Wikipedia:Reference desk/Archives/Mathematics/2022 June 12

= June 12 =

Accuracy of flat-earth approximation
I know that, for short distances, we can get away with pretending that the Earth is flat. For example, if I measure 3 meters in a horizontal line, then turn 90 degrees and measure 4 meters in another horizontal line, then the distance to my starting point will be for all intents and purposes 5 meters. What about longer distances? If I were to measure 3000 meters horizontally, and then turn 90 degrees and measure 4000 meters horizontally, then to what level of (im)precision would I need to be working to get away with saying that the distance from my starting point to my ending point is 5000 meters? If I am working, say, to the nearest centimeter, then at what distance can I no longer get away with pretending that the Earth is flat? 32.219.123.165 (talk) 05:49, 12 June 2022 (UTC)


 * You can use the Spherical law of cosines for this problem, or more simply the spherical analogue of Pythagoras' theorem, $$\cos c/R = \cos a/R \cos b/R$$, where R is the radius of the earth and a, b and c the (great circle) lengths of the sides, with c as the "hypotenuse". Applying this to a 3-4-5 triangle I calculate that for sides of 300km and 400km the error in c is about 0.02%, and for 3000km and 400km it's 2.49%. To get an error of 1% we need sides of about 1950 and 2600. AndrewWTaylor (talk) 06:44, 12 June 2022 (UTC)


 * The following table assumes a perfectly spherical Earth with a radius of 4000 miles. The side lengths, measured on the surface along a great circle, are in miles:

a    b     c       3     4     4.9999997 30   40    49.9997     300   400   499.7    3000  4000  4657    6000  8000  6401    9000 12000  3599
 * --Lambiam 08:05, 12 June 2022 (UTC)
 * Fun fact:

a           b            c    75398.22369 100530.96492      0.000000000000
 * --Lambiam 08:38, 12 June 2022 (UTC)


 * For a concrete answer to the specific question about working to the nearest centimetre, I take the Earth radius to be $6,378,137 m$, the largest of the radii commonly used in mathematical models. The hypotenuse of a flat right triangle with sides $11,539.05 m$ and $15,385.4 m$ equals $19,231.75 m$. But the hypotenuse when laid out on our orb is only $19,231.743 m$, which, rounded to the nearest centimetre, is $19,231.74 m$. An error of one 23995th part of one degree in the right angle gives an equally large discrepancy. --Lambiam 15:00, 13 June 2022 (UTC)