58 A. 0. Lane — Estimation of the Optical Angle. 



It may be shown that by exchanging a'—Q and — (/—/?) we shall only exchange 

 cos 2V and —cos 2 V. So that we need consider but the following cases: 



1. In the rhombic system (j)=o to find 2V, (a) given 6(a — 3) and <5(y—ft) with 

 y — 3^>3 — a. Lay off along MO proportional to <5(y—j3) the abscissa Ma, and the 

 ordinate ab in the same proportion to 6(a — 3). From the point b proceed in the 

 direction mb till yon strike ON at c. From c proceed | to NP, meeting the hyper- 

 bole at d, then turn at right angles and go |ON till we meet NP at e. Then 

 eN= + cos 2V and 2V corresponding may be read from the outer scale. If 

 y—3<^3—a we must interchange and 2V will be — . 



(b) Given d(y—a) and 6{a—3) to find 2V. Lay off from M an abscissa Mk pro- 

 portional to d(y — a) and as an ordinate kl in the same proportion to 6(a— 3). From 

 I proceed in the direction Ml till you strike ON, at m. Thence go || to OM till we 

 strike the zigzag MHP, at n. Turning at right angles go | to ON till we strike 

 NP, at p. pN (positive) is +cos 2V if n is on MH, —cos 2V if n is on HP. In- 

 terchange for 6(y—a) and 6(y—j3). 



2. In the monoclinic system, f being given, and 6(a'— (5) (a) and also 8(y'—3) 

 greater than it. Proceed as in l a till we get e. Thence go toward till we meet 

 a line | to OM, and at a distance yO=cos 2<j> from it, at g. From g drop upon NP 

 again, at h. hn is +cos 2T if cos 20 is +, otherwise — . Interchange as usual. 



(b) And also 6(y—a). Find pn as in l b and subtract 1— cos 20, which may be 

 read off from ON from it. 



H. Given 2V and one of the three (y—a) (y—3) (y— a)\ to find the other two 

 work backward; e.g. (a) given y—a. If joN=cos 2V erect a perpendicular cut- 

 ting PUM ai n' and n. Draw lines through n and n' || to MO, cutting ON at m 

 and m'. Draw roM and w/M. Lay off Mk=-y—a and construct an ordinate, cut- 

 ting wM and w?'M at I and /'. Then M and JeV are a— 3 and —(y—fi) [if cos 2V 

 is +]. 



(b) (a— 3) given. Proceed as before 3" till we get m and m'. Then lay off on 

 ON Oq=za—(3. Erect a perpendicular ql to ON, cutting ml at I. Then draw the 

 line HV at right angles to gl, Mk is y—a and Vk=y — 3. Of course these various 

 coordinate lines should be drawn once for all in complete sets. 



Michigan Mining School, Houghton, Mich., Sept. 14, 1889. 



