Two friends of mine are having a friendly dispute about a bit of math. The crux of the problem is this: If the earth were perfectly spherical, how much "bend" would there be in 260 miles? Or, in other words, how tall would a mountain 260 miles away have to be for its top to be visible?

The math for this is beyond me (maybe 40 years ago I knew how to figure it) so please try to keep the explanation relatively simple.

tanstaafl.

Using the diameter of the Earth at the equator of 7926.28 miles:

260 miles across the surface of the earth equates to:

(260 / (7926.28 * pi)) * 360 = 3.758863 degrees (260 miles divided by the circumference of earth in degrees - probably should have done it in radians but this still works)

Assuming the mountain is perpendicular to the earth, the distance from the centre of the earth to the tip of the mountain will be:

(7926.28 / 2) / cos (3.758863) = 3971.6839 miles (working out the length of the hypotenuse of a right angle triangle)

So the mountatain height will be

3971.6839 - (7926.28 / 2) = 8.5439 miles

Can draw a diagram if necessary...

Edit: Bah... Used the diameter in the preview on a google search which appears to be wrong. You can recalculate if you want.
http://www.google.com.au/search?sourceid...=earth+diameter

Looks like it should be 7926.41 miles here.
http://geography.about.com/library/faq/blqzdiameter.htm

Wikipedia has it in km but equates to 7926.55 miles diameter

That "edit reason" has to be one of the best I've ever seen in a forum. Sure beats my usual "fix typos"!

I just love this forum! Like a magic box of answers to anything! To top it all of you then get a whole discussion that usually follows. Who needs wikipedia!

To add to the original question, how far "off" is the shape of the earth from a 'perfect' sphere?

Refraction will let you see shorter things...

http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html

The frequent comparison I hear, which might be urban legend, is that if the earth were the size of a billiard ball; it would be smoother than a billiard ball.

edit: link

Smoothness is not roundness.

This reminds me of a fun one we had in an Algebra class-

If you had a rope that circled the Earth at the equator, how much more rope would you need in order to raise the rope's height above ground by one foot?

r=radius of earth in feet

circumfrence=pi*2*r

lg = (pi*(r+1)*2) - (pi*r*2)

lg = pi((r+1)*2 - r*2)

lg = 2pi((r+1) - r)

lg = 2pi(r+1-r)

lg = 2pi(1)

lg = 2pi

lg = 2*3.14

lg = 6.28

It's unintuitive that it doesn't matter what the radius is.

I have an obsession with geometry, and have in the past looked for a good free geometry program, and come up with some reasonable ones in the past. But this prompted me to look again, and I found a great one: GeoGebra.

Anyway, I created a diagram for this problem. It's saved in the attachment. If you assume that the earth has a circumference of 40,000km, a mountain would have to be over 7.29 miles high in order to be seen, ignoring atmospheric refraction.

Oh, and GeoGebra let me create sliders so that I can manipulate the circumference and distance. It also let me know that a 2m viewing height is nearly irrelevant.

Bitt,

You have 240 miles as the distance. The original problem was 260 miles. Changing that and the circumference and a slightly more correct km to miles conversion and it comes out as 8.5439 miles as calculated. The numbers aren't exactly the same though.

Just checking my maths is right is what I'm saying

The piece of rope one foot higher maths looks right but is certainly counter intuitive to me.

Oh, I wasn't trying to refute you or anything. I was mostly promoting GeoGebra.

No that's cool. I didn't think that. But your result of 7.29 vs 8.54 was significantly different enough for me to check the calculations.

The problem with the 1 ft higher rope, is that the expansion of area is not linear to the change in circumference.