = R (2 − φ cos φ cos( λ − λ ) − φ sin φ )= R (2 − φ cos φ [1 − λ − λ )] − φ sin φ = R (2 − φ cos φ − φ cos φ haver sin( λ − λ ) − φ sin φ )= R (2 − φ − φ ) + 4 cos φ cos φ haver sin( λ − λ )= R (4 · − cos( φ − φ )2 + 4 cos φ cos φ haver sin( λ − λ )= R (4haver sin( λ − λ ) + 4 cos φ cos φ haver sin( λ − λ ) . Kansas State University,
Distance between points on the Earth’s surface , retrievedJune 2019.
35y dividing both sides of the equation by 4 R , it seems that( d R ) = haver sin α = haver sin( φ − φ ) + cos φ cos φ haver sin( λ − λ ) . Additionally, since d = 4 R haver sin α and d R √ R − d = (cid:114) d R · (4 R − d ) = (cid:114) d R − d R , it can be deduced that d R √ R − d = (cid:114) R haver sin αR − · R haver sin α R = (cid:112) α − α = 2 (cid:112) haver sin α − α. The results retrieved above can be effectively utilized to digitally gener-ate routes between two points on the surface of an object with eccentricity(sphere, ellipsoid etc.). For example, some mapping algorithms, such as
Mov-able Type Script , use the Haversine formula to accurately calculate the dis-tance between any two points on the surface of the Earth using the shortestpossible route between them. Therefore, the Haversine formula has potentialof contributing tremendous real-life application value, hence revamping thetechniques used in the map-making industry. Movable type scripts,
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The mathematical tourist: snapshots of modern math-ematics
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Flexion and Skewnessin Map Projections of the Earth
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The Effects of Spatial Reference Systemson the Predictive Accuracy of Spatial Interpolation Methods
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Combining world map projec-tions
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Applications used
21. L A TEX22. Geogebra,
23. Google maps24. Derivative calculator,
25. Caliper - Mapping Software, GIS, and Transportation Software26. Movable type scripts,
Calculate distance, bearing and more betweenlatitude/longitude points