I would then want to know why you would want to go through LSA when 3mm falls within the 1st order limit of 8mm?
If you don't need to realistically characterize the errors of your work, there's no reason to run a statistically valid adjustment.?ÿ However, most geodetic leveling is used as the foundation for future work, and not having well-characterized errors of the original marks leaves future users without a basis to evaluate the accuracy of their own results.
What we do these days is to run levels with our electronic levels, dump the info into the program and push the button to get a lease squares adjustment, if we see anything odd we end up re-running it.?ÿ
But if I were in the field with an automatic level and the given parameters I would do the run, then a checkbook reduction of this data on a sheet in the book, in about 5 minutes I come up with a value of 11.997 for D and 10.744 for E.?ÿ
I ran A-B, adjusted E to 10.747 from 10.753
A-C, adjusted D to 11.995 from 11.999
B-C, checked in .001M, eyeball adjustment for D to 11.998 and no adjustment for E-10.741
mean 6mm at E and 3mm at D. A simple mean at that point.?ÿ
Clearly there are some issues, there is a bit over 1cm floating in there, either with the fixed Bench marks or one of the runs, for the real world you would need some error budgets established to figure out if that's an issue, for this exercise I would say the runs are good to go.?ÿ
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However, most geodetic leveling is used as the foundation for future work, and not having well-characterized errors of the original marks leaves future users without a basis to evaluate the accuracy of their own results.
How would future users have access to these LSA notes? All you are given for a certain benchmark is the elevation of the point. I doubt it very much if the ordinary users would be able to dig up these LSA notes & I would further doubt if those future users would even care to look at those LSA notes.
All you are given for a certain benchmark is the elevation of the point.
JS0768 *********************************************************************** JS0768 DESIGNATION - P 1200 JS0768 PID - JS0768 JS0768 STATE/COUNTY- CA/PLACER JS0768 COUNTRY - US JS0768 USGS QUAD - ROCKLIN (1981) JS0768 JS0768 *CURRENT SURVEY CONTROL JS0768 ______________________________________________________________________ JS0768* NAD 83(2011) POSITION- 38 49 43.37074(N) 121 11 25.70485(W) ADJUSTED JS0768* NAD 83(2011) ELLIP HT- 98.206 (meters) (06/27/12) ADJUSTED JS0768* NAD 83(2011) EPOCH - 2010.00 JS0768* NAVD 88 ORTHO HEIGHT - 127.675 (meters) 418.88 (feet) ADJUSTED JS0768 ______________________________________________________________________ JS0768 GEOID HEIGHT - -29.478 (meters) GEOID18 JS0768 NAD 83(2011) X - -2,576,657.191 (meters) COMP JS0768 NAD 83(2011) Y - -4,256,163.618 (meters) COMP JS0768 NAD 83(2011) Z - 3,977,583.218 (meters) COMP JS0768 LAPLACE CORR - 10.70 (seconds) DEFLEC18 JS0768 DYNAMIC HEIGHT - 127.588 (meters) 418.60 (feet) COMP JS0768 MODELED GRAVITY - 979,947.3 (mgal) NAVD 88 JS0768 JS0768 VERT ORDER - FIRST CLASS I JS0768 JS0768 Network accuracy estimates per FGDC Geospatial Positioning Accuracy JS0768 Standards: JS0768 FGDC (95% conf, cm) Standard deviation (cm) CorrNE JS0768 Horiz Ellip SD_N SD_E SD_h (unitless) JS0768 ------------------------------------------------------------------- JS0768 NETWORK 0.40 0.67 0.19 0.12 0.34 -0.06240227 JS0768 ------------------------------------------------------------------- JS0768 Click here for local accuracies and other accuracy information.
Those are post processed GPS information. We were talking about differential leveling not GPS derived orthometric elevation values.
JS0768.The orthometric height was determined by differential leveling and JS0768.adjusted by the NATIONAL GEODETIC SURVEY JS0768.in June 1991.
Those are post processed GPS information. We were talking about differential leveling not GPS derived orthometric elevation values.
Oops, I didn't notice that network error estimates weren't included in leveled heights.?ÿ My goof.
P 1200 is one I remember in particular because in one bluebooked project I used it to validate the ortho height of another station (G 1200) in the same level line.?ÿ In a previous adjustment the G 1200 height had been published based on GPS obs, and that bothered me.?ÿ Once we validated (withing 2.5 cm) that G 1200 and P 1200 agreed, G 1200 got published back at its leveled height.
I thought I understood what you did but came up with different numbers. I also note that you had to reuse section difference in your approach. I am down with a cold and set up a little script to do the maths. BTW, I congratulate you on your noticing that the known points do not fit together at a high level of accuracy. For my attempt to follow your description see below:
My original post was an attempt to address the utility of the least squares method using a straightforward example. Leveling was the easiest problem I could come up with. While a previous thread dealt with GPS and least squares, that implementation is rather complex.
My work was always intended for public use and therefore the idea of "good enough" or merely giving someone a sheet of paper with values on it was insufficient. I fully accept that for in-house work and quick-and-dirty surveys; to some the rigor seems unnecessary. The fact is, a leveling adjustment following the method of least squares provides a unique, supportable solution. It is not hard to write the code to do work. It is hard to incorporate all the modeling necessary to refine the observations, convert to geopotential numbers, etc. But that's beyond the scope of this problem.
FWIW, I copy below a heavily annotated script (using Matlab) I wrote to solve the network. I intentionally did not incorporate reading observation and other data from a file. The "%" is used to start a comment. I did not "prettify" the display units.
<<<< Matlab script follows >>>>
See: "Linear Algebra,Geodesy and GPS" by Strang and Borre
Published by Wellesley-Cambridge Press, 1997. ISBN 0-9614088-6-3
Data is from Example 8.1 pages: 286-288.
Text does not include portion related to the determination of the accuracy of the determinations. (Q matrix)
Known NAVD88 heights (m) from text (vertical datum inferred): A = 10.021 B = 10.321 C = 11.002
known_H = [ 10.021; 10.321; 11.002 ];
Difference in elevation (de) and section lengths (from text)
Section de (m) section length (km) 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
Create observation matrix containing de (m) and section length (km)
two column five row matrix
obs = [ 1.978 1.02; 0.732 0.97; 0.988 1.11; 0.420 1.07; 1.258 0.89 ];
Stating our problem into words.
To put it into words, our equation for the first observation is: Known height A plus (+) the difference in elevation de(1) equals (=) the new height at D. More concisely: A + de(1) = D.
N.B.The values taking adjustment are the differences in elevation.
We therefore isolate the de by moving the starting BM to the right side of the equals sign. The result: de(1) = D - A
The case without a direct connection to a known height
We treat this case similarly: E + de(5) = D
Isolating the de we move E to the right side: de(5) = D - E
Using the information provided above we create an incidence matrix (Incid)
We place under the column for the point involved in the section a one(1)
or a negative one(-1).
e.g. Our first observation, row one(1) is de(1) = D - A which becomes: -1 0 0 1 0 (note the -1 for A and +1 for D.
The annotated incidence matrix follows:
Sec A B C D E
---------------------------------
1 -1 0 0 1 0 (from A to D)
2 -1 0 0 0 1 (from A to E)
3 0 0 -1 1 0 (from C to D)
4 0 -1 0 0 1 (from B to E)
5 0 0 0 1 -1 (from E to D)
The Incid matrix without the annotations
Incid = [ -1 0 0 1 0 ;
-1 0 0 0 1 ;
0 0 -1 1 0 ;
0 -1 0 0 1 ;
0 0 0 1 -1 ];
Reduce Incid matrix to remove columns at known heights (A,B, and C)
This becomes our design matrix A
It consists of only two columns representing the two unknowns.
D E tied BM + de = preliminary H
A = [ 1 0; % 10.021 + 1.978 = 11.999
0 1; % 10.021 + 0.732 = 10.753
1 0; % 11.002 + 0.988 = 11.990
0 1; % 10.321 + 0.420 = 10.741
1 -1 ]; % + 1.258 (Not tied directly to a known BM)
Compute system's degrees of freedom (DF) from A matrix
we have five observations to determine two unknowns DF = 3
As number rows = number of observations and number of columns = number of stations use size() function to report dimensions then subtract them.
[a,b] = size(A); % dimensions of design matrix
DF = a - b; % DF = degrees of freedom
Reduced observation matrix (L) reflecting direct ties to known BMs (see above)
L = [ 11.999;
10.753;
11.990;
10.741;
1.258 ];
Create preliminary weight matrix
W = ones(5,1) ./ obs(:,2); % weight is inverse of section length
P = diag(W); % weights on diagonal of 5 x 5 matrix
Solve for heights at unknowns and residuals
Used inverse function (inv) here due to issues related to my using the latest MacOS and my old version of Matlab.
Should use X = P*A P*L, the left division operator
Solve for unknown heights and observation residuals
X = inv(A'*P*A)*(A'*P*L); % solves for unknown heights
V = A*X - L; % calculates observation residuals
% in sense computed - observed (Strang and Borre use V = L -A*X)
Solution statistics
VAR = (V'*P*V)/DF; % solve for variance of unit weight (VUW)
STDEV = sqrt(VAR); % solve for standard deviation
Q = inv(A'*P*A); % weighted design matrix
VCV = STDEV * Q; % variance-covariance matrix
Since the weights were not correct given the non unitary VUW additional work is necessary. NOT DONE HERE.
To prove the determined X is optimal
add 0.0001 m to heights at unknowns then resolve for residuals
Xtest = X + [ 0.0001; 0.0001 ]; % add a mm to computed heights (in X)
Vtest = A * Xtest - L; % variances with test X
diff_in_Vsqrd = sum(V .* V) - sum(Vtest .* Vtest);
Outputs
sprintf('nSolved heights for unknownsnD = %5.4f mtE = %5.4f m', X(1),X(2))
sprintf('nObservation and residualsn')
disp([L V])
sprintf('nVariance of Unit Weight: %5.4f', VAR )
sprintf('nStandard deviation: %5.4f', STDEV )
sprintf('nWe add a tenth millimeter to solved heights for testing.')
disp(Xtest)
sprintf('nComputed residuals using these test heights.')
disp( [L Vtest])
sprintf('nThe principle of least squares is that the correct answer is the one where the sum of squares of the errors is a minimum.')
sprintf('nWe prove our computed answer is the best by comparing the sum of variances.')
sumV_sqrd = V'*V;
sumVtest_sqrd = Vtest' * Vtest;
Solved heights for unknowns (displayed to four decimal places)
D = 11.9976 m E = 10.7445 m
Observation and residuals
11.999 -0.00138986387474738
10.753 -0.00850076737886774
11.99 0.00761013612525296
10.741 0.00349923262113272
1.258 -0.00488909649587921
Variance of Unit Weight: 0.0001
Standard deviation: 0.0075
We add a tenth millimeter to solved heights for testing.
11.9977101361253
10.7445992326211
Computed residuals using these test heights.
11.999 -0.00128986387474761
10.753 -0.00840076737886797
11.99 0.00771013612525273
10.741 0.00359923262113249
1.258 -0.00488909649587921
The principle of least squares is that the correct answer is the one where the sum of squares of the errors is a minimum.
We prove our computed answer is the best by comparing the two sums of variances.
if sumV_sqrd < sumVtest_sqrd
sprintf('nSum of V^2 squared is smaller than heights altered by 1 mm.')
else
sprintf('nSum Vtest V^2 is smaller than using computed heights.n')
end
Sum of V^2 squared is smaller than heights altered by 1 mm.
end of file
GeeOddMike ?ÿI have been following your ?ÿpost?ÿ and reading the comments and would like to add
some comments myself. #1, its to bad you had an error in labeling one of the points. It seems to have
taken away form your post (a little). #2, Your very first question was "How would you derive heights
for the unknown points..."; The very first thing I thought of was "Manual of Leveling Computation
and Adjustment by Rappleye , Special Pub. 240 USC&GS 1948.
No place in your post did you ask for something other than a Least Squares Adj.; this came later.
When you did ask for something other than LSA then I thought of the "DELL ?ÿADJ."
The Dell method is?ÿ in the references of "Surveying Theory and Practice by Davis, Foote and Kelley" ?ÿpage
212 of the 5th edition. A better place to look is in "Surveying for Civil Engineers by Prof. Kissam, 2nd edition
chapter 6, starting on page 156 (a worked example is given). Two other places you can find H.G. Dell method
are both in ASCE; Vol. 61, no. 4, April 1935 (part 1) and same journal Vol. 101, pp. 834-856, 1936. (part 2)
I have had Dr. Strang book since it was published and find it very good.
?ÿ
JOHN NOLTON
Thanks for the references. While “ancient” Rappleye’s manual is still useful.
My efforts to get those instinctively opposed to considering least squares to at least recognize its value remain unsuccessful.
I too like the Strang/Borre text. Prof Strang’s linear algebra course is available for viewing at: https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/ He is a good communicator. I had the pleasure of attending one of his lectures.
Given the inattention paid to the monumented vertical network, these monuments have diminished in numbers and not updated to reflect geophysical and other changes. This hinders efforts like the GPS on BM program as an earlier thread on efforts to resolve possible geoid modeling errors in the Scranton, PA area. It seems that many considered work on the nationwide leveling network to have been finished with NAVD88.
As for leveling adjustments, the image file below describes how one goes about determining the accuracy of a leveling project. Note that this FGCS document has much more information on the data acquisition and observation phases. See: https://www.ngs.noaa.gov/FGCS/tech_pub/Fgcsvert.v41.specs.pdf
This exercise, in my mind, is designed for a hand calculation of a least squares adjustment of the data in a classroom setting. The pedagogical value is in seeing how observations are turned into equations, how redundant observations are incorporated, and how results are obtained by applying maxima/minima principles from the calculus. Duplicating the answers from a black box removes some of the mystery from the black box.
So let's go through the exercise and see if there's anything to be learned.
Here are the observation equations, written in the form of the least-squares adjustment to be made to each difference in elevation. Note that the reciprocals of the lengths of the segments are used to weight the differences. This adjusts a shorter length more than a longer one.
We only want values for two items, the heights of D and E, yet we have five equations. Let's look for redundancies; equations that can be derived from one of the other equations. Here's how we do that.
1. We want the height of E to be the same value whether we compute it from B or from A.
2. We want the height of D to be the same value whether we compute it from C or from A.
3. We want the value of the change in height between D and E to be the same whether we compute it from D to E or from E to D. (Note: This will not be true if simple averages of AE and BE, and AD and CD are used for the heights of E and D, respectively.)
Here are the condition equations that describe these conditions:
Substituting the condition equations into the appropriate observation equations gives these new equations:
Since it is the sum of the differences squared that we want to minimize, we need an equation that sums up the differences squared:
Now for the calculus. The principle is that, when a function is minimized, its first derivative is 0. In the case of multi-variable functions (our sum function has two variables) each of the partial derivatives is 0 when the function is minimized. Here are the partial derivatives:
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We now have two equations in two unknowns, but they need to be simplified:
?ÿ
Solving these equations and substituting appropriately produces the following values:
x1 = 1.976610
x2 = 0.723499
x3 = 0.99561
x4 = 0.423499
x5 = 1.253111
D = 11.998
E = 10.744
These are the best estimates for D and E because they resolve the three readily-identifiable ambiguities in a non-arbitrary way. As to the accuracy, that requires a separate analysis and statistical assumptions that may not be met in the problem.
Well, that was a lot of work. I learned a great deal from doing it and I hope you did, too.
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E= 10.740
D (F) = 11.991
I did not use any software, I did not use least squares. I just used common sense and a calculator. It took about 4 minutes. If I average to two separate runs from E to D, I get 1.252. Calculating from B to C using 1.252 for the middle run I get .003 mm error at C. I distributed that error .001 mm for each of the three runs.
How Accurate??ÿ?ÿ?ÿ I don't know, I didn't run it but for the type of work I do, it is plenty good. If I was setting new permanent monuments, I would rerun it.
Prove your answer is the best possible??ÿ?ÿ?ÿ I can't do that. but it is good enough for me and that is all that counts.
James
I did not use any software, I did not use least squares.
That works great when you have a handful of turns and a few marks to deal with.?ÿ Try it with tens of km of leveling and dozens of marks and the error propagation is much more difficult to properly model.
