9780071086806, Chapter 9, Problem 6SP

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Chapter 9, Problem 6SP Problem 6. For the same vehicle in Problem 5, plot the stability map at least out to a lift coefficient of 1.0. Plot the data at 5-degree intervals of elevator deflection, and assume the maximum and minimum elevator deflections are +30 deg For the flight condition given in Problem 5, find the trim angle of attack and elevator deflection. Also find the trim-elevator gradient / and the maximum achievable CL for the c.g. location given. Compare your results with those for Problem 5. Should they agree? Why? Step-by-step solution Step of 2 Solution: We will use Excel and Powerpoint to plot the stability map, given the following data for the Navion (see Problem 5). =0.364 0 = 4.44 /rad = -0.683 /rad Since only the slopes for the lift and moment coefficients are available, the stability map will consist of staight lines. And the equations for these lines are simply = 0.364 + 4.44 57.3 57.3 = Step of 2 = 0 0.683 57.3 0.87 57.3 E = if and are in degrees. The resulting plot is shown below. Navion Stability Map 0.6 0 5 0.4 10 15 012 Moment Coefficient, -20 -10 -0.5 0.5 1.5 -0.2 0 10 -0.4 -0.6 Lift Coefficient, CL The point indicating the trim lift coefficient (0.41) is indicated by the white dot. The trim angle of attack and elevator deflection may be estimated by interpolating the data directly from the plot. It would appear that (0.5-0.6) deg deg which agrees with the results obtained in Problem 5. The elevator trim gradient may be estimated using the trim point and the origin indicated by the red arrow. (Actually, here any two data points along the trim line indicated by CM = 0 may be used since everything is linear.) It would appear that the elevator trim gradient is approximately (-0.5-3) (0.41-0) = -8.54 rad which agrees reasonable well with the result in Problem 5, subject to our accuracy in reading from the plot. Finally, from the figure it appears that the maximum achievable lift coefficient would be well above 1.5. It would be limited more due to a stall limit than by maximum elevator deflection, since the lines for constant elevator deflection do not have a steep negative slope. The stall limit is probably about 15 20 degrees angle of attack, although it is not revealed in this plot since the plot is based only on linear data. All these results should agree with those from Problem 5, subject to the accuracy by which we can graphically estimate the data. This is due to the fact that the plot is based only on the linear data given, and that same linear data was used in Problem 5.