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gschjetne
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I'm exploring a few topics for my IB extended essay, and since I'm into aviation, my teacher suggested I'd find the coefficient of drag for an airplane.
Finding force of net drag would be piece of cake.
[tex]F_{\Sigma drag}=\frac{Power output}{velocity}[/tex]
According to aerodynamics of an airplane at constant speed:
[tex]F_{\Sigma drag}=C_{drag} \frac{1}{2} \rho v^2 + mg \cos \alpha [/tex]
The first term, parasite drag, [itex]C_{drag} \frac{1}{2} \rho v^2[/itex], should be pretty straightforward. It is the coefficient of drag times the dynamic pressure, defined as [itex]\frac{1}{2} \rho v^2[/itex]
The second term, induced drag, on the other hand, probably needs some explanation.
At lower airspeeds, the angle of attack must be higher to generate enough lift, and lower at high airspeeds, respectively. The drawback of this, though, is that the lift generated is perpendicular to the wing chord. At high angle of attacks this force turns backwards, inducing drag, hence the name.
This can be calculated as [itex]F_{lift} \sin \alpha[/itex] which means that if the plane isn't accelerating in any direction, this equals [itex]mg \cos \alpha[/itex] as long as the aircraft is not accelerating.
Rearranging:
[tex]C_{drag} = \frac{F_{\Sigma drag}}{\frac{1}{2} \rho v^2 + mg \cos \alpha}[/tex]
I'm going to test this on a Piper PA-38-112 Tomahawk. My biggest problem is that this aircraft is not equipped with angle of attack measuring equipment. To overcome this I figured that I know that at 75 knots of indicated airspeed, or dynamic pressure (I hate all non-SI units , but that's the way things aviation works, help converting would be appreciated) the terms for induced and parasite drag are equal. This is the point where the two graphs intersect, and the net drag is at a minimum (that's why Piper chose 75kts as the climb speed)
Now, if i take several measurements of the above, varying dynamic pressure (speed), I should be able to plot this into my graphing calculator. I am no math wizard, so I would appreciate some help, perhaps some calculus would do the job to get me around the angle of attack problem?
Boy, that was a lot of LaTeX! :tongue2:
Finding force of net drag would be piece of cake.
[tex]F_{\Sigma drag}=\frac{Power output}{velocity}[/tex]
According to aerodynamics of an airplane at constant speed:
[tex]F_{\Sigma drag}=C_{drag} \frac{1}{2} \rho v^2 + mg \cos \alpha [/tex]
The first term, parasite drag, [itex]C_{drag} \frac{1}{2} \rho v^2[/itex], should be pretty straightforward. It is the coefficient of drag times the dynamic pressure, defined as [itex]\frac{1}{2} \rho v^2[/itex]
The second term, induced drag, on the other hand, probably needs some explanation.
At lower airspeeds, the angle of attack must be higher to generate enough lift, and lower at high airspeeds, respectively. The drawback of this, though, is that the lift generated is perpendicular to the wing chord. At high angle of attacks this force turns backwards, inducing drag, hence the name.
This can be calculated as [itex]F_{lift} \sin \alpha[/itex] which means that if the plane isn't accelerating in any direction, this equals [itex]mg \cos \alpha[/itex] as long as the aircraft is not accelerating.
Rearranging:
[tex]C_{drag} = \frac{F_{\Sigma drag}}{\frac{1}{2} \rho v^2 + mg \cos \alpha}[/tex]
I'm going to test this on a Piper PA-38-112 Tomahawk. My biggest problem is that this aircraft is not equipped with angle of attack measuring equipment. To overcome this I figured that I know that at 75 knots of indicated airspeed, or dynamic pressure (I hate all non-SI units , but that's the way things aviation works, help converting would be appreciated) the terms for induced and parasite drag are equal. This is the point where the two graphs intersect, and the net drag is at a minimum (that's why Piper chose 75kts as the climb speed)
Now, if i take several measurements of the above, varying dynamic pressure (speed), I should be able to plot this into my graphing calculator. I am no math wizard, so I would appreciate some help, perhaps some calculus would do the job to get me around the angle of attack problem?
Boy, that was a lot of LaTeX! :tongue2: