738 



H. F. REID ADDITIONAL NOTE ON GEOMETRY OP FAULTS 



at right angles to OA ; the angle through which the stratum must be 

 rotated to make it horizontal is 9i ; the point D on the stratum is moved 

 to D' by the rotation, and points on the axis OA are not displaced. 

 Now rotate the stratum around the vertical axis OS through an angle 9^ ; 

 D' moves to D", and A to A' . It is well known that any displacement 

 of a sphere about its center can be represented as a rotation around a 

 single axis, and that this axis is fixed if we know the displacements of 

 two points on the surface of the sphere (such as D and A in this ex- 



FiGURE I. — Combination of Finite Rotations Figure 2. — Combination of Finite Rotations 



ample). The point, R, where this axis pierces the surface of the sphere 

 must be equidistant from A and A" , and from Z) and Z)". To be equi- 

 distant from A and A'\ R must lie on a great circle passing midway 

 between A and A" ] that is, its azimuth must make an angle 9^2/2 with 

 OA. We must now find the point on this great circle which is equi- 

 distant from D and D" . It is simpler to do this by analytic than by 

 projective methods. We take the ordinary expressions for the distance 

 of R from D and D" measured on the sphere, equate them, and find 



tan 5 = sin f^l'^ (^ot 9'i/*2, 



which gives us the dip, <5^ of the axis of rotation OR. We have seen 

 that its azimuth makes an angle 9^ I '^ with OA ; its position is therefore 

 completely defined. 



To determine the amount of the rotation, s^, around OR, we solve the 

 right-angled spherical triangle RFA and find 



tan <fJ'^ 



tan <p /2 ^ 



sm 



d 



The angles ^ and <P can be found very quickly with a table of logarithms. 



