The tail sweep angle and the tail dihedral angle are tentatively considered to be the same as those of the wing. The reasons are

presented in Section $6\mathrm{.}7$:

$\backslash$Lambda $\_($h$)=10$deg,$\backslash$Gamma $\_($h$)=5$degi$\_($h$)$ C$\_($L$\_(\backslash$alpha $))=($C$\_($l$\_(\backslash$alpha $-$h$)))/(1+($C$\_($l$\_(\backslash$alpha $\_(-$h$))))/(\backslash$pi $*$AR$\_($h$)))=(6\mathrm{.}7)/(1+(6\mathrm{.}7)/(3\mathrm{.}14*18\mathrm{.}6))=6\mathrm{.}1(1)/($rad$)$

The tail angle of attack in cruise is:

$\backslash$alpha $\_($h$)=($C$\_($L$\_($h$)))/($C$\_($L$\_(\backslash$alpha $\_($l$))))=(-0\mathrm{.}121)/(6\mathrm{.}1)=-0\mathrm{.}018$rad$=-1\mathrm{.}02$deg

To calculate the tail created lift coefficient, the lifting line theory is employed as introduced in Section $5\mathrm{.}14.$ The following MATLAB m$-$file is utilized to calculate the tail lift coefficient with an angle of attack of $-1\mathrm{.}02$ deg.

CODE:

clc

clear

N $=9$; $\%($number of segments$-1)$

S $=2\mathrm{.}277$; $\%$ m$2$

AR $=18\mathrm{.}6$; $\%$ Aspect ratio

lambda $=0\mathrm{.}8$; $\%$ Taper ratio

alpha$\_($t$)$wist $=0\mathrm{.}00001$; $\%$ Twist angle $($deg$)$

a$\_($h$)=-1\mathrm{.}02$; $\%$ tail angle of attack $($deg$)$

a$\_(2)$d $=6\mathrm{.}1$; $\%$ lift curve slope $((1)/($r$)$ad$)$

alpha$\_(0)=0\mathrm{.}000001$; $\%$ zero$-$lift angle of attack $($deg$)$

b $=$ sqrt$($AR$*$S$)$; $\%$ tail span

MAC $=($ S$)/($b$)$; $\%$ Mean Aerodynamic Chord

Croot $=(1\mathrm{.}5*(1+$lambda$)*$MAC$)/(1+$lambda$+$lambda$2)$; $\%$ root chord

theta $=$ p$($i$)/(2*$N$)$:p$($i$)/(2*$N$)$:p$($i$)/(2)$;

alpha$=$a$\_($h$)+$alpha$\_($t$)$wist:$-$alpha$\_($t$)$wis$($t$)/($N$-1)$:a$\_($h$)$;

$\%$ segment's angle of attack

z $=(($b$)/(2))*$cos$($theta$)$;

c $=$ Croot $*(1-(1-$lambda$)*$cos$($theta$))$; $\%$ Mean

Aerodynamics chord at each segment

mu $=$ c $*$ a$\_(2)$d $()/()(4*$ b$)$;

LHS $=$ mu $.*($alpha$-$alpha$\_(0))/(57\mathrm{.}3)$; $\%$ Left Hand Side

$\%$ Solving N equations to find coefficients A$($i$)$:

for i$=1$:N

for j$=1$:N

B$($i$,$j$)=$sin$((2*$j$-1)*$ theta$($i$))*(1+($mu$($i$)*)/($sin$($theta$($i$)))$

$(2*$j$-1))$;

end

end

A$=($B$)/($t$)$ranspose$($LHS$)$;

for i $=1$:N

sum$1($i$)=0$;

sum$2($i$)=0$;

for j $=1$ : N

sum$1($i$)=$ sum$1($i$)+(2*$j$-1)*$ A$($j$)*$sin$((2*$j$-1)*$theta$($i$))$;

sum$2($i$)=$ sum$2($i$)+$ A$($j$)*$sin$((2*$j$-1)*$theta$($i$))$;

end

end

CL$\_($t$)$ail $=$ pi $*$ AR $*$ A$(1)$

The output of this m$-$file is:

CL$\_($t$)$ail $=-0\mathrm{.}095$

The tail is expected to generate a CL$\_($h$)$ of $-0\mathrm{.}121,$ but it generates a CL$\_($h$)$ of $-0\mathrm{.}0959.$ To increase the tail lift coefficient to the desired value, we need to increase the tail angle of attack. With trial and error and using the same m$.$ file, we find that the tail angle of attack of $-1\mathrm{.}29$ deg generates the desired tail lift coefficient.

Hence: alpha$($h$)=-1\mathrm{.}29$deg

INSTRUCTIONS:

PLEASE WRITE EXTRA CODE UNDER THE CODE PROVIDED SO THAT THE PROGRAM OUTPUTS CL LIFT AS $-0\mathrm{.}121$ INSTEAD OF $-0\mathrm{.}095$ USING TRIAL AND ERROR. Extra values are given in the image provided