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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)[reply]

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)[reply]

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)[reply]

Fun fact:

       a            b            c
   75398.22369 100530.96492      0.000000000000

--Lambiam 08:38, 12 June 2022 (UTC)[reply]

For a concrete answer to the specific question about working to the nearest centimetre, I take the Earth radius to be 6378137.0 m, the largest of the radii commonly used in mathematical models. The hypotenuse of a flat right triangle with sides 11539.05 m and 15385.40 m equals 19231.75 m. But the hypotenuse when laid out on our orb is only 19231.7433 m, which, rounded to the nearest centimetre, is 19231.74 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)[reply]