Geometrical Construction for rectifying any Arc. 473 



further bisection will be undertaken, and the point H will be 

 used to determine the final rebult. 



Draw HK parallel to OA meeting AT at K and continue 

 EH until it cuts AT at I. 



Divide KI in the ratio 1 : 2 at point P. 



Then the straight line AP will be very nearly equal in 

 length to the arc AB. 



It is easy to demonstrate that this result is true. If < AOB 



is called a and 0A = 1, then AI = 2 n tan ^, where n represents 



the number of times (in the present instance 2) that the 

 process of bisecting the angles took place. Similarly 



AK = 2 sin~. 



2 n 



Therefore 



AP = 2»{sin| r + l/3(taniL- S m^)]. 

 Expanding this one finds 



AP = 2 n { {*j2 n ) - l/6(*/2 n ) B -f l/120(a/2 n y - . . . 



+ 1/3((«/2») + l/3(*/2*) 3 4- 2/150/2") 5 + . . . 

 _(«/2-)-+-l/6(a/2 w ) 3 -l/120( a /2-) 5 +...)} 

 «=2 w {(«/2 w )-f-l/20(^/2») 5 +...} = <l + l/20(a/2-) 4 ). 



The residual error l/20(a/2 n )4 is obviously reduced to 

 1/16 by each repetition of the bisecting process, and may 

 therefore in theory be made very small indeed in a very short 

 time. Even with but two bisections as in the above diagram, 

 the error is only of the order of 1 part in 5000. A. similar 

 construction with a 90° arc would give it to 6 places of decimals 

 if the bisecting process were repeated 5 times., 



Famborougli. 

 June 8th, 1918. 



[Note. 



The method is interesting though hardly practical. 



The details seem to be these : — 



(1) The angles ABE, AEH are right angles. 



For ABC = BA = 90°-", as is seen by dropping a 



perpendicular from on AB. 



