I have the latitude and longitude of a point, say N 40 00 00, W 75 00 00. I want to find the latitude (with great precision!) of a point exactly 10 statute miles due south (so my second point is on the same meridian as my starting point).
My question: In doing this, does the fact that the earth is an oblate spheroid rather than a perfect sphere factor into it at all? Seems to me that if you walked the same distance on the two different shapes, you would wind up at slightly different latitudes, but I’m unschooled in these things.
I found a little program on the web called Geocalc which will give me a latitude, but I don’t know if it takes the flattening factor into account or even if it needs to. If it does matter, do you know of a program or set of equations I could use to get a good result?
How great is “great precision”? If you only need to know the difference between the two latitudes to about 1% or so, then you’re probably OK using the naive spherical approximation; the Earth’s oblateness is about a one-in-a-thousand effect. In particular, 10 statute miles will translate to a latitude difference of about 1.4 seconds of arc (I think).
Unfortunately, various schemes are used. The simplest is to assume a spherical earth with a radius that is some compromise between that at the equator and the poles. One standard spherical radius (used in the measurement of aviation records) is 6371.0 km.
Another approach is to use a spherical earth model with a radius appropriate for the latitudes being considered. Provided the latitudes involved in a calculation don’t vary much (as in the case mentioned in the OP), this will give very good accuracy.
And some schemes are iterative: they assume a non-spherical and thus don’t use a constant radius.
Obviously, there’s an inherent tradeoff between ease of calculation and accuracy.
The length of a degree of latitude varies from 69.407 miles at the poles to 68.703 miles at the Equator. Put another way, 10 miles will cover 8.73 arcminutes at the Equator and 8.64 minutes at the poles. The formula for interpolating between these values would involve sines and cosines, and I’d probably get it wrong, so I’ll defer to someone else to post it.
That Geocalc program probably does it right. The question is, what does it do right?
“10 statute miles south” doesn’t quite define the problem. It has different meanings for different methods.
Now you seem to want it to be along some ellipsoid rather than some map projection . If you really want to be precise you need to know some more things.
Which datum/reference frame do you want to use (i.e. what is the shape and orientation of the ellipsoid)?
Are you assuming 10 miles at a constant height above the ellipsoid, a constant mean-sea-level height, or with both points at zero height on the ellipsoid?
If the Geocalc program (there are probably dozens of different ones with the same name) doesn’t give you the option to select the ellipsoid, it is probably using something like NAD83 (which is likely fine for you). If it doesn’t ask you your altitude, it is probably assuming both points are at zero height on the ellipsoid (which may or may not be what you want).
Anyway, there are a lot of programs out there but I use this one from the Royal Netherlands Navy Hydrographic Service. It does a lot more than I even have heard of.
If I haven’t lost all my credibility already with the above series of posts, here’s what I think is the best way to do it.
If you assume a perfectly ellipsoid earth, then your north-south path lies on an ellipse whose semimajor axis is the equatorial diameter of the Earth (call it a) and whose semiminor axis is the polar radius of the earth (call it b). So essentially want to find the angle subtended by an arc of a given length, starting at a given latitude, on an ellipse with known a and b. It can be shown (e.g. here) that the arc length of the arc subtended on the ellipse between angles alpha and beta is given by
length = a * integral (sqrt(1 - e[sup]2[/sup] sin[sup]2[/sup] theta) d theta),
where e is the eccentricity of the ellipse, given by e[sup]2[/sup] = (a[sup]2[/sup] - b[sup]2[/sup])/a[sup]2[/sup], and the integral runs between alpha and beta. (Note, by the way, that alpha and beta are measured from the North Pole, not the equator as latitudes are.)
So if you know alpha (your starting latitude), a, e, and the length you want, you can in principle solve for beta (your ending latitude.) Unfortunately, the elliptic integrals aren’t nice functions like sine or cosine, so you have to look up the results in tables or use a program like Mathematica to solve this. I’ve got access to Mathematica here, so if you want send me the actual values you’re using (I’m assuming you don’t want exactly 40 N, 75 W) and I’ll plug it in to Mathematica.