A plane weighs 2280 lbs, has a wing area of 245 ft^2, and C_L of 0.15 when alpha = 0. Assume that the force of lift is equal to the force of weight, Air density = 0.002469 slug/ft^3, and to use the force of lift equation. Determine:
a). How fast (in knots) must this plane be flying to maintain cruising flight at an altitude of 1000 ft? (Manipulate the equation to solve for v)
b). Based on the chart provided below, complete the table computing the velocity (in knots) to reach 1000 ft for each of the temperatures (and resultant air densities) provided.
For wing area = 245 ft^2
c). What can you say about the impact of varying air density on the resultant velocity required to maintain cruising altitude?
From the result shown in the table in part (b), we can say that as the air density decreases, (due to increase in air temperature), the resultant velocity required to maintain the 1000 ft cruising altitude increases. Therefore, holding other factors constant, the cruising speed of the plane varies inversely proportional to the air density (and air temperature)
d). Based on the equation provided and your detailed analysis/understanding from part (a) to (c), how does changing the wing area to 230 ft^2 change your answers? (i.e., redo your analysis for part (a) and (b), put them in the second chart, and describe your results)
For wing area = 230 ft^2
e). Describe how changing the wing area from 245 to 230 ft^2 changes your velocity required to maintain the same altitude at 1000 ft.
Comparing the values obtained in parts (a) and (b), with those obtained in part (d), we can say that decreasing the wing area from 245 ft^2 to 230 ft^2 made the plane to cruise at a relatively higher speed to maintain the altitude of 1000 ft at the same conditions of air temperature and air densities.
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