1844.] 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



m 



DIAGRAM FOR DRAWING GOTHIC ARCHES, 

 By Peter Nicholson. 



To find the radius of an arc q s, concentric with a given arc A B, 

 to meet the sine B C, so that the difference of the radii A 5, may 

 divide rj s, tlie arc to be found into a given number of equal parts. 



(We are indebted to our vnhiable correspondent O.T. for the above 

 method of drawing a ciuquefoil head ; it is copied from a diagrani 

 made by Peter Niciiolson, in whose handvpriting is also the problem 

 annexed. It will be observed that the diagram does not suggest any 

 method of solving the problem, we therefore referred it to Mr. OLIVER 

 ByE-NLi who has favoured us with the following solution.) 



Let AO = r,qo=x,Co=:c, and the angle qor=:2e. 



. • . A 2 = g r := 2 .r sin 6=r—x; and x = 



1 + 2 sin »' 



Again 



C s = V .t- -c' «= i: sin 2 n e, (n being the number of equal parts into 

 which the arc q s has to be divided) ; 



And 



(A.) 



Although tlie general solution of the equation involves sonic diffi- 

 culty, yet the value of H or of 2 n 6, and therefore of .1, may be readily 

 determined in all cases, to any required degree of exactness, by pro- 

 portion; lor from the near approach of A O S to A OB, which is 

 known, two values of e may be readily selected, which wlien substi- 

 tuted in (A), will give results respectively less and greater tlian unity. 

 One or two examples will show the certainty and simplicity of this 

 mode of proceeding. 



I. Let r = 4, f = 1, and 11 = 9, i = -2500000 = cos A O B = 

 75 31' nearly. Equation (A) becomes 4 cos lb C — 2 sm 8 = 1 



— — — = 4 11'= 9 nearly. 

 18 



Let us substitute 4° 10', and 4' for 8 in (A.) then we have 



4 cos 72-2 sin 4"0'= 1-0965550 (a) 



4 cos 188—2 sin « ^ l-OOOOOOO (6) 



4 cos 75° -2 sin 4" 10' = -8899600 (c) 



f -2065950= (d-c) 



75"-72'^= 3'= IbO'-^ 0965550= (a-6) 



84'- 125= 1° 24''125, which, when 

 c 



As 206595 t 96555 : : ISO 

 added to 7-2^= 73':24'-125= 18 8. •. 8= 4" : 4'-673.i.- 

 1 



cos 73": 24'- 12 5 



= 3-6008 nearly, so that if A O be divided into 8 



equal parts, A 5 one of them, will divide the arc q, 1, 2, 3, it., into 

 9 equal parts. 



A^ajf ' ' ' C O A 9 o^ ■ ^<i 



II. Let r= 7, c = 2, and ?i= 7, f = -2857142 = cos A O Ba 

 r'io 24' 

 cos 73° 24' nearly, — r^ — = 5"' 14 = 8 nearly. 



Let us substitute 5" and 5^ 10' for 8 in equation (A,) which be 

 comes J- cos 148— 2 sin 8= 1, and we have 



i cos 70° : 0' - 2 sin 5" 0'= 1-0227589 (a) 

 I cos 14 8 - 2 sin 8 = 1-0000000 (6) 



I cos 72" 20' -2 sin 5" 10'= -8820694 (c) 



7o« • 20'-70°-2' 20' - 140' [ ^^^ = ("^"^-^'^ 

 '- •2'^-'" -^ -*" - l**^ ] -0227589 = (o) -(c) 



As 1406895 : 227589 : : 140' : 22'-647. 



. • . 70° : 22'-647 = 14 8 and 8 = 6° : 1' 617 x =. 



o 



2 n e' 



cos 14 fl~ cos 70° 22'-647 



_ S'957-2 nearly = 05. 



in. Let r=2, c = 1, and n =4. h~ -5000000 = cos A B = 

 cos 60". ^° = 7° 30' = 8 nearly. 



If 6" and 7o be substituted for 8 in equation (A,) which in this eS' 

 ample becomes 2 cos S 8 — 2 sin 8= 1, we shall have 



