<html><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; "><br><div><div>On Mar 17, 2010, at 9:56 PM, Alexander Solla wrote:</div><br class="Apple-interchange-newline"><blockquote type="cite"><span class="Apple-style-span" style="border-collapse: separate; color: rgb(0, 0, 0); font-family: Helvetica; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-border-horizontal-spacing: 0px; -webkit-border-vertical-spacing: 0px; -webkit-text-decorations-in-effect: none; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; "><span class="Apple-style-span" style="font-family: Courier; ">But your "spherical" points don't really form a basis in three-space, or even over all of two-space.</span></span></blockquote></div><br><div>I'll take this back. Lattitude and longitude is enough to "form a basis" on R^2, by taking a basis for the surface of the sphere in terms of latitude and longitude and projecting it stereographically. So if you wanted to use the normalization idea, you could use the stereographic projection formulas to turn a spherical point into a Cartesian point.</div><div><br></div><div><a href="http://en.wikipedia.org/wiki/Stereographic_projection">http://en.wikipedia.org/wiki/Stereographic_projection</a></div></body></html>