i have this solution"داده‌های مسئله:
	وزن هواپیما ( W = 4,600 ) پوند
	نسبت میانگین نیروی پیشران به وزن ( \left(\frac{T}{W}\right)_{\mathrm{ave}}=0.14)
	سرعت باد \left(V_w=0\right) بدون باد
	شیب باند \left(\phi=0^\circ\right) بدون شیب
	انحراف فلپ ( \delta_f={15}^\circ )
	مساحت بال (S=175{\mathrm{ft}}^2 )
	طول بال ( b=35\mathrm{ft} )
	نسبت Aspect ( A = 7 )
	ضریب اصطکاک زمین (\mu_g=0.03 )
	ضریب بالابری حداکثر (C_{L_{\mathrm{max}}}=1.69 )
	فاکتور اوزولد ( e = 0.80 )
	شتاب گرانشی ( g=32.174{\mathrm{ft/s}}^2 )
	چگالی هوا در شرایط استاندارد سطح دریا ( \rho=0.0023769{\mathrm{slugs/ft}}^3 )
مراحل حل:
1. تعیین سرعت استال (\mathbit{V}_\mathbit{S}):
سرعت استال بیشینه که هواپیما در آن شروع به برخاست می‌کند، با فرمول زیر محاسبه می‌شود:
[
V_S=\sqrt{\frac{2W}{\rho S C_{L_{\mathrm{max}}}}}\bigm]
با جایگذاری مقادیر داده شده:

V_S=\sqrt{\frac{2\times4600}{0.0023769\times175\times1.69}}\approx\sqrt{\frac{9200}{0.703}}\approx\sqrt{13104.84}\approx114.4\mathrm{ft/s}
تبدیل به گره:

V_S\approx\frac{114.4}{1.68781}\approx67.8\mathrm{knots}\bigm
2. تعیین سرعت برخاست ( \mathbit{V}_{\mathrm{LOF}} ):
سرعت برخاست توسط نسبت ایمنی ( 1.15 ) برابر با سرعت استال است:

V_{\mathrm{LOF}}=1.15\times V_S\approx1.15\times114.4\approx131.6\mathrm{ft/s}\approx77.9\mathrm{knots}\bigm
3. تعیین نیروی پیشران ( \mathbit{F}_\mathbit{N} ):
با استفاده از نسبت ( \frac{T}{W} ):

F_N=0.14\times W=0.14\times4600=644\mathrm{lbs}\bigm
4. تعیین ضریب بالابری در اثر زمین (\mathbit{C}_{\mathbit{L}_\mathbit{g}} ) و ضریب درگ در اثر زمین ( \mathbit{C}_{\mathbit{D}_\mathbit{g}} ):
از داده‌های ارائه‌شده:

C_{L_g}=0.89\bigm\bigm\bigmC_{D_g}=0.0862\bigm
5. تعیین نیروی درگ ( D ) و نیروی اصطکاک ( \mathbit{F}_{\mathrm{fr}} ):

D=\frac{1}{2}\rho SC_{D_g}V^2=0.5\times0.0023769\times175\times0.0862\times V^2\approx0.0178V^2\mathrm{lbs}\bigm\bigmF_{\mathrm{fr}}=\mu_g\times\left(W-L\right)=0.03\times\left(4600-0.5\times0.0023769\times175\times0.89\times V^2\right)\approx0.03\times\left(4600-0.2073\times0.89\times V^2\right)\approx138-0.0055V^2\mathrm{lbs}\bigm
6. تعیین نیروی خالص شتاب‌دهنده ( F_{\text{net}} ):

F_{\mathrm{net}}=F_N-D-F_{\mathrm{fr}}=644-0.0178V^2-\left(138-0.0055V^2\right)=506-0.0123V^2\mathrm{lbs}\bigm
7. تعیین شتاب ( a ):
با استفاده از رابطه نیوتن:

a=\frac{F_{\mathrm{net}}}{W/g}=\frac{506-0.0123V^2}{4600}\times32.174\approx3.53-0.0000854V^2{\mathrm{ft/s}}^2\bigm
8. تعیین مسافت برخاست ( \mathbit{s}_\mathbit{R} ):
با استفاده از روش انتگرال:

s_R=\int_{0}^{V_{\mathrm{LOF}}}\frac{V}{a}dV=\int_{0}^{131.6}\frac{V}{3.53-0.0000854V^2}dV\bigm
حل این انتگرال به صورت زیر است:

s_R=-\frac{1}{2\times0.0000854}\left[\ln{\left(3.53-0.0000854V^2\right)}\right]_0^{131.6}\approx1737\mathrm{ft}\bigm
9. تعیین زمان برخاست ( \mathbit{t}_\mathbit{R} ):
برای تعیین زمان برخاست، از رابطه زیر استفاده می‌کنیم:

t_R=\int_{0}^{V_{\mathrm{LOF}}}\frac{1}{a}dV\bigm
با جایگذاری مقداری از شتاب به دست آمده و حل انتگرال، تخمینی برای زمان برخاست می‌توان ارائه داد. با استفاده از روش میانگین شتاب:

a_{\mathrm{ave}}=\frac{a_{\mathrm{initial}}+a_{\mathrm{final}}}{2}\bigm10. تعیین شتاب (( a(V) )):
با استفاده از قانون دوم نیوتن:

a\left(V\right)=\frac{F_{\mathrm{net}}}{m}\bigm
که:

m=\frac{W}{g}=\frac{4,600}{32.174}\approx143\mathrm{\ slugs}\bigm
بنابراین:

a\left(V\right)=\frac{866.2}{143}\approx6.06\mathrm{\ }{\mathrm{ft/s}}^2\bigm
11. تعیین شتاب اولیه (( \mathbit{a}_{\mathrm{initial}} )):
در سرعت ( V = 0 ):

F_{\mathrm{net}}=F_N-D-F_{\mathrm{fr}}=644-0-138\approx506\mathrm{\ lbs}\bigm

a_{\mathrm{initial}}=\frac{506}{143}\approx3.54\mathrm{\ }{\mathrm{ft/s}}^2\bigm
12. تعیین نسبت شتاب‌ها (( k )):

k=\frac{a_{\mathrm{final}}}{a_{\mathrm{initial}}}=\frac{6.06}{3.54}\approx1.71\bigm
توجه: در محاسبات اولیه، نسبت شتاب‌ها به 0.7 تعیین شده بود که با محاسبه دقیق‌تر به 1.71 رسیدیم. این تغییر نشان‌دهنده تفاوت‌های ناشی از نحوه محاسبه شتاب‌ها است.
13. تعیین شتاب میانگین (( \mathbit{a}_{\mathrm{ave}})):
در محاسبات اولیه، اشتباه محاسبه شده بود که به صورت میانگین ساده شتاب‌های اولیه و نهایی:

a_{\mathrm{ave}}=\frac{a_{\mathrm{initial}}+a_{\mathrm{final}}}{2}=\frac{3.54+6.06}{2}\approx4.80\mathrm{\ }{\mathrm{ft/s}}^2\bigm
بنابراین:
\bigma_{\mathrm{ave}}=\frac{2\times870}{{22.2}^2}\approx\frac{1,740}{492.84}\approx3.53\mathrm{\ }{\mathrm{ft/s}}^2
اما این محاسبه صحیح نیست زیرا شتاب به‌طور غیرخطی با سرعت تغییر می‌کند. بنابراین، باید شتاب میانگین را به‌صورت دقیق‌تر با استفاده از انتگرال محاسبه کنیم
سپس زمان برخاست به صورت زیر محاسبه می‌شود:
t_R=\frac{V_R}{a_{\mathrm{ave}}}=\frac{125.8}{3.53}\approx35.63\mathrm{\ seconds}

" and i want you to solve numerical and step by step with this formula "The ground speed \left(V_G\right), which is the time rate of change of ground distance (s), is illustrated in Figure 9.6. It is defined as follows:
VG=ds dt=V-Vw#(9.1)
 to provide a reasonably accurate representation of jet engine thrust at low speeds:
FNFN0=1-k1V+k2V2#(9.2)
From the definitions of the lift and drag coefficients, given by Equations 7.1 and 7.5, respectively, the lift and drag forces can be expressed as follows:
L=qSCL=12ρSCLV2#(9.3)D=qSCDg=12ρSCDgV2#(9.4)
With the slats and flaps set for takeoff and the landing gear extended, the zero-lift drag coefficient can be determined from the following expression:
CD0=CD0clean +ΔCD0slats +ΔCD0flaps +ΔCD0gear #(9.5)
C_{D_g}=C_{D_0}+\lambda KC_L^2
\mu=\frac{F_{fr}}{R}
The total friction force that acts on the airplane can be expressed as follows:
Ffr=μRMW+μRNW#(9.8)
Ffr=μWcos⁡γG-L#(9.9)
0.0.8. Net Acceleration Force
For the sake of brevity, the net acceleration force acting on the airplane in the direction of motion is herein denoted as \sum F, where
 F=FN-D-Ffr-Wsin⁡γG#(9.10)
or {\ }\sum F=F_N-D-\mu\left(W\cos\funcapply\gamma_G-L\right)-W\sin\funcapply\gamma_G{\ } (from Equation 9.9)
As the runway gradient is a very small quantity, \ ^6 the small angle approximations \cos\funcapply\gamma_G\cong1 and \sin\funcapply\gamma_G\cong\gamma_G (for \gamma_G measured in radians) can be used without loss of accuracy.
Thus \sum F=F_N-D-\mu(W-L)-W\gamma_G
Substituting Equations 9.3 and 9.4 into Equation 9.12 results in the required expression for the net acceleration force:
 F=FN-12ρV2SCDg-μCL-μ+γGW#(9.13)
 The airplane's acceleration (with respect to the ground) can be expressed as follows:
a=ddtds dt=ddtV-Vw=dV dt  if dVw/dt=0#(9.14)
a=\frac{dV}{\mathrm{\ }dt}=\frac{dV}{\mathrm{\ }ds}\frac{ds}{\mathrm{\ }dt}=\frac{dV}{\mathrm{\ }ds}\left(V-V_w\right)
The ground distance measured from the start of the takeoff to the point of rotation \left(s_R\right) can be determined from the following expression:
sR=start end   ds=VwVR  V-Vwa dV#(9.16)
At any instant during the ground run, the airplane's acceleration can be obtained from Newton's second law (see Section 2.4.5):
a= Fm=g FW#(9.17)
s_R=\int_{V_w}^{V_R}\hairsp\hairsp\frac{W\left(V-V_w\right)}{g\sum\ F}\mathrm{\ }dV=\int_{V_w}^{V_R}\hairsp\hairsp\frac{WV}{g\sum\ F}\mathrm{\ }dV-\int_{V_w}^{V_R}\hairsp\hairsp\frac{WV_w}{g\sum\ F}\mathrm{\ }dV
The mean thrust FN is calculated at a reference speed, herein denoted as V^\ast, which is determined as follows:
V*2=12VR-Vw2#(9.19) or  V*=12VR-Vw=0.707VR-Vw#(9.20)
An expression for the net accelerating force ( \sum F ) can be obtained by substituting Equation 9.6 into Equation 9.13, that is,
 F=FN-12ρV2SCD0+λKCL2-μCL-μ+γGW#(9.21)
The required distance \left(s_R\right) is obtained by a summation of the two integrals that are given in Equation 9.18. For the sake of convenience, the two integral expressions will be evaluated separately, where
1  =VwVR  WVg F dV  and  2  =VwVR  WVwg F dV#(9.22)
For the first integral, the following variable substitution is made:
f=V2#(9.23)
thus df=2V\mathrm{\ }dV and dV=\frac{1}{2V}\mathrm{\ }df
hence \int_1\hairsp=\int_{V_w}^{V_R}\hairsp\frac{WV}{g\sum F}\mathrm{\ }dV=\int_{\sqrt{f_w}}^{\sqrt{f_R}}\hairsp\frac{WV}{g\sum F}\left(\frac{1}{2V}\right)df=\frac{1}{2g}\int_{\sqrt{f_w}}^{\sqrt{f_R}}\hairsp\frac{W}{\sum F}\mathrm{\ }df
Now substitute \sum F from Equation 9.21 into Equation 9.24:
1  =12gfwfR  1FNW-μ-γG-ρS2WCD0+λKCL2-μCLf df#(9.25)
To simplify the analysis, arbitrary constants A_2,B_2, and C_2 are now introduced, where
A_2=FNW-μ-γG  and  B2=ρS2WCD0+λKCL2-μCL
and {\ }C_2=\sqrt{\frac{A_2}{B_2}}
Thus \int_1\hairsp=\frac{1}{2g}\int_{\sqrt{f_w}}^{\sqrt{f_R}}\hairsp\frac{1}{A_2-B_2f}\mathrm{\ }df=\frac{-1}{2gB_2}\left[\ln\funcapply\left(A_2-B_2f\right)\right]_{\sqrt{f_w}}^{\sqrt{f_R}}
1  =-12gB2ln⁡A2-B2V2VwVR=12gB2ln⁡A2-B2Vw2A2-B2VR2#(9.27)
The second integral is now considered:
\begin{matrix}\ &\ \int_{2}\hairsp\hairsp=\frac{V_w}{g}\int_{V_w}^{V_R}\hairsp\hairsp\frac{1}{A_2-B_2V^2}\mathrm{\ } dV=\frac{V_W}{gB_2}\int_{V_w}^{V_R}\hairsp\hairsp\frac{1}{C_2^2-V^2}\mathrm{\ } dV\\\end{matrix}
(\int_{2}\ =\frac{V_w}{2gB_2C_2}\left[\ln{\left(\frac{C_2+V}{C_2-V}\right)}\right]_{V_w}^{V_R}=\frac{V_w}{2gB_2C_2}\ln{\left\{\frac{\left(C_2+V_R\right)\left(C_2-V_w\right)}{\left(C_2-V_R\right)\left(C_2+V_w\right)}\right\}})
The two results are now added to give the distance s_R :
sR=12gB2ln⁡A2-B2Vw2A2-B2VR2-VwC2ln⁡C2+VRC2-VwC2-VRC2+Vw#(9.29)
where A_2,B_2, and C_2 are defined in Equation 9.26 .
uniform acceleration:
sR=1aVwVR  V-VwdV=1aV22-VVwVwVR=12aVR-Vw2#(9.30)
the following integral expression is obtained:
sR=VwVR  V-Vwa dV=WgVwVR  V-Vw F dV#(9.31)
To integrate Equation 9.31 numerically, it is possible to divide the takeoff distance s_R into n segments (or velocity intervals) and to use the trapezoidal rule for incremental values of V :
sR=W2gi=0n  (V)i+1-Vw Fi+1+(V)i-Vw FiΔV#(9.32)
The time that the airplane takes to accelerate to the rotation speed \(\left(V_R\right)\) may be determined using similar approaches to those presented for the distance calculation (Sections 9.4.1 to 9.4.4). The time-velocity relationship is given by
\[
\mathrm{d}t=\frac{1}{a}\mathrm{d}V\emsp\left(\mathrm{\ based\ on\ Equation\ }9.15\right)
Thus\ t=\int_{\mathrm{start\ }}^{\mathrm{end\ }}\mathrm{d}t=\int_{V_w}^{V_R}\frac{1}{a}\mathrm{d}V)
t_R=\frac{1}{\bar{a}}\int_{V_w}^{V_R}\mathrm{d}V=\frac{V_R-V_w}{\bar{a}}
s_{RL}=\Delta t_{RL}\left(\frac{V_R+V_{LOF}}{2}-V_w\right)
s_g=s_R+s_{RL}
U\cong\int\left(F_N-D\right)\mathrm{d} s=\left(F_N-D\right)_{\mathrm{avg\ }}s_{L35}
U=\frac{W}{2g}\left(V_{35}^2-V_{LOF}^2\right)+h_{sc}W
By combining Equations 9.37 and 9.38, the following expression for the distance from liftoff to the 35 ft point is obtained (for zero wind):
s_{L35}=\left(\frac{W}{\left(F_N-D\right)_{avg}}\right)\left(\frac{V_{35}^2-V_{LOF}^2}{2g}+h_{sc}\right)
An alternative-and relatively straightforward-method to estimate the distance from liftoff to the 35 ft point is based on the product of the average time and the average ground speed for this segment of the takeoff:
s_{L35}=\Delta t_{L35}\left(\frac{V_{LOF}+V_{35}}{2}-V_W\right)
s35=V3522a#(9.41)
where s_{35} is the distance from brake release (V=0) to the 35 ft screen height (typical units: m,ft ); and
V_{35} is the TAS at 35 ft (equal to the ground speed as V_w=0 ) (typical units: m/s,ft/s ).
As the greatest part of the takeoff distance will be the ground run, Equation 9.21 can be used to derive a relationship for the mean acceleration. Assuming zero slope (i.e., \gamma_G=0 ), it follows that
a= Fm=gWFN-12ρV2SCD-μRCL-μRW#(9.42)
a=gFNW(1-k)
s_{35}=\frac{V_{35}^2W}{2g(1-k)F_{N_{0.7}}}
where 𝐹𝑁0.7 is the thrust evaluated at 0.7𝑉35 (typical units: N,lb ).
The speed term in Equation 9.44 is now eliminated using the lift coefficient at the screen height 𝐶𝐿35, that is,
$$
𝐶𝐿35=2𝐿𝜎𝜌0𝑉352𝑆=2𝑛𝑧𝑊𝜎𝜌0𝑉352𝑆#(9.45)

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