the direct resistance of the ballast. Taking this as the equivalent of a normal plane area of .025 sq. in., or .000174 sq. ft., and multiplying by
144
16
{\displaystyle {\frac {144}{16}}}
per sq. ft. basis, we have .00156, or resistance per sq. ft. due to ballast
=
.00156
C
ρ
V
2
{\displaystyle =.00156\ C\ \rho \ V^{2}}
or
ξ
=
.00034
−
.00156
C
ρ
C
ρ
{\displaystyle \xi ={\frac {.00034-.00156\ C\ \rho }{C\ \rho }}}
=
.00034
7
×
.078
−
.00156
{\displaystyle ={\frac {.00034}{7\times .078}}-.00156}
=
.0062
−
.0015
{\displaystyle =.0062-.0015}
ξ
=
.0047
{\displaystyle \xi =.0047}
Again ,
Two planes, weight 5.9 grams (.418 poundals).
C.g. distance from Velocity. 8 ft. altitude.
No. 3
25%
17 ft./sec. 13 ft./sec.
54 ft.37.5 ft.
or,
V
3
=
17
ft./sec.
{\displaystyle V_{3}=17{\mbox{ft./sec.}}}
V
4
=
13
ft./sec.
{\displaystyle V_{4}=13{\mbox{ft./sec.}}}
γ
3
=
8
54
=
.148
{\displaystyle \gamma _{3}={\frac {8}{54}}=.148}
γ
4
=
8
37.5
=
.213
{\displaystyle \gamma _{4}={\frac {8}{37.5}}=.213}
x
3
+
y
3
=
.148
×
5.9
=
.875
{\displaystyle x_{3}+y_{3}=.148\times 5.9=.875}
(1)
x
4
+
y
4
=
.213
×
5.9
=
1.26
{\displaystyle x_{4}+y_{4}=.213\times 5.9=1.26}
(2)
x
3
=
17
2
n
,
y
3
=
m
17
2
{\displaystyle x_{3}=17^{2}\ n,\quad y_{3}={\frac {m}{17^{2}}}}
(3)
x
4
=
13
2
n
,
y
4
=
m
13
.2
{\displaystyle x_{4}=13^{2}\ n,\quad y_{4}={\frac {m}{13^{.2}}}}
(4)
By (1) and (3)
17
2
n
+
m
17
2
=
.875
{\displaystyle 17^{2}\ n+{\frac {m}{17^{2}}}=.875}
(5)
By (2) and (4)
13
2
n
+
m
13
2
=
1.26
{\displaystyle 13^{2}\ n+{\frac {m}{13^{2}}}=1.26}
(6)
