2.2 HW2

2.2.1 Problem 1

Figure 2.18:problem 1 description

NOTE: plane B has a wing span of 150 ft = 45.72 m (the 47.72 value in the problem text is a mistake).

2.2.1.1 part a

Characteristic time (also called the time constant and given the letter \(\tau \), is the time it takes the system to reach⁴ \(63.2\%\) of its final response. The textbook on page 115 says

Two systems of the same class are dynamically similar when all the \(\pi \) functions of one are numerically equal to those of the other.

Therefore we need to find the \(\pi \) functions and equate them for each airplane to solve for the unknowns. The \(\pi \) functions are listed in page 115 as \begin{align*}{\pi _{1}} &{=\frac{m}{\rho l}}\\{\pi _{2}} &{=\frac{u_{0}t}{l}}\\{\pi _{3}} &{=\frac{u_{0}^{2}}{lg}} \end{align*}

These are now derived using Buckingham’s \(\pi \) theorem. We assume that the time constant \(\tau \) is a function of all the system parameters \(u_{0},\rho ,m,l,g,t\). The Mach number M and the Reylond’s number Re are not used since the problem says that they can be ignored. The time constant is a function of the remaining system parameters\[ \tau =f(u_{0},\rho ,m,l,g) \] In the above \(t\) in the RHS was not used as the time constant \(\tau \) itself has units of time. The system still has 6 overall variables. Since there are 3 standard dimensions given by \(M,L,T\) where \(M\) is mass and \(L\) is length and \(T\) is time, and since there are 6 variables, then we need to find \(6-3=3\) independent groups, called \(\pi _{1},\pi _{2},\pi _{3}\).

To initiate the process of using Buckingham’s \(\pi \) theorem, we start by selecting 3 repeating variables out of original 6 variables to use for finding each one of the three \(\pi \) groups. Let us select \(\left \{{u_{0},\rho ,m}\right \}\) as the three repeating variables. This choice is not unique, as we could have selected \(\left \{{u_{0},m,l}\right \}\) just as well or any other set of 3 independent variables out of the six variables to use as the three repeating variables.

The next step is to select one of non-repeating variables, one at a time, and for each one, we set up a system of equations and then match the dimensions. The non-repeating variables are \(\left \{ \tau ,l,g\right \} \).

Starting with the first variable in the above set of the non-repeating variable, which is \(\tau \), and using \(\approx \) to mean dimensionally equivalent, we set up the first equation as follows (remembering that the repeating variables are \(\left \{{u_{0},\rho ,m}\right \} \)) \begin{align} \tau & \approx u_{0}^{a}\rho ^{b}m^{c}\tag{1}\\ \lbrack T] & \approx \lbrack LT^{-1}]^{a}[ML^{-3}]^{b}[M]^{c}\nonumber \\ \lbrack T] & \approx L^{(a-3b)}T^{-a}M^{b+c}\nonumber \end{align}

Equating powers of similar dimensions gives 3 equations in 3 the three unknowns \(\left \{ a,b,c\right \} \) powers to solve for. Hence\begin{align*} \left [ T\right ] \quad 1 & =-a\\ \left [ L\right ] \quad 0 & =a-3b\\ \left [ M\right ] \quad 0 & =b+c \end{align*}

Solving the above gives \(a=-1,b=-\frac{1}{3},c=\frac{1}{3}\), therefore (1) becomes\begin{align*} \tau & \approx u_{0}^{-1}\rho ^{-\frac{1}{3}}m^{\frac{1}{3}}\\ & \approx \left ( \frac{m}{\rho }\right ) ^{\frac{1}{3}}\frac{1}{u_{0}}\\ & \approx \left ( \frac{m}{\frac{m}{l^{3}}}\right ) ^{\frac{1}{3}}\frac{1}{u_{0}}\\ & \approx \frac{l}{u_{0}} \end{align*}

Hence the first group is now found \[ \boxed{ \pi _{1}=\frac{u_{0}\tau }{l} } \] Now we find the second group \(\pi _{2}\). The next non-repeating variable to use is \(l\), therefore, as above, we set up an equation to solve for the powers \begin{align} l & \approx u_{0}^{a}\rho ^{b}m^{c}\tag{2}\\ \lbrack L] & \approx \lbrack LT^{-1}]^{a}[ML^{-3}]^{b}[M]^{c}\nonumber \\ \lbrack L] & \approx L^{(a-3b)}T^{-a}M^{b+c}\nonumber \end{align}

Equating powers of similar dimensions gives 3 equations in 3 unknowns to solve for\begin{align*} \left [ T\right ] \quad 0 & =-a\\ \left [ L\right ] \quad 1 & =a-3b\\ \left [ M\right ] \quad 0 & =b+c \end{align*}

Solving the above gives \(a=0,b=-\frac{1}{3},c=\frac{1}{3}\). Therefore (2) becomes\begin{align*} l & \approx \rho ^{-\frac{1}{3}}m^{\frac{1}{3}}\\ & \approx \left ( \frac{m}{\rho }\right ) ^{\frac{1}{3}} \end{align*}

Hence \[ \boxed{ \pi _{2}=\frac{l^{3}\rho }{m} } \] The third and final group \(\pi _{3}\) is found by selecting the last non-repeating variable which is \(g\)\begin{align} g & \approx u_{0}^{a}\rho ^{b}m^{c}\tag{3}\\ \lbrack LT^{-2}] & \approx \lbrack LT^{-1}]^{a}[ML^{-3}]^{b}[M]^{c}\nonumber \\ LT^{-2} & \approx L^{(a-3b)}T^{-a}M^{b+c}\nonumber \end{align}

Equating powers of similar dimensions gives 3 equations in 3 unknowns to solve for\begin{align*} \left [ T\right ] \quad & -2=-a\\ \left [ L\right ] \quad & 1=a-3b\\ \left [ M\right ] \quad & 0=b+c \end{align*}

Solving the above for \(a,b,c\) gives \(a=2,b=\frac{1}{3},c=-\frac{1}{3}\), therefore \begin{align*} g & \approx \left ( u_{0}^{2}\rho ^{\frac{1}{3}}m^{-\frac{1}{3}}\right ) \\ & \approx \left ( \frac{\rho }{m}\right ) ^{\frac{1}{3}}u_{0}^{2}\\ & \approx \frac{u_{0}^{2}}{l} \end{align*}

Hence\[ \boxed{ \pi _{3}=\frac{gl}{u_{0}^{2}} } \] These are the 3 non-dimensional groups that should give the same numerical values for both airplanes if the two airplanes are to be dynamically similar.

For airplane B, we are given \(\left ( u_{0}\right )_{B}=400\) knots, \(l_{B}=150\) ft\(,\) \(m_{B}=225000\) lb and altitude of B is \(20000\) ft. For airplane A \(l_{A}=100\) ft, \(m_{A}=100000\) lb. The problem asks to find \(\left ( u_{0}\right ) _{A}\) and altitude of A.

The \(\pi \) groups use \(\rho \) (density of air) and not altitude, but appendix D in the textbook contains a table to convert from altitude to corresponding air density \(\rho \) at that level. Using this table, at altitude \(20000\) ft the air density is \(\rho _{B} =1.2673\times 10^{-3}\) lb \(\frac{\text{sec}^{2}}{\text{ft}^{4}}\). Now equating the three \(\pi \) groups for both airplanes gives 3 equations \begin{align*} \left ( \pi _{1}\right ) _{A} & =\left ( \pi _{1}\right ) _{B}\\ \left ( \pi _{2}\right ) _{A} & =\left ( \pi _{2}\right ) _{B}\\ \left ( \pi _{3}\right ) _{A} & =\left ( \pi _{3}\right ) _{B} \end{align*}

Using the specific expression found for each \(\pi \) results in\begin{align*} \left ( \frac{u_{0}\tau }{l}\right ) _{A} & =\left ( \frac{u_{0}\tau }{l}\right ) _{B}\\ \left ( \frac{l^{3}\rho }{m}\right ) _{A} & =\left ( \frac{l^{3}\rho }{m}\right ) _{B}\\ \left ( \frac{gl}{u_{0}^{2}}\right ) _{A} & =\left ( \frac{gl}{u_{0}^{2}}\right ) _{B} \end{align*}

For dynamic similarity, both sides of the equations must give the same value. Therefore we substitute the known numerical values for the parameters in these equations and solve for the unknowns\begin{align*} \left ( \frac{u_{0}\tau }{100}\right ) _{A} & =\left ( \frac{400\times 20}{150}\right ) _{B}\\ \left ( \frac{100^{3}\rho }{100000}\right ) _{A} & =\left ( \frac{150^{3}\times 1.2673\times 10^{-3}}{225000}\right ) _{B}\\ \left ( \frac{100g}{u_{0}^{2}}\right ) _{A} & =\left ( \frac{150g}{400^{2}}\right ) _{B} \end{align*}

Assuming \(g\) is the same for both planes, the above reduces to\begin{align} \left ( \frac{u_{0}\tau }{100}\right ) _{A} & =53.333\tag{4}\\ 10\rho _{A} & =0.01901\,\tag{5}\\ \left ( \frac{100}{u_{0}^{2}}\right ) _{A} & =9.375\times 10^{-4}\tag{6} \end{align}

From (6)\[ \ \left ( u_{0}^{2}\right ) _{A}=\frac{100}{9.375\times 10^{-4}}=1.066\,7\times 10^{5}\text{ }\] Hence \begin{align*} \left ( u_{0}\right ) _{A} & =\sqrt{1.066\,7\times 10^{5}}\\ & = \boxed{ 326.6\text{ knot} } \end{align*}

To find the air density for airplane \(A\), using (5) above gives\begin{align*} 10\rho _{A} & =0.01901\,\\ \rho _{A} & =\frac{0.01901}{10}=1.901\times 10^{-3}\text{ lb }\frac{\text{sec}^{2}}{\text{ft}^{4}} \end{align*}

From appendix D (assuming linear relation between each entries in each row in the table) we can interpolate the altitude for \(\rho _{A}=1.901\times 10^{-3}\) lb \(\frac{\text{sec}^{2}}{\text{ft}^{4}}\) \begin{align*} \frac{7000}{y} & =\frac{1.927\times 10^{-3}}{1.901\times 10^{-3}}\\ y & =\frac{7000\times 1.901\times 10^{-3}}{1.927\times 10^{-3}}\\ &= \boxed{ 6905.6\text{ ft} } \end{align*}

2.2.1.2 part b

Using the results from part (a) and from (4) \[ \left ( \frac{u_{0}\tau }{100}\right ) _{A}=53.333 \] Where \(\left ( u_{0}\right ) _{A}=326.6\) knot found in part(a). Hence solving for \(\tau \) gives \begin{align*} \left ( \frac{326.6\tau }{100}\right ) _{A} & =53.333\\ \tau _{A} & =\frac{53.333\times 100}{326.6}\\ & = \boxed{ 16.330\text{ sec} } \end{align*}

2.2.1.3 part c

\[ \left ( C_{L}\right ) _{A}=\frac{L_{A}}{\frac{1}{2}\rho _{A}\left ( u_{0}^{2}\right ) _{A}S_{A}}\] Assuming static equilibrium then the lift is the same as the airplane weight giving\[ \left ( C_{L}\right ) _{A}=\frac{m_{A}g}{\frac{1}{2}\rho _{A}\left ( u_{0}^{2}\right ) _{A}S_{A}}\] And similarly for airplane \(B\)\[ \left ( C_{L}\right ) _{B}=\frac{m_{B}g}{\frac{1}{2}\rho _{B}\left ( u_{0}^{2}\right ) _{B}S_{B}}\] Hence the ratio is\[ \frac{\left ( C_{L}\right ) _{A}}{\left ( C_{L}\right ) _{B}}=\frac{\frac{m_{A}g}{\frac{1}{2}\rho _{A}\left ( u_{0}^{2}\right ) _{A}S_{A}}}{\frac{m_{B}g}{\frac{1}{2}\rho _{B}\left ( u_{0}^{2}\right ) _{B}S_{B}}}=\frac{m_{A}\frac{1}{2}\rho _{B}\left ( u_{0}^{2}\right ) _{B}S_{B}}{m_{B}\frac{1}{2}\rho _{A}\left ( u_{0}^{2}\right ) _{A}S_{A}}\] Substituting the numerical values found\begin{align*} \frac{\left ( C_{L}\right ) _{A}}{\left ( C_{L}\right ) _{B}} & =\frac{\left ( 100000\right ) \frac{1}{2}\left ( 1.2673\times 10^{-3}\right ) \left ( 400^{2}\right ) S_{B}}{\left ( 225000\right ) \frac{1}{2}\left ( 1.901\times 10^{-3}\right ) \left ( 326.6^{2}\right ) S_{A}}\\ & =0.444\,43\frac{S_{B}}{S_{A}} \end{align*}

The surface area of the airplanes is not given. But \(\frac{S_{B}}{S_{A}}\) can be taken as similar to \(\frac{l_{B}^{2}}{l_{A}^{2}}\) and now the above becomes\begin{align*} \frac{\left ( C_{L}\right ) _{A}}{\left ( C_{L}\right ) _{B}} & =0.444436\frac{150^{2}}{100^{2}}\\ & =0.99998\\ & \approx{ \boxed{1}} \end{align*}

2.2.2 Problem 2

Problem: Substitute the linear expressions for \(\Delta Z\) and \(\Delta M\) into the right side of (4.9,7c) and (4.9,8b) and solve the resulting equations to get the second and third components of (4.9,18).

Solution

The linearized form of \(\Delta z\) and \(\Delta M\) are given in (4.9,17) in the textbook as\begin{align} \Delta Z & =Z_{u}\Delta u+Z_{w}w+Z_{\dot{w}}\dot{w}+Z_{q}q+\Delta Z_{c}\tag{4.9,17 (c)}\\ \Delta M & =M_{u}\Delta u+M_{w}w+M_{\dot{w}}\dot{w}+M_{q}q+\Delta M_{c}\tag{4.9,17 (e)} \end{align}

(4.9,7 c) and (4.9,8 b) are given by\begin{align} \dot{w} & =\frac{\Delta Z}{m}-g\Delta \theta \sin \theta _{0}+u_{0}q\tag{4.9,7 c}\\ \dot{q} & =\frac{\Delta M}{I_{y}}\tag{4.9,8 b} \end{align}

Substituting (4.9,17 c) into (4.9,7 c) gives\[ \dot{w}=\frac{Z_{u}\Delta u}{m}+\frac{Z_{w}w}{m}+\frac{Z_{\dot{w}}\dot{w}}{m}+\frac{Z_{q}q}{m}+\frac{\Delta Z_{c}}{m}-g\Delta \theta \sin \theta _{0}+u_{0}q \] Solving for \(\dot{w}\)\begin{align} \dot{w}-\frac{Z_{\dot{w}}\dot{w}}{m} & =\frac{Z_{u}\Delta u}{m}+\frac{Z_{w}w}{m}+\frac{Z_{q}q}{m}+\frac{\Delta Z_{c}}{m}-g\Delta \theta \sin \theta _{0}+u_{0}q\nonumber \\ \dot{w}\left ( \frac{m-Z_{\dot{w}}}{m}\right ) & =\frac{Z_{u}\Delta u}{m}+\frac{Z_{w}w}{m}+\frac{Z_{q}q}{m}+\frac{\Delta Z_{c}}{m}-g\Delta \theta \sin \theta _{0}+u_{0}q\nonumber \\ \dot{w} & =\frac{Z_{u}\Delta u}{m\left ( \frac{m-Z_{\dot{w}}}{m}\right ) }+\frac{Z_{w}w}{m\left ( \frac{m-Z_{\dot{w}}}{m}\right ) }+\frac{Z_{q}q}{m\left ( \frac{m-Z_{\dot{w}}}{m}\right ) }+\frac{\Delta Z_{c}}{m\left ( \frac{m-Z_{\dot{w}}}{m}\right ) }-\frac{g\Delta \theta \sin \theta _{0}}{\left ( \frac{m-Z_{\dot{w}}}{m}\right ) }+\frac{u_{0}q}{\left ( \frac{m-Z_{\dot{w}}}{m}\right ) }\nonumber \\ & =\frac{Z_{u}\Delta u}{m-Z_{\dot{w}}}+\frac{Z_{w}w}{m-Z_{\dot{w}}}+\frac{Z_{q}q}{m-Z_{\dot{w}}}+\frac{\Delta Z_{c}}{m-Z_{\dot{w}}}-\frac{g\Delta \theta \sin \theta _{0}}{m-Z_{\dot{w}}}m+\frac{u_{0}q}{m-Z_{\dot{w}}}m\nonumber \\ & =\left ( \frac{Z_{u}}{m-Z_{\dot{w}}}\right ) \Delta u+\left ( \frac{Z_{w}}{m-Z_{\dot{w}}}\right ) w+\left ( \frac{Z_{q}+u_{0}m}{m-Z_{\dot{w}}}\right ) q-\left ( \frac{gm\sin \theta _{0}}{m-Z_{\dot{w}}}\right ) \Delta \theta +\frac{\Delta Z_{c}}{m-Z_{\dot{w}}}\tag{1} \end{align}

The above is the second component of (4.9.18) which will written in matrix form below as well. But now let us work on the second equation. Substituting (4.9,17 e) into (4.9,8 b) gives\[ \dot{q}=\frac{M_{u}}{I_{y}}\Delta u+\frac{M_{w}}{I_{y}}w+\frac{M_{\dot{w}}}{I_{y}}\dot{w}+\frac{M_{q}}{I_{y}}q+\frac{\Delta M_{c}}{I_{y}}\] The above expression contains \(\dot{w}\). But we found \(\dot{w}\) above. Hence the above becomes\begin{multline*} \dot{q}=\frac{M_{u}}{I_{y}}\Delta u+\frac{M_{w}}{I_{y}}w\\ +\frac{M_{\dot{w}}}{I_{y}}\left ( \left ( \frac{Z_{u}}{m-Z_{\dot{w}}}\right ) \Delta u+\left ( \frac{Z_{w}}{m-Z_{\dot{w}}}\right ) w+\left ( \frac{Z_{q}+u_{0}m}{m-Z_{\dot{w}}}\right ) q-\left ( \frac{gm\sin \theta _{0}}{m-Z_{\dot{w}}}\right ) \Delta \theta +\frac{\Delta Z_{c}}{m-Z_{\dot{w}}}\right ) \\ +\frac{M_{q}}{I_{y}}q+\frac{\Delta M_{c}}{I_{y}} \end{multline*} Expanding and collecting terms\begin{multline*} \dot{q}=\left ( \frac{M_{u}}{I_{y}}+\frac{M_{\dot{w}}}{I_{y}}\left ( \frac{Z_{u}}{m-Z_{\dot{w}}}\right ) \right ) \Delta u+\left ( \frac{M_{w}}{I_{y}}+\frac{M_{\dot{w}}}{I_{y}}\left ( \frac{Z_{w}}{m-Z_{\dot{w}}}\right ) \right ) w\\ +\left ( \frac{M_{q}}{I_{y}}+\frac{M_{\dot{w}}}{I_{y}}\left ( \frac{Z_{q}+u_{0}m}{m-Z_{\dot{w}}}\right ) \right ) q-\frac{M_{\dot{w}}}{I_{y}}\left ( \frac{gm\sin \theta _{0}}{m-Z_{\dot{w}}}\right ) \Delta \theta +\frac{M_{\dot{w}}}{I_{y}}\frac{\Delta Z_{c}}{m-Z_{\dot{w}}}+\frac{\Delta M_{c}}{I_{y}} \end{multline*} Factoring out the \(I_{y}\) from each term gives\begin{multline} \dot{q}=\frac{1}{I_{y}}\left ( M_{u}+\frac{M_{\dot{w}}Z_{u}}{m-Z_{\dot{w}}}\right ) \Delta u+\frac{1}{I_{y}}\left ( M_{w}+\frac{M_{\dot{w}}Z_{w}}{m-Z_{\dot{w}}}\right ) w\nonumber \\ +\frac{1}{I_{y}}\left ( M_{q}+\frac{M_{\dot{w}}\left ( Z_{q}+u_{0}m\right ) }{m-Z_{\dot{w}}}\right ) q-\left ( \frac{M_{\dot{w}}gm\sin \theta _{0}}{I_{y}\left ( m-Z_{\dot{w}}\right ) }\right ) \Delta \theta +\frac{M_{\dot{w}}\Delta Z_{c}}{I_{y}\left ( m-Z_{\dot{w}}\right ) }+\frac{\Delta M_{c}}{I_{y}}\tag{2} \end{multline} The above is the third component of 4.9.18. Putting (1) and (2) into matrix form gives

\begin{multline*} \begin{pmatrix} \dot{w}\\ \dot{q}\end{pmatrix} = \ifdefined \HCode \else \bgroup \renewenvironment{pmatrix}{ \left (\begin{BMAT}(@,45pt,45pt){c:c:c:c}{c:c} }{ \end{BMAT}\right ) } \fi \begin{pmatrix} \frac{Z_{u}}{m-Z_{\dot{w}}} & \frac{Z_{w}}{m-Z_{\dot{w}}} & \frac{Z_{q}+u_{0}m}{m-Z_{\dot{w}}} & -\frac{gm\sin \theta _{0}}{m-Z_{\dot{w}}}\\ \frac{1}{I_{y}}\left [ M_{u}+\frac{M_{\dot{w}}Z_{u}}{m-Z_{\dot{w}}}\right ] & \frac{1}{I_{y}}\left [ M_{w}+\frac{M_{\dot{w}}Z_{w}}{m-Z_{\dot{w}}}\right ] & \frac{1}{I_{y}}\left [ M_{q}+\frac{M_{\dot{w}}\left ( Z_{q}+u_{0}m\right ) }{m-Z_{\dot{w}}}\right ] & -\frac{M_{\dot{w}}gm\sin \theta _{0}}{I_{y}\left ( m-Z_{\dot{w}}\right ) }\end{pmatrix} \ifdefined \HCode \else \egroup \fi \begin{pmatrix} \Delta u\\ w\\ q\\ \Delta \theta \end{pmatrix} \\ +\begin{pmatrix} \frac{\Delta Z_{c}}{m-Z_{\dot{w}}}\\ \frac{M_{\dot{w}}\Delta Z_{c}}{I_{y}\left ( m-Z_{\dot{w}}\right ) }+\frac{\Delta M_{c}}{I_{y}}\end{pmatrix} \end{multline*}

2.2.3 Problem 3

Figure 2.19:problem 3 description

HINT: For HW2 Prob. 3 (#4.10 in the book): assume \(\phi _{0}=\theta _{0}=0\). \(u_{0}\) is non-zero (the hovercraft is moving horizontally).

Figure 2.20:Boeing Pelican ground effect vehicle. image thanks to http://www.aerospaceweb.org

Given\begin{align*} X & =Y=0\\ Z & =-mg+Z_{z}z_{E}\\ L & =L_{\phi }\phi \\ M & =M_{\theta }\theta \\ N & =0 \end{align*}

The forces and moments and angles given above are illustrated in figure 2.21.

Figure 2.21:Balance of aerodynamic forces against external loads

Using the hint, and starting from (4.9,7,8,9) in the textbook, for linear motion\begin{align} \Delta \dot{u} & =\frac{\Delta X}{m}-g\Delta \theta \cos \theta _{0}\tag{4.9,7 (a)}\\ \dot{v} & =\frac{\Delta Y}{m}+g\phi \cos \theta _{0}-u_{0}r\tag{4.9,7 (b)}\\ \dot{w} & =\frac{\Delta Z}{m}-g\Delta \theta \sin \theta _{0}+u_{0}q\tag{4.9,7 (c)} \end{align}

And for angular motion\begin{align} \dot{p} & =\frac{I_{z}\Delta L+I_{xz}\Delta N}{I_{x}I_{z}-I_{xz}^{2}}\tag{4.9,8 (a)}\\ \dot{q} & =\frac{\Delta M}{I_{y}}\tag{4.9,8 (b)}\\ \dot{r} & =\frac{I_{zx}\Delta L+I_{x}\Delta N}{I_{x}I_{z}-I_{zx}^{2}}\tag{4.9,8 (c)} \end{align}

And Euler angles\begin{align} \Delta \dot{\theta } & =q\tag{4.9,9 (a)}\\ \dot{\phi } & =p+r\tan \theta _{0}\tag{4.9,9 (b)}\\ \dot{\psi } & =r\sec \theta _{0}\tag{4.9,9 (c)} \end{align}

Starting with the set of (4.9,7) equations. Since \(X=0\) and \(Y=0\) therefore \(\Delta X=0\) and \(\Delta Y\). We need to find \(\Delta Z\) which by the problem statement \(\Delta Z=Z_{z}\Delta z_{E}\). Also using the assumption given that \(\theta _{0}=\phi _{0}=0\) then the set of (4.9.7) equations reduces to\begin{align} \Delta \dot{u} & =-g\Delta \theta \tag{4.9,7 (a1)}\\ \dot{v} & =g\phi -u_{0}r\tag{4.9,7 (b1)}\\ \dot{w} & =\frac{Z_{z}\Delta z_{E}}{m}+u_{0}q\tag{4.9,7 (c1)} \end{align}

Now considering the set of (4.9.8) equations.  Since principal body axes is used, the off diagonal terms in the inertial matrix vanish which means \(I_{xz}=I_{zx}=0\). Also, since \(N=0\) then \(\Delta N=0\). This reduces (4.9.8) to\begin{align} \dot{p} & =\frac{\Delta L}{I_{x}}\tag{4.9,8 (a1)}\\ \dot{q} & =\frac{\Delta M}{I_{y}}\tag{4.9,8 (b1)}\\ \dot{r} & =0\tag{4.9,8 (c1)} \end{align}

But in the above \(\Delta L\) and \(\Delta M\) do not yet have the gyroscopic effect. These need to be adjusted to add the gyroscopic effect before going further. Since \(h_{B}^{\prime }=\left [ 0,0,H\right ] ^{T}\) then using (4.6.2) in order to find what terms to add to each moment, we write\begin{align*} L & :qh_{z}^{\prime }-rh_{y}^{\prime }=qH\\ M & :rh_{x}^{\prime }-ph_{z}^{\prime }=-pH\\ N & :ph_{y}^{\prime }-qh_{x}^{\prime }=0 \end{align*}

Therefore, the modified moments are \[ L=L_{\phi }\phi +qH \] and \[ M=M_{\theta }\theta -ph \] Assuming \(\Delta L\equiv L_{\phi }\phi \) and \(\Delta M\equiv M_{\theta }\theta \) then the above two equations can be written as\[ L=\overset{\Delta L^{\prime }}{\overbrace{\Delta L+qH}}\] and \[ M=\overset{\Delta M^{\prime }}{\overbrace{\Delta M-ph}}\] We have to replace \(\Delta L,\Delta M\), in (4.9.8 a1,b1,c1) above with these new corrected \(\Delta L^{\prime },\Delta M^{\prime }\) giving\begin{align} \dot{p} & =\frac{\Delta L+qH}{I_{x}}\tag{4.9,8 (a2)}\\ \dot{q} & =\frac{\Delta M-ph}{I_{y}}\tag{4.9,8 (b2)}\\ \dot{r} & =0\tag{4.9,8 (c3)} \end{align}

The above now accounts for the gyroscopic effect. We now continue to the last three set of equations (4.9.9). Since \(\theta _{0}=0\) these become\begin{align} \Delta \dot{\theta } & =q\tag{4.9,9 (a1)}\\ \dot{\phi } & =p\tag{4.9,9 (b1)}\\ \dot{\psi } & =r\tag{4.9,9 (c1)} \end{align}

Summary: The final set of equations are\begin{align} \Delta \dot{u} & =-g\Delta \theta \tag{4.9,7 (a1)}\\ \dot{v} & =g\phi -u_{0}r\tag{4.9,7 (b1)}\\ \dot{w} & =\frac{Z_{z}\Delta z_{E}}{m}+u_{0}q\tag{4.9,7 (c1)} \end{align}

\begin{align} \dot{p} & =\frac{\Delta L+qH}{I_{x}}\tag{4.9,8 (a2)}\\ \dot{q} & =\frac{\Delta M-ph}{I_{y}}\tag{4.9,8 (b2)}\\ \dot{r} & =0\tag{4.9,8 (c3)} \end{align}

\begin{align} \Delta \dot{\theta } & =q\tag{4.9,9 (a1)}\\ \dot{\phi } & =p\tag{4.9,9 (b1)}\\ \dot{\psi } & =r\tag{4.9,9 (c1)} \end{align}

2.2.4 Problem 4

Figure 2.22:problem 4 description

HINT: the most useful equations are sets 4.7,2 and 4.7,3. In particular, to determine \(\dot{r}\), it is best to differentiate (4.7,3)(c).

Solution

equations (4.7,2) and (4.7,3) from the textbook are \begin{align} L & =I_{x}\dot{p}-I_{zx}\dot{r}+qr\left ( I_{z}-I_{y}\right ) -I_{zx}pq+qh_{z}^{\prime }-rh_{y}^{\prime }\tag{4.7,2(a)}\\ M & =I_{y}\dot{q}+rp\left ( I_{x}-I_{z}\right ) +I_{zx}\left ( p^{2}-r^{2}\right ) +rh_{x}^{\prime }-ph_{z}^{\prime }\tag{4.7,2(b)}\\ N & =I_{z}\dot{r}-I_{zx}\dot{p}+pq\left ( I_{y}-I_{x}\right ) +I_{zx}qr+ph_{y}^{\prime }-qh_{x}^{\prime }\tag{4.7,2(c)}\\ p & =\dot{\phi }-\dot{\psi }\sin \theta \tag{4.7,3(a)}\\ q & =\dot{\theta }\cos \phi +\dot{\psi }\cos \theta \sin \phi \tag{4.7,3(b)}\\ r & =\dot{\psi }\cos \theta \cos \phi -\dot{\theta }\sin \phi \tag{4.7,3(c)} \end{align}

Figure 2.23 which is figure 3.1 in the text book illustrating the problem with added annotations on it to help in solving the problem.

Figure 2.23:aircraft rolling pull up

We start simplifying the above equations. Since body axes is used, then all the off diagonal moments of inertial are zero. Hence \(I_{zx}=0\). Ignoring the gyroscopic effect, then \(h_{z}^{\prime }=h_{y}^{\prime }=h_{x}^{\prime }=0\). We are also told that \(\psi =0\), hence \(\dot{\psi }=0.\) The above equations reduces to\begin{align} L & =I_{x}\dot{p}+qr\left ( I_{z}-I_{y}\right ) \tag{4.7,2(a1)}\\ M & =I_{y}\dot{q}+rp\left ( I_{x}-I_{z}\right ) \tag{4.7,2(b1)}\\ N & =I_{z}\dot{r}+pq\left ( I_{y}-I_{x}\right ) \tag{4.7,2(c1)}\\ p & =\dot{\phi }\tag{4.7,3(a1)}\\ q & =\dot{\theta }\cos \phi \tag{4.7,3(b1)}\\ r & =-\dot{\theta }\sin \phi \tag{4.7,3(c1)} \end{align}

Differentiating (4.7,3) gives\begin{align} \dot{p} & =\ddot{\phi }\tag{4.7,3(a2)}\\ \dot{q} & =\ddot{\theta }\cos \phi +\dot{\theta }\dot{\phi }\sin \phi \tag{4.7,3(b2)}\\ \dot{r} & =-\ddot{\theta }\sin \phi -\dot{\theta }\dot{\phi }\cos \phi \tag{4.7,3(c2)} \end{align}

Since we are told that \(p\) is constant, then \(\dot{p}=0\). In addition, since the airplane is moving at constant speed, and since the radius of the vertical circle is constant this implies that the angular acceleration is zero or \(\ddot{\theta }=0\). The problem also says that the wings are horizontal at this moment of time, therefore the Euler angle \(\phi =0\) as can be seen from figure 2.24, which is figure 3.14 in the textbook.

Figure 2.24:aircraft rolled angle

Using the above then the set of (4.7) equations above reduces to \begin{align*} L & =I_{x}\dot{p}+qr\left ( I_{z}-I_{y}\right ) \\ M & =I_{y}\dot{q}+rp\left ( I_{x}-I_{z}\right ) \\ N & =I_{z}\dot{r}+pq\left ( I_{y}-I_{x}\right ) \\ p & =\dot{\phi }\\ q & =\dot{\theta }\\ r & =0\\ \dot{p} & =0\\ \dot{q} & =0\\ \dot{r} & =-\dot{\theta }\dot{\phi } \end{align*}

The goal is to determine \(L,M,N\). Substituting \(\dot{p}=0\) and \(r=0\) and \(\dot{q}=0\) in the first three equations above, the reduce more to \begin{align*} L & =0\\ M & =0\\ N & =-I_{z}\dot{\theta }\dot{\phi }+\dot{\phi }\dot{\theta }\left ( I_{y}-I_{x}\right ) \end{align*}

The above are the moments needed to perform the rolling. Using the values given (SI) and using \(V=R\dot{\theta }\) or \(\dot{\theta }=\frac{V}{R}\) and using \(p=\dot{\phi }\), the above becomes

\begin{align*} L & =0\\ M & =0\\ N & =-I_{z}\frac{V}{R}p+p\frac{V}{R}\left ( I_{y}-I_{x}\right ) \end{align*}

But \(p=90^{0}~\)per sec, or \(\frac{\pi }{2}\)rad/sec. Substituting numerical values the above becomes\begin{align*} L & =0\\ M & =0\\ N & =-\left ( 677\right ) \left ( \frac{152}{610}\right ) \frac{\pi }{2}+\frac{\pi }{2}\left ( \frac{152}{610}\right ) \left ( 406\right ) \end{align*}

or\begin{align*} L & =0\\ M & =0\\ N & = -106.07\,\text{ Nm} \\ & = \boxed{ -78.233\,\text{foot-lb force} } \end{align*}

2.2.5 Key solution

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