HW5_soln

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Name: HW5 Page 1 of 11 ENAE283 Homework #5 Due Saturday, 7/1/22 by 11:59PM For all homework assignments in this course, you are required to submit fully- explained solutions, indicating the sources for any numbers and equations used. Boxed areas are provided for your final answers. If you cannot use this document directly as your worksheets, please work neatly on your own paper. Include a header on each page like the ones shown, write the problem statement at the top and box your final answer at the lower right corner. Pay attention to appropriate significant figures in final answers.

Name: HW5 Page 2 of 11 (1-11) You have a choice for this portion of the HW: you can study the performance characteristics of a propeller driven airplane (Beechcraft Bonanza) or a twin jet attack aircraft (Fairchild Republic A-10). Pick one and then follow through the whole series of problems relating to that airplane. Each of these problems is most easily accomplished by employing a spreadsheet tool like Excel, since all of the calculations are repeated at different velocities and altitudes. Your Excel spreadsheet or MATLAB arrays should contain the following calculations. 𝑉𝑉 ∞ � 𝑓𝑓𝑓𝑓 𝑠𝑠 � 𝐶𝐶 𝐿𝐿 = 𝑊𝑊 𝑞𝑞 ∞ 𝑆𝑆 𝐶𝐶 𝐷𝐷 = 𝐶𝐶 𝐷𝐷 , 0 + 𝐶𝐶 𝐷𝐷 , 𝑖𝑖 = 𝐶𝐶 𝐷𝐷 , 0 + 𝐶𝐶 𝐿𝐿 2 / 𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 𝐶𝐶 𝐿𝐿 / 𝐶𝐶 𝐷𝐷 𝑇𝑇 𝑅𝑅 = 𝑊𝑊 𝐶𝐶 𝐿𝐿 / 𝐶𝐶 𝐷𝐷 [ 𝑙𝑙𝑙𝑙 ] 𝑃𝑃 𝑅𝑅 = 𝑇𝑇 𝑅𝑅 𝑉𝑉 ∞ �𝑓𝑓𝑓𝑓 ∙ 𝑙𝑙𝑙𝑙 𝑠𝑠 � 𝑃𝑃 𝐴𝐴 �𝑓𝑓𝑓𝑓 ∙ 𝑙𝑙𝑙𝑙 𝑠𝑠 � 𝑇𝑇 𝐴𝐴 [ 𝑙𝑙𝑙𝑙 ] 𝜋𝜋𝐶𝐶 = 𝑃𝑃 𝐴𝐴 − 𝑃𝑃 𝑅𝑅 𝑊𝑊 � 𝑓𝑓𝑓𝑓 𝑠𝑠 � The only variable that changes is the density between Sea Level and altitude. Besides changing the thrust/power required curves via the 𝑞𝑞 ∞ term, this also affects the power available from the engine which is assumed to be proportional to density, so 𝑃𝑃 𝐴𝐴 , 𝑆𝑆𝐿𝐿 𝜌𝜌 𝑆𝑆𝐿𝐿 = 𝑃𝑃 𝐴𝐴 , 𝑎𝑎𝑎𝑎𝑎𝑎 𝜌𝜌 𝑎𝑎𝑎𝑎𝑎𝑎 → 𝑃𝑃 𝐴𝐴 , 𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑃𝑃 𝐴𝐴 , 𝑆𝑆𝐿𝐿 � 𝜌𝜌 𝑎𝑎𝑎𝑎𝑎𝑎 𝜌𝜌 𝑆𝑆𝐿𝐿 � For PR at altitude, you can either rerun the above calculations with a new density value, or multiply both 𝑉𝑉 ∞ and 𝑃𝑃𝜋𝜋 𝑆𝑆𝐿𝐿 by the altitude correction factor �𝜌𝜌 𝑆𝑆𝐿𝐿 𝜌𝜌 𝑎𝑎𝑎𝑎𝑎𝑎 ⁄ . Note: If you don’t multiply 𝑉𝑉 ∞ by the correction factor, your plot will be incorrect. See below.

Name: HW5 Page 3 of 11 Can Prop Plane - Beechcraft Bonanza V-tailed, Single Engine Private Plane The characteristics of this airplane are as follows: wing area = 180 ft 2 ; aspect ratio = 6.2; Oswald efficiency factor = 0.92; weight = 3000 lb; zero-lift drag coefficient = 0.027; single piston engine power = 350 hp (at sea level); propeller efficiency = 0.81; specific fuel consumption = 0.42 lb fuel/hp.hr; fuel capacity = 45 gallons; maximum gross weight = 3400 lb; C Lmax = 1.2 on take-off; wings are 4 ft off the ground during take-off roll; C Lmax = 1.8 on landing (with flaps). 1. Find a picture of this airplane online, copy and paste it into your homework and cite. 2. Plot the power required versus velocity curve at sea level.

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Name: HW5 Page 4 of 11 3. Determine the maximum velocity at sea level. From PR=PA max , 𝑉𝑉 𝑚𝑚𝑎𝑎𝑚𝑚 = 293 𝑓𝑓𝑓𝑓 / 𝑠𝑠 4. Draw the power required versus velocity curve at 14,000 ft altitude. See above 5. Determine the maximum velocity at 14,000 ft altitude. 𝑉𝑉 𝑚𝑚𝑎𝑎𝑚𝑚 = 283 𝑓𝑓𝑓𝑓 / 𝑠𝑠 6. Calculate the maximum rate of climb at sea level and at 14,000 ft altitude. 𝜋𝜋 . 𝐶𝐶 . 𝑚𝑚𝑎𝑎𝑚𝑚 = ( 𝑃𝑃 𝐴𝐴 − 𝑃𝑃 𝑅𝑅 ) 𝑚𝑚𝑎𝑎𝑚𝑚 / 𝑊𝑊 𝜋𝜋 . 𝐶𝐶 . 𝑆𝑆𝐿𝐿 = 42.3 𝑓𝑓𝑓𝑓 𝑠𝑠 = 2,538 𝑓𝑓𝑓𝑓 𝑚𝑚𝑚𝑚𝑚𝑚 𝑎𝑎𝑚𝑚𝑎𝑎 𝜋𝜋 . 𝐶𝐶 . 14𝑘𝑘 = 21.8 𝑓𝑓𝑓𝑓 𝑠𝑠 = 1,308 𝑓𝑓𝑓𝑓 / 𝑚𝑚𝑚𝑚𝑚𝑚 7. Estimate the absolute ceiling of this airplane (assuming rate of climb varies linearly with altitude). From the R.C. calculations above, 𝜋𝜋 . 𝐶𝐶 . ( ℎ ) = 42 . 3−21 . 8 0−14 , 000 ℎ + 42.3 [ 𝑓𝑓𝑎𝑎 𝑠𝑠 ] Extrapolating the 𝜋𝜋 . 𝐶𝐶 . trend from above to find when 𝜋𝜋 . 𝐶𝐶 . = 0, ℎ 𝑎𝑎𝑎𝑎𝑠𝑠 ≈ 28,888 𝑓𝑓𝑓𝑓 8. Calculate the range and endurance of this aircraft at sea level. For propeller-driven aircraft, 𝜋𝜋 𝑚𝑚𝑎𝑎𝑚𝑚 = 𝜂𝜂 𝑐𝑐 � 𝐶𝐶 𝐿𝐿 𝐶𝐶 𝐷𝐷 � 𝑚𝑚𝑎𝑎𝑚𝑚 ln � 𝑊𝑊 0 𝑊𝑊 1 � � 𝐶𝐶 𝐿𝐿 𝐶𝐶 𝐷𝐷 � 𝑚𝑚𝑎𝑎𝑚𝑚 𝑜𝑜𝑐𝑐𝑐𝑐𝑜𝑜𝑜𝑜𝑠𝑠 𝑤𝑤ℎ𝜋𝜋𝑚𝑚 𝐶𝐶 𝐷𝐷 , 0 = 𝐶𝐶 𝐷𝐷 , 𝑖𝑖 = 𝐶𝐶 𝐿𝐿 2 𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 → 𝐶𝐶 𝐿𝐿 = 0.696 𝑎𝑎𝑚𝑚𝑎𝑎 𝐶𝐶 𝐷𝐷 = 𝐶𝐶 𝐷𝐷 , 0 + 𝐶𝐶 𝐷𝐷 , 𝑖𝑖 = 0.054 → � 𝐶𝐶 𝐿𝐿 𝐶𝐶 𝐷𝐷 � 𝑚𝑚𝑎𝑎𝑚𝑚 = 12.88 𝜋𝜋 𝑚𝑚𝑎𝑎𝑚𝑚 = 3.81 × 10 6 𝑓𝑓𝑓𝑓 = 822.6 𝑚𝑚𝑚𝑚 𝐶𝐶 𝐿𝐿 3 2 / 𝐶𝐶 𝐷𝐷 𝑚𝑚𝑎𝑎𝑚𝑚 = 12.167 𝐸𝐸 𝑚𝑚𝑎𝑎𝑚𝑚 = 𝜂𝜂 𝑐𝑐 � 𝐶𝐶 𝐿𝐿 3 2 𝐶𝐶 𝐷𝐷 � 𝑚𝑚𝑎𝑎𝑚𝑚 � 2 𝜌𝜌𝑆𝑆 � 1 �𝑊𝑊 1 − 1 �𝑊𝑊 0 � = 2.93 × 10 4 𝑠𝑠 = 8.15 ℎ𝑜𝑜 9. Estimate the sea level take-off distance on a paved runway. 𝑆𝑆 𝐿𝐿𝐿𝐿 = 1.44 𝑊𝑊 2 /( 𝑔𝑔𝜌𝜌𝑆𝑆𝐶𝐶 𝐿𝐿 , 𝑚𝑚𝑎𝑎𝑚𝑚 { 𝑇𝑇 − [ 𝐷𝐷 + 𝜇𝜇 𝑅𝑅 ( 𝑊𝑊 − 𝐿𝐿 )]} 𝑉𝑉 𝑎𝑎𝑎𝑎 where 𝑉𝑉 𝑎𝑎𝑎𝑎 = 0.7( 𝑉𝑉 𝐿𝐿𝐿𝐿 ) = 0.7 � 1.2 � 2 𝑊𝑊 𝑚𝑚𝑎𝑎𝑚𝑚 𝜌𝜌𝑆𝑆𝐶𝐶 𝐿𝐿 , 𝑚𝑚𝑎𝑎𝑚𝑚 � = 115.1 𝑓𝑓𝑓𝑓 / 𝑠𝑠 so 𝑆𝑆 𝐿𝐿𝐿𝐿 = 543.1 𝑓𝑓𝑓𝑓 (670 ft w/o ground effect) 10. Estimate the sea level landing roll distance assuming the plane is landing with a weight of 3000 lb and that lift is zero after touchdown. 𝑆𝑆 𝐿𝐿 = 1.69 𝑊𝑊 2 /( 𝑔𝑔𝜌𝜌𝑆𝑆𝐶𝐶 𝐿𝐿 , 𝑚𝑚𝑎𝑎𝑚𝑚 [ 𝐷𝐷 + 𝜇𝜇 𝑟𝑟 ( 𝑊𝑊 − 𝐿𝐿 )] 0 . 7𝑉𝑉 𝑇𝑇 where 0.7 𝑉𝑉 𝑇𝑇 = 0.7(1.3 𝑉𝑉 𝑠𝑠𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ) = 0.7 � 1.3 � 2 𝑊𝑊 𝜌𝜌𝑆𝑆𝐶𝐶 𝐿𝐿 , 𝑚𝑚𝑎𝑎𝑚𝑚 � = 80.3 𝑓𝑓𝑓𝑓 𝑠𝑠 so 𝑆𝑆 𝐿𝐿 = 495.7 𝑓𝑓𝑓𝑓 11. Look up some interesting fact about this airplane and include that in your homework.

Name: HW5 Page 5 of 11 Jet Plane - Fairchild Republic A-10 Twin Jet Attack Aircraft The characteristics of this airplane are as follows: wing area = 47 m 2 ; aspect ratio = 6.5; Oswald efficiency factor = 0.88; Dry weight = 86 kN; zero-lift drag coefficient = 0.08; static thrust of each jet engine = 40 kN (at sea level); assume the engine thrust varies directly with free stream density; thrust-specific fuel consumption = 1.1 N of fuel per N of thrust per hour; fuel capacity = 49000 N; maximum gross weight = 135kN; C Lmax = 0.8 on take-off; wings are 5 ft off the ground during take-off roll; C Lmax = 2.8 on landing (with flaps). 1. Find a picture of this airplane online, copy and paste it into your homework and cite. 2. Plot the power required versus velocity curve for SLF at sea level. Use gross weight. 3. Determine the maximum velocity at sea level. 𝑉𝑉 𝑚𝑚𝑎𝑎𝑚𝑚 = ~185 𝑚𝑚 / 𝑠𝑠 4. Plot the power required versus velocity curve at 6 km altitude. 5. Determine the maximum velocity at 6 km altitude. 𝑉𝑉 𝑚𝑚𝑎𝑎𝑚𝑚 = ~153 𝑚𝑚 / 𝑠𝑠 6. Calculate the maximum rate of climb at sea level and at 6 km altitude. 𝜋𝜋 . 𝐶𝐶 . 𝑚𝑚𝑎𝑎𝑚𝑚 = ( 𝑃𝑃 𝐴𝐴 − 𝑃𝑃 𝑅𝑅 )/ 𝑊𝑊 2 𝜋𝜋 . 𝐶𝐶 . 𝑆𝑆𝐿𝐿 = 40.1 𝑚𝑚 𝑠𝑠 = 144 𝑘𝑘𝑚𝑚 ℎ 𝑎𝑎𝑚𝑚𝑎𝑎 𝜋𝜋 . 𝐶𝐶 . 14𝑘𝑘 = 9.50 𝑚𝑚 𝑠𝑠 = 34.2 𝑘𝑘𝑚𝑚 / ℎ𝑜𝑜 7. Estimate the absolute ceiling of this airplane (assuming rate of climb varies linearly with altitude). From the R.C. calculations above, 𝜋𝜋 . 𝐶𝐶 . ( ℎ ) = 40 . 1−9 . 5 0−6 , 000 ℎ + 40.1 [ 𝑚𝑚 𝑠𝑠 ] Extrapolating the 𝜋𝜋 . 𝐶𝐶 . trend from above to find when 𝜋𝜋 . 𝐶𝐶 . = 0, ℎ 𝑎𝑎𝑎𝑎𝑠𝑠 ≈ 7,860 𝑚𝑚

Name: HW5 Page 6 of 11 8. Calculate the range and endurance of this aircraft at sea level. � 𝐶𝐶 𝐿𝐿 1 2 𝐶𝐶 𝐷𝐷 � 𝑚𝑚𝑎𝑎𝑚𝑚 is needed for this calculation, and at that condition 𝐶𝐶 𝐷𝐷 , 0 = 3 𝐶𝐶 𝐷𝐷 , 𝑖𝑖 So 𝐶𝐶 ( 𝐷𝐷 , 0 ) = 3 𝐶𝐶 𝐿𝐿 2 𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 → 0.08 = 3 ( 𝐶𝐶 𝐿𝐿 2 ) 𝜋𝜋 (0.88)(6.5) → 𝐶𝐶 𝐿𝐿 = 0.692241 𝐶𝐶 𝐷𝐷 = 0.08 + (0.692241) 2 𝜋𝜋 (0.88)(6.5) = 0.032 + 1 3 (0.032) = 0.106667 � 𝐶𝐶 𝐿𝐿 1 2 𝐶𝐶 𝐷𝐷 � 𝑚𝑚𝑎𝑎𝑚𝑚 = 0.692241 1 2 0.106667 = 7.8 𝜋𝜋 𝑚𝑚𝑎𝑎𝑚𝑚 = 2 � 2 𝜌𝜌𝑆𝑆 1 𝑐𝑐 𝑎𝑎 � 𝐶𝐶 𝐿𝐿 1 2 𝐶𝐶 𝐷𝐷 � 𝑚𝑚𝑎𝑎𝑚𝑚 �𝑊𝑊 0 1 2 − 𝑊𝑊 1 1 2 � = 2 � 2 (1.225)(47) 3600 1.1 (7.8) � (135000) 1 2 − (86000) 1 / 2 � = 706. 𝑘𝑘𝑚𝑚 𝐸𝐸 𝑚𝑚𝑎𝑎𝑚𝑚 = 1 𝑐𝑐 𝑎𝑎 � 𝐶𝐶 𝐿𝐿 𝐶𝐶 𝐷𝐷 � 𝑚𝑚𝑎𝑎𝑚𝑚 ln � 𝑊𝑊 0 𝑊𝑊 1 � Where at 𝐶𝐶 𝐿𝐿 𝐶𝐶 𝐷𝐷 𝑚𝑚𝑎𝑎𝑚𝑚 , 𝐶𝐶 𝐷𝐷 , 0 = 𝐶𝐶 𝐷𝐷 , 𝑖𝑖 → 0.08 = 𝐶𝐶 𝐿𝐿 2 𝜋𝜋𝜋𝜋𝐴𝐴𝑅𝑅 = 𝐶𝐶 𝐿𝐿 2 𝜋𝜋 ( 0 . 88 )( 6 . 5 ) → 𝐶𝐶 𝐿𝐿 = 1.199 And 𝐶𝐶 𝐷𝐷 = 𝐶𝐶 𝐷𝐷 , 0 + 𝐶𝐶 𝐷𝐷 , 𝑖𝑖 = 2 𝐶𝐶 𝐷𝐷 , 0 = 0.16 so � 𝐶𝐶 𝐿𝐿 𝐶𝐶 𝐷𝐷 � 𝑚𝑚𝑎𝑎𝑚𝑚 = 1 . 199 0 . 16 = 7.49375 𝐸𝐸 𝑚𝑚𝑎𝑎𝑚𝑚 = 3600 1 . 1 (7.49375) ln � 135000 86000 � = 11100 𝑠𝑠 = 3.07 ℎ𝑜𝑜 9. Estimate the sea level take-off distance on a paved runway. 𝑆𝑆 𝐿𝐿𝐿𝐿 = 1.44 𝑊𝑊 2 /( 𝑔𝑔𝜌𝜌𝑆𝑆𝐶𝐶 𝐿𝐿 , 𝑚𝑚𝑎𝑎𝑚𝑚 { 𝑇𝑇 − [ 𝐷𝐷 + 𝜇𝜇 𝑅𝑅 ( 𝑊𝑊 − 𝐿𝐿 )]} 𝑉𝑉 𝑎𝑎𝑎𝑎 where 𝑉𝑉 𝑎𝑎𝑎𝑎 = 0.7( 𝑉𝑉 𝐿𝐿𝐿𝐿 ) = 0.7 � 1.2 � 2 𝑊𝑊 𝜌𝜌𝑆𝑆𝐶𝐶 𝐿𝐿 , 𝑚𝑚𝑎𝑎𝑚𝑚 � = 0.7 � 1.2 � 2(135000) (1.225)(47)(0.8) � = 64.3 𝑚𝑚 / 𝑠𝑠 So, using aerodynamic forces at this velocity (and modifying 𝐷𝐷 to include ground effect via eq. 6.99) 𝐶𝐶 𝐷𝐷 = 𝐶𝐶 𝐷𝐷 , 0 + 𝜙𝜙 𝐶𝐶 𝐿𝐿 2 𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 = 𝐶𝐶 𝐷𝐷 , 0 + ( 16ℎ 𝑙𝑙 ⁄ ) 2 1 + ( 16ℎ 𝑙𝑙 ⁄ ) 2 𝐶𝐶 𝐿𝐿 2 𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 = 0.08 + (16(1.524) (17.48) ⁄ ) 2 1 + (16(1.524) (17.48) ⁄ ) 2 (0.8) 2 𝜋𝜋 (0.88)(6.5) 𝐶𝐶 𝐷𝐷 = 0.103525 𝐷𝐷 = 𝑞𝑞 ∞ 𝑆𝑆𝐶𝐶 𝐷𝐷 = 1 2 (1.225)(64.3) 2 (47)(0.103525) = 12,321 𝑁𝑁 𝐿𝐿 = 𝑞𝑞 ∞ 𝑆𝑆𝐶𝐶 𝐷𝐷 = 1 2 (1.225)(64.3) 2 (47)(0.8) = 95,217 𝑁𝑁 𝑆𝑆 𝐿𝐿𝐿𝐿 = 1.44(135000) 2 /((9.81)(1.225)(47)(0.8){80000 − [12,321 + 0.02(135000 − 95,217)]} 𝑉𝑉 𝑎𝑎𝑎𝑎 𝑆𝑆 𝐿𝐿𝐿𝐿 = 868 𝑚𝑚 (887 𝑚𝑚 𝑤𝑤𝑚𝑚𝑓𝑓ℎ 𝑚𝑚𝑜𝑜 𝑔𝑔𝑜𝑜𝑜𝑜𝑜𝑜𝑚𝑚𝑎𝑎 𝜋𝜋𝑓𝑓𝑓𝑓𝜋𝜋𝑐𝑐𝑓𝑓 ) 10. Estimate the sea level landing roll distance assuming the plane is landing with its full gross weight of 135 kN with i) C Lmax and that ii) spoilers make C Lmax =0 after touchdown. 𝑆𝑆 𝐿𝐿 = 1.69 𝑊𝑊 2 /( 𝑔𝑔𝜌𝜌𝑆𝑆𝐶𝐶 𝐿𝐿 , 𝑚𝑚𝑎𝑎𝑚𝑚 [ 𝐷𝐷 + 𝜇𝜇 𝑟𝑟 ( 𝑊𝑊 − 𝐿𝐿 )] 0 . 7𝑉𝑉 𝑇𝑇

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Name: HW5 Page 7 of 11 where 0.7 𝑉𝑉 𝑇𝑇 = 0.7(1.3 𝑉𝑉 𝑠𝑠𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ) = 0.7 � 1.3 � 2 𝑊𝑊 𝜌𝜌𝑆𝑆𝐶𝐶 𝐿𝐿 , 𝑚𝑚𝑎𝑎𝑚𝑚 � = 32.53 𝑚𝑚 𝑠𝑠 𝐷𝐷 = 𝑞𝑞 ∞ 𝑆𝑆𝐶𝐶 𝐷𝐷 = 𝑞𝑞 ∞ 𝑆𝑆 �𝐶𝐶 𝐷𝐷 , 0 + 𝐶𝐶 𝐿𝐿 2 𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 � = 1 2 (1.225)(32.3) 2 (47) � 0.08 + (2.8) 2 𝜋𝜋 (0.88)(6.5) � = 15,506 𝑁𝑁 so with ground effect and 𝐿𝐿 = 0 after touchdown 𝑆𝑆 𝐿𝐿 = 1.69(86000) 2 /((9.81)(1.225)(47)(2.8)[15,506 + (0.4)(86000)] 0 . 7𝑉𝑉 𝑇𝑇 𝑆𝑆 𝐿𝐿 = 158 𝑚𝑚 empty or 280 m fully fueled. 11. Look up some interesting fact about this airplane and include that in your homework.

Name: HW5 Page 8 of 11 12. Consider a wing-body combination, having a planform area of 3 m 2 and a chord length of 1 m, flying at a standard altitude of 3.0 km and a speed of 70 m/s. At an absolute angle of attack of zero, the moment about its center of gravity is -35 N·m. When the vehicle pitches up 10° from this condition, the lift force and moment about the center of gravity are 6,000 N and +30 N·m, respectively. Is this wing-body statically stable? (Explain why or why not, with numbers. That the zero-lift moment is negative is sufficient to conclude that it can’t be longitudinally balanced even if it is stable. Strictly speaking, we should look at the slope if we’re interested specifically in stability (although points will not be deducted if you stopped after the above reasoning). From Appendix A, 𝜌𝜌 ∞ = 0.90926 𝑘𝑘𝑘𝑘 𝑚𝑚 3 At 𝛼𝛼 𝑎𝑎 = 0°, we have zero lift and therefore the moment about the aerodynamic center is the same as the moment about the center of gravity: 𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑎𝑎 = 𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑘𝑘 ( 𝛼𝛼 𝑎𝑎 = 0) = 𝑀𝑀 𝑎𝑎𝑘𝑘 1 2 𝜌𝜌 ∞ 𝑉𝑉 ∞ 2 𝑆𝑆𝑐𝑐 = − − 35 1 2 (0.90926)(70) 2 (3)(1) = − 0.005237 At 𝛼𝛼 𝑎𝑎 = 10°, 𝐶𝐶 𝐿𝐿 ( 𝛼𝛼 𝑎𝑎 = 10°) = 𝐿𝐿 1 2 𝜌𝜌 ∞ 𝑉𝑉 ∞ 2 𝑆𝑆 = 6,000 1 2 (0.90926)(70) 2 (3) = 0.8978 𝑎𝑎 𝑤𝑤𝑎𝑎 = 0.8978 10° = 0.08978 𝑝𝑝𝜋𝜋𝑜𝑜 ° 𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑘𝑘 ( 𝛼𝛼 𝑎𝑎 = 10°) = 𝑀𝑀 𝑎𝑎𝑘𝑘 1 2 𝜌𝜌 ∞ 𝑉𝑉 ∞ 2 𝑆𝑆𝑐𝑐 = − 30 1 2 (0.90926)(70) 2 (3)(1) = 0.004489 It is clear from inspection of the moment coefficients that the moment coefficient has a positive derivative with respect to angle of attack, indicating that the wing-body is unstable. (A slight increase in α will make the moment more positive, tending to increase α even further; and vice versa for a decrease in α .) We can calculate this derivative explicitly if we wish: 𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑘𝑘 , 𝑤𝑤𝑎𝑎 = 𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑎𝑎 + 𝑎𝑎 𝑤𝑤𝑎𝑎 𝛼𝛼 𝑎𝑎 �ℎ − ℎ 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑎𝑎 � 0.004489 = − 0.005237 + 0.08978(10°) �ℎ − ℎ 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑎𝑎 � ℎ − ℎ 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑎𝑎 = 0.01083 𝜕𝜕𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑘𝑘 , 𝑤𝑤𝑎𝑎 𝜕𝜕𝛼𝛼 𝑎𝑎 = 𝑎𝑎 𝑤𝑤𝑎𝑎 ( ℎ − ℎ 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑎𝑎 ) = 9.726 × 10 −4 > 0 Positive sloping moment coefficient indicates longitudinal statically unstable behavior. (A slight increase in 𝛼𝛼 makes the moment more positive , tending to increase 𝛼𝛼 even further, and vice versa for a decrease in 𝛼𝛼 ).

Name: HW5 Page 9 of 11 13. We wish to add a tail to the wing-body from problem #2 such that the static margin will be 0.5. If the tail has a lift slope of 0.08 per degree and can be placed a maximum of 3 m from the vehicle's center of gravity, what is the smallest tail planform area that can achieve this? Neglect downwash from the wings (i.e., ε ≡ 0). We want static margin, ℎ − ℎ 𝑛𝑛 = 0.5 𝜕𝜕𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑘𝑘 𝜕𝜕𝛼𝛼 𝑎𝑎 = −𝑎𝑎 𝑤𝑤𝑎𝑎 ( ℎ − ℎ 𝑛𝑛 ) = − 0.08978(0.5) = − 0.04489 𝑝𝑝𝜋𝜋𝑜𝑜 ° 𝜕𝜕𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑘𝑘 𝜕𝜕𝛼𝛼 𝑎𝑎 = 𝑎𝑎 𝑤𝑤𝑎𝑎 �ℎ − ℎ 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑎𝑎 − 𝑉𝑉 𝐻𝐻 𝑎𝑎 𝑎𝑎 𝑎𝑎 � − 0.04489 = 0.08978 � 0.01083 − 𝑉𝑉 𝐻𝐻 0.08 0.08978 � 𝑉𝑉 𝐻𝐻 = 0.5733 𝑉𝑉 𝐻𝐻 = 𝑙𝑙 𝑎𝑎 𝑆𝑆 𝑎𝑎 𝑐𝑐𝑆𝑆 = 0.5733 = 3 𝑆𝑆 𝑎𝑎 (1)(3) → 𝑆𝑆 𝑎𝑎 = 0.5733 𝑚𝑚 2

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Name: HW5 Page 10 of 11 14. For zero elevator deflection, we desire for the vehicle from problem #2 to be statically stable with an equilibrium angle of attack of 3°. Still neglecting downwash, what tail setting angle is necessary to achieve this? The equilibrium angle of attack is where the moment coefficient is zero. 𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑘𝑘 = 𝐶𝐶 𝑀𝑀 , 0 + 𝜕𝜕𝐶𝐶 𝑀𝑀 , 0 𝜕𝜕𝛼𝛼 𝑎𝑎 𝛼𝛼 𝜋𝜋 = 0 𝐶𝐶 𝑀𝑀 , 0 = − 𝜕𝜕𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑘𝑘 𝜕𝜕𝛼𝛼 𝑎𝑎 𝛼𝛼 𝜋𝜋 = 0.04489(3) = 0.1347 𝐶𝐶 𝑀𝑀 , 0 = 𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑎𝑎 + 𝑉𝑉 𝐻𝐻 𝑎𝑎 𝑎𝑎 𝑚𝑚 𝑎𝑎 0.1347 = − 0.005237 + 0.5733(0.08)( 𝑚𝑚 𝑎𝑎 ) 𝑚𝑚 𝑎𝑎 = +3.05°, 𝑚𝑚 𝑎𝑎 𝑚𝑚𝜋𝜋𝑎𝑎𝑠𝑠𝑜𝑜𝑜𝑜𝜋𝜋𝑎𝑎 𝑎𝑎𝑜𝑜𝑤𝑤𝑚𝑚𝑤𝑤𝑎𝑎𝑜𝑜𝑎𝑎 𝑓𝑓𝑜𝑜𝑜𝑜𝑚𝑚 𝑤𝑤𝑚𝑚𝑚𝑚𝑔𝑔 − 𝑙𝑙𝑜𝑜𝑎𝑎𝑏𝑏 𝑧𝑧𝜋𝜋𝑜𝑜𝑜𝑜 𝑙𝑙𝑚𝑚𝑓𝑓𝑓𝑓 𝑙𝑙𝑚𝑚𝑚𝑚𝜋𝜋

Name: HW5 Page 11 of 11 15. Suppose we now wish to operate the vehicle from problem #3 at an angle of attack of 5°. If the elevator control effectiveness is 0.04, what elevator deflection angle will trim the vehicle for this flight condition? 𝐶𝐶 𝐿𝐿 , 𝑤𝑤𝑎𝑎 = 𝑎𝑎 𝑤𝑤𝑎𝑎 𝛼𝛼 𝑤𝑤𝑎𝑎 = 008978(5°) = 0.4489 At equilibrium, 𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑘𝑘 = 𝐶𝐶 𝑀𝑀 , 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑎𝑎 + 𝐶𝐶 𝐿𝐿 , 𝑤𝑤𝑎𝑎 �ℎ ℎ 𝑎𝑎𝑎𝑎 , 𝑤𝑤𝑤𝑤 � − 𝑉𝑉 𝐻𝐻 �𝑎𝑎 𝑎𝑎 𝛼𝛼 𝑎𝑎 + 𝜕𝜕𝐶𝐶 𝐿𝐿 , 𝑎𝑎 𝜕𝜕𝛿𝛿 𝜋𝜋 𝛿𝛿 𝜋𝜋 � = 0 0 = − 0.005237 + 0.4489(0.01083) − 0.5733[0.08(5° − 3.05°) +). 04 𝛿𝛿 𝜋𝜋 ] So 𝛿𝛿 𝜋𝜋 = − 3.92°