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No Redundant Observations – No Chi Sq. Test
Posted by Alya on May 30, 2019 at 7:10 amhello,
I am alya and I am a student. I have a question regarding Star*net. It’s my first time hands on this software and I have no experience how to interpret the output produced, so recently I was given a task to run my traversing observation using this software and do I have to change any setting in the project option or I should I just stick with the default setting? I hope someone would mind helping me on this problem and share opinions on how to know whether it passes upper bound or level pass
thank you so much,
Norman_Oklahoma replied 5 years, 4 months ago 4 Members · 10 Replies 
10 Replies

Hi,
Starnet is a software that uses least square to adjust all your data to what parameters you enter. Least Squares works when you have redundant observations, meaning having more than one measurements to the station or even just closing your traverse by sitting on the last point and shooting your first point.
It seems you ran your traverse by angles and no distances measured or key into your data.
What is shown in your summary is that you had performed 6 angles observation but did not close onto your first point nor did you have something to hold true like a control point or bearing. using the C or B function. Without any of this inputs, the software is unable to calculate and tell you what is mathematically true as it is unable to calculate.
If you have measured distances in your traverse maybe you could enter your angles and distance using the M or T observation types. If you have a control point that you started with or at least hold 2 station bearings as fixed. It will give Starrnet something to work with

Sireath is quite right. You appear to have only angle data. At a minimum you also need distances, a starting coordinate, and a bearing basis. If you post your data file we might be able to offer more detailed advise.

thank you so much sireath on your explanation, I kinda get it but I will try to see first what I can to correct this.but what I understand from your explanation is that I’m lack of measurement data such as distance and bearing, but for my case I don’t have those data because I have angle reading from one stn to another stn plus the coordinate of 5 control point, so how can I fix this? I am totally lost

@normanoklahoma, thanks so much i would be happy to do that. I’ll attach the input data that I have key in here.

Your data file looks, in large part, correct as to form. But without distances the angle only data is insufficient to compute positions for the instrument points. Therefore you need to add “! !” to the end of your instrument point coordinates, to “fix” their positions. Then you have both the instrument points and the tied points fixed in place, from which the angles between them can be computed. Then the angular observation data becomes redundant, and the software can evaluate how well the angles fit.
I had a little trouble with the data until I remembered that most of the world outside of the US uses the coordinate order Easting / Northing (xy), instead of the Northing / Easting that is common in the US.

Hey there Mark, thanks so much for dropping by and answering the question. but I got a question to ask, based on what I understand, the purpose we assign “! !” is to show the coordinates are fixed in position. therefore the reason why I didn’t put “! !” on my CP station because it is not fixed in position plus the coordinate value that I put there is the only assumed coordinate value to make it easier to compute coordinates. the task given to me was meant to find the Coordinate of the CP1,CP2 and CP3 based on the angle measure at the station of unknown coordinate from known stations which is Z1,Z2, Z3, Z4, Z5?
this is the manual that i refer to perform the task

Since the advent of EDMs, particularly reflectorless EDMs, angle only resections are not very common. If you can access a point to set a target you can read a distance easily enough. So an angle only resection is mostly an academic exercise. There are guys here who are more attuned to these theoretical mathematical problems. Perhaps one of those will chime in.
Generally, you need a minimum of 3 angle only observations to known points to perform a resection. You’ll notice that the StarNet manual example has 6. If you want to get by with only 2, you will need to feed in distances as well. You have a network of observations, so maybe it can be done. But it’s beyond my ability.
StarNet will solve for CP1, CP2, & CP3 with those points “free” (ie/ with asterisks instead of exclamation points). But it reports no redundant observations, therefore no chisquare testing. Notably you have no observation to given point Z4. Have you left something out of your data?

Norman thank you so much for sharing your knowledge, basically during performing the observation we were reminded not to take the distance measurement instead we were only being asked to take the angle observation, I think I understand better now after reading all response given and maybe we were asked to analyze the result produced and understand why it can be like that. for your information there are 5 control points however I only use 4 of them excluding Z4, that is why there is no observation done on Z4.
If I may ask, let say if I have 3 control point with two of them fixed value, and the other one has tolerance, how should I write control point with tolerance e.g CP3 (5249.105 ?Ã± 0.005, 3338.575 ?Ã± 0.005) inside the data text file, should I write the tolerance value or how? compared to the fixed control point which we have to put the exclamation mark.

.ORDER EN
#KNOWN POINT(Z)
#
C Z1 5234.160 3340.430 0.005 0.005
C Z2 5253.702 3338.575 0.005 0.005
C Z3 5271.359 3347.541 0.005 0.005
C Z4 5249.105 3378.303 0.005 0.005
C Z5 5244.494 3368.823 0.005 0.005
#
#SENARAI UNKNOWN NAK CARI
#C CP1 5229.805072 3349.840092 **
#C CP2 5272.662487 3366.418797 **
#C CP3 5287.161369 3347.443319 **
C CP1 5229 3349 * *
C CP2 5272 3366 * *
C CP3 5287 3347 * *
#AT CP1
A CP1Z5Z3 552603
A CP1Z3Z2 220421
#AT CP2
A CP2Z3Z2 301812
A CP2Z2Z1 214340
#AT CP3
A CP3Z1Z5 340908
A CP3Z1Z3 075331
#E 
If we change the seed coordinates of CP1, CP2, & CP3 to something close to the known correct answer, but not the exact correct answer, we seethat StarNet does solve for the correct coordinates. Note that under these conditions of zero redundancy those seed coordinates can’t be too far off before the solution will fail to converge.
In the example above I’ve also changed the standard error of Z1 thru Z5 to 0.005 0.005. Since the errors are all the same, and the observations are near perfect, that change doesn’t affect your result at all.
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