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Is drag coefficient lowest at zero angle of attack?
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The drag coefficient of a symmetric airfoil is lowest when its angle of attack is zero. I'm not sure if this is true in general.
aerodynamics airfoil drag angle-of-attack
asked 6 hours ago
simple jacksimple jack
103
New contributor
simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
The drag coefficient of a symmetric airfoil is lowest when its angle of attack is zero. I'm not sure if this is true in general.
aerodynamics airfoil drag angle-of-attack
asked 6 hours ago
simple jacksimple jack
103
New contributor
simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome to Av.SE!
$endgroup$
– Ralph J
6 hours ago
add a comment |
$begingroup$
The drag coefficient of a symmetric airfoil is lowest when its angle of attack is zero. I'm not sure if this is true in general.
aerodynamics airfoil drag angle-of-attack
asked 6 hours ago
simple jacksimple jack
103
New contributor
simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The drag coefficient of a symmetric airfoil is lowest when its angle of attack is zero. I'm not sure if this is true in general.
aerodynamics airfoil drag angle-of-attack
aerodynamics airfoil drag angle-of-attack
asked 6 hours ago
simple jacksimple jack
103
New contributor
simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
simple jacksimple jack
103
New contributor
simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
simple jacksimple jack
103
New contributor
simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
simple jacksimple jack
103
asked 6 hours ago
simple jacksimple jack
103
103
New contributor
simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Welcome to Av.SE!
$endgroup$
– Ralph J
6 hours ago
add a comment |
$begingroup$
Welcome to Av.SE!
$endgroup$
– Ralph J
6 hours ago
$begingroup$
Welcome to Av.SE!
$endgroup$
– Ralph J
6 hours ago
$begingroup$
Welcome to Av.SE!
$endgroup$
– Ralph J
6 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Obviously, drag should be smallest for symmetrical airfoils at zero angle of attack.
However, most airfoils have camber, and then the lowest drag is at positive lift coefficients in case of positive camber. Where that point is exactly depends on many parameters; in case of laminar airfoils even local imperfections can have a noticeable effect. Generally, the lowest drag can be found at the angle of attack where the stagnation point is exactly at the center of the leading edge, where the local curvature is highest. A deviation from this point will force the flow on one side to negotiate this point of highest curvature all by itself, resulting in a suction peak which will increase the losses in the boundary layer.
This is a theoretical drag polar (calculated with XFOIL) of an airfoil with a 20% camber flap at different settings and a Reynolds number of 1.5 million. The laminar bucket is clearly visible and produces a range of lift coefficients with nearly identical drag. The small waviness at the lower end of the laminar bucket is an artificial result of smoothing the plot.
What is obvious is how camber shifts the area of minimum drag up and down. If you use the right side of the plot to find the angle of attack of minimum drag, you will find that this is not constant but goes down as flap angles go up. For the 0° flap polar it is at about -2° AoA. This is caused by the induced angle of attack which increases with the lift coefficient.
The 6-series NACA airfoils were the first to be systematically designed with the pressure distribution in mind, and the design lift coefficient is where the condition of the ideal stagnation point location is met. This is indicated by the digit right after the hyphen in the airfoil name: Divide this digit by 10 and you have the lift coefficient of minimum drag.
Example: The $63_1-412$ airfoil has its lowest drag at a lift coefficient of 0.4.
If you want to know the angle of attack with the lowest drag of a whole airplane, this is a very different matter and needs to include the drag due to lift, which is of course smallest at the zero lift polar point.
answered 1 hour ago
Peter KämpfPeter Kämpf
162k12411656
$endgroup$
add a comment |
$begingroup$
Yes, for a symmetrical lift generating airfoil this is true.
The drag coefficient is computed by dividing the wetted area $A_w$ of the airfoil by its frontal area $A_f$ :
$$ c_d = frac{A_w}{A_f} $$
For non-symmetrical airfoils, the lowest drag coefficient is found at the angle of attack were the frontal area is at its smallest. For almost all the airfoils this is at 0 degrees AoA.
edited 5 hours ago
simple jack
103
answered 6 hours ago
BrilsmurfffjeBrilsmurfffje
3,41621536
$endgroup$
$begingroup$
It is not true for the Clark-Y airfoil.
$endgroup$
– simple jack
5 hours ago
$begingroup$
Frontal area is smallest? April 1st was two weeks ago!
$endgroup$
– Peter Kämpf
2 hours ago
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
Obviously, drag should be smallest for symmetrical airfoils at zero angle of attack.
However, most airfoils have camber, and then the lowest drag is at positive lift coefficients in case of positive camber. Where that point is exactly depends on many parameters; in case of laminar airfoils even local imperfections can have a noticeable effect. Generally, the lowest drag can be found at the angle of attack where the stagnation point is exactly at the center of the leading edge, where the local curvature is highest. A deviation from this point will force the flow on one side to negotiate this point of highest curvature all by itself, resulting in a suction peak which will increase the losses in the boundary layer.
This is a theoretical drag polar (calculated with XFOIL) of an airfoil with a 20% camber flap at different settings and a Reynolds number of 1.5 million. The laminar bucket is clearly visible and produces a range of lift coefficients with nearly identical drag. The small waviness at the lower end of the laminar bucket is an artificial result of smoothing the plot.
What is obvious is how camber shifts the area of minimum drag up and down. If you use the right side of the plot to find the angle of attack of minimum drag, you will find that this is not constant but goes down as flap angles go up. For the 0° flap polar it is at about -2° AoA. This is caused by the induced angle of attack which increases with the lift coefficient.
The 6-series NACA airfoils were the first to be systematically designed with the pressure distribution in mind, and the design lift coefficient is where the condition of the ideal stagnation point location is met. This is indicated by the digit right after the hyphen in the airfoil name: Divide this digit by 10 and you have the lift coefficient of minimum drag.
Example: The $63_1-412$ airfoil has its lowest drag at a lift coefficient of 0.4.
If you want to know the angle of attack with the lowest drag of a whole airplane, this is a very different matter and needs to include the drag due to lift, which is of course smallest at the zero lift polar point.
answered 1 hour ago
Peter KämpfPeter Kämpf
162k12411656
$endgroup$
add a comment |
$begingroup$
Obviously, drag should be smallest for symmetrical airfoils at zero angle of attack.
However, most airfoils have camber, and then the lowest drag is at positive lift coefficients in case of positive camber. Where that point is exactly depends on many parameters; in case of laminar airfoils even local imperfections can have a noticeable effect. Generally, the lowest drag can be found at the angle of attack where the stagnation point is exactly at the center of the leading edge, where the local curvature is highest. A deviation from this point will force the flow on one side to negotiate this point of highest curvature all by itself, resulting in a suction peak which will increase the losses in the boundary layer.
This is a theoretical drag polar (calculated with XFOIL) of an airfoil with a 20% camber flap at different settings and a Reynolds number of 1.5 million. The laminar bucket is clearly visible and produces a range of lift coefficients with nearly identical drag. The small waviness at the lower end of the laminar bucket is an artificial result of smoothing the plot.
What is obvious is how camber shifts the area of minimum drag up and down. If you use the right side of the plot to find the angle of attack of minimum drag, you will find that this is not constant but goes down as flap angles go up. For the 0° flap polar it is at about -2° AoA. This is caused by the induced angle of attack which increases with the lift coefficient.
The 6-series NACA airfoils were the first to be systematically designed with the pressure distribution in mind, and the design lift coefficient is where the condition of the ideal stagnation point location is met. This is indicated by the digit right after the hyphen in the airfoil name: Divide this digit by 10 and you have the lift coefficient of minimum drag.
Example: The $63_1-412$ airfoil has its lowest drag at a lift coefficient of 0.4.
If you want to know the angle of attack with the lowest drag of a whole airplane, this is a very different matter and needs to include the drag due to lift, which is of course smallest at the zero lift polar point.
answered 1 hour ago
Peter KämpfPeter Kämpf
162k12411656
$endgroup$
add a comment |
$begingroup$
Obviously, drag should be smallest for symmetrical airfoils at zero angle of attack.
However, most airfoils have camber, and then the lowest drag is at positive lift coefficients in case of positive camber. Where that point is exactly depends on many parameters; in case of laminar airfoils even local imperfections can have a noticeable effect. Generally, the lowest drag can be found at the angle of attack where the stagnation point is exactly at the center of the leading edge, where the local curvature is highest. A deviation from this point will force the flow on one side to negotiate this point of highest curvature all by itself, resulting in a suction peak which will increase the losses in the boundary layer.
This is a theoretical drag polar (calculated with XFOIL) of an airfoil with a 20% camber flap at different settings and a Reynolds number of 1.5 million. The laminar bucket is clearly visible and produces a range of lift coefficients with nearly identical drag. The small waviness at the lower end of the laminar bucket is an artificial result of smoothing the plot.
What is obvious is how camber shifts the area of minimum drag up and down. If you use the right side of the plot to find the angle of attack of minimum drag, you will find that this is not constant but goes down as flap angles go up. For the 0° flap polar it is at about -2° AoA. This is caused by the induced angle of attack which increases with the lift coefficient.
The 6-series NACA airfoils were the first to be systematically designed with the pressure distribution in mind, and the design lift coefficient is where the condition of the ideal stagnation point location is met. This is indicated by the digit right after the hyphen in the airfoil name: Divide this digit by 10 and you have the lift coefficient of minimum drag.
Example: The $63_1-412$ airfoil has its lowest drag at a lift coefficient of 0.4.
If you want to know the angle of attack with the lowest drag of a whole airplane, this is a very different matter and needs to include the drag due to lift, which is of course smallest at the zero lift polar point.
answered 1 hour ago
Peter KämpfPeter Kämpf
162k12411656
$endgroup$
Obviously, drag should be smallest for symmetrical airfoils at zero angle of attack.
However, most airfoils have camber, and then the lowest drag is at positive lift coefficients in case of positive camber. Where that point is exactly depends on many parameters; in case of laminar airfoils even local imperfections can have a noticeable effect. Generally, the lowest drag can be found at the angle of attack where the stagnation point is exactly at the center of the leading edge, where the local curvature is highest. A deviation from this point will force the flow on one side to negotiate this point of highest curvature all by itself, resulting in a suction peak which will increase the losses in the boundary layer.
This is a theoretical drag polar (calculated with XFOIL) of an airfoil with a 20% camber flap at different settings and a Reynolds number of 1.5 million. The laminar bucket is clearly visible and produces a range of lift coefficients with nearly identical drag. The small waviness at the lower end of the laminar bucket is an artificial result of smoothing the plot.
What is obvious is how camber shifts the area of minimum drag up and down. If you use the right side of the plot to find the angle of attack of minimum drag, you will find that this is not constant but goes down as flap angles go up. For the 0° flap polar it is at about -2° AoA. This is caused by the induced angle of attack which increases with the lift coefficient.
The 6-series NACA airfoils were the first to be systematically designed with the pressure distribution in mind, and the design lift coefficient is where the condition of the ideal stagnation point location is met. This is indicated by the digit right after the hyphen in the airfoil name: Divide this digit by 10 and you have the lift coefficient of minimum drag.
Example: The $63_1-412$ airfoil has its lowest drag at a lift coefficient of 0.4.
If you want to know the angle of attack with the lowest drag of a whole airplane, this is a very different matter and needs to include the drag due to lift, which is of course smallest at the zero lift polar point.
answered 1 hour ago
Peter KämpfPeter Kämpf
162k12411656
answered 1 hour ago
Peter KämpfPeter Kämpf
162k12411656
answered 1 hour ago
Peter KämpfPeter Kämpf
162k12411656
answered 1 hour ago
Peter KämpfPeter Kämpf
162k12411656
162k12411656
add a comment |
add a comment |
$begingroup$
Yes, for a symmetrical lift generating airfoil this is true.
The drag coefficient is computed by dividing the wetted area $A_w$ of the airfoil by its frontal area $A_f$ :
$$ c_d = frac{A_w}{A_f} $$
For non-symmetrical airfoils, the lowest drag coefficient is found at the angle of attack were the frontal area is at its smallest. For almost all the airfoils this is at 0 degrees AoA.
edited 5 hours ago
simple jack
103
answered 6 hours ago
BrilsmurfffjeBrilsmurfffje
3,41621536
$endgroup$
$begingroup$
It is not true for the Clark-Y airfoil.
$endgroup$
– simple jack
5 hours ago
$begingroup$
Frontal area is smallest? April 1st was two weeks ago!
$endgroup$
– Peter Kämpf
2 hours ago
add a comment |
$begingroup$
Yes, for a symmetrical lift generating airfoil this is true.
The drag coefficient is computed by dividing the wetted area $A_w$ of the airfoil by its frontal area $A_f$ :
$$ c_d = frac{A_w}{A_f} $$
For non-symmetrical airfoils, the lowest drag coefficient is found at the angle of attack were the frontal area is at its smallest. For almost all the airfoils this is at 0 degrees AoA.
edited 5 hours ago
simple jack
103
answered 6 hours ago
BrilsmurfffjeBrilsmurfffje
3,41621536
$endgroup$
$begingroup$
It is not true for the Clark-Y airfoil.
$endgroup$
– simple jack
5 hours ago
$begingroup$
Frontal area is smallest? April 1st was two weeks ago!
$endgroup$
– Peter Kämpf
2 hours ago
add a comment |
$begingroup$
Yes, for a symmetrical lift generating airfoil this is true.
The drag coefficient is computed by dividing the wetted area $A_w$ of the airfoil by its frontal area $A_f$ :
$$ c_d = frac{A_w}{A_f} $$
For non-symmetrical airfoils, the lowest drag coefficient is found at the angle of attack were the frontal area is at its smallest. For almost all the airfoils this is at 0 degrees AoA.
edited 5 hours ago
simple jack
103
answered 6 hours ago
BrilsmurfffjeBrilsmurfffje
3,41621536
$endgroup$
Yes, for a symmetrical lift generating airfoil this is true.
The drag coefficient is computed by dividing the wetted area $A_w$ of the airfoil by its frontal area $A_f$ :
$$ c_d = frac{A_w}{A_f} $$
For non-symmetrical airfoils, the lowest drag coefficient is found at the angle of attack were the frontal area is at its smallest. For almost all the airfoils this is at 0 degrees AoA.
edited 5 hours ago
simple jack
103
answered 6 hours ago
BrilsmurfffjeBrilsmurfffje
3,41621536
edited 5 hours ago
simple jack
103
edited 5 hours ago
simple jack
103
edited 5 hours ago
simple jack
103
103
answered 6 hours ago
BrilsmurfffjeBrilsmurfffje
3,41621536
answered 6 hours ago
BrilsmurfffjeBrilsmurfffje
3,41621536
answered 6 hours ago
BrilsmurfffjeBrilsmurfffje
3,41621536
3,41621536
$begingroup$
It is not true for the Clark-Y airfoil.
$endgroup$
– simple jack
5 hours ago
$begingroup$
Frontal area is smallest? April 1st was two weeks ago!
$endgroup$
– Peter Kämpf
2 hours ago
add a comment |
$begingroup$
It is not true for the Clark-Y airfoil.
$endgroup$
– simple jack
5 hours ago
$begingroup$
Frontal area is smallest? April 1st was two weeks ago!
$endgroup$
– Peter Kämpf
2 hours ago
$begingroup$
It is not true for the Clark-Y airfoil.
$endgroup$
– simple jack
5 hours ago
$begingroup$
It is not true for the Clark-Y airfoil.
$endgroup$
– simple jack
5 hours ago
$begingroup$
Frontal area is smallest? April 1st was two weeks ago!
$endgroup$
– Peter Kämpf
2 hours ago
$begingroup$
Frontal area is smallest? April 1st was two weeks ago!
$endgroup$
– Peter Kämpf
2 hours ago
add a comment |
simple jack is a new contributor. Be nice, and check out our Code of Conduct.
simple jack is a new contributor. Be nice, and check out our Code of Conduct.
simple jack is a new contributor. Be nice, and check out our Code of Conduct.
simple jack is a new contributor. Be nice, and check out our Code of Conduct.
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