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#1
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Lift-to-Drag Ratio?
Guys-
If we have higher L/D, we need higher thrust to maintain lift equal weight. Am I reading this correctly? In this case, you would need higher horsepower? Toks Desalu PP-SEL Dyin' to fly! |
#2
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"Toks Desalu" wrote in
news:ikurb.154486$Tr4.407177@attbi_s03: Guys- If we have higher L/D, we need higher thrust to maintain lift equal weight. Am I reading this correctly? In this case, you would need higher horsepower? I'm not entirely sure I understand what you are trying to say. The higher the L/D ratio, the *less* thrust is needed to maintain the same lift. Look at 1920's aircraft. Very anemic engines, so they have high lift wings (and usually biplanes - double wings to boot). Not very good for going fast (because parasitic drag goes up with speed), but great for getting enough lift to become airborne without much HP. OTOH, look at an F-16 fighter. Virtually no "lift" from the airfoil design at all. The lift comes from Angle of Attack (AOA), which causes induced drag. So how does it get airborne (not to mention fly so well at supersonic speeds)? LOTS and LOTS of thrust! ----------------------------------------------- James M. Knox TriSoft ph 512-385-0316 1109-A Shady Lane fax 512-366-4331 Austin, Tx 78721 ----------------------------------------------- |
#3
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In article ,
"James M. Knox" wrote: "Toks Desalu" wrote in news:ikurb.154486$Tr4.407177@attbi_s03: Guys- If we have higher L/D, we need higher thrust to maintain lift equal weight. Am I reading this correctly? In this case, you would need higher horsepower? I'm not entirely sure I understand what you are trying to say. The higher the L/D ratio, the *less* thrust is needed to maintain the same lift. Look at 1920's aircraft. Very anemic engines, so they have high lift wings (and usually biplanes - double wings to boot). Not very good for going fast (because parasitic drag goes up with speed), but great for getting enough lift to become airborne without much HP. OTOH, look at an F-16 fighter. Virtually no "lift" from the airfoil design at all. The lift comes from Angle of Attack (AOA), which causes induced drag. So how does it get airborne (not to mention fly so well at supersonic speeds)? LOTS and LOTS of thrust! How can people write something as incorrect as AOA generated lift causes induced drag that "airfoil design" wouldn't???? -- Alan Baker Vancouver, British Columbia "If you raise the ceiling 4 feet, move the fireplace from that wall to that wall, you'll still only get the full stereophonic effect if you sit in the bottom of that cupboard." |
#4
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Toks,
L/D is what it says, Lift divided with Drag Lift is at one G = weight (+ some down force from stabilator) So a smaller Nr of drag, the higher L/D ratio! Jan Carlsson www.jcpropellerdesign.com "Toks Desalu" skrev i meddelandet news:ikurb.154486$Tr4.407177@attbi_s03... Guys- If we have higher L/D, we need higher thrust to maintain lift equal weight. Am I reading this correctly? In this case, you would need higher horsepower? Toks Desalu PP-SEL Dyin' to fly! |
#5
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Alan Baker wrote in
: OTOH, look at an F-16 fighter. Virtually no "lift" from the airfoil design at all. The lift comes from Angle of Attack (AOA), which causes induced drag. How can people write something as incorrect as AOA generated lift causes induced drag that "airfoil design" wouldn't???? Lack of sleep? Rushed? G Still, it's not that incorrect. I didn't actually say what you paraphrased... that lift due to airfoil design didn't cause induced drag. Increasing the angle of attack (until stall) increases lift, with a corresponding increase in induced drag. The alternative is a higher lift airfoil that produces the same amount of lift at a much lower airspeed, where lower parasitic drag will be encountered. Please feel free to give him a much more detailed (and hopefully more specifically accurate) explanation. But remember that he is looking for a a basic understanding... not Reynolds numbers. ----------------------------------------------- James M. Knox TriSoft ph 512-385-0316 1109-A Shady Lane fax 512-366-4331 Austin, Tx 78721 ----------------------------------------------- |
#6
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"James M. Knox" wrote in message ...
Alan Baker wrote in : OTOH, look at an F-16 fighter. Virtually no "lift" from the airfoil design at all. The lift comes from Angle of Attack (AOA), which causes induced drag. .... Please feel free to give him a much more detailed (and hopefully more specifically accurate) explanation. But remember that he is looking for a a basic understanding... not Reynolds numbers. In supersonic flight the passage of the leading edge of the airfoil through the air generates a shock wave. Ahead of the shock wave the air pressure is ambient, of course because the shock wave hasn't arrived there yet. At the shock wave the pressure is higher than ambient, which why it is a wave. Behind the shock wave the pressure is less than ambient because some of that air has been pushed forward into the shock wave. Now, there actually are two shock waves generated, one over the top of the wing and and one under the bottom. Both trail back at some angle relative to the wing. For various angles of attack the angle between the wing and the upper shock wave will be different from the angle between the wing and the lower shock wave. There is also a difference in the pressure behind the wave over the wing and the wave under the wing. The lift is approximately the product of this pressure differential and the area of the wing, e.g. (Po - Pu) * Ah = Lift where Po is the pressure over the wing, Pu under the wing and Ah is the cross sectional area of the wing in a plane parallel to the direction of motion. Now, there is a region of reduced pressure behind the wing which is somewhere between Po and Pu in value, call that Pb. Remembering that ahead of the wing is the shock wave (Pw) and ahead of that ambient air (Pa) then to a first approximation: [(Pa - Pw) + (Pw - Pb)] * Av = (Pa -Pb) * Av = Drag, where Av is the cross-sectional area of the wing perpendicular to the direction of motion. Now of course the pressures in these regions are not constant so to get a more exact answer you would integrate all the pressures over the entire surface of the aircraft (not just the wings) to arrive at one force which is the net superposition of lift and drag. At least that is the way I remember it from aeronautics. -- FF |
#7
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Toks Desalu wrote:
Guys- If we have higher L/D, we need higher thrust to maintain lift equal weight. Am I reading this correctly? In this case, you would need higher horsepower? Toks Desalu PP-SEL Dyin' to fly! A simple way to look at it is: You need thrust to overcome drag. You need lift to overcome gravity. So, the more lift you have for a given amount of drag the less thrust you need to achieve a given speed. Or: To maintain straight and level flight you need "X" amount of lift to hold up the wieght of your aircraft. A wing with a higher L/D will give you "X" amount of lift with less drag (L/D is just "amount of lift" per "amount of drag"). Because you now have less drag you don't need as much power to maintain a given speed. Regards Martin Morgan |
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