Answer:
a) at 2° => 0.22
6° => 0.66
10° => 1.1
b) at 2° => 0.363
6°=>0.803
10° => 1.243
c) at 2° => 0.0275
6° => 0.4675
10° => 0.9075
Explanation:
We are given:
Lift curve slope = 0.11 per degree
a) to find symmetric airfoil, we use the formula;
= 0.22
=0.66
= 1.1
b) to find cambered airfoil, with a zero lift angle of attack at -1.3°, we use:
Making C_L subject of the formula, we have:
At a = 2 => 0.11(2+1.3) = 0.363
At a = 6 => 0.11(6+1.3) = 0.803
At a = 10 => 0.11(10+1.3)= 1.243
c) to calculate for a eflexed airfoil with a zero lift angle of attack of 1.75 degrees, we use:
At a= 2 => 0.11(2-1.75) = 0.0275
At a = 6 => 0.11(6-1.75) = 0.4675
At a = 10 => 0.11(10-1.75) = 0.9075