The diagram shows osculating circles to the ellipse at points A, B and C. At A the curvature is $\frac{2}{3}$, at B it is $\frac{1}{12}\approx 0.083$ and at C it is $0.288$.

A perfect sphere has constant curvature everywhere on the surface whereas the curvature on other surfaces is variable. For example on a rubgy ball the curvature is greatest at the ends and least in the middle. Measuring curvature at a point using curves through that point on the surface will not work. Which plane curve should we use? At the '2' on the rugby ball, the curve in one direction, going between the B and the E, has greater curvature than the curve along the length of the ball. Gauss proved that, taking the curvatures in all directions at a point on a surface, the product of the maximum and minimum curvatures at the point is constant when the surface is distorted provided that lengths in the surface are unchanged.

One method used to measure the Gaussian curvature of a surface at a point is to take a small circle of radius $r$ on the surface with centre at that point and to calculate the circumference or area of the circle. If the circumference is $2\pi r$ and the area is $\pi r^2$ the surface is flat and is said to have zero curvature. If the circumference is less than $2\pi r$ and the area is less than $\pi r^2$ the surface has positive curvature; if the circumference is greater than $2\pi r$ and the area is greater than $\pi r^2$ the surface has negative curvature.

There are three distinct geometries for the three types of surfaces and each of these geometries has its own trigonometry with the results in each trigonometry having counterparts in the other two trigonometries. Euclidean geometry is the geometry of surfaces with zero curvature. Spherical geometry, also known as elliptic geometry, is the geometry of surfaces with positive curvature. Hyperbolic geometry is the geometry of surfaces with negative curvature. See the article How Many Geometries Are There? .

What about the curvature of the surface of a cylinder. Clearly the top and bottom are flat but what about the surface, often called the curved surface, where the label is wrapped around as in the illustration? Along the length of the cylinder the curvature is zero and in other directions there is positive curvature so the product of the maximum and minimum curvatures is zero making the Gaussian curvature zero. To see this another way, unwrap the label and you can flatten it out, draw a small circle of radius $r$ anywhere on the label and it has a circumference of $2\pi r$ and an area of $\pi r^2.$ Wrap the label around the can again and the circumference and area of the circle you have drawn do not change as the paper takes the shape of the surface of the cylinder. So the surface of the cylinder has zero Gaussian curvature.

In order to find the areas of circles on spheres we shall use a result discovered by Archimedes of Syracuse who invented calculus type methods almost two thousand years before Newton and Leibniz. Archimedes made many discoveries and inventions including the Archimedean Screw pump, still in use today, depicted on this stamp, and you can find out more about him on the St Andrew's website. The result we shall use is depicted in the diagram on the right and said to have been on Archimede's tombstone.

The cylinder shown has radius $r$, height $2r$ and area $4\pi r^2$, the same as the surface area of the sphere inside. Moreover the area of a strip on the surface of a sphere cut off by two parallel planes is equal to the area cut off by the planes on the circumscribing cylinder.

In this cross section, A and B are planes at a small distance dy apart, cutting the sphere and circumscribing cylinder. The 'triangle' with sides dy, dx and dc is so small that we can treat the arc dc as if it were a straight line and the hypotenuse of this triangle. This tiny triangle is similar to the triangle shown with sides $x,y$ and $r$ so that ${\displaystyle \frac{\mathrm{dy}}{\mathrm{dc}}=\frac{x}{r}.}$ The area of the strip cut off by planes A and B on the surface of the sphere is $2\pi \mathrm{xdc}$ because this strip has circumference $2\pi x$ and height dc. The area of the band on the cylinder is $2\pi \mathrm{rdy}$. As we have shown that $\mathrm{xdc}=\mathrm{rdy}$ it follows that these areas are equal.

Now we study the curvature of the surface of a sphere of unit radius. Take any point N and draw a circle centre N radius $r$ on the surface of the sphere where the radius is an arc in the surface of the sphere subtending an angle of $r$ radians at the centre of the sphere. The circumference of the circle centre N radius $r$ is $2\pi \sin r$. Because $\sin r<r$ for all non-zero values of $r$ we see that this circumference $2\pi \sin r$ is less that $2\pi r$ showing that the surface has positive curvature.. By Archimedes' method the area of this circle is $2\pi r(1-\cos r).$ You can verify that this is less than $\pi r^2$, and hence that the curvature of the surface of the sphere is positive, by drawing the graphs of $y=\cos x$ and $y=1-\frac{x^2}{2}$ on the same axes and showing that $\cos r>1-\frac{r^2}{2}.$

Tie a long rope to a post at this point and, keeping the rope taught, walk around a circle with the post as centre. As you have to walk uphill and downhill as well as around the post you walk farther than you would walk on flat ground showing that the curvature at that point is negative.

Although the curvature is concentrated at 16 points, the block shown with a hole through it is analagous to the torus (or doughnut shaped solid) shown in yellow.

We investigate the curvature at the red spot at the edge of the hole in the block. The angle deficiency at the red spot is $-{90}^o$. The circumference of a circle of radius $r$ with centre at the red spot on the block is $\frac{5}{4}\times 2\pi r$ so the curvature at that point is negative.