DR. S. CROMPTON ON A PORTRAIT OF TYCHO BRAKE. T7 



Draw HK parallel to Cy, meeting the vertical line ^L 

 at K. 



As the triangles KH^ and CLy are similar, 

 Ky_Cy_C6_B6_6E, 

 H/ CL CL g\A LA/ 



But Hy=LAy by construction. 



Therefore K/=6E^ = C0. 



Consequently the line H K produced will pass through 

 O, and the vertical projection of the tangent to the face- joint 

 at G is Og'. 



In the same way it may be shown that the tangent to a 

 face-joint at any other point of AGD passes through 

 O. Q.E.D. 



Let the angle BC6 = d = angle of obliquity of the arch. 



Then B6 = /' tan ^, where r = Q,h, the radius of the soffit. 



E,6=B6tan(/) = CO. 



Therefore CO = r tan ■& tan ^. 



This is the same value for the excentricity as that found 

 by Buck, and is perfectly general, depending, as it does, 

 only on the radius of the cylinder, the angle of skew- 

 back for that cylinder, and the obliquity of the arch. The 

 simplest method, however, of obtaining the point O is that 

 shown above by means of a geometrical construction. 



XI. On a Portrait of Tycho Brake. 

 By Samuel Crompton, M.D. 



Eead October sist, 1876. 



Dr. Crompton exhibited a portrait in oil of Tycho Brahe, 

 and gave some particulars respecting the known portraits 



