NGS quotes locations on data sheets rounded to the nearest 100 thousandth of a second. A second is 1/3600th of a degree. A 100 thousandth of a second is 1/3600/100000 = 2.778 * 10^(-9). So the equivalent of NGS accuracy would seem to be 9 or 10 decimal places in degrees. As an aside, that 5th decimal place in NGS seconds is about 3/10ths of a millimeter.
However, since the NGS numbers are rounded to 5 decimal places in seconds, the sixth decimal place is also considered. Considering this, NGS accuracy is 5 millionths of a second (remember rounding rules, over or under 5 in the last place.) Thus NGS accuracy is really 5/3600/1000000 = 1.389 * 10^(-9).
The additional analysis doesn't really change the conclusion. If you're matching NGS data sheet accuracy, then you need 10 decimal places in degrees to be sure.
Looked at another way, a second of arc is about 101 feet (purists will point out that this number is variable from equator to pole, but it's close enough for what we're doing here.) So, a minute of arc is about 6060 feet (compare to a nautical mile) and a degree is 363,600 feet. To get accuracy better than 1/10th of a foot, then, you need 7 decimal places in degrees.
The chart says that 7 decimal places is Waldo on a page, but it's better than that. Overall, though, the chart is a pretty good guide.
In the end, it's professional judgement, based on mathematical truisms and the situation at hand that is the determining factor.
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XKCD ?????ÿ
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Historic boundaries and conservation efforts.
It depends on your equipment and how you use it, but surveyors, as opposed to geodesists, don't take much stock in significant figures. Every bearing is reported to the nearest second and every distance to the nearest 1/100 of a foot, equipment capabilities, methodology, and usefulness be dammed.?ÿ
".. but it's close enough for what we're doing here.) So, a minute of arc is about 6060 feet..."
Im not being critical.?ÿ Just pointing out the " it depends" perspective always is in play behind the scenes....kind of like gravity.?ÿ Always there, instantaneously acting across all distances and time
Fun discussion!
latitude: 1second=100ft; 0.0001"=0.01'- approximately?ÿ
longitude: varies depending on latitude
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The distance along a meridian does change as the Latitude increases from the equator to the pole.
This can be found in most basic geodesy books. In Frederick Pearson II work for the Naval Surface Weapons Center, report
# TR-3624 of 1977?ÿ "Map Projections Equations" on page 80 he gives the following (for WGS-72 ellipsoid)
geodetic latitude(in degrees)..../ distance in meters for 1' (one minute). Just divide by 60 and you have for 1 second of lat.
25.00 Deg....1846.224 meters
30.00 Deg....1847.549
35.00 Deg. ...1849.018
40.00?ÿ Deg....1850.585
45.00?ÿ Deg....1852.204
50.00?ÿ Deg. ...1853.825
55.00?ÿ ?ÿDeg....1855.399
60.00?ÿ ?ÿDeg....1856.878 meters
I can calculate this for the GRS80 ellipsoid if you want. I did this for Dr. Herb Stoughton for a project he was doing for the island of
Guam.
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JOHN NOLTON
I use 100 or 101 ft for lat, 100 cos(lat) for lon unless I want to be picky, and then I go to INVERSE.
Sorry Bill93 but your equation of?ÿ 100 x cos (lat) for long. is NOT CORRECT.
Use 101.45 and even better would be?ÿ 101.57?ÿ The 101.57 would give you a very close value at a Lat. of 35 Degrees.
and fit between Lat. of 20 Deg. to 50 Deg. not bad.?ÿ The equation to use for a difference in Longitude (being at the same latitude)
is?ÿ d= a (diff. of long. in radians) cos(lat.)/sqrt(1-e^2 sin^2(lat.)?ÿ very simple. example using this equation at a Lat. of 35 Deg. you
get a d=25.3578 meters = 83.1947 us ft. and using a corrected equ. of yours of 101.57 you get 83.2013 usft. or a difference?ÿ
of 0.0066 usft.?ÿ Using your incorrect equ. you would have a difference of 1.2795 usft. Now at 50 Deg. Lat. the error will be 0.0515 usft
and at 20 Deg. Lat. you will be in error of 0.0752 usft?ÿ(that is if you use 101.57 times the cos) and not 100 cos.
If you do a geodetic inverse you will get the same value as using the above equation I gave you?ÿAS it should be.
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JOHN NOLTON
Note: the 101.57 is a mean of the bottom of the correct equation above for 20 Deg. and 50 Deg.
the 101.45 comes from the first part of the equation which is [a(difference Long. in radians)]. 1 sec in radians = 0.0000048481368110953599+
times a (6378137 for GRS 80)=101.45 usft approx.
e^2 is the square of the 1st eccentricity of the ellipsoid and for GRS80 = 0.0066943800229034157495749485862893 +
round where you like.
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JOHN NOLTON
I had Dr Stoughton for my first surveying class.?ÿ He was tough, curmudgeonly, and knows his stuff.?ÿ I learned more in one class than I thought was possible. He was also fair and acknowledged your effort when you did the work to show the right answers.
I didn't claim it was "correct", but have found it useful at times for estimation. And rather than look up e^2 and do the more complicated calculation, inverse is handier.
@bill93?ÿ It looks like you don't read very well also. Your equation of [100 cos(Lat.)] = long. distance is very crude.
As I have pointed out if you would use?ÿ101.57 cos(Lat.) will give you a much better approximate solution
between latitude of 20 Deg. to 50 Deg.?ÿ Remembering 101.57 should not be to difficult??
Remember:?ÿ
A little learning is a dang'rous Thing;
Drink deep, or taste not the Pierian Spring:
There shallow Draughts intoxicate the Brain,
And drinking largely sobers us again.
by Alexander Pope (1688-1744)
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JOHN NOLTON
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I make a point of reading XKCD and Pearls Before Swine regularly. As most people have little sense of the magnitudes represented by differences in geographic coordinates, many have adopted a reasonable approximation like 1? = 30 meters (based on (2*pi*r)/(360*60*60) ) and divide this value by tens as they work their way to decimal parts of a second.
Of course an accurate calculation requires more effort. As Mr Nolton and others note, the difference in length varies as a function of latitude.
This reminds me of the famous debate among geodesists in the early 18th century about whether the Earth was oblate or prolate. This led to the famous Cassini surveys in Lapland and what is now Ecuador.
See: https://www.fig.net/resources/proceedings/fig_proceedings/fig_2002/Hs4/HS4_smith.pdf
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As for the matter of NGS data sheets, five-decimal place seconds are used for geodetic coordinates as that magnitude is required to obtain three-decimal place metric values. Anyone imagining these five-decimal place seconds represent the accuracy of the point should reconsider.
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For those unfamiliar with the NGS interactive INVERSE tool (I used the 3D version), I provide screen captures below:
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Surely NGS doesn't just invent the numbers in the digits beyond, say, the third decimal place. On the other hand, maintaining sub-millimeter accuracy of the location of a point on the earth's surface for years seems pretty far-fetched as well.
The accuracy question is settled on the data sheet. Position accuracy is described by error ellipses. With a given probability, the point is likely to lie somewhere within the ellipse. I look at the error ellipse as the modern-day incarnation of the old mariner's "cocked hat." A snippet from AI2383 is illustrative:
Rules of thumb have guided artists, physicists, astronomers, statisticians, and surveyors for millenia. We need to know that our particular one serves its purpose well, but I think we're trying to put too fine a point on some pretty good ones.
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Once you have an error ellipse or +/- you can give all the digits you want without implying false accuracy.
123.45678901234 +/-0.02@ 95% or 987.654321 with sd=0.01?ÿ is unambiguous.
NGS didn't complicate the production of data sheets by trying to represent accuracy with the number of digits.
Bill93 better captured what I meant than I did.
The following NGS datasheet extracts show a third-order intersection station (a courthouse spire) and a Cooperative CORS station. I can assure you that their positional accuracies differ greatly. Both are shown to the same five-decimal place seconds. Accuracy is provided by other information on the datasheet.?ÿ
JW1494?ÿ DESIGNATION -?ÿ CUMBERLAND ALLEGANY CO CTHSE
JW1494?ÿ PID ?ÿ ?ÿ ?ÿ ?ÿ -?ÿ JW1494
JW1494?ÿ STATE/COUNTY-?ÿ MD/ALLEGANY
JW1494?ÿ COUNTRY ?ÿ ?ÿ -?ÿ US
JW1494?ÿ USGS QUAD ?ÿ -?ÿ CUMBERLAND (2016)
JW1494
JW1494 ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ *CURRENT SURVEY CONTROL
JW1494?ÿ ______________________________________________________________________
JW1494* NAD 83(1995) POSITION- 39 39 02.95827(N) 078 46 00.38401(W) ?ÿ ADJUSTED
JW1494?ÿ ______________________________________________________________________
JW1494?ÿ GEOID HEIGHT?ÿ ?ÿ -?ÿ ?ÿ ?ÿ ?ÿ -32.707 (meters) ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ GEOID18
JW1494?ÿ LAPLACE CORR?ÿ ?ÿ - ?ÿ ?ÿ ?ÿ ?ÿ -5.63?ÿ (seconds)?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ DEFLEC18
JW1494?ÿ HORZ ORDER?ÿ ?ÿ ?ÿ -?ÿ THIRD
JW1494
JW1494.The horizontal coordinates were established by classical geodetic methods
JW1494.and adjusted by the National Geodetic Survey in April 1998.
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DF6303?ÿ DESIGNATION -?ÿ CUMBERLAND COOP CORS ARP
DF6303?ÿ CORS_ID ?ÿ ?ÿ -?ÿ ALLG
DF6303?ÿ PID ?ÿ ?ÿ ?ÿ ?ÿ -?ÿ DF6303
DF6303?ÿ STATE/COUNTY-?ÿ MD/ALLEGANY
DF6303?ÿ COUNTRY ?ÿ ?ÿ -?ÿ US
DF6303?ÿ USGS QUAD ?ÿ -?ÿ EVITTS CREEK (2016)
DF6303
DF6303 ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ *CURRENT SURVEY CONTROL
DF6303?ÿ ______________________________________________________________________
DF6303* NAD 83(2011) POSITION- 39 39 13.88871(N) 078 43 50.50328(W) ?ÿ ADJUSTED
DF6303* NAD 83(2011) ELLIP HT- ?ÿ 175.744 (meters)?ÿ ?ÿ ?ÿ ?ÿ (06/??/19) ?ÿ ADJUSTED
DF6303* NAD 83(2011) EPOCH ?ÿ -?ÿ 2010.00
DF6303?ÿ ______________________________________________________________________
DF6303?ÿ GEOID HEIGHT?ÿ ?ÿ -?ÿ ?ÿ ?ÿ ?ÿ -32.788 (meters) ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ GEOID18
DF6303?ÿ NAD 83(2011) X?ÿ -?ÿ ?ÿ 960,973.378 (meters) ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ COMP
DF6303?ÿ NAD 83(2011) Y?ÿ - -4,822,646.747 (meters) ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ COMP
DF6303?ÿ NAD 83(2011) Z?ÿ -?ÿ 4,048,582.199 (meters) ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ COMP
DF6303
DF6303?ÿ Network accuracy estimates per FGDC Geospatial Positioning Accuracy
DF6303?ÿ Standards:?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ
DF6303 ?ÿ ?ÿ ?ÿ ?ÿ FGDC (95% conf, cm) ?ÿ ?ÿ Standard deviation (cm) ?ÿ ?ÿ CorrNE
DF6303?ÿ ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ Horiz?ÿ Ellip ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ SD_N ?ÿ SD_E ?ÿ SD_h?ÿ ?ÿ ?ÿ (unitless)
DF6303?ÿ -------------------------------------------------------------------
DF6303?ÿ NETWORK?ÿ ?ÿ 0.14 ?ÿ 0.44 ?ÿ ?ÿ ?ÿ ?ÿ ?ÿ 0.06 ?ÿ 0.05 ?ÿ 0.23?ÿ ?ÿ ?ÿ 0.00968500
DF6303?ÿ -------------------------------------------------------------------
DF6303
DF6303
DF6303.The coordinates were established by GPS observations
DF6303.and adjusted by the National Geodetic Survey in June 2019.
Agreed. But if you were using NGS Inverse to compute the ellipsoidal distance between DF6303 and JW1494, how many decimal places would you enter from each station? How many decimal places would you include in the computed distance and how would you state the computed distance?
how many decimal places would you enter
I usually compute with all the decimals available.?ÿ The only place for rounding is in the final answer.
Any rounding or truncation is equivalent to adding a small error, which is acceptable if and only if that error is insignificant with respect to the error the number is already expected to have.
The improvement by not rounding inputs is usually insignificant, but it's non-zero and essentially free.