Table of flight distance calculations struck me
Table of Contents?Table of Contents 2Introduction 3Aim 3Relevant Topics 3Method and calculations 4Table 1: To show the cartesian coordinates that will be used throughout this investigation. 4Fig 1: to demonstrate the measure in which the degrees relate to the axis and angles on earth’s sphere 4Converting Latitude and Longitude to Cartesian Coordinates 5Fig 2: to serve as visual representation of the 5action of three axis components to locate a point. 5Fig.
3: To show the use of axes and angles in order to convert latitude and longitude into Cartesian coordinates 6Converting my points into Cartesian Coordinates 6Equations + Calculations 7Calculation 1: Moscow (a) – Barcelona (b) (with explanation) 7Fig. 4: to demonstrate the relationship between the determined angle and the distance on the surface. 8Calculation 2: Barcelona (b) – London Luton (c) 9Analysis of results 9Table 2: to show and compare results with actual distance. 9Figures 5 & 6: To demonstrate the actual distance on a map and compare with theoretical results 10Conclusion 11Further Research 12Bibliography 12?Introduction In learning about vectors in class, I have begun to wonder about the multiple applications of vectors in three-dimensional geometry, specifically in the geometry of three dimensional space, for instance in measuring distances on a curved surface by combining pre-existing knowledge of spherical geometry.
Vectors introduced a new means of measurement which exposed multiple different possibilities. It occured to me that this method could be used by mapping software in order to calculate distances for practical things, such as flight paths. To investigate the maths behind of flight distance calculations struck me as I was flying back home, to Barcelona, from school in the UK. The distance in kilometers, 1190, was displayed on the flight status screens, and I realized that this seemingly very complicated measurement could be achieved by using the information covered in the SL Maths syllabus. In this investigation I will attempt to use vectors to calculate the distances between my hometown airport, Moscow Sheremetyevo, El Prat (Barcelona), and London Luton, using Latitude and Longitude as coordinates. Aim To calculate distances on the earth using vectors and euclidean geometry from specific cartesian coordinates.
This will require me to convert the latitude and longitude to cartesian coordinates which could then me used as vectors. These vectors should theoretically create an angle, which could then be used to determine the distance on the curved surface of a sphere using the earth’s circumference. Because the earth isn’t perfectly spheroid, the accuracy of the findings should fall within an error of 0.3% of the actual distance. Relevant Topics Vectors, Mapping software, Great circle theorem,Haversines. Method and calculationsTable 1: To show the cartesian coordinates that will be used throughout this investigation.
Moscow, Sheremetyevo AirportBarcelona, El PratLondon Luton55.9736° N, 37.4125° E41.
3120° N, 2.0959° E51.8763° N, -0.3717° W All coordinates taken from data provided by Google Maps.Fig 1: to demonstrate the measure in which the degrees relate to the axis and angles on earth’s sphereFigure 1 – Original image from http://modernsurvivalblog.com/wp-content/uploads/2013/09/definition-of-latitude-longitude.
jpg , Annotated by author using Microsoft Power Point. The latitude is measured as degrees from the equator, which in this scenario will be denoted as , and seen as the height on the z-axis. The longitude, as seen on the right, is moves along the x or y axis and is denoted as .Converting Latitude and Longitude to Cartesian Coordinates The equation for latitude and longitude to be converted into Cartesian Coordinates is x=cossiny=sinsinz=cosWith being the longitude and, denoting the colatitude. As latitude, (), is equal to (90°-), sin(90-)=cos()Therefore, the final equation used to convert the latitude and longitude into a position vector points isx=coscosy=sincosz=sinFig 2: to serve as visual representation of the action of three axis components to locate a point. Figure 2 – created by author using Microsoft Word.
In order to convert the latitude and longitude into cartesian coordinates, the earth must be re-imagined as a three dimensional space. For the sake of the conversion, the x-axis will be on 0° N, 0° E, the y-axis will run perpendicular at 0° N, 90° E, and the z-axis will go through 90° N. Fig. 3: To show the use of axes and angles in order to convert latitude and longitude into Cartesian coordinatesFigure 3 – created by author using Microsoft word. – Latitude- LongitudeConverting my points into Cartesian CoordinatesTherefore, the three coordinate points will be derived using; A: Moscow, Sheremetyevo Airport: = 55.9736 = 37.
4125B: Barcelona, El Prat: = 41.3120= 2.0959C: London Luton: = 51.8763= -0.3717 **due to the fact that Luton is west of the meridian, the longitude is negative Equations + CalculationsCalculation 1: Moscow (a) – Barcelona (b) (with explanation)To get the distance on the sphere, I will first find the angle between the two points on the surface. The formula for vector angles: ab|a||b|= cos()The dot product gives the cosine angle between the two vectors multiplied by the magnitude of both vectors. The dot product can be determined using the equation below; ab=(x?x?)+(y?y?)+(z?z?)The absolute values of both |a| and |b| would both be 1, since both are unit vectors, therefore in this case the dot product gives simply the cosine angle of the two vectors. Therefore the angle in between vector A and vector B is (0.
333621+0.009338+0.547125=0.892673cos()=ab|a||b|cos()=0.892673cos-1=26.7888 Fig. 4: to demonstrate the relationship between the determined angle and the distance on the surface. Figure 4 – created by author using Microsoft Word Now in order to get the distance on the earth’s circumference (c); c=2rc=2(6371.
008) c=40030.226.7888360=x40030.2x = 2978.78Therefore the overall distance between Moscow Sheremetyevo and Barcelona El Prat is 2978.78 km. Calculation 2: Barcelona (b) – London Luton (c) (0.
7506230.617248)+(0.027470-0.004010)+(0.6601580.786679) = 0.463321-0.000110+0.
519332 =0.982543 cos-1(0.982543)=10.721510.7215360 40030.2 = 1192.
18Analysis of resultsTable 2: to show and compare results with actual distance. Calculated DistanceActual Distance*Error:Moscow to Barcelona, 2978.783004.250.84%Barcelona to London Luton,1192.181191.850.03%Table 2 – comparing actual distance and theoretical calculations to determine error.
*actual distance calculated using Google Maps % Error = |(experimental-Theoretical)(Theoretical)|100 Figures 5 & 6: To demonstrate the actual distance on a map and compare with theoretical resultsFigure annotated by author in Microsoft Word, original image courtesy of Google MapsFigure annotated by author in Microsoft Word, original image courtesy of Google MapsConclusion The two distances discovered are accurate, with a percentage error less than 1%. This means that the method can be deemed accurate in order to calculate distances on a sphere using vectors. The average error between the two results is 0.44%, close to the predicted error of 0.30% due to the fact that the earth isn’t perfectly spheroid and in fact slightly resembles an ellipsis. Furthermore, another possible explanation for the error could be that, in limiting myself to only 6 significant figures, the accuracy of the final calculation was compromised. This error becomes more and more prominent with the amount of such steps in the calculation, which is why I attempted to minimize the amount of steps by simplifying and combining some equations.
This also helped minimize possibilities of errors. For example modifying the initial equation to accommodate latitude instead of colatitude, or when performing the step which required me to find the absolute value. I realized that removing the radius component from the initial equation x=rcossiny=rsinsinz=rcos would allow me to perform the calculations using a unit vector to begin with, which meant taking out an extra step such as finding its absolute value to satisfy the equation cos()=ab|a||b| One way to make my calculations more accurate would have been to use an existing number for the circumference of the earth rather than obtaining a value by using the radius. For example, in calculation 1, if the circumference used was 40,070, the final result would be 2981.74, which means the error of my calculations would be decreased by 22%. In this investigation I wanted to see if I could measure distances on the earth’s surface using my knowledge of vectors and angles. However, I have also discovered that there are other ways of both using my calculations or completely different methods in order to come to the same conclusion. For instance, the Havershine’s formula, which gives the error of 0.
3% rather than the 0.5% in my calculations, also due to the slightly elliptical form of the earth. In fact, when calculated with the Haversine formula, the Moscow-Barcelona distance was predicted to be 3015 km, a 0.35% error from the Google Maps Data, which is more accurate than that which was acquired by using vectors. Because this investigation concerned itself also with airline travel and navigation, it is important to note that another way of determining distance between two cartesian coordinates is using rhumb lines. Rhumb lines are easier to navigate on a sphere, as the azimuth – or bearing – will remain the same, due to crossing meridians at exactly the same angle.
Rhumb line paths, however, are longer than great circle paths. The distance for Moscow-Barcelona data points using rhumb lines would be 3043, 0.9% longer than the Haversine formula distance. Overall, this investigation allowed me to explore a topic and find an explanation for something which interested me, using the maths which I have already covered in class, which both gave me a sense of the vast applications of my course, as well as introduce me to some new, more complex concepts. Further ResearchThe methods used in these calculations could be repeated to measure the distance on any kind of curved surface.
Furthermore, if the vectors are not in fact assumed to be of the same magnitude (due to earth being treated as a sphere in this instance), the distance on a curved surface could be measured more accurately. Calculations such as these have also been the basis of many computerized technologies which use similar methods to produce answers. This way, the conclusions drawn from this particular investigation could lead to a better understanding of the computational process, as well as other methods which measure distances in a similar way.
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