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successive sets of twenty flights : thus I find cos 29 to he positive for eighteen out of thirty, and %l sin 20 to be negative for nineteen out of the thirty. § 38. A very interesting test-case is represented in the accompanying diagram, (fig. 6)—a circular boundary of semi- Fig. 6. circular corrugations. In this case it is obvious from the symmetry that the time-integral of kinetic energy of component motion parallel to any straight line must, in the long run, be equal to that parallel to any other. But the Boltzmann-Maxwell doctrine asserts, that the time-integrals of the kinetic energies of the two components, radial and transversal, according to polar coordinates, would be equal. To test this, I have taken the case of an infinite number of the semicircular corrugations, so that in the time-integral it is not necessary to include the times between successive impacts of the particle on any one of the semicircles. In this case the geometrical construction would, of course, fail to show the precise point Q at which the free path would cut the diameter AB of the semicircular hollow to which it is approaching ; and I have evaded the difficulty in a manner thoroughly suitable for thermodynamic application such as the kinetic theory of gases. I arranged to draw lots for 1 | out of the 199 points dividing AB into 200 equal parts. This was done by taking 100 cards *, 0, 1 ..... 98, 99, to represent distances from the middle point, and, by the toss of a coin, determining on which side of the middle point it was to be (plus or minus for head or tail, frequently changed to avoid possibility of error by bias). The draw for one of the hundred numbers (0 . . . . 99) was taken after very thorough shuffling of the cards in each case. The point of entry having been found, a large-scale geometrical construction was used to determine the successive points of impact and the inclination 0 of the emergent path to the diameter AB. The inclination of the entering path to the diameter of the semicircular hollow struck at the end of the flight, has the same value 0, If we call the diameter of the large circle unity, the length of each flight is sin 0. Hence, if the velocity is unity and the mass of the particle 2, the time-integral of the whole kinetic energy is sin 0; and it is easy to prove that the time-integrals of the components of the velocity, along and perpendicular to the line from each point of the path to tho centre of the large circle, are respectively 0 cos 0, and sin 0—0 cos 0. The excess of the latter above the former is sin 0 — 20 cos 6. By summation for 143 flights we have found, 2 sin (9 = 121*3 ; 220cos6= 108’3 ; whence, 2 sin 0 — 220 cos 0=13*0. This is a notable deviation from the Boltzmann-Maxwell doctrine, which makes 2 (sin 0 — 0 cos 0) equal to 20 cos 0. We have found the former to exceed the latter by a difference which amounts to 10*7 of the whole 2 sin 0. Out of fourteen sets of ten flights, I find that the time-integral of the transverse component is less than half the w hole in twelve sets, and greater in only two. This seems to prove beyond doubt that the deviation from the Boltzmann-Maxwell doctrine is genuine; and that the time-integral of the transverse component is certainly smaller than the time-integral of the radial component. * I had tried numbered billets (small squares of paper) drawn from a bowl, but found this very unsatisfactory. The best mixing |