Suppose you measured a distance to be 25,897.89 feet using a total station whose 95% uncertainty was 5 mm + 2ppm. How would you report the distance and its uncertainty?
My answer is 25,897.89 +/- 0.0542 feet. The answer in this publication http://www.californiasurveyors.org/members/Exam%20Guide.pdf , where the problem appears on page 322, is 25,897.89 +/- 0.068 feet.
The difference is trivial, to be sure, but the source of the difference may be important.
I interpret the stated uncertainty as two different standard errors. Thus, to calculate their combined effect, I take the square root of the sum of the squares of their individual effects. The source above merely adds the distance uncertainties of the two together.
This source, http://www.dot.ca.gov/hq/row/landsurveys/LSITWorkbook/03.pdf , presents a similar problem on page 16 and accepts two different answers on page 18. It then chooses what I think is the wrong one to compute an error ellipse.
Here is a source I agree with: http://www.wfps.org/files/AMUW09/Handouts/Sunday%20-%20Greenfeld%20(60).pdf . See page 13 for the work. Another is in the May, 2014 issue of Professional Surveyor, where Dr. Charles Ghilani discusses similar uncertainties in GPS measurements. The web copy is incomplete, so we have to look at a print copy to see his full discussion.
So how should the uncertainty be interpreted and presented?
Us older surveyors, who think in terms of 4 cups of coffee in the morning, don't measure the cups. We just drink it.
More or less.
Nate
> Suppose you measured a distance to be 25,897.89 feet using a total station whose 95% uncertainty was 5 mm + 2ppm. How would you report the distance and its uncertainty?
>
> My answer is 25,897.89 +/- 0.0542 feet. The answer in this publication http://www.californiasurveyors.org/members/Exam%20Guide.pdf , where the problem appears on page 322, is 25,897.89 +/- 0.068 feet.
> So how should the uncertainty be interpreted and presented?
Well, if the distance was obtained by direct measurement and the stated uncertainty of the EDM was +/-(5mm + 0.000002 x D) 2-sigma, then I'd say that the uncertainty is +/-(0.016 ft + 0.052 ft.) = +/- 0.068 ft. 2-sigma.
On the other hand, if the nature of the measurement process is such that the scale uncertainty component of 0.000002 x D is independent of some bounded error that has an uncertainty of +/-0.016 ft., then the two errors combine as root sum of squares.
The practical problem would be that one would have to know some facts about the performance of the EDM, such as that the bounded error component of +/-5mm is distance independent, in order to justify treating the two components as independent. Failing that, one is left with the bare statement of uncertainty that does not treat them as completely independent.
Footnote:
Probably the best, most exhaustive treatment of EDM errors that I've seen is given in J.M. Rüeger's "Electronic Distance Measurement". In Section 13.5 - Accuracy Specifications of EDM Instruments, the author states:
[pre]
Most manufacturers state the accuracy of their instruments in the following form
s = +/-(A = Bd), (13.63)
where A in MM, B in ppm, d = distance (in km)
The variance follows as
s^2 = (A + Bd)^2 (13.64)
Most manufacturers consider the Eq. (13.63) as a
standard deviation at 66% confidence level, but use
it as a tolerance in their predelivery testing procedures.
[/pre]
So the first source is consistent with Rueger. Some manufacturers' specs put the two components in parentheses as Rueger did. Perhaps that is a clue to proper interpretation and use.
I wonder where the two-standard-error interpretation came from. Its users are learned practitioners, but they disagree with the only definitive source I've seen.
Thanks for the reference and the enlightenment.
I hear you, Nate. This old teacher has 9 working days left before his retirement begins. Your reflection applies to education, too. I can't precisely measure how much they've learned, but a bunch of PhD's think that they can.
> I wonder where the two-standard-error interpretation came from. Its users are learned practitioners, but they disagree with the only definitive source I've seen.
I'd say that it came from wishfulness if that is how they are choosing to interpret a manufacturer's specification. In the expression s = +/-(A + Bd), the term A is a composite of multiple effects that may include:
- phase resolution of the EDMI
- average effect of short periodic errors (e.g.: cyclic errors)
- average effect of non-linear distance-dependent errors, and
- accuracy of a present addition constant.
The term B is either:
- the range of the typical frequency drift of the main oscillator within the specified range of operating temperatures or
- the maximum error which may be caused by the limited step interval of the "ppm dial".
But the point is that the expression, s = +/-(A + Bd), is a test value quoted by the manufacturer to explicitly describe the standard error of a range measured by its product. So, whatever factors were reflected, the value was chosen to describe actual results.
Math Teacher, did you see that the distance was in feet and the s.d. of the measurement was given in mm + 2ppm.
You might want to change the distance to meters and then calculate the error using the given 5mm + 2ppm. Then convert that back to feet.
It looks like they are trying to make sure a prospective surveyor knows feet and meters and how to convert back and forth. (?)
JOHN NOLTON
Tombstone, AZ
> My answer is 25,897.89 +/- 0.0542 feet. The answer in this publication http://www.californiasurveyors.org/members/Exam%20Guide.pdf , where the problem appears on page 322, is 25,897.89 +/- 0.068 feet.
(25,897.89 X (2/1000000))+0.016 = 0.068
Watch you significant figures; I would argue the California answer down to showing the nearest hundredth... 5mm is greater than 0.01.
95% isn't applied to the value... I am guessing there is assumption that the standard deviation is 2ppm. Guess I could have looked at the link.
I'm sure that the units conversion was part of the problem. In the published solution, 0.016 feet was used as the equivalent of one millimeter, so I used that number, to be consistent.
The stastical question is whether each of the two components is a separate standard error or if the linear combination of the two is the total standard error. Kent's source states the latter while equally prestigious sources use the former with no further discussion.
The example measurement is one of a series of measurements to determine the area of a tract, so the accumulation of the errors is also being illustrated.
That publication has a wealth of information, specifically for California exam-takers, but likely useful to many others.
Right, and that's the published answer. But, if you consider the two error components, 5mm and 2ppm, to each be standard errors, then the total standard error is sqrt( 0.016^2 + (25,897.89/500,000)^2 ) = 0.0542. I left the ten-thousandths place for the benefit of anyone who wanted to duplicate the calculation.
Note that a smaller standard error would accept fewer measurements at any confidence level, so there may be some importance beyond some arcane mathematics. From my viewpoint, I just want to know what the correct assumption is.
> Kent's source states the latter while equally prestigious sources use the former with no further discussion.
It would be a grave mistake to think that the New Jersey Institute of Technology is on a par with the University of New South Wales and in particular that J.M. Rueger's "Electronic Distance Measurement", now into at least the 4th Ed. at Springer-Verlag, isn't in a completely different universe of expertise than a bunch of Star*Net screenshots with annotations scrawled on them.
Readers of this message board will recall postings by one former NJIT faculty member that can most charitably be described as self-referential opinion delivered with great certainty, but contrary to the bulk of the modern surveying literature and thought on the subject of measurements.
Math Teacher, you DO NOT consider each component separately. You might find this in Bomford's book on Geodesy. There is another reference around 1973 that was in Journal of Geophysical Research (I believe). If I have time in the next several days I will look for it IF you really want it.
JOHN NOLTON
Tombstone, AZ.
Thanks, John. It's probably not worth the effort, but it might be helpful.
Part of my problem is the conflicting information from references that should be reliable. Take page 10 of this one http://www.ngs.noaa.gov/PUBS_LIB/EstablishmentofCalibrationBaseLines_final.pdf for example. It's the NGS CBL procedures, revised less than a year ago.
The tolerance for CBLs clearly separates the two and defines the allowable error in terms of added variances. I may be mixing apples with oranges, but one interpretation for CBL measure and a different one for EDM specifications seems like an invitation to confusion and errors.
All I'm after here is a better understanding of how my profession is used by yours. If I offended anyone, I apologize.
Math Teacher you did not offend me nor do I think you offended anyone on the board for wanting to understand how this problem is solved.
I wrote up some information for you but got timed out. Can you go to my profile and send me an e-mail with your phone number. I will then go over everything I put into the message.
Thanks
JOHN NOLTON
Tombstone, AZ.
I didn't see an email address in your profile.
Math Teacher, I checked it and its there. If you don't find it we can work something out.
Thanks
JOHN NOLTON
That evidently is true, confirmed by the reference to Rueger. The NGS guidelines for establishing CBL's specify that the two components be considered as two separate standard errors.
So, manufacturers' specifications and specifications for field measurements seem to be apples and oranges. As is always true, confirming the meaning of the spec before applying the statistics should be the rule.
