Let there be described on Q_q_ a semicircle cutting the ray CR at B, from which BV is drawn perpendicular to Q_q_; and let the proportion of the refraction for this section be, as before, that of the line N to the semi-diameter CQ.

Then as N is to CQ so is VC to CD, as appears by the method of finding the refraction which we have shown above, Article 31; but as VC is to CD, so is VB to DS.  Then as N is to CQ, so is VB to DS.  Let ML be perpendicular to CL.  And because I suppose the eyes R_r_ to be distant about a foot or so from the crystal, and consequently the angle RS_r_ very small, VB may be considered as equal to the semi-diameter CQ, and DP as equal to CL; then as N is to CQ so is CQ to DS.  But N is valued at 156,962 parts, of which CM contains 100,000 and CQ 105,032.  Then DS will have 70,283.  But CL is 99,324, being the sine of the complement of the angle MCL which is 6 degrees 40 minutes; CM being supposed as radius.  Then DP, considered as equal to CL, will be to DS as 99,324 to 70,283.  And so the elevation of the point I by the refraction of this section is known.

[Illustration]

41.  Now let there be represented the other section through EF in the figure before the preceding one; and let CM_g_ be the semi-ellipse, considered in Articles 27 and 28, which is made by cutting a spheroidal wave having centre C. Let the point I, taken in this ellipse, be imagined again at the bottom of the Crystal; and let it be viewed by the refracted rays ICR, I_cr_, which go to the two eyes; CR and cr being equally inclined to the surface of the crystal G_g_.  This being so, if one draws ID parallel to CM, which I suppose to be the refraction of the perpendicular ray incident at the point C, the distances DC, D_c_, will be equal, as is easy to see by that which has been demonstrated in Article 28.  Now it is certain that the point I should appear at S where the straight lines RC, rc, meet when prolonged; and that this point will fall in the line DP perpendicular to G_g_.  If one draws IP perpendicular to this DP, it will be the distance PS which will mark the apparent elevation of the point I. Let there be described on G_g_ a semicircle cutting CR at B, from which let BV be drawn perpendicular to G_g_; and let N to GC be the proportion of the refraction in this section, as in Article 28.  Since then CI is the refraction of the radius BC, and DI is parallel to CM, VC must be to CD as N to GC, according to what has been demonstrated in Article 31.  But as VC is to CD so is BV to DS.  Let ML be drawn perpendicular to CL.  And because I consider, again, the eyes to be distant above the crystal, BV is deemed equal to the semi-diameter CG; and hence DS will be a third proportional to the lines N and CG:  also DP will be deemed equal to CL.  Now CG consisting of 98,778 parts, of which CM contains 100,000, N is taken as 156,962.  Then DS will be 62,163.  But CL is also determined, and contains 99,324 parts, as has been said in Articles 34 and 40.  Then the ratio of PD to DS will be as 99,324 to 62,163.  And thus one knows the elevation of the point at the bottom I by the refraction of this section; and it appears that this elevation is greater than that by the refraction of the preceding section, since the ratio of PD to DS was there as 99,324 to 70,283.