Take a compass and draw a circle on a globe. Then take the same compass, opened to the same width, and draw a circle on a flat piece of paper. Which circle has more area?
If the circle is small compared to the radius of the globe, then the two circles will be approximately equal because a small area on a globe is approximately flat.
To get an idea what happens for larger circles, let’s a circle on the globe as large as possible, i.e. the equator. If the globe has radius r, then to draw the equator we need our compass to be opened a width of √2 r, the distance from the north pole to the equator along a straight line cutting through the globe.
The area of a hemisphere is 2πr². If we take our compass and draw a circle of radius √2 r on a flat surface we also get an area of 2πr². And by continuity we should expect that if we draw a circle that is nearly as big as the equator then the corresponding circle on a flat surface should have approximately the same area.
Interesting. This says that our compass will draw a circle with the same area whether on a globe or on a flat surface, at least approximately, if the width of the compass sufficiently small or sufficiently large. In fact, we get exactly the same area, regardless of how wide the compass is opened up. We haven’t proven this, only given a plausibility argument, but you can find a proof in [1].
Note that the width w of the compass is the radius of the circle drawn on a flat surface, but it is not the radius of the circle drawn on the globe. The width w is greater than the radius of the circle, but less than the distance along the sphere from the center of the circle. In the case of the equator, the radius of the circle is r, the width of the compass is √2 r , and the distance along the sphere from the north pole to the equator is πr/2.
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[1] Nick Lord. On an alternative formula for the area of a spherical cap. The Mathematical Gazette, Vol. 102, No. 554 (July 2018), pp. 314–316
This broke something in my brain – so will put this out here and wait for community illumination —
Slice a sphere of radius R in half, (along equator, say) and place it on a table, the area of the planar circle is clearly πR^2. The area of the hemisphere is half that of the full sphere (4πR^2) …so 2πR^2.
The area of the hemisphere and the planar circle which it projects down to are clearly unequal.
I think I understand the source of the confusion, and so I added one more paragraph to the post to explain. The key is that the compass width is not the radius of the circle on the sphere.
Illumination Unlocked. Thanks John.