1904] The Longitudinal Stability of Aerial Gliders. 

 Then we have from Langley's results, 



/(O = 0-56, f(a!) = 1, 



f(*") = 0-28, fia!') = 2-5; 



= -V2f(vi); 



ill 



for equilibrium 



therefore, if we put r)i = a, we have rj. 2 = 2a. Let & 2 be assumed = 2a 2 . 

 Substituting in the expressions for X w , etc., we have 



mX u = 3-6KSV, 

 mX v = 1KSV, 

 mXfl = 4KSV«, 

 and therefore, 



A = 2ft 2 , 



D = 2721 



V 



B = 680 " 

 E = 297000 



mQ u = 4KSVa, 

 mG v = 0-7KSVo» 

 mGce = llKSVft 2 , 



C = 156ft -22800ft 2 /V 2 



V 2 



H 



2.7.108^-1-4.10^ 



2-9.10 10 — -l-3.10 n ^- 



This is positive if V 2 > 590ft. 



Ex. 5. — Let us now take the case of two equal square planes inclined 

 at a small angle to each other. 



Let 



a = 10°, 



ft 



-5 C 



ft = 5°, 



and let 2ft be the breadth of either plane. 

 For these angles we have 



f (ocx- Pi) = 0-44, 

 /(«2-ft) = 015, 

 = 0'45, 

 <f> (oc 2 - p 2 ) = 0-55, 



For equilibrium we must have 



- Pi) {li + ^ («i - Pi)} =/(a 2 - ft) + (a 2 - ft)} 



whence 



/' ("i-ft) = 1'7, 

 /' (a 2 -ft) = 1-8, 

 <t>'(*i-Pi) = 0-6, 

 <£' (ai-ft) = 0-58. 



therefore, if we put ^ 2 



3^i + ^2 = 0*28ft; 



- 4ft, we have 171 = l*4ft. Let & 2 be assumed 



