When the objects are at rest, and the Kaleidoscope in motion, a new series of appearances is presented. Whatever be the direction in which the Kaleidoscope moves, the object seen by direct vision must always be stationary, and it is easy to determine the changes which take place when the Kaleidoscope has a progressive motion over the object. A very curious effect, however, is observed when the Kaleidoscope has a rotatory motion round the angular point, or rather round the common section of the two mirrors. The picture created by the Instrument seems to be composed of two pictures, one in motion round the centre of the circular field, and the other at rest. The sectors formed by an odd number of reflexions are all in motion in the same direction as the Kaleidoscope, while the sector seen by direct vision, and all the sectors formed by an even number of reflexions, are at rest. In order to understand this, let =M=, Fig. 10, be a plane mirror, and =A= an object whose image is formed at _a_, so that =_a_ M = A M=. Let the mirror =M= advance to =N=, and the object =A=, which remains fixed, will have its image _b_ formed at such a distance behind =N=, that =_b_ N = A N=; then it will be found that the space moved through by the image is double the space moved through by the mirror; that is, =_a b_ = 2 M N=. Since =M N = A M - A N=, and since =A M = _a_ M=, and =A N = _b_ N=, we have =M N = _a_ M - _b_ N=; and adding =M N= or its equal =_b_ M + _b_ N= to both sides of the equation, we obtain =2M N = _a_ M - _b_ N + _b_ N + _b_ M=; but =-_b_ N + _b_N = 0=, and =_a_ M + _b_ M = _a b_=; hence =2M N = _a b_=. This result may be obtained otherwise, by considering, that if the mirror =M= advances _one inch_ towards =A=, one inch is added to the distance of the image _a_, and one subtracted from the distance of the object; that is, the difference of these distances is now two inches, or twice the space moved through by the mirror; but since the new distance of the object is equal to the distance of the new image, the difference of these distances, which is the space moved through by the image, must be two inches, or twice the space described by the mirror. [Illustration: FIG. 10.] Let us now suppose that the object A advances in the same direction as the mirror, and with twice its velocity, so as to describe a space =A α = 2 M N = _a b_=, in the same time that the mirror moves through =M N=, the object being at α when the mirror is at =N=. Then, since =A α = _a b_= and =_b_ N = A N=, the whole =α N= is equal to the whole =_a_ N=, that is, =_a_= will still be the place of the image. Hence it follows, _that if the object advances in the same direction as the mirror, but with twice its velocity, the image will remain stationary_. [Illustration: FIG. 11.]