# Maths

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How many parallels of latitude have a radius of 2500km?

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Assuming the earth is a perfect sphere, and with a radius of 6400 km, there are exactly two parallels with a radius of 2500 km, located at approximately
cos-1(2500/6400)
=67° N or S.

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Show that the points (20°N,20°E) and (60°N,160°W) lie on the same great circle.Find their great circle distance apart.

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There is at least one great circle that passes through any two non-coincident points lying on a sphere.

If the two points are diametrically located, there is an infinite number of great circles that pass through the two points.

The proof is relatively simple. The two points and the centre of the sphere form a plane that bisects the sphere into two hemispheres.

As for the distance, we are into spherical trigonometry.
There are various formulas to calculate the central angle subtended at the centre. The great circle distance is therefore the radius multiplied by this angle (in radians).

There are three formulas
1. spherical cosine formula.
This formula does not apply to small distances between the two points (say less than 10 km) because numerical round-off errors are magnified by the difference of cosine of small angles.
2. Haversine formula works well for almost all cases except for points which are, or almost, diametrically placed (which is our case in point).
3. A more complicated formula applies to all cases, and that's the Vincenty formula.

Here are the results, using R=6371 km.
(1) 11119.49 km
(2) 9623.09 km (rejected)
(3) 11119.49 km

The respective formulas are, using
R=radius of the earth (assumed a sphere)
L =difference in longitude of the two points (in radians)
D=great circle distance
(1)
R*acos(sin(p1)*sin(p2)+cos(p1)*cos(p2)*cos(L))
(2)
2*R*asin(sqrt(sin((p1-p1)/2)^2+cos(p1)*cos(p2)*sin(L/2)^2))
R*atan2(sqrt((cos(p2)*sin(L))^2+(cos(p1)*sin(p2)-sin(p1)*cos(p2)*cos(L))^2),
sin(p1)*sin(p2)+cos(p1)*cos(p2)*cos(L))

Reference:
https://en.wikipedia.org/wiki/Great-circle_distance

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