Hi Guys
I tried to replicate the way StarNet inflates angular standard deviations if centering errors are set in the adjustment. I used the formulae from the Starnet help file Appendix B and all works well for HZ Slope etc. but I can't get the following
formula to work:
I = Horizontal Instrument Centering Error
T =Horizontal Target Centering Error
V =Vertical Centering Error at Instrument and at Target
d = Horizontal distance from instrument to target
s = Slope distance from instrument to target
e = Elevation difference from instrument to target
Since slope distances and zenith angles are both “sloped” observations, their total standard errors are each inflated by both horizontal and vertical centering errors.
[tex]Total Zenith StdErr = Sqrt( ZenithStdErr^2 + (e/s)^2 * (I^2 + T^2) + 2*(d/s)^2 * V^2 ) [/tex]
I think the part of the formula where the distance dependent inflation of the angle standard deviation happens is missing. I get a constant ZenithStdErr if I have a constant zenith angle and decreasing slope distance.
Any ideas what is happening here?
Thank You!
PS:
My StarNet results are
I=0.000m
T=0.000m
V=0.005m
ZenithStdErr=6"
SlopeDist____Zenith____StdErr
100.0000___90-00-00.00____15.77
97.5000____89-00-00.00____16.12
95.0000____88-00-00.00____16.47
92.5000____87-00-00.00____16.85
90.0000____86-00-00.00____17.24
87.5000____85-00-00.00____17.66
85.0000____84-00-00.00____18.09
82.5000____83-00-00.00____18.54
80.0000____82-00-00.00____19.02
77.5000____81-00-00.00____19.53
75.0000____80-00-00.00____20.07
72.5000____79-00-00.00____20.64
70.0000____78-00-00.00____21.25
67.5000____77-00-00.00____21.89
65.0000____76-00-00.00____22.58
62.5000____75-00-00.00____23.33
60.0000____74-00-00.00____24.12
57.5000____73-00-00.00____24.99
55.0000____72-00-00.00____25.92
52.5000____71-00-00.00____26.94
50.0000____70-00-00.00____28.06
47.5000____80-00-00.00____30.83
45.0000____80-00-00.00____32.48
42.5000____80-00-00.00____34.33
40.0000____80-00-00.00____36.41
37.5000____80-00-00.00____38.77
35.0000____80-00-00.00____41.48
32.5000____80-00-00.00____44.60
30.0000____80-00-00.00____48.25
27.5000____80-00-00.00____52.57
25.0000____80-00-00.00____57.77
22.5000____80-00-00.00____64.12
20.0000____80-00-00.00____72.07
17.5000____80-00-00.00____82.30
15.0000____80-00-00.00____95.94
12.5000____80-00-00.00___115.06
10.0000____80-00-00.00___143.76
> I think the part of the formula where the distance dependent inflation of the angle standard deviation happens is missing. I get a constant ZenithStdErr if I have a constant zenith angle and decreasing slope distance.
> Any ideas what is happening here?
Basically, you just reduce the vertical centering error to an equivalent standard error of the zenith angle and add that root-sum-of-squares to the instrumental component of the standard error of the zenith angle.
Kent
I tried this before and got the following results:
ZenithStdErr=6"
Hor. Centering Errors=0.000m
Vert. Centering error=0.005m
SlopeDist(m)___Zenith°___STDErr "_____STDErr Starnet "
100____________90__________12______________16
95_____________89__________12______________16
90_____________88__________13______________17
85_____________87__________14______________18
80_____________86__________14______________19
75_____________85__________15______________20
70_____________84__________16______________22
65_____________83__________17______________23
60_____________82__________18______________25
58_____________81__________19______________26
56_____________80__________19______________26
54_____________79__________20______________27
52_____________78__________21______________28
50_____________77__________21______________29
48_____________76__________22______________30
46_____________75__________23______________31
44_____________74__________24______________32
42_____________73__________25______________34
40_____________72__________26______________35
38_____________71__________28______________37
36_____________70__________29______________39
34_____________80__________31______________43
32_____________80__________33______________45
30_____________80__________35______________48
28_____________80__________37______________52
26_____________80__________40______________56
24_____________80__________43______________60
22_____________80__________47______________66
20_____________80__________52______________72
18_____________80__________58______________80
16_____________80__________65______________90
14_____________80__________74______________103
12_____________80__________86______________120
10_____________80__________103_____________144
This difference is way too big for short distances, no idea how StarNet calculates these values.
> I tried this before and got the following results:
> ZenithStdErr=6"
> Hor. Centering Errors=0.000m
> Vert. Centering error=0.005m
>
> SlopeDist(m)___Zenith°___STDErr "_____STDErr Starnet "
> 100____________90__________12______________16
Okay so over 100m at Z=90deg an instrumental uncertainty of 0-00-06 in the zenith angle gives an uncertainty (s.e.) of +/-0.0029m in the height difference. A vertical centering uncertainty (s.e.) of +/-0.005m over 100m is equivalent to an uncertainty of +/-0-00-10 in the height difference.
Taking just the height differences, SQRT(0.0029^2 + 0.005^2)= 0.0058m
arcsin (0.0058/100) = 0-00-12, which checks the value you posted.
But Star*Net is applying the same +/-0.005m uncertainty to both the instrument height and the target height, so the net uncertainty would be, SQRT (0.005^2 + 0.005^2 _ 0.0029^2) = 0.0076m
arcsin (0.0076/100) = 0-00-16, which checks the Star*Net value you posted.
Thank You so much Kent!
When I saw your calculation I realised that I used the centering errors in metres
in the StarNet formula instead of centering error(m) divided by slope distance(m).
I should have spent some time thinking the whole thing through instead of just
staring at my Excel results and not seeing what went wrong.
Dear all,
Is there any error in this formula?
Total Zenith StdErr = Sqrt(ZenithStdErr2 + (e/s)2 * (I2 + T2) + 2*(d/s)2 * V2)
because the result of this equation doesn't make any sense.
Thank you all.
