PALMER: THE HYPERBOLIC SPIRAL. 167 
point Q, and striking a spiral arc across FH, which is drawn with 
any assumed radius, known in terms of OA. (2) By setting at E 
of 6’, and striking across FH of 6’. (3) Setting to R, and striking 
Fig. 9. 
across arc CM of 6”. (4) Setting to M, and striking an arc across 
FH of @. 
(1) Set at Q,and strike arc at W. Draw WO and find Z HOW 
in terms of 6, and r and 1’, 
Bi tS Kas 
Mae ya ORL eNO = ZR ON Salt OO =O oa EL OW: 
Then gt Ze LOW , and from above ules ae ESL 
2 r 20 
2r—tr’ 
And Z How—( : jin part of 4 6. 
(2) Set the instrument at E, where CO of @” cuts arc AB, and 
strike arc SX, finding X. XO determines some fraction of 6, which 
can be found in terms of r, r’, and r”. 
The Z EOA, through which the instrument turns, =(0—0")= 
Z FOX. Hence the fraction of 6, determined, Z HOX=6’—(0—@’), 
which =(6@'-+-6’)—6. And the ratio of this’ to 0, or the part ex- 
pressed as a fraction of 6, is 
0’ +6" I I 
( 6 )— or (tor) —1 
(3) Setthrough Rand get point S. To find relation of Z MOS 
fone A ROR== (70) COS. So, 7 SOM] 8 — (0%) = 
(20"—6’). Or 
_(20"—#) 
oe 6 
ES 
part of Z 0, or =| yo =) part. 
3 r 
