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Is drag coefficient lowest at zero angle of attack?

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Is drag coefficient lowest at zero angle of attack?



Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How do insects decrease aircraft performance?How to draw NACA 6-Series Airfoils?How can the zero-lift drag coefficient (parasitic drag) be calculated?What is the relation between the Lift Coefficient and the Angle of Attack?Is it possible to fly horizontally with zero angle of attack?How to find trim condition of a sectional airfoil without knowing the angle of attack?What is the effect of flow separation on lift, pressure distribution and drag?How can the zero-lift drag coefficient (parasitic drag) be calculated?Do negative angles-of-attack create lift?How do you calculate the lift coefficient of an airfoil at zero angle of attack?Calculating induced drag approximation using XFoil generated parasitic dragDoes speed or angle of attack generally have the greatest impact on total induced drag?What's the theoretical background of the critical 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.










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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.










share|improve this question







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.







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  • $begingroup$
    Welcome to Av.SE!
    $endgroup$
    – Ralph J
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1












1








1





$begingroup$


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.










share|improve this question







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






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simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







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simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 6 hours ago









simple jacksimple jack

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103




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simple jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.
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  • $begingroup$
    Welcome to Av.SE!
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    – Ralph J
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$begingroup$
Welcome to Av.SE!
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2 Answers
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3












$begingroup$

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.



flap polar



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.






share|improve this answer









$endgroup$




















    0












    $begingroup$

    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 $A_f$ :



    $$ c_d = fracA_wA_f $$



    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.






    share|improve this answer











    $endgroup$












    • $begingroup$
      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
      1 hour ago











    Your Answer








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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    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.



    flap polar



    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.






    share|improve this answer









    $endgroup$

















      3












      $begingroup$

      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.



      flap polar



      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.






      share|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        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.



        flap polar



        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.






        share|improve this answer









        $endgroup$



        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.



        flap polar



        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.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        Peter KämpfPeter Kämpf

        162k12411656




        162k12411656





















            0












            $begingroup$

            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 $A_f$ :



            $$ c_d = fracA_wA_f $$



            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.






            share|improve this answer











            $endgroup$












            • $begingroup$
              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
              1 hour ago















            0












            $begingroup$

            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 $A_f$ :



            $$ c_d = fracA_wA_f $$



            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.






            share|improve this answer











            $endgroup$












            • $begingroup$
              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
              1 hour ago













            0












            0








            0





            $begingroup$

            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 $A_f$ :



            $$ c_d = fracA_wA_f $$



            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.






            share|improve this answer











            $endgroup$



            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 $A_f$ :



            $$ c_d = fracA_wA_f $$



            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.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 4 hours ago









            simple jack

            103




            103










            answered 5 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
              1 hour 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
              1 hour 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
            1 hour ago




            $begingroup$
            Frontal area is smallest? April 1st was two weeks ago!
            $endgroup$
            – Peter Kämpf
            1 hour ago










            simple jack is a new contributor. Be nice, and check out our Code of Conduct.









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