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Therefore the whole line AH will represent the time along AD, DB. Similarly the line AC or AF will represent the time along AC; and FH being by construction equal to 3/2 of CB, it will represent the time along CB in the medium; and in consequence the whole line AH will represent also the time along AC, CB. Whence it appears that the time along AC, CB, is equal to the time along AD, DB. And similarly it can be shown if L and K are other points in the curve CDE, that the times along AL, LB, and along AK, KB, are always represented by the line AH, and therefore equal to the said time along AD, DB.
Cb. Cerebellum. M. Midbrain. P. Pons. B. Bulb. I-XII. Cranial nerves. *The Bulb*, or medulla oblongata, is, properly speaking, an enlargement of the spinal cord within the cranial cavity. It is somewhat triangular in shape, and lies immediately below the cerebellum. It contains important clusters of cell-bodies, as well as the nerve fibers that pass from the spinal cord to the brain.
And one will easily see the truth of this, namely that CB being larger than 2/3 of AB, the waves excited beyond the plane AB will have no common tangent if about the centres K one then draws circles having radii equal to 3/2 of the lengths LB to which they correspond. For all these circles will be enclosed in one another and will all pass beyond the point B.
But there could be no position of the spheroid which would have the same relation to these three sections except that in which the axis was also the axis of the solid angle C. Consequently I saw that the axis of this angle, that is to say the straight line which traversed the crystal from the point C with equal inclination to the edges CF, CA, CB was the line which determined the position of the axis of all the spheroidal waves which one imagined to originate from some point, taken within or on the surface of the crystal, since all these spheroids ought to be alike, and have their axes parallel to one another.
For if the angle DAQ or CBA is such that in the triangle ACB, CB is equal to 2/3 of AB, or is greater, then AN cannot form one side of the triangle ANB, since it becomes equal to or greater than AB: so that the portion of wave BN cannot be found anywhere, neither consequently can AN, which ought to be perpendicular to it. And thus the incident ray DA does not then pierce the surface AB.
The piece C, then, of the wave AC, will in a certain space of time have advanced as far as the plane AB following the straight line CB, which may be imagined as coming from the luminous centre, and which consequently will cut AC at right angles.
The piece C of the wave AC, will in a certain space of time advance as far as the plane AB at B, following the straight line CB, which may be supposed to come from the luminous centre, and which in consequence is perpendicular to AC. Now in this same space of time the portion A of the same wave, which has been hindered from communicating its movement beyond the plane AB, or at least partly so, ought to have continued its movement in the matter which is above this plane, and this along a distance equal to CB, making its own partial spherical wave, according to what has been said above.
Which wave is here represented by the circumference SNR, the centre of which is A, and its semi-diameter AN equal to CB.
It was stated, in Article 26, that the straight line CS, which in the preceding figure is CH, being the axis of the crystal, that is to say being equally inclined to the three sides CA, CB, CF, the angle GCH is 45 degrees 20 minutes. This is also easily calculated by the same spherical triangle.
If one considers further the other pieces H of the wave AC, it appears that they will not only have reached the surface AB by straight lines HK parallel to CB, but that in addition they will have generated in the transparent air, from the centres K, K, K, particular spherical waves, represented here by circumferences the semi-diameters of which are equal to KM, that is to say to the continuations of HK as far as the line BG parallel to AC. But all these circumferences have as a common tangent the straight line BN, namely the same which is drawn from B as a tangent to the first of the circles, of which A is the centre, and AN the semi-diameter equal to BC, as is easy to see.
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