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Date: Fri, 4 Nov 1994 09:53:00 -0700
From: Paul Kairis
Subject: let's do math

The problem is found in sea navigation. Given two points A,B with
LON and LAT coordinates, find the travel distance from A to B. The
distance is actually the great circle passing A and B.

The solution is needed in the form of a double integral where
the first int covers the LONs and the second the LATs. I guess
the diff. part is to form the vector.

My calculus books are not with me, can you help? Thanks

-Paul
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From: Dr. Ken
Date: Fri, 4 Nov 1994 12:15:29 -0500 (EST)

Paul!

I wanted to let you know that we're working on your problem.  The first
thing my colleague Ethan noticed, however, is that you probably don't want
to look at a double integral at all, but instead a single integral.

The reason for this is that when you find the length of a curve in three
space, you integrate the number one over the region in question, i.e. the
curve you want the area of.

I think that there's an easier way to handle your problem, though.  Use the
distance formula to find the real distance (through the crust of the earth)
between the two points.  Then you've reduced the problem to a
two-dimensional Euclidean Geometry problem, namely finding the length
of an arc of a circle cut off by a segment with a certain length.

I hope this helps, and let us know if it doesn't!

-Ken
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Date: Fri, 4 Nov 1994 10:40:39 -0700
From: Paul Kairis

Ken, thanks for your input. What you have suggested  is quite true. I am
interested in the symbolic representation though, e.g.

Here're my notes:
-----------------

Let dS = sqrt[ dx^2 + dy^2 + dz^2 ]

where x = r cosf cosg
y = r cosf sing
z = r sinf

then,

latB   lonB
/     /
S = Radius  |     |  [1+cos^2(f)] df dg
/     /
latA  lonA

This is what I came up with but it is not right!

-Paul
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From: Dr. Ken
Date: Mon, 7 Nov 1994 12:02:16 -0500 (EST)

Hello there!

I think the main problem is that you're trying to use a double integral
to find the length of a curve, but the double integral will give you the
AREA of the curve.  And that's Zero.  I'm not quite sure how you got
that particular integral, but that's why it doesn't work.

-Ken "Dr." Math

Note: There's a complete solution of this type of problem, available for
http://mathforum.org/dr.math/problems/longandlat.html   .
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Associated Topics:
High School Calculus

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