THE LONGITUDINAL STRENGTH OF RIGID AIRSHIPS. 157 
The contraction of that girder is: 
Ngo OSES 
is aEp akEp (25) 
Qe nee 2 
T = Ay—— sin ¢@ sec’ @ sin 0 (26) 
at+a 72 
Since the total shearing force must be equal to the sum of the vertical components of 
the tension in the wires, we have: 
YQ=2T sin ¢ sin 6 (27) 
te BQ=Ay 32 S44 tan’ sin’ 0 (28) 
where the 2 sign comprises all the panels on both sides of the ship. Substituting the value 
of Ay as determined by (28), in (26) we obtain: 
Ta 2% sin @ sec’ @ sin 6 LO (29) 
Coma an Dias ee 
a ate sin’ 
When the section of the ship is a regular polygon as here assumed, the angle @ is constant 
and we get: 
T aa  cosec¢ sin é@d LO 
Saar (29") 
a+a aa ane 
sin’ 0 
Sn Ga 
If, moreover, a and a are the same in all the panels, we get: 
Tp = cosec ¢ sin 6 zQ (29”) 
> sin? 6 
The expressions (29) and (29’) differ somewhat from that usually given, because ordi- 
narily no account is taken of Av, the contraction of the girders. In fact, the usual formula 
as given for a ship with equidistant longitudinals but varying sectional area of the wires is: 
T = Mo cosec b sin 9 ZQ mn 
Za, sin’ @ (29") 
We assume here that the conditions underlying (29”) hold good, in which case, since 
g, DQ, and y sin?@ are constant for the given section, T is proportional to sin 9. The 
compression in the girders is found by multiplying the expression for T with cos@. From 
(29") we obtain: 
JP cot @ sin 6 ZO (30) 
D sin? 0 
which, again, is proportional to sin @. It must be noted that @ is the inclination to the 
horizon of the panel of which the girder in question forms the lower boundary. 
