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| | #1 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Anyone care to tell me how you'd explain the aerodynamics to a commercial applicant why when the aircraft is slow, it needs a higher AoA to maintain altitude? Thanks!!
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| | #2 |
| Old Skool Join Date: Mar 2006 Location: KRST
Posts: 1,819
| Bust out the lift equation. Same weight, same wing area, same coefficient of lift, only thing that changed is speed (velocity) of the air mass.
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| | #3 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Yeah, I actually tried to explain that. Didnt work.. Any other non-mathematical thoughts on how to explain it??
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| | #4 |
| Old Skool Join Date: Mar 2006 Location: KRST
Posts: 1,819
| Don't shy away from the equation, it can answer so many questions about aerodynamics. Lets look it at and break it down L = Cl * A * .5 * r * V^2 Cl = coefficeint of lift, a fixxed number for a given airfoil and AoA A = area of the wing (generally a fixed number) r = air density (in slugs, fixed at a given altitude and temp) V = velocity of the airflow
__________________ Aircraft without engine(s) prohibited... -KMIA 10-9 Last edited by Dugie8; August 13th, 2006 at 21:08. |
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| | #5 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| I am listening..
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| | #6 |
| Old Skool Join Date: Mar 2006 Location: KRST
Posts: 1,819
| grrr, fish has hit this on the head, and I officially know I need to stop drinking so much!
__________________ Aircraft without engine(s) prohibited... -KMIA 10-9 Last edited by Dugie8; August 13th, 2006 at 21:07. |
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| | #7 |
| Senior Member | Hey Merit, I answered you in a PM, but I'll put it here too. Sorry I didn't get to this earlier, but I was cross country with a student all weekend. This is the same thing that I PM'd you, just a cut and paste, but I wanted to correct something in a couple of previous posts on this thread. AoA doesn't affect the velocity term of the lift equation, it affects the coefficient of lift. Here's that PM: Yeah, man. No problem. It's all about generating lift. The equation for lift is: L=1/2Cl rho V^2 S where L is lift, rho is the symbol for density, Cl is the coefficient of lift, V is velocity and S is the surface area of the wing. If you ignore putting out flaps, slats, slots, etc. you don't really have a way to change the surface area of the wing, so you can't control S as a pilot. Also, you don't have any control over the density of the air, except by climbing or descending, but ignore that for the purposes of this discussion. So you can't control rho, either. The things that you CAN control are airspeed, and Cl. Airspeed control is pretty obvious, so let's look at how you control the coefficient of lift for a minute. Cl is kind of a nebulous term that is actually made up of a number of factors. The big ones are the shape of the airfoil, the shape of the wing, and the angle of attack of the wing. Increasing the angle of attack, (by pulling back on the stick or yoke) increases Cl, and in turn increases the lift produced by the wing. That is, until you increase angle of attack past the "critical" angle of attack, and then lift drops off sharply. (That's what a stall is). So back to slow flight. From the equation above, we see that as V (velocity) gets smaller lift would get smaller also, if everything else remains the same. To keep flying level, however, we need lift to stay the same. If we decrease velocity, we need to increase the coefficient of lift to compensate. We do this by increasing the angle of attack.
__________________ Dude, what are you trying to do? Land the airplane or adjust the field elevation? |
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| | #8 |
| Old Skool Join Date: Mar 2006 Location: KRST
Posts: 1,819
| ahhhh, grrrr, I can't this right in my head. here is a great link http://selair.selkirk.bc.ca/aerodyna...ift/Page6.html
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| | #9 | |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Quote:
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| | #10 |
| Old Skool Join Date: Mar 2006 Location: KRST
Posts: 1,819
| someday, I will learn to stick to my guns fish and I are both correct, it all depends on how you want to show it to your student. http://www.grc.nasa.gov/WWW/K-12/FoilSim/index.html Download and run the foilsimII program. Put the probe on the upper part of the airfoil and do nothing but increase the AoA, you will see a rise in velocity. That rise is what causes the increase in Cl. Think about from an "engergy conservation" point of view. Simply increasine the AoA at slower speeds doesn't magically make the wing produce more lift, something needs to happen to the airflow over the wing.
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| | #11 | |
| Old Skool Join Date: Mar 2006 Location: KRST
Posts: 1,819
| Quote:
Not really, and we are getting away from how to explain this to a student and into how to design a wing. Cl is just a way to quantify what is happening to a wing at different angles of attack and puts a number on it to make it work (no such thing as unitless variables in physics, but we have one here!). If you play with that foilsim program you start to see that for a given lift amount the velocity over the top of the wing can be a wide range of values and still produce the same lift, at first glance that shouldn't work. Don't forget that lift is the result of pressure differential between the upper and lower surfaces. I just ran some quick numbers, and this will be hard to reproduce since it varies on probe placement. for 1250 pounds of lift for the default airfoil, it took 100 mph at 4 degrees AoA, that translated to 129 mph over the top of the wing near the leading edge. At 50 mph, it was 84 mph over the top of the wing at 16 degrees AoA. At first that doesn't make sense, until you start looking at the velocities on the lower surfaces at slower speeds the differnce is speed is greater (faster on top, slower on bottom) resulting in greater pressure differntial. This is what Cl is showing, in a way. Clear as mud yet?
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| | #12 |
| Old Skool Join Date: Apr 2002 Location: Denver, CO
Posts: 2,078
| Pretty much just like that. When the aircraft is slow, it needs a higher AoA. You said you tried the lift equation, well that should at least demonstrate that AoA and airspeed are the two main things taht we use to create lift. If we have less of one, we need more of the other. You can simplify the lift equation by saying someting like "if you have 2000 pounds of weight then you need 2000 pounds of lift to maintain alititude. If you are getting 1600 pounds from airspeed, than you need to get 400 from AoA. If you slow your airspeed and only get 1000 pounds from airspeed, then you must increase your AoA to get the other 1000 or you'll have less than 2000 total and will start to descend" Dosn't have to be real ratios or anything - just a simple mathmateic exmple that shows how if one goes down, the other must go up, or else the total goes down.
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| | #13 |
| Old Skool | When AoA is increased the thrust vector in angled upward in relation to relative wind, and its the verticle component of thrust that compensates for the reduced lift created by the increased AoA. The L/D ratio graph can illustrate that. ![]()
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| | #14 |
| Senior Member | Alright, DesertDog, now you're intentionally confusing the poor guys!! He's not being serious here, note the smiley in his last post. Ok, let's simplify it down to student level. When you increase airspeed and keep AoA the same, the wing produces more lift. When you increase AoA and keep airspeed the same, you increase lift. So you have two ways to increase lift, by either increasing airspeed, or by increasing angle of attack. So back to slow flight... If you decrease airspeed, you have to increase AoA in order to compensate and keep lift the same.
__________________ Dude, what are you trying to do? Land the airplane or adjust the field elevation? |
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| | #15 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| I cant really coin the whole thrust vector idea. Maybe you're referring to the center of pressure (CP). The L/D drag graph illustrates different coefficients of drag (CD), lift (CL), CLmax, and their effect on gliding the aircraft over a maximum range.
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| | #16 | |
| Old Skool | Quote:
My point remains the same. Now leaving out such things as load factors and centrifigul force, lets assume that weight always pulls towards the center of the earth. Lift is at a 90 degree angle to the chord of the wing. With an increased angle of attack drag has a slight defelction of force towards the ground and lift has a slight deflection towards the rear of the aircraft. Thrust is also deflected upwards in relation to relative wind. The deflected thrust is a major contributor to keeping the aircraft aloft along with the additional lift. I could totally illustrate this, but I suck at describing it obviously. Besides you know I like to talk out of my ass sometimes. I am pretty sure I am right on this one, just not explaining it very good.
__________________ www.flywhiteair.com http://www.myspace.com/desertdog71 Following message is for SkyCougar. ![]() Took my chances on a big jet plane, Never let them tell you that they're all the same. | |
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| | #17 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Riighht.. (I still love you D Dog in a pilot brother kinda way)
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| | #18 | |
| Old Skool | Quote:
In a low-power, high-speed descent, lift is once again less than weight --- because drag is supporting part of the weight. When you drop a bomb and it reaches terminal velocity it has no lift so why does it stop accellerating? Because Drag equals weight. But that is not the question you asked now that I have re-read it dammit. Use the Vg Diagram and Newtons 2nd and 3rd laws. Force occurs when mass is accelerated therefore in order to generate force (lift) the airfoil needs to be accelerated to the point where you generate the additional effect of downwash and the resultant lift due to the whole opposite and equal thing. The FSDO is gonna eat my alive come CFI Oral time.... LMAO!!!!!!
__________________ www.flywhiteair.com http://www.myspace.com/desertdog71 Following message is for SkyCougar. ![]() Took my chances on a big jet plane, Never let them tell you that they're all the same. | |
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| | #19 | |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| Quote:
2 - Only for a transition phase; 3 - We're gettin' there; 4 - Correct; 5 - The Vg diagram addresses velocity of and aircraft and load factor; and 6 - Yes, most likely they will. DOMINOS PIZZA?? I gotta run!
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| | #20 |
| Old Skool | 5. Yes it does, but it also illustrates that lift will not support weight at certain airspeeds.
__________________ www.flywhiteair.com http://www.myspace.com/desertdog71 Following message is for SkyCougar. ![]() Took my chances on a big jet plane, Never let them tell you that they're all the same. |
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| | #21 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| You're correct. I guess I just never thought of the Vg diagram like that.
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| | #22 | |
| Old Skool | Quote:
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__________________ www.flywhiteair.com http://www.myspace.com/desertdog71 Following message is for SkyCougar. ![]() Took my chances on a big jet plane, Never let them tell you that they're all the same. | |
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| | #23 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| I still think the Vg diagram doesnt accurately depict what we're discussing though... [snicker] ![]()
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| | #24 |
| Old Skool | You said for a Commercial Pilot Student not an Aeronautical Engineer
__________________ www.flywhiteair.com http://www.myspace.com/desertdog71 Following message is for SkyCougar. ![]() Took my chances on a big jet plane, Never let them tell you that they're all the same. |
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| | #25 |
| Old Skool Join Date: Dec 2005 Location: Winchestertonfieldville
Posts: 6,608
| I meant NASA engineer.. what this isnt the NASA forum? Ahh crap... too much pizza for me. ![]()
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