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    <p>I have recently done something similar to this using the "Haversine Formula" on WGS-84 data, which is a derivative of the "Law of Haversines" with very satisfying results. </p> <p>Yes, WGS-84 assumes the Earth is an ellipsoid, but I believe you only get about a 0.5% average error using an approach like the "Haversine Formula", which may be an acceptable amount of error in your case. You will always have some amount of error unless you're talking about a distance of a few feet and even then there is theoretically curvature of the Earth... If you require more rigidly WGS-84 compatible approach checkout the "Vincenty Formula."</p> <p>I understand where <strong>starblue</strong> is coming from, but good software engineering is often about trade offs, so it all depends on the accuracy you require for what you are doing. For example, the result calculated from "Manhattan Distance Formula" versus the result from the "Distance Formula" can be better for certain situations as it is computationally less expensive. Think "which point is closest?" scenarios where you don't need a precise distance measurement.</p> <p>Regarding, the "Haversine Formula" it is easy to implement and is nice because it is uses "Spherical Trigonometry" instead of a "Law of Cosines" based approach which is based on two-dimensional trigonometry, therefore you get a nice balance of accuracy over complexity.</p> <p>A gentlemen by the name of <strong><em>Chris Veness</em></strong> has a great website at <a href="http://www.movable-type.co.uk/scripts/latlong.html" rel="noreferrer">http://www.movable-type.co.uk/scripts/latlong.html</a> that explains some the concepts you are interested in and demonstrates various programmatic implementations; this should answer your x/y conversion question as well.</p>
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    1. POConverting from longitude\latitude to Cartesian coordinates
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    1. POConverting from longitude\latitude to Cartesian coordinates
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    1. CO0.5% error - 0.5% of what? In the context of this question it could be the radius of the earth, so 0.5% could be 30 km :)
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    2. COChecked out your link. The 0.5% quote is for error in the great-circle distance between two points so not strictly relevant to this question. I would think when converting lat-long to Cartesian coordinates with the origin at the centre of the earth, the errors from assuming a spherical earth could be significant. It's not clear what the questionner wants to **do** with the Cartesian coordinates. Either it's just more convenient to work in them for some bizarre reason, or perhaps it's some requirement for data export? If the latter, accuracy would be important.
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