Does anyone know what the equation is to reduce the raw data to an elevation?
my occupy point EL= 10.687'
Here is my foresight point
LS,HI5.035,HR4.885
SS,OP11411,FP11413,AR237.21207,ZE93.08328,SD324.908,--TWP
The elevation of the foresight point is supposed to be -6.9722
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I though the equation to reduce it was: Occ EL + HI- Sin(Zenith Angle) x Slope Distance - HR= FS Pt EL
but this is not working for me.?ÿ
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Note*
This is in the San Francisco bay area so that is why there is negative elevations. Thanks!
You need to use Cos rather than Sin when using a Zenith angle.
I calculated an elevation of -6.974 not taking into account curvature and refraction.
You need to use Cos rather than Sin when using a Zenith angle.
I calculated an elevation of -6.974 not taking into account curvature and refraction.
Right on!
Appling curvature & refraction, I get -6.9718.
I get -6.974'
While I am hesitant to involve myself in quibbles over thousandths of a foot when dealing with zenith angles and slope distances, I see neither of the two "complete" solutions (including curvature and refraction) offered agree. Neither show their work! All values shown are in feet.
Elevation of starting point: 10.687
UP offset from marker: 5.035
Resulting H.I.: 15.722
Using the provided zenith angle (assumed to be DMS) and slope distance (assuming atmospheric corrections and prism offsets were applied) the difference in elevation to the reflector (center?): -17.811
The UP offset from the prism to the unknown point provided is: -4.885.
Uncorrected for earth curvature height of unknown is -6.974.
Using the curvature correction in?ÿ https://www.ngs.noaa.gov/PUBS_LIB/ResultsOfLevelingRefractionTests_by_NGS_TR_NOS92_NGS22.pdf?ÿ equation 1 on printed page 6: Cc = s^2 / 2 * r where s is distance in meters and r is mean earth radius. Whalen used:r = 6,363,000. Converting to units of feet Cc = 0.002. Note that the computation of the curvature correction is merely a function of the distance. Its algebraic sign in this case should be negative as the earth's surface is negatively increasing with respect to the height of instrument.
This means we have a third answer! Mine is -9.976.?ÿ
Note that I did not apply a correction for refraction. This is a negative vertical angle. We have no meteorological data to compute it. As the angle of refraction bends to the normal from low to high refractive index we need more information than merely the distance.?ÿ
While I recognize that the thousandths of a foot discrepancies are well within the errors in the components of the problem, who can enlighten us/me about the correct solution? Please provide your work.
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The combined curvature and refraction correction formula I have always used, which I believe I got from either Brinker, or Moffat, Bouchard, and Wolf (I would have to hike through a bunch of snow to find the book) is:
((distance/1000)^2)*.0206
This product is then subtracted from the foresight target height.
The combined correction for a distance of 324.908 = 0.0021746
Therefore the Target Height would be 4.88282
So from the beginning
Starting Elevation 10.687, + HI 5.035 = 15.722
ZA (DDD.MMSS) 93.08328?ÿ = (DDD.DDDD) = 93.14244444?ÿ
Cos ZA 93.142444444 = -0.0548185094
-0.0548485094 * SD 324.908 = -17.8109722
Elevation at head of instrument 15.722 + Elevation Difference (-17.8109722) - Target Height 4.88282 = Foresight Point Elevation 6.9717922
Assumptions were made based upon the information contained in the question stating the occupied point elevation was " 10.687' "
My answer did not include the correction for curvature and refraction as I felt it was more important to correct the erroneous trigonometric formula being used.
I, of course, failed in my oblique attempt to get those on this thread to think about how realistic the values in the original post and responses are. The original post indicates he was attempting to check a value at the ten-thousandths of ?ÿa foot. He does not specify the equipment used but provides measurements at the thousandths of a foot and a non-reciprocal vertical angle reported to the tenth of a second.?ÿ
Without knowledge of the procedures and instrument specifications, it is hard to imagine meeting this accuracy. Looking at the specifications for an arbitrarily chosen high quality total station (see below), I wonder...
For a discussion of accuracies when performing work at high levels of accuracy see:?ÿ http://adsbit.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1979IAUS...89..373D&db_key=AST&page_ind=4&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES
