Steven Dutch, Natural and Applied Sciences, University
of Wisconsin - Green Bay

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**Note:** All formulas on this page are in terms of the primitive circle.

The proof here is not a completely rigorous one but it should convince you that the basic property of the stereographic projection is true. A plane (blue) cuts the sphere to form a small circle with diameter PQ. P and Q project onto the projection plane at P' and Q'. |

- Let the angular diameter of the small circle be 2a and its angular distance from the point of tangency be 2u. Then the angles measured from V, the projection point, are a and u, respectively. (This is a well-known property of circles. If you're not familiar with it and want proof, consider the angles in triangle VOQ.)
- Triangles VOP and POQ are both isoceles. Their base angles are therefore equal and of course their angles sum to 180 degrees. Thus angle VPO = a + u, and OPQ = 90 - a.
- From the above results, angle QPP' = 90+u
- Triangle VCQ' is a right triangle, therefore angle CQ'V=90-u and angle VQ'P=90+u
- Although not shown for reasons of clarity, angle PQV is easy to find. We know the other angles of triangle PQV, so angle PQV = 180 - a - (a+u) - (90-a) = 90-a-u. Note that this is equal to angle PP'Q'.

Now, the surface containing point V and circle PV is a cone. It's not a circular cone because the section cut by plane PV is oblique to the cone axis. A section perpendicular to the cone axis would have an elliptical cross-section. But note that plane CP' cuts the cone at the same angles, just in the opposite direction. Since plane PV cuts the cone in a circle, plane CP', with the same obliquity, should also cut the cone in a circle. Thus circle PV projects onto the plane as a circle, P'V'.

The key to constructing stereographic projections is constructing any circle given its center and radius. Assume the circle has radius r and its center is angular distance a from the point of tangency. It the circle is a unit circle, than CP'=tan((a+r)/2) and CQ'=tan((a-r)/2). |

The center of the circle on the sphere will not project as the center of the circle on the plane. The distance to the center of the projected circle is (CP'+CQ')/2 = tan((a+r)/2) + tan((a-r)/2).

Since tan u + tan v = sin(u + v)/(cos u cos v) we can let u = (a+r)/2 and v = (a-r)/2.
We can then write:

Radius of center =

- sin(((a+r)/2) + ((a-r)/2))/(cos((a+r)/2) cos ((a-r)/2) =
- sin a /(cos((a+r)/2) cos((a-r)/2)

We can also make use of the fact that cos u cos v = (cos(u+v) + cos(u-v))/2 to continue:

- sin a /(cos((a+r)/2) cos((a-r)/2) =
- sin a /(cos(((a+r)/2)+((a-r)/2)) + cos(((a+r)/2)-((a-r)/2))/2) =
- 2sin a /(cos a + cos r) =

The formula for the radius of the projected circle differs only in the middle sign, which is negative instead of positive. The radius of the projected circle is (CP'- CQ')/2 = tan((a+r)/2) - tan((a-r)/2). Proceeding as above, we obtain radius = sin r /(cos a + cos r)

Thus:

**Distance to center of projected circle = sin a /(cos a + cos r)**

**Radius of projected circle = sin r /(cos a + cos r)**

For great circles, the radius is 90 degrees and cos r = 0. Thus the distance and radius formulas for the projected circle become:

- Distance = tan a
- Radius = 1/cos a

If we want to construct a stereonet in the standard orientation with the great circles like meridians of longitude, then the circle that represents longitude w will have its pole at longitude w plus or minus 90. Thus the distance to the pole is -1/tan(w) and the center of the circle lies on the equator. The minus sign in the distance formula simply means that the center of the circle lies on the opposite side of the sphere from the meridian. The radius of the projected circle is 1/cos(w).

The formulas suggest the simple construction at left. Linear dimensions are in red,
angles in purple. It is useful to introduce angle u, since 1/tan w = tan(90-w) = tan u. Measure angle u from the prime meridian as shown and draw SPC. Point C is the center of the great circle. Alternatively, if the primitive circle is marked with degrees, simply measure off angles 2u and 2w as shown and construct SPC. |

For the great circles very close to the prime meridian, point C may be so far away it
is impossible to plot. It's useful to have some formula that relates distance AO and angle
w. (AO is sometimes called the *sagitta*, from the Latin word for arrow, because arc
NAS and line AO somewhat resemble a bow and arrow.) The formula is simple and follows
directly from the construction of the projection: **AO = tan (w/2)**.

For the small circles of the stereonet, a always equals 90 because the circles are centered on the pole. Thus, a circle of latitude l has radius 90-l. The distance to its center is 1/cos r = 1/sin l. Its radius is tan r = 1/tan l.

The simple construction at left can be used to plot a small circle of latitude l. Draw OP making angle l with the equator. Draw PC perpendicular to OP. C is the center of the circle. |

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*Created 1 March 1999, Last Update
14 December 2009*

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