106 Dr. G. H. Bryan and Mr. W. E. Williams. [Jan. 7, 



mYe = 2 [2KSi ( - u m + - ft) sin ft 



- KSi (vrn + ufyf (aj - ft) sin ft], 



mGu = 2 [ - 2KSi (it - ^)/(*i - ft){i?i + ( ai - ft)} 



- KS^x/' (ai - ft){^ + ( ai - ft)} 



-KS^/( ai -ft) O^'fa-ft)]. 



Since in the steady motion G = 0, 



i.e., 2 (u* + «*) 2KSa/(% - ft) + (a 2 - ft)} = 0, 

 wG„ = v2 [ - KSi/' ( ai - ft) {p 1 + atf fa - ft)} 



-W(«i-A)flif(«i-Al 



WlG, = U? [KSi/' ( ai - ft) + («! - ft)} 



+ KS 1 /( ai -ft)fl 1 f( ai -ft)], 



* = 2 [ - KSi /' (a x - ft) ( - v m + fife) + (*i - ft)} 



- KSi/(«i - ft) ( - v m + t£) ( ai - ft)]. 



4. Numerical Calculations for Particular Cases. Single Latnince. 



We now consider the stability of certain particular systems, 

 beginning with a single plane lamina. 



Since the lamina is falling steadily under gravity, its plans must 

 necessarily be horizontal. 



Take the axis of y parallel to this plane, then 6 = 0, /? = 0, and, 

 therefore, Y = ; also, for equilibrium, rj + a<j> (a) = 0. 



The coefficients, therefore, become 



mX u = 2KSuf(a) + K$vf(ot), 



mX v = 2KSfl/(a)-KSw/>), 



mX§ = - 2KS (urj - v£) f{a) + KS/'(a) (w? + «f ), 



mG tt = - KSfl/(a) a <£' (a), 



mG„ = KS«/(a)a^>'(a), 



mGg = - KSa/(a) (vrj + wf) (a). 



Ex. 1.— Consider a square plane, balanced so as to fly at an angle 

 of 10°, with the centre of gravity in the plane, and suppose its radius 

 of gyration given by k 2 = j& 2 . 



Professor Langley's results give for this particular angle 



f(a) = 0-3, f(a) = 1-6, 



and from Joesel's formula 



<j>(a) = 0-49, <f>'(a) = - 0'59, .'. r) = 0-49a. 



