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Unformatted text preview: LSGI 3341 Geodetic Control Assignment 1 Chong Wai Ming 06701301d October 6 th , 2007 1a) Prove: 2 1 e b a ′ + = b e ε = ' Eq.1 2 / 1 2 2 ) ( b a = ε Eq.2 Substitute Eq.2 into Eq.1 b b a e 2 / 1 2 2 ) ( ' = 2 2 2 2 ' b a e b = 2 2 2 2 ' b e b a + = ) 1 ' ( 2 2 2 + = e b a ) 1 ' ( 2 2 + = e b a ∴ 2 ' 1 e b a + = 1b) Prove: 2 1 e a b = a e ε = Eq.3 Substitute Eq.2 into Eq.3 a b a e 2 / 1 2 2 ) ( = 2 / 1 2 2 ) ( b a ae = 2 2 2 2 b a e a = 2 2 2 2 e a a b = ) 1 ( 2 2 2 e a b = ∴ 2 1 e a b = 1c) Prove: 2 ' 1 ' e e e + = From Eq.1 b e ε = ' ' e b ε = Eq.4 From Eq.3 a e ε = e a ε = Eq.5 Substitute Eq.2 into Eq.1 b b a e 2 / 1 2 2 ) ( ' = 2 2 2 2 ' b b a e = 1 ' 2 2 2 = b a e Substitute Eq.4 and Eq.5 1 ) ' ( ) ( ' 2 2 2 = e e e ε ε 2 2 2 ) ' ( ) ( 1 ' e e e ε ε = + 2 2 2 ) ' ( ) ( 1 ' ε ε e e e = + 2 2 ) ' ( 1 ' e e e = + Eq.6 e e e ' 1 ' 2 = + ' 1 ' 2 e e e = + ∴ 2 ' 1 ' e e e + = 1d) Prove: 2 1 ' e e e = Substitute Eq.2 into Eq.1 a b a e 2 / 1 2 2 ) ( = Substitute Eq.4 and Eq.5 e e e e ε ε ε 2 / 1 2 2 ) ) ' ( ) (( = 2 2 2 2 ) ( ) ' ( ) ( e e e e ε ε ε = 2 2 2 ) ( ) ' ( 1 e e e ε ε = 2 2 2 1 ) ( ) ' ( e e e = ε ε 2 2 1 ) ' ( e e e = ε ε 2 2 2 1 ' e e e = ) 1 ( ' 2 2 2 e e e = ∴ 2 1 ' e e e = 2) Prove: a) The equipotential surface of the earth’s gravity fields are not parallel to each other. The potential different between p to p’ equal the potential different between q to q’. dWq dWp = , Hence the work done from p to p’ equal to the work done from q to q’ Assume the point mass at point p and q are the same. q q p p dS g m dS g m = Since the gravity is not uniform along the equipotential surface q p g g ≠ q p dS dS ≠ ∴ Since the distance between p to p’ and q to q’ are not equal, therefore the equipotential surface A and B are not parallel as well. b) The gravity vector is vertical to the equipotential surface. Both A and B are points on an equipotential surface, thus have the same potential energy by the gravity potential, Wa = Wb = Constance. Therefore, the work done from point A to point B on the same equipotential is equal to zero. AB dW = dS F AB • = dS g • = p q p’ A B θ Equipotential surface A Equipotential surface B q’ AB dW θ COS dS g * * = Since g and dS ≠ Therefore = θ COS 90 = θ The gravity vector is perpendicular to the equipotential surface at any point on the surface. 3) geoid undulation It is the height different from the geoid surface to the reference ellipsoid surface. It can be determined by using ellipsoid height minus and the orthometric height, which can also be find out easily by using GPS to determine the orthometric height and minus the local height datum referenced by the mean sea level....
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 Spring '07
 KADY
 Geodesy, Equator Plane

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