MAE 365. Flight Dynamics. Homework 1 Alain Islas1 West Virginia University, Mechanical and Aerospace Engineering Faculty 401 Evansdale Drive, Morgantown, West Virginia 26506-6070, United States. a
Abstract
In this work there are solved 4 exercises from the textbook, that result fundamental to the correct understanding of the equations that describe the flight dynamics of an aircraft. There is also presented a method to estimate the moments of inertia of different types of airplanes, always keeping in mind the importance that these values represent during the flight maneuvers. Keywords: Moment Moment of inertia, inertia, Linear momentum, Angular Momentum Momentum
1.
Problem lem 1
Problem 1.8 from the textbook. Use sketches and drawings whenever appropiate to document your answer. 1.1. 1.1.
Solu Solutio tion n
First we decide to consider every aircraft the simplest as possible in order to calculate the moments of inertia in an easier way. Let’s refer to the next figure, where we changed the entire airplane geometry to simple euclidean geometric bodies.
[email protected] (Alain Islas) email:
[email protected]
HW 1 MAE
365
January 25th, 2018
MAE 365 Spring Semester, 2018 West Virginia University We changed the fuselage by a hollow cylinder, and the wings for simple plates. There is also showed the C.G. and the reference frame within the aircraft body as the XYZ origin. Now, it is our goal to describe the full moments of inertia tensor of each aircraft. So let’s recall it
I xx xx I xy xy I xz xz I = I yz yz I yy yy I yz yz I zx zx I zy zy I zz zz
Now, because as we’ve seen in class, it is usual for an airplane to have a plane of symmetry. In this case let the XZ plane be that plane, and because the moment of inertia tensor is symmetric (I (I xy xy = I yx yx & I yz yz = I zy zy ). Furthermore, if for this exercise we consider the XYZ frame striclty aligned with the principal axes of the aircraf, then we have that I xx 0 0 xx 0 I = 0 I yy yy 0 0 I zz zz
So our initial goal then reduces to describe only 3 different moments of inertia for every single aircraft. aircraft. Let’s start the adventure! First we’re going to describe all the 3 different I xx xx for the 4 different euclidean bodies (The fuselage, the wings and the empenagge). Starting with the hollow cylinder then
I xx xx =
V
2
2
ρ y + z dV
We want to compute the above integral, but we notice that it results easier to calculate it in cylindrical coordinates. Then
MAE 365 WVU
P a g e|2
MAE 365 Spring Semester, 2018 West Virginia University
I xx xx =
V
ρ r 2 cos2 θ + r + r 2 sin2 θ rdrdθdx
where y = r cos θ, z = r sin θ, x = x and dV = rdrdθdx1 . Then, because the x -axis -axis is aligned with the C.G., solving this integral results in
I xx xx =
h/2 h/2
2π
R2
h/2 h/2 0 h/2 h/2 2π
R1 R2
h/2 h/2 0
R1
· −
=
−
2
2
2
2
ρ r cos θ + r + r sin θ rdrdθdx
ρr2 rdrdθdx
where R1, R2 and h are the inner radius, outer radius and the length of the fuselage, respectively. Now because of the assumption of constant mass distribution we’ve discussed in class then we have2
ρ
h/2 h/2
2π
R2
h/2 h/2 0
R1
−
2
r · rdrdθdx = rdrdθdx = ρ ρ
h/2 h/2
2π
R2
− h/2 h/2
dx
−
0
h/2 h/2
= ρ x
dθ
R1
r4 4
2π
θ
h/2 h/2
0
−
= ρ (h) (2π (2π )
R42
r 3dr R2 R1
R14
4
Now introducing the hollow cylinder density as ρ = m/ = m/ (πh (R22 − R12)). )). Then m R24 − R14 = (2πh (2 πh)) πh (R22 − R21 ) 4 2 2 2 m ( R R + R12 ) − 2 1 ) (R2 + R = 2 2 πh 2 4 πh ( R 2 − R1 ) m = R22 + R + R12 2
Then we have m = R22 + R + R12 (1) 2 Now let’s compute the same moment of inertia of the wing. Take a look to the next figure Tube I xx
1 2
Take caution with this arrangement of the axes, is not the usual one See Fubini’s theorem
MAE 365 WVU
P a g e|3
MAE 365 Spring Semester, 2018 West Virginia University
We observe that the axis of rotation doesn’t pass through the C.G of the wing, so let’s recall the parallel axis theorem, which states that I = I C.G. + md2 C.G. + md where I C.G. C.G. is the moment of inertia as if the axis of rotation is aligned with the C.G. and d and d is the distance from the rotation axis to the C.G. So we’re going to first calculate the value of I I xx xx as if it were aligned with the axis of rotation. Then I xx xx =
2
2
y + z dV =
V
a/2 a/2
b/2 b/2
c/2 c/2
a/2 a/2
−
b/2 b/2
2
ρ y + z dxdydz
c/2 c/2
−
2
−
where a1 , b1 and c1 are the width, depth and height of the plate, respectively. So continuing
= ρ
a1 /2
a1 /2
−
dx
b1 /2
c1 /2
b /2
2
y dy
− 1
c /2
a1 /2
b1 /2
dz + ρ + ρ
− 1
dx
a1 /2 3 c1 /2
−
y 3 b /2 z = ρ (a1 ) (c1 ) + (a ( a1 ) (b1 ) 3 b /2 3 b3 c3 = ρ (a1 ) 1 (c1) + (a ( a1 ) (b1 ) 1 12 12 1
− 1
b /2
dy
− 1
c1 /2
c /2
z 2 dz
− 1
c /2
− 1
Now we can introduce the plate density as ρ = m/ = m/ (a1 b1c1). So we have m b3 c3 = (a) 1 (c1 ) + (a ( a1 ) (b1 ) 1 a1 b1 c1 12 12 m 2 = b1 + c + c21 12
MAE 365 WVU
P a g e|4
MAE 365 Spring Semester, 2018 West Virginia University Then let d1x be the distance from the axis of rotation to the C.G. of the wing, then wing I xx
m 2 = b1 + c + c21 + md + md21x 12
m 2 = b1 + c + c21 + 12 1 2d21x (2) 12 Because the horizontal stabilizers have the same shape as the wings, the moment of tube inertia is going to be equal to the previous formula. Now let’s calculate the value for I yy . Please refer to the next figure wing I xx
I yy yy =
V
2
ρ z + x
2
dV
rdrdθdx
Because it results easier to compute this integral in cylindrcal coordinates. Then I yy yy =
V
2
2
ρ r sin θ + x + x
2
Now because the axis is aligned with the C.G., solving the above results easy
V
2
2
2
ρ r sin θ + z + z rdrdθdx = rdrdθdx = =
h/2 h/2
h/2 h/2 0
−
2π
R2
R1
h/2 h/2
2π
h/2 h/2 0
−
3
2
R2
ρr sin θdrdθdx + θdrdθdx +
R1
h/2 h/2
h/2 h/2 0
−
ρ r2 sin2 θ + z + z 2 rdrdθdx 2π
R2
R1
ρx2 rdrdθdx
where R1, R2 and h are the inner radius, outer radius and the length of the fuselage, respectively. Now, considering constant mass distribution as we’ve previously discussed in class then
MAE 365 WVU
P a g e|5
MAE 365 Spring Semester, 2018 West Virginia University
= ρ
h/2 h/2
2π
dx
h/2 h/2 h/2 h/2
−
0
= ρ x
h/2 h/2
−
= ρ (h) (π ) = ρ
R2
h/2 h/2
2π
R2
− − − − − πh R24 4
sin2 θdθ
θ 2 R42
r3 dr + dr + ρ ρ
R1 2π
h/2 h/2
x2dx
0 3 h/2 x h/2
−
sin2θ sin2θ 2 0 R14 h3 + 4 12 πh3 4 R1 + R22 12
r4 4
R2
+ ρ
3 R21
R1
(2π (2π )
dθ
R1 2π
θ
h/2 h/2
0
−
R22
rdr r2 2
R2 R1
2
R21
Now introducing the hollow cylinder density as ρ = m/ = m/ (πh (R22 − R12)). )). Then m πh πh3 4 4 = R2 − R1 + R22 − R21 2 2 πh (R2 − R1 ) 4 12 m πh 2 πh 3 2 2 2 2 2 − − = R R R + R + R + R R 2 1 2 1 2 1 2 2 4 12 πh (R 2 − R1 ) m = 3 R22 + R + R12 + h + h2 12
So m = 3 R22 + R + R12 + h + h2 12 tube Now, if we try to calculate the I zz value, it will result result in the following following integral integral
Tube I yy
tube I zz
=
h/2 h/2
2π
R2
h/2 h/2 0
R1
−
ρ r 2 cos2 θ + z + z 2 rdrdθdx
which actually drives us to ρ
h/2 h/2
h/2 h/2
−
dx
2π
0
2
sin θdθ
R2
R1
(3)
3
r dr + dr + ρ ρ
h/2 h/2
h/2 h/2
2
x dx
−
2π
R2
0
dθ
R1
rdr
tube whose result is just the same as I yy . Then
m = 3 R22 + R + R21 + h + h2 (4) 12 Now, why we didn’t find the other values for the vertical and horizontal stabilizers?. tube I zz
There are two reasons for that, the first one is that it would be a long job that we don’t want to get involved with, indeed because we used volume integrals, then the moments of inertia for the wings can be handled as for generic volumes, the only thing we need to keep in mind are the respective dimensions of each one, how are they aligned with respect to the MAE 365 WVU
P a g e|6
MAE 365 Spring Semester, 2018 West Virginia University axis of rotation and of course the distance between their C.G’s and the axis of rotation, so the second reason lies in the fact that we’ve already calculated them! In the next figure we can view the distances d1x (The one we were talking about when wing computing I xx ) d2x , d3x
The same idea happens with respect to the y -axis, -axis, being d1y , d2y , d3y the distances between the C.G.’s of the wing, horizontal stabilizer and vertical stabilizer respectively and the y -axis. -axis. Also d1z , d2z , d3z are the distances of the same components with respect to the z -axis. -axis.
Now in the above figure we have all the dimensions of the the wing and the stabilizers. So we can summarize the previous results in:
MAE 365 WVU
P a g e|7
MAE 365 Spring Semester, 2018 West Virginia University For the fuselage :
mf 2 I xx R2 + R + R12 xx = 2
mf I yy 3 R22 + R + R12 + h + h2 yy = 12
mf I zz 3 R22 + R + R12 + h + h2 zz = 12
For the wing :
mw 2 + c21 + 12d 12d21x I xx b1 + c xx = 12
mw 2 + c21 + 12 1 2d21y I yy a1 + c yy = 12
mw 2 + b21 + 12d 12 d21z I zz a1 + b zz = 12
For the horizontal stabilizer :
mhs 2 I xx b2 + c + c22 + 12 1 2d22x xx = 12
mhs 2 I yy a2 + c + c22 + 12d 12 d22y yy = 12
mhs 2 I zz a2 + b + b22 + 12 1 2d22z zz = 12
mvs 2 I yy = a3 + c + c23 + 12d 12 d23y yy 12
mvs 2 I zz = a3 + b + b23 + 12d 12d23z zz 12
For the vertical stabilizer :
mvs 2 I xx = b3 + c + c23 + 12 1 2d23x xx 12
where m where m f is the mass of the fuselage, m fuselage, m w is the mass of the wing, m wing, m hs is the mass of the horizontal horizontal stabilizer stabilizer and mvh is the mass of the vertical stabilizer. So for the entire aircraft we have that mf 2 mw 2 mhs 2 I xx R2 + R + R12 + b1 + c + c21 + 12 1 2d21x + b2 + c + c22 + 12d 12 d22x xx = 2 6 6 m vs 2 + b3 + c + c23 + 12d 12 d23x 12
I yy yy =
mf m w 2 m hs 2 3 R22 + R + R12 + h + h2 + a1 + c + c21 + 12d 12d21y + a2 + c + c22 + 12d 12d22y 12 6 6 mvs 2 + a3 + c + c23 + 12 1 2d23y 12
mf m w 2 m hs 2 I zz 3 R22 + R + R12 + h + h2 + a1 + b + b21 + 12d 12 d21z + a2 + b + b22 + 12d 12 d22z zz = 12 6 6 mvs 2 + a3 + b + b23 + 12d 12 d23z 12
MAE 365 WVU
P a g e|8
(5)
(6)
(7)
MAE 365 Spring Semester, 2018 West Virginia University Now we this information, we are going to guess the probable values for each of the constants described previosuly. Then for the Hughes H-4 we suppose that mf = 45, 45, 000 kg a1 = 7 m b1 = 38 m c1 = 2 m d1x = 23 m d1y = 10 m d1z = 25 m R9 = 9 m
mw = 25, 25, 000 kg a2 = 5 m b2 = 16 m c2 = 1.5 m d2x = 12 m d2y = 30 m d2z = 25 m R1 = 8 m
mhs = 8, 000 kg a3 = 4 m b3 = 1.5 m c3 = 14 m d3x = 10 m d3y = 30 m d3z = 35 m h = 65 m
mvs = 5, 000 kg
Substituting these values into Eqs. (5), (6) and (7) we obtain I xx 38.97 × 106 kg · m2 xx = 38. I yy 41.7 × 106 kg · m2 yy = 41. I zz 71.44 × 106 kg · m2 zz = 71. Now, let’s consider the next aircraft. For the Boeing 747-400 we suppose that mf = 75, 75, 000 kg a1 = 3 m b1 = 29 m c1 = 1.5 m d1x = 16 m d1y = 8 m d1z = 15 m R9 = 6 m
mw a2 b2 c2 d2x d2y d2z R1
= 30, 30, 000 kg = 2.5 m = 10 m = 1m = 9m = 24 m = 18 m = 5m
mhs = 15, 15, 000 kg a3 = 3 m b3 = 1 m c3 = 10 m d3x = 7 m d3y = 24 m d3z = 25 m h = 70 m
Substituting these values into Eqs. (5), (6) and (7) we obtain I xx 25.11 × 106 kg · m2 xx = 25. I yy 58.79 × 106 kg · m2 yy = 58. I zz 54.54 × 106 kg · m2 zz = 54. In the same fashion, for the Airbus A380-800 we suppose that MAE 365 WVU
P a g e|9
mvs = 10, 10, 000 kg
MAE 365 Spring Semester, 2018 West Virginia University
mf = 120, 120, 000 kg a1 = 3.5 m b1 = 32 m c1 = 2 m d1x = 18 m d1y = 10 m d1z = 17 m R9 = 7 m
mw a2 b2 c2 d2x d2y d2z R1
= 40, 40, 000 kg = 23 m = 12 m = 1.5 m = 10 m = 26 m = 20 m = 6m
mhs = 30, 30, 000 kg a3 = 4 m b3 = 1.5 m c3 = 12 m d3x = 9 m d3y = 26 m d3z = 26 m h = 73 m
mvs = 15, 15, 000 kg
Substituting these values into Eqs. (5), (6) and (7) we obtain I xx 45.99 × 106 kg · m2 xx = 45. I yy 114.89 × 106 kg · m2 yy = 114. I zz 123.58 × 106 kg · m2 zz = 123. Finally, for the An-225-Mriya we suppose that mf = 150, 150, 000 kg a1 = 4 m b1 = 35 m c1 = 2.5 m d1x = 18 m d1y = 10 m d1z = 15 m R9 = 8 m
mw a2 b2 c2 d2x d2y d2z R1
= 50, 50, 000 kg = 3m = 13 m = 2m = 12 m = 25 m = 23 m = 7m
mhs = 30, 30, 000 kg a3 = 5 m b3 = 1.5 m c3 = 14 m d3x = 10 m d3y = 25 m d3z = 25 m h = 84 m
Substituting these values into Eqs. (5), (6) and (7) we obtain I xx 63.53 × 106 kg · m2 xx = 63. I yy 143.08 × 106 kg · m2 yy = 143. I zz 173.58 × 106 kg · m2 zz = 173.
MAE 365 WVU
P a g e|10
mvs = 25, 25, 000 kg
MAE 365 Spring Semester, 2018 West Virginia University 2.
Problem lem 2
Problem 1.9 from the textbook. Use sketches, drawings whenever appropiate to document your answer. 2.1. 2.1.
Solu Solutio tion n
To answer this question, let’s recall once again the definition of a moment of inertia I =
V
ρr2 dV
where r where r is is the distance from any point within the body to the axis of rotation. Now if we take a look to the F-111, we’ll see that at low-subsonic velocities its wings are opened. This may result in an increase of the moments of inertia because the distance from the axis of rotation is bigger than if it were flying at high-subsonic velocities, where the aircraft closes its wings. In other words, let’s first look at the next figure
We observe that when the wings are opened (low-subsonic velocity) the distance from the axis of rotation to a differential mass element dm is dm is r, r , whilst when the wings are closed (high-subsonic velocity) the distance from the axis of rotation to a differential mass element dm is r . It follows then
I w.o w.o =
MAE 365 WVU
v
2
I w.c w.c =
r dm
P a g e|11
V
r 2 dm
MAE 365 Spring Semester, 2018 West Virginia University Now because d because dm m = dm = dm and r > r we have then
I w.o w.o > I w.c. w.c. This same idea can be extended to any axis of rotation, so every moment of inertia is going to change because of the distance between the wings and the axis of rotation is going to increase or decrease respectively.
MAE 365 WVU
P a g e|12
MAE 365 Spring Semester, 2018 West Virginia University 3.
Problem lem 3
Go over Student Sample Problem 1.1 from the textbook. Make sure you fully understand it!. Next, re-do the process and explain with your own words the process leading to the demonstration that the initial vectorial CAME relationship with respect to X , Y , Z .
d dt
dr r × ρA dV = dt V
¯
¯
V
r × ρA g dV + ¯
¯
S
r × F dS
with F = F A + F T T
¯
¯
¯
¯
¯
can be reduced to d dt
V
r × ¯
dr ρA dV = M dt ¯
¯
with M = M A + M T T . Of course, the key is NOT to copy the textbook. Essentially you are asked to re-do it using your own words. ¯
3.1. 3.1.
¯
¯
Solu Solutio tion n
First we look that r = r p + r, so we have
¯
d dt
¯
¯
d (r p + r ) × ρA (r p + r) dV = dt V
¯
¯
¯
¯
V
(r p + r ) × ρA gdV + ¯
¯
¯
S
(r p + r ) × F dS ¯
¯
¯
Computing the LHS of the above equation then d dt
dr p r p × ρA dV + dt V
¯
¯
dr r p × ρA dV + dt V
¯
¯
dr p r × ρA dV + dt V
¯
¯
dr r × ρA dV dt V
¯
¯
Now recalling Leibniz integral rule , we know that there’s a chance to apply differentiation differentiation inside the integral operation, just when the limits of integration are constants with respect to the variable we are differentiating with. In this case, the derivative is with respect to time and the integral is with respect to spatial coordinates. So then if we apply this rule to the first three terms, then we have
V
d dr p r p × ρA dt dt ¯
¯
dV + +
V
d dr dV + + r p × ρA dt dt
¯
¯
V
d dr p r × ρA dt dt ¯
¯
d dV + + dt
dr r × ρA dV dt V ¯
¯
Also we note that ρ that ρA is constant in time due to the constant mass distribution assumption ( ) we stated in the first class of the course. So, any time derivative of the form dtd ρA ddt is just equal to
d ρA dt
MAE 365 WVU
d(∗) dt
. Expanding all this terms then
P a g e|13
∗
MAE 365 Spring Semester, 2018 West Virginia University
V
d2 r p dr p dr d2 r × ρA dV + r p × ρA 2 dV dV + r p × ρA dt2 dt dt V V dt V 2 dr dr p d r p d dr × ρA + dV + r × ρA 2 dV + r × ρA dV dt dt dt V dt V dt V
dr p dr p × ρA dV + dt dt
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
Now, because the self cross product of a vector is equal to the zero vector and recalling the rigid body assumption we made in the first class then any time derivative of the vector r is just equal to zero ¯
0 dr p d r p × ρA dV + dt dt V
¯
¯
0 0 2 d2 r p dr p d d r r × ρA dV + r × r p × ρA dV + ρA 2 dV p dt2 dt dt V V dt V 0 r dr d d2 r p d dr p × + ρA dV + r × ρA 2 dV + r × ρA dV dt dt dt V dt V dt V
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
So we are only left with the second, sitxth and seventh terms of the LHS d2r p dV + r p × ρA dt2 V
¯
¯
d2 r p d dV + r × ρA dt2 dt V
¯
¯
dr r × ρA dV dt V ¯
¯
Indeed we want to get rid of the first & second terms and only mantain the third one to complete the demonstration, so let’s continue working. Because the vector r p has nothing to do with the volume integral of the aircraft (just remember remember that r p measures the distance between the X the X , Y , Z frame and the C.G.) we can take it out of the integral
¯
¯
d2 r p dV = r p × r p × ρA dt2 V
¯
¯
d2 r p ρA 2 dV dt V 2 d r p = r p × m 2 dt
¯
¯
¯
¯
Now, we also know that the cross product is anticomm anticommutativ utative, e, so we can write the second term as d2 r p r × ρA dV = dt2 V
¯
¯
− V
d2r p ρA 2 × r dV dt
¯
¯
0 d2r p = − 2 × ρ A r dV dt V
¯
MAE 365 WVU
P a g e|14
¯
MAE 365 Spring Semester, 2018 West Virginia University Taking advantage of the C.G. property, then it was possible to cancel the above term. So finally, the LHS can be written as d2 r p r p × m 2 dt
¯
¯
d + dt
dr r × ρA dV dt V ¯
¯
Now, focusing on the RHS we have then
V
(r p + r ) × ρA gdV + ¯
¯
¯
S
(r p + r ) × F dS = ¯
¯
¯
r p × ρA g dV + ¯
V
¯
+
S
V
r × ρA g dV ¯
¯
r p × F dS + + ¯
¯
× r
¯
S
F dS ¯
Once again, because r p has nothing to do with the volume integral of the aircraft then the RHS can be written as
¯
r p
r p
¯
× × × ρA g dV + ¯
V
¯
mg +
ρA g dV + r p
r
¯
V
F dS + ¯
¯
S
V
¯
¯
× × ×
r × ρA g dV + ¯
¯
+ F dS + ¯
S
r
¯
S
r
¯
S
F dS ¯
F dS ¯
Matching both the LHS and the RHS then d2 r p r p × m dt2 ¯
¯
d + dt
dr r × ρA dV = r p × mg + dt V ¯
¯
¯
¯
F dS + ¯
S
V
r × ρA g dV ¯
¯
+
× r
¯
S
F dS ¯
Now recalling the conservation of linear momentum equation in vectorial form (as seen in the first class of the course also) d2 r p m 2 = m g + dt We can express the above equation as
¯
¯
r p ¯
×
0
−
r p d2 m 2 dt ¯
r p × ¯
0 m + F dS + g S ¯
¯
d dt
¯
dr r × ρA dV = dt V ¯
¯
V
r × ρA gdV ¯
+
Almost there! MAE 365 WVU
S
F dS
P a g e|15
¯
× r
¯
S
F dS ¯
MAE 365 Spring Semester, 2018 West Virginia University
d dt
dr r × ρA dV = dt V ¯
¯
r × ρA g dV + ¯
V
¯
× r
¯
S
F dS ¯
Now we notice that the first term of the RHS can be re-arranged as
V
r × ρAg dV = ¯
¯
− V
ρAg × r dV ¯
¯
Applying same idea as in the previous situations, we observe that the g vector has nothing to do with the volume integral, then ¯
− V
Finally!
d dt
ρA g × rdV = ¯
¯
× S r
MAE 365 WVU
¯
g
¯
¯
¯
V
r × ρA ¯
¯
r
¯
S
F dS ¯
dr dV = M dt ¯
F dS = M A + M T T ¯
¯
×
dr r × ρA dV = dt V
d dt where M =
¯
0 ρ r dV A V
− ×
¯
P a g e|16
¯
MAE 365 Spring Semester, 2018 West Virginia University 4.
Problem lem 4
Go over Student Sample Problem 1.2 from the textbook. Work out and write down the process leading to the demonstration that the vectorial relationship
V
ˆ , r = x ˆ with ω = P = P ˆi + Q jˆ + R + Rk = xˆi + y jˆ + z + z k
r × ω (ω · r ) ρA dV ¯
¯
¯
¯
¯
¯
will be reduced to the following ’scalarized’relationship 2 2 ˆ + I yz + RQ (I zz {I xy xy P R + I yz R − Q − I xz xz P Q + RQ zz − I yy yy )}i
2 2 ˆ + {(I xx + I xz QR + I I yz xx − I zz zz ) P R + I xz P − R − I xy xy QR + yz P Q} j
ˆ + {(I yy + I x Q2 − P 2 + I + I xz yy − I xx xx ) P Q + I xz QR − I yz yz P R}k 4.1. 4.1.
Solu Solutio tion n
First we start calculating the dot product between the vectors ω and r ¯
ˆ · xˆi + y jˆ + z ˆ ω · r = P ˆi + Q jˆ + R + Rk + z k ¯
¯
= P x + Qy + Qy + + Rz Rz
¯
which is a scalar number. Now let’s compute the cross product between the same vectors
i
ˆ
ω × r = x ¯
j
ˆ
k ˆ
y z P Q R
¯
= (Ry − Qz ) i + (P (P z − Rx) j + (Qx − P y) k − Rx) ˆ
ˆ
ˆ
Plugging both calculations into the original integral then
V
r × ω (ω · r ) ρAdV = ¯
¯
¯
¯
V
− Rx) (P x + Qy + Qy + + Rz Rz ) (Ry − Qz ) i + (P (P z − Rx) j + (Qx − P y) k ρA dV ˆ
ˆ
ˆ
With these results, then we can split the last integral into 3 new ones (each of them along the 3 different directions)
V
(P x + Qy + Qy + + Rz Rz ) (Ry − Qz ) ρA dV i + ˆ
V
+ MAE 365 WVU
− Rx) (P x + Qy + Qy + + Rz Rz ) (P z − Rx) ρA dV j
P a g e|17
V
ˆ
(P x + Qy + Qy + + Rz Rz ) (Qx − P y) ρAdV k ˆ
MAE 365 Spring Semester, 2018 West Virginia University Now, if we expand the terms of the x -direction -direction integral then we have
− − − − − −
− P Qxz PRxy + PRxy + QRy QRy 2 + R2 yz − Qxz Q2 yz QRz 2 ρA dV i
V
ˆ
PRxy − P Qxz Qxz + QR + QR y2 − z 2 + R2
V
PR
V
xyρ A dV − P Q
V
xzρ AdV + QR
V
Q2 yz ρAdV i
ˆ
y
2
2
2
z ρA dV + R − Q
2
V
yzρ yz ρA dV i
ˆ
Recalling the moments of inertia formulas I xx xx = I yy yy = I zz zz =
V V V
2
2
y + z ρA dV
I xy xy =
x2 + z 2 ρA dV
I xz xz =
x2 + y 2 ρAdV
I yz yz =
V V V
xyρ A dV xzρ AdV yzρ yz ρAdV
we note that we can re-write the above integrals as the following
P RI xy + QR xy − P QI xz xz + QR
2
2
2
y − z ρA dV + R − Q
V
Now we notice that it is possible to express the third term as QR
2
2
y − z ρA dV = QR
V
= QR
V
V
2
I yz yz i
−
ˆ
x2 + y 2 − x2 − z 2 ρA dV
x2 + y 2 ρA dV
= QR [I zz zz − I yy yy ]
x2 + z 2 ρA dV
V
Finally, the CAME equation for the x -direction -direction can be expressed as
2
P RI xy + QR (I zz xy − P QI xz xz + QR zz − I yy yy ) + R − Q
2
I yz yz i
(8)
ˆ
In similar fashion we can expand the terms of the y -direction -direction integral as
− − 2
2
2
P 2
+ P Qyz + P + P R z 2 − x2 R2 xz + P
− − − 2
P xz + P + P Qyz Qyz + P + P Rz − P Rx − QRxy
V V
2
P
R
2
V
xzρ A dV + P Q
V
R xz ρA dV j
ˆ
QRxy ρA dV j
yzρ yz ρA dV + P R
ˆ
2
V
z
x
2
ρA dV − QR
V
xyρ A dV j
Now writing them in terms of the moments of inertia coefficients then we have MAE 365 WVU
P a g e|18
ˆ
MAE 365 Spring Semester, 2018 West Virginia University
P 2 − R2 I xz + P QI yz + P R (I xx xz + P yz + P xx − I zz zz ) − QRI xy xy j
(9)
ˆ
Following the same procedure as the previous ones, for the z -direction integral
2
V V
PQ
2
2
2
− P xy − P Qy − P Ryz P Qx + Q xy + xy + QRxz QRxz − Ryz ρA dV k PQ x V
x
− − − −
2
2
2
y + Q2
y
2
ˆ
P 2 xy + xy + QRxz QRxz − Ryz k − P Ryz 2
ρAdV + Q
2
P
V
ˆ
xyρ AdV + QR
V
xzρ AdV − P R
V
yzρ yz ρA dV k ˆ
Once again, writing them down in terms of the moments of inertia coefficients we have
MAE 365 WVU
2 2 P Q (I yy + QRI xz yy − I xx xx ) + Q − P I xy xy + QRI xz − P RI yz yz k
P a g e|19
ˆ
(10)
REFERENCES
MAE 365 Spring Semester, 2018 West Virginia University
References [1] M. R. R. Napolitano Napolitano.. Aircraft Dynamics: From modeling to simulation . 1st. Ed. John Wiley & Sons. 2012. U.S.A [2] Aviastar. Retrieved on January 24th, 2018 from: General General Dynamics F-111. http://www.aviastar.org/pic http://www.aviastar.org/pictures/usa/genera.f-111.gif tures/usa/genera.f-111.gif
MAE 365 WVU
P a g e|20
