# Math Help - Using the Pythagorean theorem to determine distance on a curved surface

1. ## Using the Pythagorean theorem to determine distance on a curved surface

I am working through the book "Geometry - A High School Course" by Serge Lang, and the only problem I have gotten stuck on so far is in chapter 3, which concerns the Pythagorean theorem. The question is in two parts, the first of which I have no trouble with, but the second of which puzzles me:

a. A ship travels 6 km due South, 5 km due East, and then 4 km due South. How far is it from its starting point? (The answer is NOT 15 km!) You may assume that the path of the ship lies in a plane.
b. If actually the ship starts at the North Pole, on earth, what would the actual distance be?

For part a, I got the answer $5\sqrt{5}$ km for the distance. But I am confused by the second part, which seems to add the curvature of the Earth into the equation. Earlier in the book there was a section on arcs which involved a couple of questions calculating latitude and longitude (and for which we were told to use 40,000 km for the circumference of the Earth). But there is nothing in it about relating distance on the Earth's surface in two dimensions to its central angles, which is the only thing I can think of doing.

I would really like to work this out for myself, but I keep getting confused with diagrams. Is it perhaps a simpler problem than I'm making it out to be? Thanks a lot!

2. You will have to define things better than that. A couple of thought question...

Is "actual distance" the distance achieved by boring a hole through the Earth? If that is not what "actual distance" means, then it's quite possible that 5 miles = 5 miles.

Normally, one would measure "actual distance" as the distance actually travelled. Why would the curvature of the Earth make any difference?

3. Originally Posted by HyperKaehler
I am working through the book "Geometry - A High School Course" by Serge Lang, and the only problem I have gotten stuck on so far is in chapter 3, which concerns the Pythagorean theorem. The question is in two parts, the first of which I have no trouble with, but the second of which puzzles me:

a. A ship travels 6 km due South, 5 km due East, and then 4 km due South. How far is it from its starting point? (The answer is NOT 15 km!) You may assume that the path of the ship lies in a plane.
b. If actually the ship starts at the North Pole, on earth, what would the actual distance be?

For part a, I got the answer $5\sqrt{5}$ km for the distance. But I am confused by the second part, which seems to add the curvature of the Earth into the equation. Earlier in the book there was a section on arcs which involved a couple of questions calculating latitude and longitude (and for which we were told to use 40,000 km for the circumference of the Earth). But there is nothing in it about relating distance on the Earth's surface in two dimensions to its central angles, which is the only thing I can think of doing.

I would really like to work this out for myself, but I keep getting confused with diagrams. Is it perhaps a simpler problem than I'm making it out to be? Thanks a lot!
Hi HyperKaehler,

You calculations in Part (a) seem to be correct.

For Part (b), you travel 6 miles South. When you make your turn to the East, no matter how far you travel, you will still be the 6 miles from your starting point at the N. Pole.

So, when you move South again for another 4 miles, you are actually only 10 miles from the N. Pole.

4. Ah! Thank you so much. I was overthinking again as usual.

5. Originally Posted by HyperKaehler
I am working through the book "Geometry - A High School Course" by Serge Lang, and the only problem I have gotten stuck on so far is in chapter 3, which concerns the Pythagorean theorem. The question is in two parts, the first of which I have no trouble with, but the second of which puzzles me:

a. A ship travels 6 km due South, 5 km due East, and then 4 km due South. How far is it from its starting point? (The answer is NOT 15 km!) You may assume that the path of the ship lies in a plane.
b. If actually the ship starts at the North Pole, on earth, what would the actual distance be?

For part a, I got the answer $5\sqrt{5}$ km for the distance. But I am confused by the second part, which seems to add the curvature of the Earth into the equation. Earlier in the book there was a section on arcs which involved a couple of questions calculating latitude and longitude (and for which we were told to use 40,000 km for the circumference of the Earth). But there is nothing in it about relating distance on the Earth's surface in two dimensions to its central angles, which is the only thing I can think of doing.

I would really like to work this out for myself, but I keep getting confused with diagrams. Is it perhaps a simpler problem than I'm making it out to be? Thanks a lot!
Hi HyperKaehler,

For small distances on the earth's surface, such as those in this question
allow calculations to be performed on a flat plane.

You can gain a perspective by finding the angle of the ship's movement.

Earth circumference =40,000 km.

The ship sails 10 km south from the pole, which is $\frac{10}{40,000}=\frac{1}{4000}$

of the Earth's circumference, so the angle it moves through is $\frac{360^o}{4000}=\frac{9^o}{100}=0.09^o$

This is one-tenth of a degree, for which the attachment gives a perspective.
At this scale, we don't distinguish between the arc and a straight distance.

Earth radius = $\frac{40,000}{2{\pi}}\ km$

Hence, if we use the cosine rule to calculate the straight distance,
comparing it with the arc length

$L^2=(2)\frac{40,000^2}{4{\pi}^2}-(2)\frac{40,000^2}{4{\pi}^2}cos(0.1^o)=99.999979\ km^2$

$L=9.999989\ km$