2010-2 - (Agm-10v LF) Page 1 av 11 NTNU Geomatics Exam:...

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(Agm-10v – LF) Page 1 av 11 NTNU Geomatics Exam: TBA4231 APPLIED GEOMATICS Date: 6.12.2010 Time: 0900 – 1300 Solution, version 12.1.2011. For the theory questions check also the curriculum. For some of the questions the answers are in keyword form (some more text is required to obtain the top score). Grades, included the project assignment counting 30% of the final grade: 3A-3B-13C-3D-1E. Question no 1: Total station etc 1.1 (a) Explain what a bearing is. See the textbook, chapter 6 (2010). The horizontal angle from the northern part of a parallel of the north axis going through a point (G2 on the figure) to the vector from the same point to another point (T1). Clockwise. Numerical value between 0 and 400 gon (b) Verify with your own calculations that the bearing from G2 to T1 is φ G2-T1 = 247,6998 gon. 12 21 1655,90 tan 1780,10 G GT G EE NN ϕ  ==  −−  => ϕ G2-T1 = +47,6998 gon From the signs of E and N (and looking at the figure) we find the bearing to be in the 3.rd quadrant: ϕ G2-T1 = 247,6998 gon (c) Calculate the bearing from T1 to T2, φ T1-T2 , by using all the measurements in point T1 towards the known control points. NB: Two of the bearings, those to G1 and G3, were given in the exam text. Some of the students did not see that, and they did some unnecessary calculations. Remember to ”turn” the bearing from G2 to T1 which is calculated in (b),:to the bearing from T1 to G2 : 200 47,6998 TG ϕϕ =− = , see the table below. Bearing to the Gi point Measured hor.angle, β Bearing to T2 Weight To G1: φ TØ1-G1 = 77,5524 gon β 2 = 239,035 gon 316,5874 gon 2 or 1 To G2: φ TØ1-G2 = 47,6998 gon β 1 = 268,891 gon 316,5908 gon 2 or 1 To G3: φ TØ1-G3 = 117,8638 gon β 3 = 198,727 gon 316,5908 gon 4 or 2 The weights are relative numbers, the weights 1, 1 and 2 are used here. The observation of unit weight is therefore an angle measured 2 times. 3 suggestions in the table above, where one of the horizontal angles ( β 3 ) is measured 4 times compared with 2 times for the other 2 β angles. Therefore: Weighted mean must be used to calculate the mean value of the measurements, where measurement/suggestion no 3 have the double weight compared with the 2 others. () (316,5908 1 316,5874 1 316,5908 2) / 1 1 2 316,5900 TT =⋅ + + + + = gon (Arithmetic mean: (316,5874 316,5908 316,5908) / 3 316,5897 =+ + = gon) G2-T1 G2 T1
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(Agm-10v – LF) Page 2 av 11 1.2 (a) Calculate the standard deviation of the bearing in (1.1.c) from point T1 to T2, s T1-T2 . Starting point is the 3 calculated suggestions of the bearing T1-T2 in question 1.2.c. Alternatively you can start with the given standard deviation, not correct but you get some score. The weighted mean value is 316,58996 gon. We have to calculate the corrections, v i , of the 3 measurements (the difference between mean value and the actual value): 11 21 2 2 3 316,58996 316,5874 0,00255 gon 316,58996 316,5908 0,00087 gon 316,58996 316,5908 0,00084 gon GG TT v v v ϕϕ −− =−= = + =− = = p i v i = 0,00 mgon => OK The number of observations: n = 3 The number of unknowns: e = 1 (the bearing T1-T2) The standard deviation of the unit weight, defined in 1.1.c, the weights p i are 1, 1 and 2: 22 2 0 1 0,87 1 2,55 1 2 0,84 2,08 mgon 31 pvv ne σ ⋅+ ⋅⋅ + == = The standard deviation of the weighted mean of 3 suggestions/measurements, the formula can be
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2010-2 - (Agm-10v LF) Page 1 av 11 NTNU Geomatics Exam:...

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