Solution of Chapter 2
?.1
lf the slope of the C* vs Cscurve is -0.15 ancl the pitching monrent atzero lift is equal to 0.0g, determine tfre trim iift coefficient. lf the center of gravity of the airplane is located at X.o / c = 0.3, determine the stick fixed neL:tral point.
SP-lul-q'!:
dc_
Given:
---u
=
-6,15
Cr.n
= 0.Ct8 @ C,- = 0
X6gic=0.3
Finci: The trim litt coefficient and the stick fixed neutial point.
r^
umlUL-o t,gto.
t.,rflcg .A dC,_ '*'t
Lrrncg
'lrltn
,a
,
= um@C'- = 0 *
=+
9ct ..
*L
d6t
C,.n = S cg Therefore 0 = 0.08 - O,iS
t\.,*
0.08 n f^ t-L_-. =oJ€=u'cr
Inm
hler"rti"al Point
df:. x_('g
dC,.E
x..^ ul' t_
YXdC
"NP .'cg "-m
f'lF'*. v.u ! v. rv,
._ ne-1_n1q\
C
i0
l:
v
"Nlrf
2.2
,r ar
-
?
For the data shown in Figure P2.2, determine the following.
n\ Tha eiink fivod nar rirel nnini
b) If we wish to fly the airpiane
at a veiocity of 125 ft/sec, then what would be the trln iift coefficient and what would be the elevator angle for trim?
u. t3
= 2,750 tb
S = '180 fi2
\4,/
0.10
0.0 5
C
n"q
0.0
-0.05
-0. 1i)
-u. t3
Figure P2.2.
Scilution:
Given: W = 2,750 lb, S = 180 i12, c.g. = 0.25Find: The neutrai point, trim lift coeificient and
125 itlsec.
ttn
^cg -lr^NP v
--_- = v ccdC,
measure dcm I CC; f rom graph P2.2 irr Ot/
n
-tiin
I -
L
it
^l^,,^+.,r
:lg'/d.LUl
rnnlr rur JilYl cnood ILJ u 1r dl lult; in frimil fnr flinhi li JyqtEu r-
Y
_'_fr_= 0,2F - (-0,15) = 0.40 e 4
o
=
Lvt
2r'
(125 fUsec)
= (0.s) (2.977 x to-3; slug/ft3
/
= 18.6 lb/ft-
/^ -Wv,
UU
L.
I nrn
2750 lbs
(18.6 lb/ft4) (180 ft2)
C, - 0.82 =+ 6e = -10.5o trim arim
2.,)
P2"3. The canard and wing are
Analyze the canard-vring combination shown in Figure glo;"tti."lly siniiiar anl are made from the same airfoil section"
AR*
-
ARw, $c
=
S*,6. = 0'45
1l*
lf the slope of the C* vs Cscurve is -0.15 ancl the pitching monrent atzero lift is equal to 0.0g, determine tfre trim iift coefficient. lf the center of gravity of the airplane is located at X.o / c = 0.3, determine the stick fixed neL:tral point.
SP-lul-q'!:
dc_
Given:
---u
=
-6,15
Cr.n
= 0.Ct8 @ C,- = 0
X6gic=0.3
Finci: The trim litt coefficient and the stick fixed neutial point.
r^
umlUL-o t,gto.
t.,rflcg .A dC,_ '*'t
Lrrncg
'lrltn
,a
,
= um@C'- = 0 *
=+
9ct ..
*L
d6t
C,.n = S cg Therefore 0 = 0.08 - O,iS
t\.,*
0.08 n f^ t-L_-. =oJ€=u'cr
Inm
hler"rti"al Point
df:. x_('g
dC,.E
x..^ ul' t_
YXdC
"NP .'cg "-m
f'lF'*. v.u ! v. rv,
._ ne-1_n1q\
C
i0
l:
v
"Nlrf
2.2
,r ar
-
?
For the data shown in Figure P2.2, determine the following.
n\ Tha eiink fivod nar rirel nnini
b) If we wish to fly the airpiane
at a veiocity of 125 ft/sec, then what would be the trln iift coefficient and what would be the elevator angle for trim?
u. t3
= 2,750 tb
S = '180 fi2
\4,/
0.10
0.0 5
C
n"q
0.0
-0.05
-0. 1i)
-u. t3
Figure P2.2.
Scilution:
Given: W = 2,750 lb, S = 180 i12, c.g. = 0.25Find: The neutrai point, trim lift coeificient and
125 itlsec.
ttn
^cg -lr^NP v
--_- = v ccdC,
measure dcm I CC; f rom graph P2.2 irr Ot/
n
-tiin
I -
L
it
^l^,,^+.,r
:lg'/d.LUl
rnnlr rur JilYl cnood ILJ u 1r dl lult; in frimil fnr flinhi li JyqtEu r-
Y
_'_fr_= 0,2F - (-0,15) = 0.40 e 4
o
=
Lvt
2r'
(125 fUsec)
= (0.s) (2.977 x to-3; slug/ft3
/
= 18.6 lb/ft-
/^ -Wv,
UU
L.
I nrn
2750 lbs
(18.6 lb/ft4) (180 ft2)
C, - 0.82 =+ 6e = -10.5o trim arim
2.,)
P2"3. The canard and wing are
Analyze the canard-vring combination shown in Figure glo;"tti."lly siniiiar anl are made from the same airfoil section"
AR*
-
ARw, $c
=
S*,6. = 0'45
1l*