FORMULAE TO DETERMINE THE NUMERICAL QUANTITIESIN THE CASES OF THE CIRCLE ELLIPSE AND CYCLOID.
Circle. Put the radius of the circle CF = r zz 50 Fithe height of the key AV ~ GF = LY — n — 5 , theabsciss to the extrados VD zz m zz 70.
Then u 3 — n (m + n — r) v —. ii 2 r ; and where m,n, and r are given quantities, v may be found by thesolution of a cubic equation.* v zz 14.5273.
' KD = (2 + A) — 74.2248.
GH = HI - - 13.6385.
Also put the angle the intrados makes with itsordinate EF, which is equal to the angle the extradosmakes with its ordinate DK — <p. Then Tang. L <p
= — = 2.7277, which found in a table Nat. Tang.
gives the angle <p = 69° 527 Or secant L <p — Y
— 2.90546, which found in a table of Nat. secants,gives the angle 9 = AFE = GFH = VKL = LYK= 69°.52' as before.
GH = HI = 11 Tang. 9 = 13.6385, and HF = KY= n sec. <p = 14.5273 as before.
When CD = 0 ; that is, when YD = r + n, thecubic equation becomes t? — %n 2 v — ri 2 r ; from which
v = 12.31 and KD - (2 + ~) - 68.197- The
sec. <p = — = 2.462, which gives the angle 66°.2'.
GH = n Tang. <p =11.2479, HF = n sec. <p = ^/u 2 + ifzz 12.31. Put the radius of curvature, at any point of