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Distance between two sets of coordinates... update

  1. #1
    Marcus Fox
    Guest

    Distance between two sets of coordinates... update

    Sorry for making a new post, but it seems that the original has moved down
    the page due to the date and wouldn't be as visible to new contributors.
    Much more new information here.

    Was using the Haversine formula to find the distance between two sets of
    coordinates, but it seemed to break down where we didn't use North or West
    coordinates, as Excel didn't like negative dd:mm:ss, as I would have has to
    use for east and south. So I converted them to decimal degrees multiplying
    by 24/-24 depending on a condition of N/S or E/W. The coordinates are
    originally entered as they are on maps in dd:mm:ss format.

    Lat1 is in B2, long1 is in B3, lat2 is in B5 and long 2 is in B6. The
    letters N/S for latitude are in C2 and C5 and E/W in C3 and C6.

    The formula is -

    =6371*ACOS(COS(RADIANS(90-(IF(C2="S",B2*(-24),B2*24))))*COS(RADIANS(90-(IF(C
    5="S",B5*(-24),B5*24))))+SIN(RADIANS(90-(IF(C2="S",B2*(-24),B2*24))))*SIN(RA
    DIANS(90-(IF(C5="S",B5*(-24),B5*24))))*COS(RADIANS((IF(C3="E",B3*(-24),B3*24
    ))-(IF(C6="E",B6*(-24),B6*24)))))

    Seems to work ok, until I realised that the earth isn't actually
    spherical, - we can assume it is for the purposes of calculation, to get a
    reasonably accurate approximation, as the above formula gives the radius of
    curvature, but since the radius of curvature varies with latitude, say we
    used a formula for an elipsoid, as the polar radius of the earth is 6357 km
    and the equatorial radius 6378, would

    Radius =
    6378*(1-(1-6357^2/6378^2)^(1/2)^2)/(1-(1-6357^2/6378^2)^(1/2)^2SIN^2(lat))^(
    3/2)

    do instead of the figure of 6371 for the radius at the start of the first
    equation, but I do notice I need to use a figure for the latitude, and since
    I have two different latitudes as two different points, how do I integrate
    it into my equation?

    Marcus



  2. #2
    Marcus Fox
    Guest

    Re: Distance between two sets of coordinates... update


    "Marcus Fox" <[email protected]> wrote in
    message news:[email protected]...
    > Sorry for making a new post, but it seems that the original has moved down
    > the page due to the date and wouldn't be as visible to new contributors.
    > Much more new information here.


    OK, I knew that spherical trigonometry wouldn't be the easiest subject for
    one to get their head around, but was hoping someone would have some ideas,
    LOL.

    Marcus



  3. #3
    WhiteStarLine
    Guest

    Re: Distance between two sets of coordinates... update

    Hi Marcus,

    If you really, really need this precision, I wouldn't approach it this
    way. I use Excel as a quick satellite viewing angle predictor and
    factor in zonal harmonics as it makes quite a difference to the local
    observer's altitude / azimuth.

    But firstly, why not simply use standard spherical trigonometry, as
    suggested when you posted this to a satellite newsgroup. Lets get the
    distance from Canberra to London:

    Canberra
    Latitude 1 -35.45
    Longitude 1 149.09
    London
    Latitude 2 51.50
    Longitude 2 -0.17

    Theta = Longitude 1 - Longitude 2
    Theta 149.26

    Dist = sin(deg2rad(Latitude1)) * sin(deg2rad(Latitude2)) +
    cos(deg2rad(Latitude1)) * cos(deg2rad(Latitude2)) *
    cos(deg2rad(Theta))
    Dist -0.889746961

    Distance

    10559.81698 miles
    16994.3781 kilometres
    9170.145065 nautical miles

    Now, if you still want to factor in a 3% oblated spheroid, then move to
    ECI vectors. Get the instantaneous position of each lat / lon on an
    earth centred inertial system using the vernal equinox as your
    reference. Factor in the meridional radius of curvature at latitude
    (lat) as N = a[cos^2(lat)+(1-e^2)sin^2(lat)]^(0.5).

    Use transformational matrices to get the relative XYZ vector. I can
    send you the Excel spreadsheet if you want.


  4. #4
    Dana DeLouis
    Guest

    Re: Distance between two sets of coordinates... update

    > COS(RADIANS(90-...etc
    > SIN(RADIANS(90-...etc


    Hi. Just a note on your equations. When studying Spherical Geometry, you
    will often have equations like you did above as
    Cos(90-x). Just note that these equations reduce to just Sin(x).
    And Sin(90-x) reduces to just Cos(x).
    I have seen some complicated equations when working with earth as a
    spheroid.
    For example, given WhiteStarline's data, with an SemimajorAxis of 6378.137
    and Earth's eccentricity of 0.081819, a distance of 16,990.7867 km vs a
    distance of 16994.3781 km with a fixed Radius. (This depends on Radius used
    also).
    Good luck.
    --
    HTH. :>)
    Dana DeLouis
    Windows XP, Office 2003


    "Marcus Fox" <[email protected]> wrote in
    message news:[email protected]...
    > Sorry for making a new post, but it seems that the original has moved down
    > the page due to the date and wouldn't be as visible to new contributors.
    > Much more new information here.
    >
    > Was using the Haversine formula to find the distance between two sets of
    > coordinates, but it seemed to break down where we didn't use North or West
    > coordinates, as Excel didn't like negative dd:mm:ss, as I would have has
    > to
    > use for east and south. So I converted them to decimal degrees multiplying
    > by 24/-24 depending on a condition of N/S or E/W. The coordinates are
    > originally entered as they are on maps in dd:mm:ss format.
    >
    > Lat1 is in B2, long1 is in B3, lat2 is in B5 and long 2 is in B6. The
    > letters N/S for latitude are in C2 and C5 and E/W in C3 and C6.
    >
    > The formula is -
    >
    > =6371*ACOS(COS(RADIANS(90-(IF(C2="S",B2*(-24),B2*24))))*COS(RADIANS(90-(IF(C
    > 5="S",B5*(-24),B5*24))))+SIN(RADIANS(90-(IF(C2="S",B2*(-24),B2*24))))*SIN(RA
    > DIANS(90-(IF(C5="S",B5*(-24),B5*24))))*COS(RADIANS((IF(C3="E",B3*(-24),B3*24
    > ))-(IF(C6="E",B6*(-24),B6*24)))))
    >
    > Seems to work ok, until I realised that the earth isn't actually
    > spherical, - we can assume it is for the purposes of calculation, to get a
    > reasonably accurate approximation, as the above formula gives the radius
    > of
    > curvature, but since the radius of curvature varies with latitude, say we
    > used a formula for an elipsoid, as the polar radius of the earth is 6357
    > km
    > and the equatorial radius 6378, would
    >
    > Radius =
    > 6378*(1-(1-6357^2/6378^2)^(1/2)^2)/(1-(1-6357^2/6378^2)^(1/2)^2SIN^2(lat))^(
    > 3/2)
    >
    > do instead of the figure of 6371 for the radius at the start of the first
    > equation, but I do notice I need to use a figure for the latitude, and
    > since
    > I have two different latitudes as two different points, how do I integrate
    > it into my equation?
    >
    > Marcus
    >
    >




  5. #5
    Dana DeLouis
    Guest

    Re: Distance between two sets of coordinates... update

    > COS(RADIANS(90-...etc
    > SIN(RADIANS(90-...etc


    Hi. Just a note on your equations. When studying Spherical Geometry, you
    will often have equations like you did above as
    Cos(90-x). Just note that these equations reduce to just Sin(x).
    And Sin(90-x) reduces to just Cos(x).
    I have seen some complicated equations when working with earth as a
    spheroid.
    For example, given WhiteStarline's data, with an SemimajorAxis of 6378.137
    and Earth's eccentricity of 0.081819, a distance of 16,990.7867 km vs a
    distance of 16994.3781 km with a fixed Radius. (This depends on Radius used
    also).
    Good luck.
    --
    HTH. :>)
    Dana DeLouis
    Windows XP, Office 2003


    "Marcus Fox" <[email protected]> wrote in
    message news:[email protected]...
    > Sorry for making a new post, but it seems that the original has moved down
    > the page due to the date and wouldn't be as visible to new contributors.
    > Much more new information here.
    >
    > Was using the Haversine formula to find the distance between two sets of
    > coordinates, but it seemed to break down where we didn't use North or West
    > coordinates, as Excel didn't like negative dd:mm:ss, as I would have has
    > to
    > use for east and south. So I converted them to decimal degrees multiplying
    > by 24/-24 depending on a condition of N/S or E/W. The coordinates are
    > originally entered as they are on maps in dd:mm:ss format.
    >
    > Lat1 is in B2, long1 is in B3, lat2 is in B5 and long 2 is in B6. The
    > letters N/S for latitude are in C2 and C5 and E/W in C3 and C6.
    >
    > The formula is -
    >
    > =6371*ACOS(COS(RADIANS(90-(IF(C2="S",B2*(-24),B2*24))))*COS(RADIANS(90-(IF(C
    > 5="S",B5*(-24),B5*24))))+SIN(RADIANS(90-(IF(C2="S",B2*(-24),B2*24))))*SIN(RA
    > DIANS(90-(IF(C5="S",B5*(-24),B5*24))))*COS(RADIANS((IF(C3="E",B3*(-24),B3*24
    > ))-(IF(C6="E",B6*(-24),B6*24)))))
    >
    > Seems to work ok, until I realised that the earth isn't actually
    > spherical, - we can assume it is for the purposes of calculation, to get a
    > reasonably accurate approximation, as the above formula gives the radius
    > of
    > curvature, but since the radius of curvature varies with latitude, say we
    > used a formula for an elipsoid, as the polar radius of the earth is 6357
    > km
    > and the equatorial radius 6378, would
    >
    > Radius =
    > 6378*(1-(1-6357^2/6378^2)^(1/2)^2)/(1-(1-6357^2/6378^2)^(1/2)^2SIN^2(lat))^(
    > 3/2)
    >
    > do instead of the figure of 6371 for the radius at the start of the first
    > equation, but I do notice I need to use a figure for the latitude, and
    > since
    > I have two different latitudes as two different points, how do I integrate
    > it into my equation?
    >
    > Marcus
    >
    >




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