(3)
| Fig. 40. |
Again, consider the point of the solid which was initially at A′ in the figure. This is displaced relatively to A′ through a space δψ perpendicular to the plane of the meridian, whilst A′ itself is displaced through a space cos θ δψ in the same direction. Hence
ζ = δφ + cos θ δψ.
(4)
To find the component displacements of a point P of the body, whose co-ordinates are x, y, z, we draw PL normal to the plane yOz, and LH, LK perpendicular to Oy, Oz, respectively. The displacement of P parallel to Ox is the same as that of L, which is made up of ηz and −ζy. In this way we obtain the formulae
δx = ηz − ζy, δy = ζx − ξz, δz = ξy − ηx.
(5)
The most general case is derived from this by adding the component displacements λ, μ, ν (say) of the point which was at O; thus
| δx = λ + ηz − ζy, δy = μ + ζx − ξz, δz = ν + ξy − ηx. |