If) 
the tanpfent line to the helix traced on the soffit thiou 2 ’h G. 
The tangent to the helix lies in the plane G^'H, so that 
by drawing making the angle 0 with gh and setting off 
equal to L/i^, the vertical trace of the tangeut to the 
helix in G is found. Hence the vertical trace of the tan- 
gent plane to tlie helical surface at G passes through H and 
is parallel to gC, the vertical projection of the generator 
at G. 
Draw HK parallel to Cg', meeting the vertical line g'L 
at K. 
As the triangles KH^' and CL^' are similar, 
Kg' ^ Cg' _Ch hE, 
Bg’ " CL “CL'^L ~ Lh, 
But H^'=L/q by construction, 
Therefore K^'=r6E^ = CO. 
Consequently the line HK produced will pass through 
O and the vertical projection of the tangent to the face 
joint at G is 0^'. 
In the same way it may be shown that the tangent to a 
face joint at any other point of AGD passes through 0. 
Q.E.D. 
Let the angle BC6=rSr= angle of obliquity of the arch. 
Then B6=r tan where r — the radius of the soffit. 
Kh=:Bh tan 0 = CO. 
Therefore CO = r tan 3” tan 
This is the same value for the eccentricity as that found 
by Buck, and is perfectly general, depending, as it does, 
only on the radius of, the cylinder, the angle of skewback 
for that cylindei', and the obliquity of tlie arch. The sim- 
plest method, however, of obtaining the point 0 is that 
shown above by means of a geometrical construction. 
