2009-2 - (Gm-09v Sol) Page 1 of 11 NTNU Geomatics Exam:...

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(Gm-09v – Sol) Page 1 of 11 NTNU Geomatics Exam: TBA4230 GEOMATICS Date: 8.12.2009 Time: 0900 – 1300 Solution, version 28.11.2010 Question no. 1: GPS, adjustment by parameters, etc. Here method 1 is used: The calculations are carried out in the local coordinate system. NB: Only calculations in the ground plane (horizontal plane) are done (no estimations of heights). Results from the estimation in the ground plane ( x , y ): () 2 3,84 mgon pvv == T VPV Figure 1.1 Observations and points The normal equations: 2,145 0,354 1,910 0,450 719372 1,029 0,400 0,740 1450000 2,717 1,581 0 2,219 1450000 C C D D x y x y   =  T XA P F the Q matrix of the estimated coord. 1.1 (a) State the number of excess measurements (degree of freedom) for the determination of the new points C and D on figure 1.1. The number of excess measurements, 5 baselines, 2 unknown points (NB: No calculations of heights): n - e = 5 2 – 2 2 = 6 With 2 known points (A and B) we also can choose to have 2 additional unknowns in the adjustments: Unknown scale and rotation. Normally this is done when GPS baselines calculated in WGS84 are used as observations in an adjustment in another datum: n - e = 5 2 – 2 2 – 2 = 4 (b) The observation with the unit weight is a plane bearing (as GPS is used). Calculate the standard deviation of the unit weight for the adjustment of the network in figure 1.1. Standard deviation of the unit weight, where n - e = 6 is used here, a plus if you use n - e = 4: 0 3,84 0,80 mgon 6 pvv ne σ = (With n – e = 4 we obtain 0,96 mgon) (c) Calculate the standard deviation of the estimated x coordinates of C and D. Calculate the covariance between the x coordinates of C and D. Calculate the covariance between the y coordinates of C and D. Interpret the results. x y B A D C
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(Gm-09v – Sol) Page 2 of 11 We know we can obtain the co-factor matrix of X from the normal equations: = 1 XX QN  , see the values of the elements above (NB: Symmetrical matrix, only the upper triangle is shown). Known connections give the variance/covariance matrix: 2 0 1,373 0,227 1,222 0,288 0,659 0,256 0,474 1,739 1,012 1,420 σ   =⋅ =  XX XX CQ The answers of the questions can also be found directly from the matrix above, or using the definitions directly (on element form): 22 0 0 0,80 2,145 1,373 1,17 mm 0,80 2,717 1,32 mm CC DD q q σσ = = = = = 1,4 and 1,6 mm with n - e = 4 0 0,80 1,910 1,222 CD q = = mm 2 0 0,80 0,740 0,474 q = = YY mm 2 Interpret: Lower standard deviation on C compared to D. Theoretically Ok, as there are 3 baselines starting or ending in point C, and only 2 in point D. The co-variances show correlations, as expected. Large values, mainly caused by the baselines in the triangle B-C-D. The signs are positive (+), as a positive/negative uncertainty in C also will give a positive/negative uncertainty in D, ”they are moving in the same direction”.
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This note was uploaded on 02/03/2012 for the course CIVIL AND TBA5100 taught by Professor Arnfinn during the Fall '11 term at Norwegian Univ. of Science & Technology.

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2009-2 - (Gm-09v Sol) Page 1 of 11 NTNU Geomatics Exam:...

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