Class 25: Even More Corrections and Survey

Networks Project Planning

21 April 2008

Deadlines

• Reading assignments (2) due 30 April 2008– Sample of what I expect is posted to class

wiki page

• Extra credit due 23 April 2008

• No lab assignments/homework will be accepted after today.

Textbook

• EM 1110-1-1003, Chapters 4-10

• Geodesy text, Chapter 11

Does distance matter in GPS?

Article available on NGS web site

http://www.ngs.noaa.gov/CORS/Articles/

Plumb line• Indicates the average direction of gravity

between the point of suspension and the plumb bob.

• It is perpendicular to every each level surface it intersects.– Equipotential surfaces are NOT parallel

• Effected by mass of the earth– Oceans are a mass deficiency– Mountains a mass concentration

Astro-Geodetic

Deflection of the VerticalAngle between the plumb line and a normal to the ellipsoid is the deflection of the vertical.

City of Corpus Christi, TX

Ft. Davis, TX

Denali, AK

Computation of Deflection

• Has a North South component ξ (xi) and East West component η (eta)

• Computed with respect to astronomical azimuth (A)– ξ = θ cos A and η = θ sin A

• Positive ξ indicates astronomic latitude will fall north of the geodetic latitude

• Positive η indicates astronomic longitude will fall east of geodetic longitude

http://www.exploretheline.com/

Mason-Dixon LineMonument Latitude () Xi (ξ) Lat. Seconds (Φ)

50 N39-43-15.36789 5.06 20.42

54 N39-43-14.50754 5.94 20.45

83 N39-43-11.68275 7.00 18.68

91 N39-43-20.87172 7.91 19.78

Near Warfordsburg

N39-43-20.87172 -2.15 18.72

MARPENN N39-43-21.29309 -1.83 19.46

= Φ - ξ

Distance Corrections

• Refraction – may be overcome by reciprocal vertical/zenith angles– Otherwise:

k is coefficient of atmospheric refraction

R is mean earth radius

hA is ellipsoid height of instrument station

Curvature Correction

• Accounts for the fact that plumb lines are not parallel at different locations on earth’s surface.

Geodetic Distance

• Computed from level distance (LD) – LD * [ R / (R + h) ]– LD * [ R / (R + H + N) ]

• Use R = 6,371,000 m– H = Orthometric Height– h = ellipsoid height– N = geoid height

Planning GPS Control Surveys

• Your plan will be developed in accordance with your client’s goals and the relevant standards.

• EM 1110-1-1003 describes work for the USACE. – These differ from FGCS standards and

specifications.

Issues in Any Plan• Project Datum

– Mostly will be NAD 83 (but which version?)

• Intended accuracy– Horizontal, Vertical or both

• Which height system (ellipsoidal or NAVD 88)• More accuracy = More $$$$

• Monumentation

• Equipment needed (and available)

Project Planning/Standards and Specifications

• See class page for links to relevant documents.

Meeting standards

• We must have redundant observations in order to evaluate their precision.

• We must have ties to fixed control (more than one) in order to attach our observations to this framework as well as to verify the relationship of the fixed points.

• When our observations are more accurate than the fixed network, we degrade them to fit the network.

Level Data Reduction and Adjustment

5

Leveling summary

Difference in elevation for section (edge)

Section length kilometers

A to D = 1.978 1.02

A to E = 0.732 0.97

C to D = 0.988 1.11

B to E = 0.420 1.07

E to D = 1.258 0.89

Populating the design matrix

• Height difference (i to j) is equal to the observed difference and its residual/variance.

• Line 1. D – A = 1.978 + variance

• Line 2. E – A = 0.732 + v

• Line 3. D – C = 0.988 + v

• Line 4. E – B = 0.420 + v

• Line 5. D – E = 1.258 + v

Design Matrix (Free)Nodes -> A B C D E

Section 1 -1 0 0 1 0

Section 2 -1 0 0 0 1

Section 3 0 0 -1 1 0

Section 4 0 -1 0 0 1

Section 5 0 0 0 1 -1

“From” station gets a “-1” “To” station gets a “1”

Note diagonal elements!

Observations/Distances

Section Height Diff. (m)

Distances (KM)

1.978 1.02

0.732 0.97

0.988 1.11

0.420 1.07

1.258 0.89

Account for Known BMs• Review diagram

• Solve for heights where direct connection to unknown

5

Updated Matrices

D E

1 0

0 1

1 0

0 1

1 -1

Height D.E Sum

10.021 +1.978 11.999

10.021 +0.732 10.753

11.002 +0.988 11.990

10.321 +0.420 10.741

+1.258 1.258

Rearrange to put BMs on right side

• D = 11.999 + v

• E = 10.753 + v

• D = 11.990 + v

• E = 10.741 + v

• D – E = 1.258 + v

Why not just mean heights?D E

11.999 10.753

11.990 10.741

Mean: 11.9945 Mean: 10.747

How do we account for the difference in elevation from E to D? How accurate is our result?

Matlab Un-weighted Result

Calculate heights

Calculate residuals

Weight Matrix• We assume that error will accumulate as a

function of distance.

• Weights assigned as 1/distKM.

Section Length Weight

A to D 1.02 0.98

A to E 0.97 1.031

C to D 1.11 0.901

B to E 1.07 0.935

E to D 0.89 1.124

Weight Matrix

• P used for weight matrix (also W)

• Diagonal Matrix assumes no correlations

• Applied to both design matrix and observation matrix

• X = ATPA-1*ATPL

D = 11.9976

E = 10.7445

Differences in ResultsSimple Mean Least Squares Un-

weightedLeast Squares Weighted

D = 11.994 5 11.997 1 11.997 7

E = 10.747 0 10.744 4 10.744 5

Residual Residual (weighted)

-0.0019 -0.0014

-0.0086 -0.0085

0.0071 0.0076

0.0034 0.0035

-0.0052 -0.0049