reynolds number

Morgans schreef:


"Brian Whatcott" wrote

I could even go a little further: if you get yourself in a situation
when you have to deploy your considerable engineering skills in
evaluating Re, it is because you forgot to use your even more
considerable judgment is selecting well-liked, useful, relevent airfoils.

:-)
Amen!

Ya define the mission, and how fast you think you will go and look at
the list of airfoils used on airplanes of similar speed and mission.

And that's the lines along which I was thinking, until this Reynolds
thing crossed my way. My gratitude to all who responded, I have received
a powerful lot to put under my thinking cap for the next couple of
weeks/months/years

KA

First, I'd like to say that I am no professional aerodynamicist, and the following text may contain errors. I'd be happy if someone points them out!

About Reynolds numbers:

Let's say you're a pioneer of flight, and you want to build an airplane that can fly. How big a wing do you need? To answer this, you could simply build a wing, put it in your pioneer-wind tunnel, and measure the forces on it at the speed you think your airplane will be able to attain with the pioneer-engine you have available. Let's say the wing wasn't big enough. Now, what do you do? You have to build another one! And another one... etc, until you get it right. You don't wan't a too big wing (weight!), and not a too small (can't fly!).

Anyway, this quickly grows old. Someone said: Why can't we predict how much lift a given wing will have without building the darned thing every time?

So someone set out to build lots of different wings, and tried to see if there was some kind of pattern to the amount of lift different wings gave. Let's see, this one over here is 1m wide, and gives 1N at this speed, but that one over there gave 2N at the same speed. Hmm, why.. oh, it's twice as wide, but identical otherwise! Okay, lift seems to increase linearly with wingspan! And people built lots of wings, and thought hard, and came up with a formula:

Lift = chord * span * CL * Velocity * Velocity * airdensity / 2

Whe
chord = Width of wing, from leading edge to trailing edge
span = Length of wing, from tip to tip
velocity = The relative forward speed of the wing in relation to the air
air density = The weight of air, per volume (1.2kg / cubic meter)
CL = Lift Coefficient, (varies with angle of attack, and reynolds number!)

Anyway, this worked pretty well.

The was just one problem! When you build a 30 cm (1') wing, and it gives 10N of lifting force, you'd expect the same kind of wing, but 3m (10'), to give 100N (if it has the same chord), at the same speed! But, this did not occur! Rather, it turned out, that when you take a small wing, and make it many times bigger, or run it many times faster through the air, the above formula doesn't work anymore.

When does it work? Does speed have to be the same? Does chord have to be the same? No! It works when the Reynolds number is the same!

So, if you build a small wing, and run it in a wind tunnel with a different medium (perhaps water), and this makes the reynolds number the same as it will be when your real aircraft flies its real mission, you can expect the lift equation above to 'work'.

So, the Lift Equations allows the lifting force from a wing to be predicted (very handy when constructing aircraft). But the lift equation only works when supplied with a CL measured at the same Reynolds number as you'll be flying at! So you need the Reynolds number in order to be able to use the lift equation!

Anders wrote:
First, I'd like to say that I am no professional aerodynamicist, and the
following text may contain errors. I'd be happy if someone points them
out!

About Reynolds numbers:

Let's say you're a pioneer of flight, and you want to build an airplane
that can fly. How big a wing do you need? To answer this, you could
simply build a wing, put it in your pioneer-wind tunnel, and measure
the forces on it at the speed you think your airplane will be able to
attain with the pioneer-engine you have available. Let's say the wing
wasn't big enough. Now, what do you do? You have to build another one!
And another one... etc, until you get it right. You don't wan't a too
big wing (weight!), and not a too small (can't fly!).

Anyway, this quickly grows old. Someone said: Why can't we predict how
much lift a given wing will have without building the darned thing
every time?

So someone set out to build lots of different wings, and tried to see
if there was some kind of pattern to the amount of lift different wings
gave. Let's see, this one over here is 1m wide, and gives 1N at this
speed, but that one over there gave 2N at the same speed. Hmm, why..
oh, it's twice as wide, but identical otherwise! Okay, lift seems to
increase linearly with wingspan! And people built lots of wings, and
thought hard, and came up with a formula:

Lift = chord * span * CL * Velocity * Velocity * airdensity / 2

Whe
chord = Width of wing, from leading edge to trailing edge
span = Length of wing, from tip to tip
velocity = The relative forward speed of the wing in relation to the
air
air density = The weight of air, per volume (1.2kg / cubic meter)
CL = Lift Coefficient, (varies with angle of attack, and reynolds
number!)

Anyway, this worked pretty well.

The was just one problem! When you build a 30 cm (1') wing, and it
gives 10N of lifting force, you'd expect the same kind of wing, but 3m
(10'), to give 100N (if it has the same chord), at the same speed! But,
this did not occur! Rather, it turned out, that when you take a small
wing, and make it many times bigger, or run it many times faster
through the air, the above formula doesn't work anymore.

When does it work? Does speed have to be the same? Does chord have to
be the same? No! It works when the Reynolds number is the same!

So, if you build a small wing, and run it in a wind tunnel with a
different medium (perhaps water), and this makes the reynolds number
the same as it will be when your real aircraft flies its real mission,
you can expect the lift equation above to 'work'.

So, the Lift Equations allows the lifting force from a wing to be
predicted (very handy when constructing aircraft). But the lift
equation only works when supplied with a CL measured at the same
Reynolds number as you'll be flying at! So you need the Reynolds number
in order to be able to use the lift equation!




Building on your response, I should mention that lift coefficients are
sometimes based on cross section area perpendicular to the wind, rather
than on chord times span. The former gives bigger numbers than the
latter method. So a flat board perpendicular to the wind was sometimes
given as Cl = 1 to 1.2, whereas a flat board at zero incidence might be
given as span times chord and make the contrast between the angle of
attack of a flat board for AoA = 0 versus AoA = 90 look even look bigger
than it already is.

Brian W

On Sat, 27 Jun 2009 17:43:27 -0400, "Morgans"
wrote:



"Brian Whatcott" wrote

I could even go a little further: if you get yourself in a situation when
you have to deploy your considerable engineering skills in evaluating Re,
it is because you forgot to use your even more considerable judgment is
selecting well-liked, useful, relevent airfoils.

:-)
Amen!

Ya define the mission, and how fast you think you will go and look at the
list of airfoils used on airplanes of similar speed and mission. That makes
airfoil choice a real choice.

....and then you look at mark langford's web site and pick the aerofoil
according to thickness. ...from the list of excellent aerofoils
developed for the KR2S on his web site. :-)

you dont have to be correct or competent to design an aeroplane.
you stand the chance of designing a damn site better one though if you
are. having experience and attitude sometimes gets you there.
competence can mire you in decisions and see you achieve nothing.

Why is the Wittman W8 tailwind still a standout in the efficiency
figures?

Stealth Pilot

"Stealth Pilot" wrote

Why is the Wittman W8 tailwind still a standout in the efficiency
figures?

Seems to me that it has a few things that keep it on top.

See if you think I am on the right track.

The shapes used in the fuselage and anything that is sticking out in the wind
are all good aerodynamic tradeoffs of slippery and light.

The basic shape of the fuselage is good for contributing to the lift of the
aircraft, more than most other designs. Probably the most important feature of
the design, in my eyes.

Attention is always on making structures easy to build light and no extra weight
is there that does not contribute to lightness.

The airfoil and fuselage are light enough and slippery enough to be powered by a
small engine, so extra engine weight and fuel weight does not have to be carried
around, which allows the structures to be built more lightly. It is sort of a
good circle that keeps weight down, versus the other circle that keeps growing
the weight of the aircraft.

How did I do? g
--
Jim in NC

On Tue, 21 Jul 2009 00:40:21 -0400, "Morgans"
wrote:


"Stealth Pilot" wrote

Why is the Wittman W8 tailwind still a standout in the efficiency
figures?

Seems to me that it has a few things that keep it on top.

See if you think I am on the right track.

The shapes used in the fuselage and anything that is sticking out in the wind
are all good aerodynamic tradeoffs of slippery and light.

The basic shape of the fuselage is good for contributing to the lift of the
aircraft, more than most other designs. Probably the most important feature of
the design, in my eyes.

Attention is always on making structures easy to build light and no extra weight
is there that does not contribute to lightness.

The airfoil and fuselage are light enough and slippery enough to be powered by a
small engine, so extra engine weight and fuel weight does not have to be carried
around, which allows the structures to be built more lightly. It is sort of a
good circle that keeps weight down, versus the other circle that keeps growing
the weight of the aircraft.

How did I do? g

pretty damn good.