MAE175__Discussion_5
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University of California, Irvine *
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Course
175
Subject
Aerospace Engineering
Date
Apr 3, 2024
Type
Pages
19
Uploaded by CoachGerbilPerson1025
MAE 175: Discussion 5
Jordi Ventura Siches
February 9, 2024
Jordi Ventura Siches
University of California, Irvine - Henry Samueli School of Engineering
MAE 175: Discussion 5
1 of 17
Problem 2.6 (Nelson)
An airplane has the following pitching moment characteristics at the
center of gravity position:
x
cg
/
¯
c
= 0
.
3
.
C
m
cg
=
C
m
0
+
dC
m
cg
dC
L
C
L
+
C
m
δ
e
δ
e
where
C
m
0
= 0
.
05,
dC
m
cg
dC
L
=
−
0
.
1,
C
m
δ
e
=
−
0
.
01
/
deg,
dC
m
cg
dC
L
=
x
cg
¯
c
−
x
np
¯
c
If the airplane is loaded so that the center of gravity position moves to
x
cg
/
¯
c
= 0
.
10, can the airplane be trimmed during landing,
C
L
= 1
.
0?
Assume that
C
m
0
and
C
m
δ
e
are unaffected by the center of gravity travel
and that
δ
e
max
=
±
20
º
.
Jordi Ventura Siches
University of California, Irvine - Henry Samueli School of Engineering
MAE 175: Discussion 5
2 of 17
Problem 2.7 (Nelson): Solution
Solution
:
We first need to compute the new value for
dC
mCG
dC
L
. In order to do so, we
first find the position for the neutral point by using the data from the
previous scenario:
dC
m
cg
dC
L
|
{z
}
−
0
.
1
=
x
cg
¯
c
|{z}
0
.
3
−
x
np
¯
c
→
x
np
¯
c
= 0
.
4
Now we can find the new slope
dC
m
cg
dC
L
=
x
cg
¯
c
|{z}
0
.
1
−
x
np
¯
c
|{z}
0
.
4
=
−
0
.
3
Jordi Ventura Siches
University of California, Irvine - Henry Samueli School of Engineering
MAE 175: Discussion 5
3 of 17
Problem 2.7 (Nelson): Solution
Solution
:
Then,
dC
m
cg
dC
L
=
−
0
.
3. Trim implies
C
m
,
cg
= 0, thus
0 = 0
.
05
−
0
.
3
C
L
|{z}
1
.
0
−
0
.
01
·
δ
e
(deg)
We isolate
δ
e
and get
δ
e
=
0
.
05
−
0
.
3
0
.
01
=
−
25
◦
Since
−
25
◦
is out of the bounds, the aircraft can not be trimmed in this
conditions.
Jordi Ventura Siches
University of California, Irvine - Henry Samueli School of Engineering
MAE 175: Discussion 5
3 of 17
Problem 2.7 (Nelson)
The pitching moment characteristics of a general aviation airplane with
the landing gear and flaps in their retracted position are given by
(a) Where is the stick fixed neutral point located?
Jordi Ventura Siches
University of California, Irvine - Henry Samueli School of Engineering
MAE 175: Discussion 5
4 of 17
Problem 2.7 (Nelson): Solution (a)
Solution:
To find the slope, select 2 points from any of the curves (they have
the same slope). For
δ
e
= 0
◦
:
C
L
= 0
.
4
,
→
C
m
,
cg
= 0;
C
L
= 1
.
2
,
→
C
m
,
cg
=
−
0
.
2
and the slope is
dC
m
,
cg
dC
L
=
−
0
.
2
−
0
1
.
2
−
0
.
4
=
−
0
.
25
Therefore
x
np
¯
c
=
x
cg
¯
c
−
dC
m
,
cg
dC
L
= 0
.
5
Jordi Ventura Siches
University of California, Irvine - Henry Samueli School of Engineering
MAE 175: Discussion 5
5 of 17
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