Show how to distribute nine points around the surface of a sphere in such a way that each point is equidistant from its four nearest neighbors.
(In reply to
Solution by Jer)
I get a latitude for the smaller circle as 48.19° (and a resulting distance between points as : 70.53°):
The arc along the equator between a point and the longitude of the point on the smaller circle is 60°, refer to the latitude of the smaller circle as lat, and the angle between the meridian of the point on the smaller circle and the equator is 90°.
Call the unknown arc distance from the point on the equator to the point on the smaller circle d.
cos(d) = cos(60°)*cos(lat) + sin(60°)*sin(lat)*cos(90°) = cos(60°)*cos(lat)
The distance between two successive points on the smaller circle should be the same:
cos(d) = sin(lat)^2 + cos(lat)^2 * cos(120°)
cos and sin reversed as the sides of the triangle are the complement of the latitude
We need to solve
cos(60°)*cos(lat) = sin(lat)^2 + cos(lat)^2 * cos(120°)
to find the latitude.
Wolfram Alpha solves this as 114.59 * 0.42053 = 48.189° rather than 36.9°,
By the first formula this gives 70.5283° between points. The second formula gives the same, so this solves the equation.
Later in MATLAB:
clc, clearvars
syms lat
eqn=cosd(60)*cosd(lat)==sind(lat)^2+cosd(lat)^2*cosd(120);
solve(eqn,lat)
gives
(180*acos(2/3))/pi
>> (180*acos(2/3))/pi
ans =
48.1896851042214
PS: I've always been upset about the lack of teaching of spherical trig in the curricula I've had both in high school and college. But it was fun learning on my own, although I never got into haversines.
Edited on June 4, 2021, 7:11 am
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Posted by Charlie
on 2021-06-04 06:49:40 |
re: Solution -- Using spherical trig
