Subject: Re: how to determine distance from line?
Date: Sep 09, 2004 @ 08:23
Author: aletheiak ("aletheiak" <aletheiak@...>)
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>bravo
> SUBJECT WAS:
> "Decimal digits for Latitude and Longitude: 0.001" Lat/Long"
>
> "aletheiak" asks,
> "could you show more details of your computations?"
>
> Be careful what you ask for. :)
>sci.geo.cartography
> ======================================
>
> *** Dave Patton's post to comp.info.gis and
> with light editing + some info ***monuments:
> A)
> Official NAD27 coordinates for two Canada/USA border
> M02300 MONUMENT 194 48 59 56.85 117 04 17.79<http://www.internationalboundarycommission.org/coordinates/
> M02310 MONUMENT 195 48 59 57.42 117 01 39.71
>
>
>the name of the surveyors
> B)
> The Canada/USA border is defined as being straight lines
> between border monuments.
> C)
> The NAD83/91 coordinates shown on this plaque
>
> Land Surveyor's Associaton of Washington (LSAW)
> Historical Society Monument #5
>
> STATE PLANE COORDINATES: DEMEYER [?]
> NORTHING: 751827.675 EASTING: 2552579.905see the text running down the right column in daves report
> LATITUDE: N 48º 59' 57.110"
> LONGITUDE: W 117º 01' 56.715"
> ELEVATION: 5932.34 FEET, 1808.182 METERS
> DATUM: NAD 83/91
>
> <http://www.confluence.org/photo.php?visitid=8945&pic=5>
>
> [Lat/Long represent what? No obvious "center" mark.
> Isn't there usually an inscribed reference point?==================================================
> Assume center of plaque until someone
> who really knows speaks up. RCMcC]
>
> What is the 'best' way for a 'recreational user' to
> accurately calculate the distance(north or south)
> between the plaque coordinates and the boundary line?
>
> The distance involved is small, so using typical
> 'recreation-grade' mapping software will not be
> sufficiently accurate.
>
> I'm more interested in learning how to do this myself,
> rather than just having someone who has the facilities
> tell me the answer
>
> Dave Patton
>
>
>right & i surmise that the difference is not merely negligible but
> Dave notes that the border is defined
> by the line of sight between monuments.
> There are 3 candidates for the "line of sight":
> (1) great circle (shortest spherical distance,
> constantly varying azimuth along the path),
> (2) rhumb line (constant azimuth along the path,
> greater distance than great circle),
> and (3) WGS-84 ellipsoidal geodesic path
> (more accurate version of great circle
> for earth's non-spherical shape),
> (4) Other?
> For the short distances here,
> there is little practical difference among them.
> My GCB program calculates all three.ok
> Hand calculations for the plane geometry triangle
> agree well with the fancy computer spherical
> triangle stuff for these distances.
>
> But, how to find the north-south distance from
> a 3rd point to an intermediate point on the line
> isn't obvious.
> a formula that can be applied.what do you mean specifically by the above comment
>
> ==========================
>
> RCMcC
> Angles and Distance greatly distorted.
>117.072683º
> [194] -- 3212 m -- I -- [195]
> \ | /
> 2946 m 267.048 m
> \ /
> [LSAW 5]
>
> ^ North (up) South (down) V
> <- West East ->
>
> ==========================================
>
> PROPOSED RCMcC SOLUTION
>
> Given (all NAD-27 converted to NAD-83 with CORPSCON )
>
> Monument #194 = Point 1 = Lat1/Long1 = N 48.999056º W
> Monument #195 = Point 2 = Lat2/Long2 = N 48.999215º W117.028771º
> LSAW#5 plaque = Point 3 = Lat3/Long3 = N 48.999197º W117.032421º
>not sure i want all that
> TBD
> Where: Great Circle between Points 1 & 2 (GC1-2)
>
> Find intermediate Point I on GC1-2 = LatI/LongI
> that is due north or south of Point 3
> where LongI = Long3 = W 117.032421º
> LatI = TBD from formula from
>
> Ed Williams' Aviation Formulary
>
> <http://williams.best.vwh.net/avform.html>
>
> "Latitude of point on GC"
>
> [edited]
>
> latI = atan(
> ( sin( lat1 ) * cos( lat2 ) * sin( lonI - lon2 )
> - sin( lat2 ) * cos( lat1 ) * sin( lonI - lon1 ) )
> / ( cos( lat1 ) * cos( lat2 ) * sin( lon1 - lon2 ) )
> )
>
> Result:
> Point I = LatI/LongI = N 48.999202º W 117.032421º
> = 0.024 sec lat or 7.473 cm = 2.942 in. due north
> of Point 3 on LSAW #5 plaque {center?]
>
> ============================
>
> I have a draft fortran 77 program
> with all the excruciating details
> which works this problem for these three points.
> If this is something BP folks do occasionally,
> it could be generalized to allow variable input
> and worldwide distances, and to take care
> of the special cases where the formula fails.
> I can email the source code and executable file
> to interested parties. Someone can do
> a modern fancy GUI translation.
>
> This approach is all subject to correction,
> of course.
>
> Cheers, 73,
>
> Ron McC.
> w2iol@a...
>
> Ronald C. McConnell, PhD
>
> WGS-84: N 40º 46' 57.6" +/-0.1"
> W 74º 41' 22.1" +/-0.1"
> FN20ps.77GU31 +/-
> V +5058.3438 H +1504.2531
>
> http://home.earthlink.net/~rcmcc
>
> "The first day or so,
> we all pointed to our countries.
> The third or fourth day,
> we were pointing to our continents.
> By the fifth day,
> we were only aware of one Earth."
>
> -Prince Sultan Bin Salmon Al-Saud,
> Saudi Arabian astronaut