22 Rev. 0. Fisher — Oblique and Orthogonal Sections. 



Then we have by spherical trigonometry, in the triangle E D B 

 which is right-angled at J), 



cos E£=cos BD cos ED, 



or sin ^=sin a sin /3 (1) 



Also the position of the plane of with reference to the horizontal 

 line ^ C is the angle CAF=90°—GF. 



Let CAF=(j). 

 Then we have in the triangle E D B, right-angled at D, 



sin D E=tan D B cot DEB. 

 And in the triangle E G F, right-angled at G, 



sin E G=tan GF cot D E B. 

 But E G=90° and sin E G=l 



.'. eliminating DEB 



tan 0=:cos /3 tan a (2) 



The direction in which the dimension of the trace will be the same 

 as that of the original surface will be at right angles to the plane 

 E AB, and therefore to the line A F. 



The direction upon tlie orthogonal section of the cylinder which 

 will be most distorted in the trace will be that in which the plane 

 E B F cuts it. Hence it will make the angle C B F or E B D with 

 the horizon. Call it ■v/r. 



From the right-angled triangle E D B. 



sin D B=tan E D cot E B D 

 or cos a=cot ^ cot -v^ 



.*. cot -v|r=cos a tan /3. (3) 



(Hence cot yfr tan ^=sin a sia /3 



=sin e.) 



Equation (2) shows how the obliquity of the trace is inverted 

 when the cutting plane is inclined in the opposite direction to the 

 horizon, as upon opposite sides of a railway cutting ; for in that case 

 we must substitute 180° — /3 for /3, and then, 

 tan <f)'=i — cos /3 tan a, 

 whence 4>' =1S0°—(f). 

 If the axis of the prism should not be horizontal, but lie in a plane 

 inclined at an angle j to the horizon, it is obvious that our demon- 

 stration will be rendered applicable to this case also by writing 

 /3 — 7 for /3. The angle a will not however lie in a horizontal plane. 



Fig. 2. 



Knowing the strike of a surface, our equations, 

 cot yfr =cos a tan /3, 

 tan 9=:cos /S tan a, 

 and sin ^:=sin a sin jS, 

 enable us to delineate its trace on any plane. 



