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    Re: The magical maths of Google Maps
    From: Gary LaPook
    Date: 2015 Oct 8, 03:00 -0700

    
    If a surveyor started at the equator and walked straight north, setting up his 
    theodolite from time to time to determine his latitude, when he got to the 
    place where he determined that his latitude was exactly one degree north when 
    he then looks at the distance he measured with his chain he will find that he 
    walked 59.701 nautical miles not the expected 60 NM exactly. 
    Gl
    
    
    ------------------------------
    On Thu, Oct 8, 2015 2:12 AM PDT Gary LaPook wrote:
    
    >I think I can see what your are missing. When we learn celnav we work with 
    the assumption that the earth is a perfect sphere and a degree on the 
    surface, as measured at the center of the earth, is everywhere 60 nautical 
    miles. This assumption is appropriate given the level of accuracy that is 
    obtainable at sea with celnav. Based on this assumption we also accept that 
    straight down at our position points to the center of the earth so the 
    horizon, from which we measure our altitudes, is exactly at right angles from 
    a line from out feet to the center of the earth. If this were true then, when 
    we take a noon latitude, we would be determining our geocentric latitude, the 
    angle between the equator and our position as measured by a person at the 
    center of the earth. This is the simplified diagram that we have been trained 
    on. 
    >However, there are two other types of latitude, geodetic and astronomic. 
    >What we are actually determining is the last type, the astronomic latitude. 
    Because of the oblateness of the earth and because of local deflection of the 
    vertical due to non uniform density, the horizon is not at right angles to a 
    line from out feet to the center of the earth but is at a right angle to the 
    local field of gravity as it is deflected from the tangent to the surface of 
    the oblate earth. As to deflection of the vertical, Frank has posted before 
    about this effect in the vicinity of Bermuda but is is usually very small, 
    maybe a few tenths of a minute, is not charted so it can not be corrected for 
    by a navigator, If you could correct for the local deflection of the vertical 
    then you would be able to determine the geodetic latitude, the latitude as 
    measured from a horizon that is tangent to the surface of the oblate earth at 
    our position but because the deflection of the vertical is very small we 
    ignore it (in celnav) and treat our astronomic
     latitude as though it were geodetic latitude. Why do we do this?
    >Because charts show the positions of places on earth by their geodetic 
    latitude, not their geocentric latitude. That is the reason for the 
    meridional parts, to allow creating a chart in which the latitude determined 
    with a sextant will correspond to the latitude of the shoreline. (ignoring 
    deflection of the vertical.) That is the reason that the length of a degree 
    on latitude varies because it is based on the oblate earth (the Clarke 
    spheroid of 1866 in Bowditch). If, instead, the shoreline was placed on a 
    chart with a gratical  of geocentric latitude then it would be off by about 
    11 minutes of latitude at latitude 45 north and south so you would be 
    crashing into reefs all the time. 
    >So forget the simplified diagram of geocentric latitude, now you know the true story.
    >gl      From: Lu Abel 
    > To: garylapook---.net 
    > Sent: Wednesday, October 7, 2015 12:07 PM
    > Subject: [NavList] Re: The magical maths of Google Maps
    >   
    >I have no doubt of these numbers, Gary -- and at the same time they seem 
    counterintuitive.   If I were to fit a perfect sphere to the curvature of the 
    earth at the equator it would be - as you pointed out earlier - about 11 nm 
    greater in diameter than a perfect sphere that matched the curvature of the 
    earth at the poles.  A greater diameter should imply a greater surface 
    distance for one degree than for a smaller diameter.   What am I missing?
    > 
    >
    >
    >      From: Gary LaPook 
    > To: luabel{at}ymail.com 
    > Sent: Tuesday, October 6, 2015 10:42 PM
    > Subject: [NavList] Re: The magical maths of Google Maps
    >   
    >That made me curious so I turned to table 6 of Bowditch which gives the 
    length of a degree of latitude  for each degree from 0 to 90. This length 
    varies from 59.701 NM at the equator to 60.313 NM at the pole based on the 
    Clarke Spheroid of 1866. I added all the distances up and found that half 
    way, at 45 degrees of latitude the distance is 8.3 NM short of the 2,700 NM 
    it would be on a spherical earth and then the remaining distance to the pole 
    is 9.2 NM greater so the total I came up with was 5,400.5915 NM but I assume 
    the discrepancy from the perfect 5,400 NM is probably due to rounding. So to 
    test Google Earth it doesn't do to measure the distance from the pole to the 
    equator since you will get 5400 either way but you should measure from the 
    pole to 45 degree latitude.
    >gl
    >gl    
    >
    >  From: Frank Reed 
    > To: garylapook---.net 
    > Sent: Tuesday, October 6, 2015 10:12 AM
    > Subject: [NavList] Re: The magical maths of Google Maps
    >   
    >David Pike, you wrote:
    >"From this I conclude that the Google Maps projection is close to standard 
    Mercator’s.". You’ll find you’ve got a “Globe” presentation, which you can 
    tilt until the line between Boston and London is straight while the distance 
    remains constant. You would appear to have hit the view in with the plane of 
    the great circle."
    >Yes, that's right. The only thing you have to double-check with an 
    earth-based mapping product is that they are not being too clever by 
    including ellipsoidal corrections to the distance calculation. If that 
    happens then it's not quite a great circle but awafully close. I haven't 
    checked, but I assume that Google Maps products use the standard spherical 
    great circle calculation.
    >Frank Reed
    >Conanicut Island USA
    >
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    >
    >View and reply to this message: 
    >http://fer3.com/arc/m2.aspx/magical-maths-Google-Maps-LaPook-oct-2015-g32992
    

       
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