MAE175__Discussion_5

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University of California, Irvine *

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175

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Aerospace Engineering

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Apr 3, 2024

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MAE 175: Discussion 5 Jordi Ventura Siches February 9, 2024 Jordi Ventura Siches University of California, Irvine - Henry Samueli School of Engineering MAE 175: Discussion 5 1 of 17
Problem 2.6 (Nelson) An airplane has the following pitching moment characteristics at the center of gravity position: x cg / ¯ c = 0 . 3 . C m cg = C m 0 + dC m cg dC L C L + C m δ e δ e where C m 0 = 0 . 05, dC m cg dC L = 0 . 1, C m δ e = 0 . 01 / deg, dC m cg dC L = x cg ¯ c x np ¯ c If the airplane is loaded so that the center of gravity position moves to x cg / ¯ c = 0 . 10, can the airplane be trimmed during landing, C L = 1 . 0? Assume that C m 0 and C m δ e are unaffected by the center of gravity travel and that δ e max = ± 20 º . Jordi Ventura Siches University of California, Irvine - Henry Samueli School of Engineering MAE 175: Discussion 5 2 of 17
Problem 2.7 (Nelson): Solution Solution : We first need to compute the new value for dC mCG dC L . In order to do so, we first find the position for the neutral point by using the data from the previous scenario: dC m cg dC L | {z } 0 . 1 = x cg ¯ c |{z} 0 . 3 x np ¯ c x np ¯ c = 0 . 4 Now we can find the new slope dC m cg dC L = x cg ¯ c |{z} 0 . 1 x np ¯ c |{z} 0 . 4 = 0 . 3 Jordi Ventura Siches University of California, Irvine - Henry Samueli School of Engineering MAE 175: Discussion 5 3 of 17
Problem 2.7 (Nelson): Solution Solution : Then, dC m cg dC L = 0 . 3. Trim implies C m , cg = 0, thus 0 = 0 . 05 0 . 3 C L |{z} 1 . 0 0 . 01 · δ e (deg) We isolate δ e and get δ e = 0 . 05 0 . 3 0 . 01 = 25 Since 25 is out of the bounds, the aircraft can not be trimmed in this conditions. Jordi Ventura Siches University of California, Irvine - Henry Samueli School of Engineering MAE 175: Discussion 5 3 of 17
Problem 2.7 (Nelson) The pitching moment characteristics of a general aviation airplane with the landing gear and flaps in their retracted position are given by (a) Where is the stick fixed neutral point located? Jordi Ventura Siches University of California, Irvine - Henry Samueli School of Engineering MAE 175: Discussion 5 4 of 17
Problem 2.7 (Nelson): Solution (a) Solution: To find the slope, select 2 points from any of the curves (they have the same slope). For δ e = 0 : C L = 0 . 4 , C m , cg = 0; C L = 1 . 2 , C m , cg = 0 . 2 and the slope is dC m , cg dC L = 0 . 2 0 1 . 2 0 . 4 = 0 . 25 Therefore x np ¯ c = x cg ¯ c dC m , cg dC L = 0 . 5 Jordi Ventura Siches University of California, Irvine - Henry Samueli School of Engineering MAE 175: Discussion 5 5 of 17
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