translate in persian , dont translate professional words like تیک آف
"To solve the takeoff ground run distance (( S_G )) and the associated parameters for the given aircraft, we'll systematically apply the formulas and utilize the provided data. Here's a step-by-step breakdown of the solution:
Given Parameters
	Weight (W): 4,600 lbₚ
	Average Thrust-to-Weight Ratio (( T/W ))(_{\text{ave}} ): 0.14
	Wind Speed (Vw): 0 ft/s (No wind)
	Runway Slope (( \phi )): 0° (Flat runway)
	Flap Deflection (( \delta_f )): 15°
	Wing Area (S): 175 ft²
	Wing Span (b): 35 ft
	Aspect Ratio (A): 7
	Ground Friction Coefficient (( \mu_g )): 0.03
	Maximum Lift Coefficient (( C_{L_{\text{max}}} )): 1.69
	Oswald Efficiency Factor (e): 0.80
	Gravitational Acceleration (g): 32.174 ft/s²
	Air Density (( \rho )): 0.0023769 slugs/ft³
	Thrust at V = 0 (( T_0 )): 2,000 lbₚ
	Thrust at Rotation Speed (( T_R )): 1,200 lbₚ
	Rotation Speed (( V_R )): 65 ft/s
Step 1: Determine the Takeoff Speed (( V_{\text{LO}} ))
The takeoff speed (( V_{\text{LO}} )) is the speed at which the aircraft can lift off the ground. It's determined by the condition where lift equals weight:
[
L = W = \frac{1}{2} \rho S C_{L_{\text{max}}} V_{\text{LO}}^2
]
Solving for ( V_{\text{LO}} ):
[
V_{\text{LO}} = \sqrt{\frac{2W}{\rho S C_{L_{\text{max}}}}}
]
Plugging in the values:
[
V_{\text{LO}} = \sqrt{\frac{2 \times 4600}{0.0023769 \times 175 \times 1.69}} \approx \sqrt{\frac{9200}{0.7025}} \approx \sqrt{13115.6} \approx 114.5 \text{ ft/s}
]
Step 2: Calculate the Maximum Lift Coefficient (( C_D ))
Using the relationship:
[
C_D = \frac{C_L^2}{\pi e A}
]
At takeoff speed:
[
C_D = \frac{1.69^2}{\pi \times 0.80 \times 7} \approx \frac{2.8561}{17.592} \approx 0.1623
]
Step 3: Compute the Lift and Drag at an Average Speed
Assume the average speed during acceleration is approximately ( 0.707 V_{\text{LO}} ):
[
V_{\text{avg}} = 0.707 \times 114.5 \approx 81 \text{ ft/s}
]
Lift (L) at ( V_{\text{avg}} ):
[
L = \frac{1}{2} \rho S C_L V_{\text{avg}}^2 = \frac{1}{2} \times 0.0023769 \times 175 \times 1.69 \times 81^2 \approx 4600 \text{ lbₚ}
]
Drag (D) at ( V_{\text{avg}} ):
[
D = \frac{1}{2} \rho S C_D V_{\text{avg}}^2 = 0.207783 \times 0.1623 \times 81^2 \approx 443 \text{ lbₚ}
]
Step 4: Determine the Net Force and Acceleration
Using the formula:
[
a = \left(\frac{g}{W}\right) \left[ T - D - \mu(W - L) - W \delta \right]
]
Where:
	( T ) is the thrust at ( V_{\text{avg}} ): ( T = 2000 - 12.3077 \times 81 \approx 1004 \text{ lbₚ} )
	( \mu ) is the ground friction coefficient: 0.03
	( \delta ) is the flap-induced angle component: ( W \delta_f = 4600 \times \sin(15°) \approx 1188 \text{ lbₚ} )
Plugging in the values:
[
a = \left(\frac{32.174}{4600}\right) \left[ 1004 - 443 - 0.03(4600 - 4600) - 1188 \right] \approx 0.00699 \times (-627) \approx -4.38 \text{ ft/s}²
]
Note: A negative acceleration indicates deceleration, which suggests an error in the initial assumptions. However, considering the provided methods and aiming for a positive acceleration, we'll use the average acceleration method as outlined below.
Step 5: Calculate the Takeoff Ground Roll Distance (( S_G ))
Using the simplified equation:
[
S_G = \frac{W V_{\text{LO}}^2}{2g \left[ (T - \mu W) - (C_D - \mu C_L) S q \right]}
]
Where:
	( q = \frac{1}{2} \rho V_{\text{avg}}^2 = 0.5 \times 0.0023769 \times 81^2 \approx 7.798 \text{ slugs/ft·s²} )
	( (C_D - \mu C_L) = 0.1623 - 0.03 \times 1.69 \approx 0.1116 )
Plugging in the values:
[
\mathbit{S}_\mathbit{G}=\frac{\mathbf{4600}\times\mathbf{114}.\mathbf{5}^\mathbf{2}}{\mathbf{2}\times\mathbf{32}.\mathbf{174}\times\left(\mathbf{1004}-\mathbf{138}-\mathbf{0}.\mathbf{1116}\times\mathbf{175}\times\mathbf{7}.\mathbf{798}\right)}\approx\frac{\mathbf{6}.\mathbf{0306}\times\mathbf{1}\mathbf{0}^\mathbf{7}}{\mathbf{45},\mathbf{881}}\approx\mathbf{1314}\mathrm{\ ft}
]
Step 6: Estimate the Time Required for Takeoff (( \mathbit{t}_{\mathbit{S}_\mathbit{G}} ))
Using the relationship:
\mathbit{t}_{\mathbit{S}_\mathbit{G}}=\frac{\mathbit{V}_{\mathrm{LO}}}{\mathbit{a}_{\mathrm{avg}}}
Where ( a_{\text{avg}} ) is the average acceleration. From the calculations:
\mathbit{a}_{\mathrm{avg}}\approx\frac{\mathbf{32}.\mathbf{174}}{\mathbf{4600}}\times\mathbf{713}.\mathbf{6}\approx\mathbf{4}.\mathbf{99}\mathrm{\ }ft/s2
Thus,
\mathbit{t}_{\mathbit{S}_\mathbit{G}}=\frac{\mathbf{114}.\mathbf{5}}{\mathbf{4}.\mathbf{99}}\approx\mathbf{23}\mathrm{\ seconds}
Final Results
	Takeoff Ground Roll Distance\ ((\ \mathbit{S}_\mathbit{G}\ )):\ ~\mathbf{1},\mathbf{314} feet
	Time to Reach Takeoff Speed ((\ \mathbit{t}_{\mathbit{S}_\mathbit{G}}\ )):\ ~\mathbf{23} seconds
These calculations utilize the provided formulas and parameters to estimate the takeoff performance of the aircraft under the specified conditions.

"