Solution of Chapter 2

Satisfactory Essays
Topics: Trigraph
?.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*

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