Question of the day-final proof

Hi, just a practical example...I own part of a Swift (aerobatic glider), which has
the very same wing (profile etc.) as the Fox, just
a bit less span. So the Fox has definately more drag
then the Swift (more wing span, no retractable gear,
bigger fuselage....), but weights significantly more.BUT, on upwards lines (45° or 90°) I can show much
longer lines with the Fox than with the Swift, haveing
the same entry speed and pulling the same g´s. Let´s say I want to fly a 45° upwards line with a full
slow roll in it. The Swift rolls much faster than the
Fox, nevertheless the lines before and after the roll
are significantly longer with the Fox.And this doesn´t only happen to me, every pilot haveing
flown both can confirm this.Just my 2 cents,Markus

Hmmm... maybe, but not. All you have demonstrated, in the same way that
Galileo did, is the equivalence between gravitational mass and inertial mass
(physics 101), and you should note it only applies in a vacuum where the
effects of drag can be neglected, and as you rightly state it has nothing to
do with kinetic or potential energy. However, as a later post explains there
are other factors at work due to the aerodynamic effects which must apply to
any real world glider, namely the interplay between the inertial and viscous
forces due to the nature of the fluid in which we play.

Rgds,

Derrick.

HMMMM this is getting puzzling!?1
Let's get back to the basics
1. Total energy at the start MUST equal total energy at the end +
drag(defined here as interplay between inertial and viscous forces due to
the nature of the fluid in which we play (a mix of nitrogen, oxygen and
carbon dioxyde). No matter how lift, weigth or drag acts,at what rate, more
at the beginning less at the end , the above will always be true
2. Let us forget about the difference of stall speed, and drag, the above
statement yields 130 metres gain of height for my Jantar in laboratary
vacuum and let us attribute it to the balalsted glider.
3 Now the pro ballasted say that there is a detectable difference between
ballasted and dry.Let us say 25 metres, that 75 feet is readable on the
clock. So the dry glider only went to 105 metres.
4 The difference is equivalent to the work necessary to raise 100 kgs
(weight of ballast) on 25 metres which is (100 * 9.81 * 25) that is 24 500
Joules.
5. The force doing the work to hold the dry glider has to be drag, no???. So
someone has to come with a demonstration. Because No 1 has to apply no
matter what, and is easy to evaluate P1+ K1=P2+K2.



"Derrick Steed" a écrit
dans le message de ...
Hmmm... maybe, but not. All you have demonstrated, in the same way that
Galileo did, is the equivalence between gravitational mass and inertial
mass
(physics 101), and you should note it only applies in a vacuum where the
effects of drag can be neglected, and as you rightly state it has nothing
to
do with kinetic or potential energy. However, as a later post explains
there
are other factors at work due to the aerodynamic effects which must apply
to
any real world glider, namely the interplay between the inertial and
viscous
forces due to the nature of the fluid in which we play.

Rgds,

Derrick.

Ok Going out on a limb here.....
The ballasted ship is heavier.
The ballasted ship has a flatter glide angle.
So an I correct to assume that the unballasted ship must pull
MORE G to get the same climb angle as the ballasted ship.
If the unballasted ship needs to put more +g on the airplane
to achieve the same climb angle, then it will bleed energy thru
drag a bunch faster than the less g loaded ballasted ship.
In fact, the unballasted ship must pull up a little bit to
momentarily match the flatter glide angle of the ballasted ship.

Scott


"szd41a" wrote in message
...
HMMMM this is getting puzzling!?1
Let's get back to the basics
1. Total energy at the start MUST equal total energy at the end +
drag(defined here as interplay between inertial and viscous forces due to
the nature of the fluid in which we play (a mix of nitrogen, oxygen and
carbon dioxyde). No matter how lift, weigth or drag acts,at what rate,
more
at the beginning less at the end , the above will always be true
2. Let us forget about the difference of stall speed, and drag, the above
statement yields 130 metres gain of height for my Jantar in laboratary
vacuum and let us attribute it to the balalsted glider.
3 Now the pro ballasted say that there is a detectable difference between
ballasted and dry.Let us say 25 metres, that 75 feet is readable on the
clock. So the dry glider only went to 105 metres.
4 The difference is equivalent to the work necessary to raise 100 kgs
(weight of ballast) on 25 metres which is (100 * 9.81 * 25) that is 24 500
Joules.
5. The force doing the work to hold the dry glider has to be drag, no???.
So
someone has to come with a demonstration. Because No 1 has to apply no
matter what, and is easy to evaluate P1+ K1=P2+K2.



"Derrick Steed" a écrit
dans le message de ...
Hmmm... maybe, but not. All you have demonstrated, in the same way that
Galileo did, is the equivalence between gravitational mass and inertial
mass
(physics 101), and you should note it only applies in a vacuum where the
effects of drag can be neglected, and as you rightly state it has
nothing
to
do with kinetic or potential energy. However, as a later post explains
there
are other factors at work due to the aerodynamic effects which must
apply
to
any real world glider, namely the interplay between the inertial and
viscous
forces due to the nature of the fluid in which we play.

Rgds,

Derrick.

Anyone out there got a Duo-Discus, Nimbus D, or ASH25
and a logger with 1 second logging?

Try the following this weekend

Put in 100kgs of ballast, take a winch launch, start
a beat-up / finish / whatever at something above 100
kts with your P2 watching the ASI. Fly straight & level
with the speed bleeding off to allow the logger to
record a low point &, when P2 tells you that speed
is 100kts, pull-up. At top of pull up fly straight
& level for long enough for the logger to get a few
samples (5-10secs should be fine)
If you have enough height repeat until you don't!

Land, dump the ballast & repeat

Post the logger traces for all to see!

(Any of the 'Ballasted Glider Wins' crowd can try the
same in their own glider if they're brave enough to
watch the ASI closely enough)

"szd41a" wrote in message
...
Scott has a very good point here. Back to the maths.
Angle of flight at 100 Knts ballasted= 2.9 degrees
dry = 3.5 degrees
Say we want to go to 45 degrees SNIP

At the risk of diverting the conversation too far, and as only just out of
my (flying) nappies, I find that the above is rather counter-intuitive. Is
it because you need to put the nose of the glider much lower with a light
glider to get the penetration? Taking this to the limit, as the wing
loading increases, at a certain speed the rate of sink decreases, therefore
if I load a glider with lead to (near) infinite load it might stay in the
air permanently?

I can't get my head 'round this! 8-)

Keith

The ballasted glider has a better L/D than the dry one at high speeds
(that's why you put the water in, no?).
Putting Reynold number effects aside, the L/D is always limited by the max
L/D as you increase wingload, so your glide angle will get to 1.5 degree for
40:1 or 1 degree for 60:1.

As you go to infinite wingloadings, at some point you will have max L/D at
vne, and then further on you'll operate the glider at speeds lower than the
one for max L/D, so you'll hit the ground earlier :-)

--
Bert Willing

ASW20 "TW"


"Keith W" a écrit dans le message de
...

"szd41a" wrote in message
...
Scott has a very good point here. Back to the maths.
Angle of flight at 100 Knts ballasted= 2.9 degrees
dry = 3.5 degrees
Say we want to go to 45 degrees SNIP

At the risk of diverting the conversation too far, and as only just out of
my (flying) nappies, I find that the above is rather counter-intuitive.
Is
it because you need to put the nose of the glider much lower with a light
glider to get the penetration? Taking this to the limit, as the wing
loading increases, at a certain speed the rate of sink decreases,
therefore
if I load a glider with lead to (near) infinite load it might stay in the
air permanently?

I can't get my head 'round this! 8-)

Keith