Lift-to-Drag Ratio?

[Toks Desalu]

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!

[James M. Knox]

"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
-----------------------------------------------

[Alan Baker]

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."

[Jan Carlsson]

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!
>
>

[James M. Knox]

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
-----------------------------------------------

[Fred the Red Shirt]

"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

[Martin Morgan]

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