Re: Calculating Miles per Degree Longitude - Check my Math Please

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Thanks, Goran,

You are correct of course, the Earth is not a perfect sphere, however the
assumption of a sphere yields close enough results for my purposes.

private const double MilesPerDegLat = 69.04;

What you have seems to be the radius at the equator.

Not according to the information I've seen at
http://www.hypernews.org/HyperNews/get/trails/SAR/291/1.html
------------------------------------
Conceptually and practically, latitude is the same no matter where you go on
earth; however, in reality it varies from 69.41 statute miles per minute at
the poles to 68.70 statute miles per minute at the equator due to the earth
bulging slightly from its rotational spin.
------------------------------------

So, what I've done is assume an equally distributed gradient (again, an
assumption which might not be strictly correct, but close enough for my
purposes) and taken the average of 69.41 and 68.70 to arrive at an average
distance of approximately 69.04. Mathematically speaking, this comes out to
69.055, but I did see the figure 69.04 referenced somewhere, which is why
I've used it. Again, in the context of the accuracy which I need, I don't
think this deviation is significantly altering my results.

Thanks!

Joseph Geretz

"Göran Andersson" <guffa@xxxxxxxxx> wrote in message
news:eNTRv7lZHHA.4008@xxxxxxxxxxxxxxxxxxxxxxx
The earth is not spherical, so if you want a more exact value, you should
use the formula for an ellipse rather than a circle.

As it's not spherical, the radius varies with the latitude. What you have
seems to be the radius at the equator. The difference is only 0.19%, so
again it depends on how exact you want the value to be.

Also, you should add the altitude over sea level to the radius if you want
a more exact value, but at five miles over sea level (Mount Everest) the
difference is only 0.12%


Joseph Geretz wrote:
I need to calculate miles per degree longitude, which obviously depends
on latitude since lines of longitude converge at the poles. Via Google,
I've come up with the following calculation:

private const double MilesPerDegLat = 69.04;
private const double EarthRadiusMiles = 3963.1676;
private static double PiDiv180 = Math.PI / 180;

double MilesPerDegLon = MilesPerDegLat * Math.Cos(Latitude * PiDiv180)

I'm not fluent at all in the trigonometric branches of mathematics and
I'd appreciate it if you could check my math for me.

I basically got this from:

http://answers.google.com/answers/threadview?id=577262

I'm using the formula which Steven expressed as: 69.1703234283616 *
COS(Lat*0.0174532925199433) with just one difference. I am using the
value 69.06 for miles per degree latitude. Steven is using 69.17. I
understand that the Earth is not a perfect sphere and I'm under the
impression that the value 69.06 is the correct average to use. On the
other hand, my calculations will be exclusively confined to North
America. Should I be using a value of 69.17?

I'm also curious about how this formula works without taking into account
the Earth's radius. Again, I'm not fluent by any means in trigonometry,
but I'd been assuming that the Earth's radius would play a role in the
calculation. Also, I'm curious as to why the formula proposed directly
prior to the final solution doesn't work:

(pi/180) * R * cosA where R is the radius of the earth in miles and A is
the degree latitude.

For some reason this didn't work for me, but perhaps I encoded it
incorrectly. If this should work and is deemed to be more precise than
the claculation I've implemented I'll happily substitute it.

Thanks for your help!

- Joseph Geretz -


--
Göran Andersson
_____
http://www.guffa.com


.

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