v at angle A down from horizontal
D = drag
vertical component of D = D sin A
L = lift
vertical component of L= L cos A
vertical force down = weight = W
so
W = D sin A + L cos A
also horizontal
D cos A = L sin A
so
tan A = D/L = CD/CL
Lacking thrust, a glider can’t maintain level flight at constant speed in still air. It can, however, glide at a constant speed at a fixed downward angle. Find this angle in terms of the lift and drag coefficients CL and CD. Recall that lift is perpendicular to the glider’s velocity, and drag is anticipated to its velocity. Turbulent-flow for laws are appropriate here.
5 answers
Do you mean "parallel" where you wrote "anticipated" ?
You do not have to invoke turbulent flow to answer this question. That statement will only confuse students, and should not have been made. The values of Cl and Cd do depend upon whether there is turbulent flow. In airplane and glider flow, there will be turbulent flow within boundary layers, whether the free stream flow is turbulent or not.
The "angle of attack" of the wing equals the descent angle of the plane, and that angle is arctangent (Drag)/(Lift) = Cd/Cl
You do not have to invoke turbulent flow to answer this question. That statement will only confuse students, and should not have been made. The values of Cl and Cd do depend upon whether there is turbulent flow. In airplane and glider flow, there will be turbulent flow within boundary layers, whether the free stream flow is turbulent or not.
The "angle of attack" of the wing equals the descent angle of the plane, and that angle is arctangent (Drag)/(Lift) = Cd/Cl
Sorry, I meant antiparallel....
Thanks both of you very much...then the answer will be theta= arctan (CD/CL)... I was doing it wrong, I thought that the angle of the wing and the angle of the force were different and I had theta=(pi/2) -arctan(CD/CL)
Actually the angle of attack is zero, but the lift force is tilted forward (from vertical) by the descent angle.
This does not change my answer. I agree with Damon, who derived it more carefully.
The resultant of the lift force and the drag force is vertical, balancing the weight. There is no net horizontal force, and the speed remains constant.
This ignores fuselage lift and drag, and assumes both are all due to the wings. Lift and drag coefficients are usually applied to the wings only.
This does not change my answer. I agree with Damon, who derived it more carefully.
The resultant of the lift force and the drag force is vertical, balancing the weight. There is no net horizontal force, and the speed remains constant.
This ignores fuselage lift and drag, and assumes both are all due to the wings. Lift and drag coefficients are usually applied to the wings only.
1 answer