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Convert Longitude/Latitude dependent upon the location on the Earth

  1. Mar 30, 2017 #1
    I am planning to plot all the observations made by a satellite in two respective grid boxes. One grid located at the equator and one grid located at the mid latitudes...

    Each grid box will be a 10° x 10° box. However, due to the shape of the earth the area of each box will indeed not be the same size. Is there some formula to convert latitudinal and longitundal degrees dependent on location of earth? To some useful unit (km).

  2. jcsd
  3. Mar 30, 2017 #2


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    The north-south distances will be one to one everywhere. A simple conversion of the east-west distances will be proportional to the latitude.

    For example, at the equator, 10 degrees in the east-west direction is 25,000 miles / 36 = 694 miles. This distance is essentially zero at the pole, so that it's a linear relationship as you move north and south. Therefore at 80 degrees north, 10 degrees of longitude would be a ninth of 694 or about 77 miles.

    Note that these are simple approximations and do not take into account things like datums, etc.
  4. Mar 30, 2017 #3
    Hi Borg:

    I am pretty certain that the longitude linear distance for 10 degrees is the same as the 10 degree distance at the equator times the cosine of the latitude. This is based on a spherical Earth. Some small and a bit complicated corrections would be needed for the Earth's approximately ellipsoidal shape. Smaller and even more complicated corrections are required if you want to take the actual shape of the Earth into consideration.

    The longitudinal distance at the poles is zero. That is the grid "boxes" between 80 degrees and 90 degrees will be triangles.

  5. Mar 30, 2017 #4


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    @Buzz Bloom I think that we're saying generally the same thing. The distance between two latitudes along a longitudinal line is the same for every degree of distance. It's the distance between two longitudinal lines that changes depending on the latitude.
  6. Mar 30, 2017 #5

    jim mcnamara

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    Hmm. @Borg I do not see it quite your way.

    As you know - The Earth is not a perfect sphere. So, Geodesy has a little complexity.
    Parallel distances across longitude A to longitude B is more interesting.
    @Anthony LaRosa
    What exactly do you need? How much accuracy? The usual choice is probably a great circle distance between two points given a lat long for each one.
    See Vincenty formula for special case of ellipsoids:

    You can often use haversines because if the N/S differences are small a spherical approximation is probably okay:

    It gets more interesting if you are doing some kind of surveying, where accuracy needs to greater than locating the nearest McDonalds restuarant.

    You get to decide.
  7. Mar 30, 2017 #6


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    I must be wording things in a way that is confusing. Yes, I agree that the distance between each degree of latitude is mostly the same (ignoring the spheroid issue). Stated your way, I've been trying to say that the distance between each degree of longitude varies proportionally to the latitude.
  8. Mar 30, 2017 #7


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    This is the real question. Most of this depends on how accurate the OP wants to be. A simple conversion gets your nuke on the right block. Taking the rest into account lets you pick whose window you want to put it through. :oldtongue:
  9. Mar 30, 2017 #8


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    @Borg. You keep saying that it is proportional to the latitude. This is not true. As BuzzBloom said, the distance between two longitude lines depends on the cosine of the latitude. So, while 10 degrees of longitude is 694 miles at the equator, at 80 degrees latitude it is 694 * cos(80) = 120 miles, not the 77 miles you quoted. At 10 degrees latitude, your linear approximation would say 10 degrees of longitude is 694 * 8/9 = 617 miles, while the correct answer is at 10 degrees latitude, 10 degrees of longitude = 694 * cos(10) = 683 miles. These differences are large, and will not get you "on the right block".
  10. Mar 30, 2017 #9


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    Thanks for clarifying my mistake. It has been a long time since I've had to do those calculations.
  11. Mar 31, 2017 #10


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    Ugh. What was I thinking yesterday? :oldruck:
  12. Apr 6, 2017 #11


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    Find out about WGS84 here; https://en.wikipedia.org/wiki/World_Geodetic_System#WGS84
    This code converts latitude, longitude and height to x, y, z coordinates in metres.
    You can then map that to your grid box.
    Code (Text):

    Sub wgs84(_
    Byval latit As Double,_
    Byval lng As Double,_
    Byval h As Double,_
    Byref x As Double,_
    Byref y As Double,_
    Byref z As Double )

    Const a As Double = 6378137 ' Defined semi-major axis, metres
    Const ba As Double = .9933056200098587 ' square of b to a ratio bb/aa
    Const ee As Double = 6.694379990141316D-03 'square of first eccentricity f(2-f)

    Dim As Double sinlat = Sin(latit)
    Dim As Double RofC = a / Sqr(1 - ee * sinlat * sinlat) ' radius of curvature
    Dim As Double term = (RofC + h) * Cos(latit)

    x = term * Cos(lng) ' x-axis is Atlantic Ocean positive
    y = term * Sin(lng) ' y-axis is Indian Ocean positive
    z = (h + (RofC * ba)) * sinlat ' z-axis is North Pole positive

    End Sub
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