for resolving Transcendental Equations. 435 



IJ. Moreover, let the index circle instead of being as before con- 

 centric with the axis C be now described about the center /, and 

 the index arm with its vernier be connected with that axis as 

 we before supposed to be with the axis C. 



(17.) Put u for the arc read off on this circle or the angle XIJ, 

 also let e = CB, BD = BE = 1 ; then since the angle GCB is twice 

 XIJ, or = 2 u, we have 



MJ = DE + EJ- DEM = JDE + BG + KJ - DEM, 

 = 2 u + e . sin 2 u + KL . tan u - DEM. 



DEM is the part of the thread included between its point of at- 

 tachment to the wheel, and the zero of the scale — it is therefore 

 an arbitrary constant. If we put c to represent it, and observe 

 that KL = CL- CG - GK= CI . cotan u - CB . cos 2 u - 1 



= b . cotan u - e . cos 2 u — 1, (putting b for the constant distance CI) 

 we shall have for the expression of MJ, 



MJ =2M + <?.sin2w — c + {b . cotan u — e . cos 2k-1j. tan u 

 = (2 u + b - c) + {e - 2) . tan u. 



If therefore we put 3IJ=2A +b — c; and take e = 2p + 1, the 

 relation between A and u becomes 



u + p . tan w = A, 

 which is the equation proposed. 



(18.) In order then to adopt the above construction to any given 

 case of this equation we must set the slider so that the distance 

 BC shall = Sip + 1. That is to say, the zero of the scale of the 

 divisions on the slider must commence at a distance from C equal 

 to the unit or radius of the excentric + that of the thread, and 

 its graduation must be into parts double of the corresponding parts 

 of p. In like manner the graduation of the vertical scale M J must 

 begin at the point where the revolving straight edge intersects 



Vol. IV. Part III. 3 K 



