Thought Toys · Shape & space · Exhibit 134
Every triangle you were ever taught has angles adding to exactly 180°. Draw one on a globe — three straight-as-possible paths joined corner to corner — and the sum comes out bigger. The overshoot is not a mistake in the drawing and not a distortion of the map. It is the triangle's area, measured in the world's own radius, in one line. Which means a creature living on the surface could measure how curved its world is without ever leaving it.
A triangle on a globe — drag a corner ● the triangle ● its three angles ● the far side
Keyboard: focus the globe, then ← → ↑ ↓ to slide the ringed corner (hold Shift for bigger strides), and , . to ring a different corner — the same thing dragging does.
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The disc is a globe seen from outside. The amber triangle's sides are great circles — the straightest path there is on a sphere, the route a long-haul flight approximates. At each corner the cyan arc is the interior angle, measured exactly as a surveyor would measure it standing on that spot: the angle between the two paths as they leave. The three angles are stacked end to end in the bar beside the globe, against a mark at 180°. The bar always sticks out past the mark — by a hair for a tiny triangle, by a lot for a big one.
How far it sticks out is the whole story. Take the angle sum, subtract 180°, and what is left — the excess — is the triangle's area, in units of the world's radius squared. Not approximately. The third box under the picture is the area worked out by a completely different route, one that never looks at an angle, and it matches the excess to the last digit on every drag.
Now shrink the triangle. As it gets smaller the excess shrinks too, and faster than the triangle does: once it is small, halving the size cuts the excess to a quarter, which is how an area behaves and not how a length does. (For a big triangle the ratio is a little more than four, because a sphere's triangles are not scale copies of one another; it settles onto four exactly as they shrink.) By the time the triangle is a garden the sum is 180° to within a twentieth of a degree. That is why nobody at a survey of a building plot ever noticed the world was round. Flat geometry is not wrong; it is the small-triangle limit of the curved kind.
The octant is the case you can check by counting. Three corners at right angles to each other — the North Pole and two points a quarter-turn apart on the equator — give three angles of 90°, a sum of 270°, an excess of 90°, and a triangle that is visibly one eighth of the sphere. One eighth of the sphere's area, 4πR², is πR²/2; and 90° is π/2. The formula is not being clever.
The radius slider makes the last point. Blow the world up to twice the size and the triangle's area quadruples, but not one angle changes, because uniformly enlarging a shape leaves every angle in it alone. So a surveyor who measures the three angles and the area of a triangle in their own field has, in the ratio excess ÷ area, the number 1/R² — the curvature of their world, read off the ground, with no need to look up or step off. On a sphere that ratio is exactly 1/R² everywhere; on a lumpier world it is the average curvature inside the triangle, and a small enough triangle reads the curvature at one spot.
Extend each side of the triangle all the way round into a full great circle. Three great circles cut the sphere into eight triangles, and they come in four opposite pairs of equal area. Each pair of circles bounds two orange-segment shapes, the lunes; take, for each pair, the lune that contains the triangle. Each of the three covers a fraction of the sphere equal to its angle over 360°. Add the three lunes up and you have covered the hemisphere once, plus the triangle two extra times. Rearranging that one sentence is the theorem.
The rule, exactly. For a triangle with great-circle sides on a sphere of radius R,
with interior angles A, B, C in radians,
area = (A + B + C − π) · R²
(Thomas Harriot 1603, published by Albert Girard 1629). The page computes each angle from the tangent
directions of the two sides leaving that corner; the area shown in the third box comes from the
solid-angle identity tan(E/2) = |a·(b×c)| /
(1 + a·b + b·c + c·a)
(Van Oosterom & Strackee 1983), which uses the corners as vectors and no angle at all — taken
with the two-argument arctangent, so it stays right when the denominator turns negative for a triangle
bigger than a quarter of a hemisphere. For a triangle with
geodesic sides on any curved surface the same statement is the Gauss–Bonnet theorem: angle sum
− π equals the integral of the Gaussian curvature over the inside.
Checked before this page was written (improve/verify/134-spherical-excess.js):
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