Did we decide if ballasted pullups are higher????????
Scott Correa
I'm sure some logger equipped pullups were made.
Who wins?? Wet or dry.
I still think wet pullups go higher, but I can't prove it.
Scott.
Lars Peder Hansen
fereaks, the lot:
Off course wet pullups go higher. In the end, a pullup is conversion from
kinetic energy (speed) to potential energy (altitude). Soaring is all about
trading one of these for the other.
Picture two identical gliders (please, no Discus/Duo Discus (sorry, couldn't
resist)) both at 100 feet, 250 Km/h. Their kinetic energy is derived from
their speed (equal in this setup) and mass. So the heavy one has the most
kinetic energy. -Which one can then obtain the highest potential energy ?
Elementary, my dear Watson.
Happy soaring,
Lars Peder
Replace the obvious by a dot to respond via e-mail
"Scott Correa" <scorrea-r...@socket.net> skrev i en meddelelse
news:vn6dg83...@corp.supernews.com...
Alan Baker
"Lars Peder Hansen" <l...@post1TheObvioustele.dk> wrote:
> OK, here we go, just to exite trolls, believers, non-believers, math
> fereaks, the lot:
>
> Off course wet pullups go higher. In the end, a pullup is conversion from
> kinetic energy (speed) to potential energy (altitude). Soaring is all about
> trading one of these for the other.
> Picture two identical gliders (please, no Discus/Duo Discus (sorry, couldn't
> resist)) both at 100 feet, 250 Km/h. Their kinetic energy is derived from
> their speed (equal in this setup) and mass. So the heavy one has the most
> kinetic energy. -Which one can then obtain the highest potential energy ?
Except that potential energy is proportional to both altitude *and* mass.
IOW, double the mass, and you double the kinetic energy at a given
speed, but you also double the potential energy of the change in
altitude.
> Elementary, my dear Watson.
If you think about it a moment, the correct answer is "Elementary, my
dear Galileo" (think dropping balls of different masses, and then
reversing the experiment).
>
> Happy soaring,
> Lars Peder
>
> Replace the obvious by a dot to respond via e-mail
>
>
>
>
> "Scott Correa" <scorrea-r...@socket.net> skrev i en meddelelse
> news:vn6dg83...@corp.supernews.com...
> > OK people, what was the verdict.
> > I'm sure some logger equipped pullups were made.
> > Who wins?? Wet or dry.
> > I still think wet pullups go higher, but I can't prove it.
> >
> > Scott.
> >
> >
>
>
--
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."
Mike Borgelt
<l...@post1TheObvioustele.dk> wrote:
>OK, here we go, just to exite trolls, believers, non-believers, math
>fereaks, the lot:
>
>Off course wet pullups go higher. In the end, a pullup is conversion from
>kinetic energy (speed) to potential energy (altitude). Soaring is all about
>trading one of these for the other.
>Picture two identical gliders (please, no Discus/Duo Discus (sorry, couldn't
>resist)) both at 100 feet, 250 Km/h. Their kinetic energy is derived from
>their speed (equal in this setup) and mass. So the heavy one has the most
>kinetic energy. -Which one can then obtain the highest potential energy ?
>Elementary, my dear Watson.
>
>Happy soaring,
>Lars Peder
>
>Replace the obvious by a dot to respond via e-mail
>
>
Dear God!
Didn't anyone pay attention to physics lessons in high school?
Your TE probe doesn't know the mass of the glider and seems to work
well regardless of how much water you are carrying doesn't it?
Mike Borgelt
mm
"Lars Peder Hansen" <l...@post1TheObvioustele.dk> wrote in message
news:3f735449$0$24692$edfa...@dread14.news.tele.dk...
> OK, here we go, just to exite trolls, believers, non-believers, math
> fereaks, the lot:
>
> Off course wet pullups go higher. In the end, a pullup is conversion from
> kinetic energy (speed) to potential energy (altitude). Soaring is all
about
> trading one of these for the other.
> Picture two identical gliders (please, no Discus/Duo Discus (sorry,
couldn't
> resist)) both at 100 feet, 250 Km/h. Their kinetic energy is derived from
> their speed (equal in this setup) and mass. So the heavy one has the most
> kinetic energy. -Which one can then obtain the highest potential energy ?
> Elementary, my dear Watson.
>
> Happy soaring,
> Lars Peder
>
> Replace the obvious by a dot to respond via e-mail
>
>
Ralph Jones
What Alan, Mike and m said. Strictly from a potential/kinetic energy
standpoint, it's a wash. Not APPROXIMATELY a wash --- EXACTLY a wash.
However, there at least two more factors.
First, the wet ship has to pitch up to a higher angle of attack as it
starts the pullup, in order to achieve a given upward acceleration.
This will extract a penalty in the form of energy removed by induced
drag.
Once the change of direction is completed, both pilots can reduce
their angle of attack to the zero-lift point, giving an ascending
parabola which will maximize the altitude reached. In this condition
the only drag is parasite drag, which will be the same for both ships;
consequently the wet ship has an edge here because its larger mass
will decelerate less for a given drag force.
So the answer to the question depends on which of these effects is
larger. It shouldn't be hard to settle with an empirical test: just
have two ships pull up simultaneously in line-abreast formation.
rj
Pete Brown
Lars Peder Hansen wrote:
> OK, here we go, just to exite trolls, believers, non-believers, math
> fereaks, the lot:
OK I'll bite
>
> Off course wet pullups go higher.....
> Picture two identical gliders (please, no Discus/Duo Discus (sorry, couldn't
> resist)) both at 100 feet, 250 Km/h. Their kinetic energy is derived from
> their speed (equal in this setup) and mass. So the heavy one has the most
> kinetic energy. -Which one can then obtain the highest potential energy ?
> Elementary, my dear Watson.
>
> Happy soaring,
> Lars Peder
The heavier one does have more energy. Then again, it
requires more energy to lift a heavier object. And the
additional work required to lift a heavier object to the
same height as a lighter one is directly proportional to the
difference in mass.
--
Peter D. Brown
http://home.gci.net/~pdb/
http://groups.yahoo.com/group/akmtnsoaring/
Lars Peder Hansen
The heavy and the light glider will get to the exact same altitude, provided
they slow down to the same speed. I know you can argue about minute
differences in CG, G, drag, but that is academic. My popular belief was
simply wrong. I forgot to realize that mass is also part of the potential
energy equation, as many folks have been kind enough to point out, using
even good ol' Galileo and his findings..
It's amazing the various kinds of response you get from RAS: Many point out
facts in a friendly and understandable manner, others simply say "you dont
know what you are talking about!"
Wonder what would have happened if Galileo had been able to publish his
findings on the Internet. -Now that was one thread I would have liked to
follow ;-)
Happy and safe soaring (and pullups) to all,
Lars Peder
"Lars Peder Hansen" <l...@post1TheObvioustele.dk> wrote in message
news:3f735449$0$24692$edfa...@dread14.news.tele.dk...
Lars Peder Hansen
>> Dear God!
>
> Didn't anyone pay attention to physics lessons in high school?
Actually, I skipped a lot of physics & math lessons in high school, to go
soaring with a friend from my class. ( I guess you figured that out by now
;-)
We had a math teacher who eventually figured out the connection between blue
skies with cumulus, and the two of us being absent from class. One day he
called the gliderport, just to let us know what chapters to read for next
week. -I was the unlucky one who picked up the phone...
Happy soaring, and safe pullups to exactly the same altitude no matter what
mass you fly at,
Lars Peder
Derrick Steed
I think you may be forgetting something else. For the same speed of travel,
the light glider is using energy faster than the heavy glider. It seems
paradoxical, but it is true - look at the loaded and unloaded polars to see
that the heavy glider is either: a) travelling faster for the same sink
rate, or b) sinking less for the same speed of travel (provided that the
speed of travel in both instances is above the crossover point for the
polars, the speed this occurs at depends on the wing loading).
So, as they zoom upwards during the pull up: a) the light glider is using up
its energy faster than the heavy glider, and it has less to start with, and
b) the heavy glider is using up its energy slower, and it has more to start
with.
The heavy glider, of the same type, goes higher.
Rgds,
Derrick.
Maule Driver
"Lars Peder Hansen" <l...@post1TheObvioustele.dk
> Wonder what would have happened if Galileo had been able to publish his
> findings on the Internet. -Now that was one thread I would have liked to
> follow ;-)
>
Couldn't be any worse than sending a copy to the Vatican and being welcomed
to the inquisition...
Makes what usenet rudeness seem like warm kisses from a comely troll.
szd41a
The polar for a given glider is reprentative of gliding flight
characteristics. Once your nose is pointing up, you are coasting, you are
not gliding anymore....so forget about the polar......and at the risk of
becoming rude let me add this.........LIFT IS NOT HELPING YOU ANYMORE it may
actually pull you away from optimal trajectory. Being the one who started
this whole discussion, I am starting to think that I missed a good occasion
to shut the F...up! But what the hell!!!! it all started in front of the
club house, we're all drinking beer, and I throw The QUESTION!!!!! Immediate
answer from everybody the ballasted is actually going wayyyyy
higher!!!!!!!! I said that I doubt it, and did a littlle maths (you don,t
have to own an eng. degree......which I do ), checked the drag, the only
force that could make a difference and came to conclusion the the light
glider has an edge, contrary to popular belief. So I came back to my
buddies, thinking I was bearing the Good News. I was looked down at ( I am
joking ok!!) So short of doing a real test, which is not easy to do
"scientifically" I though of posting a message on this forum
here!!!!!!!!hell broke loose (I am joking again) but yet I am amazed to see
that the "Non-silent" majority is pro-ballasted. One can easily realise how
your intuition can mislead you on this ......but yet we have some
experimented pilots who are doing the maths to prove their intuition, and
they do it in a much scientific way. So I can't wait to see the new sets of
equations, and the coasting polar (this is not a joke). we are actally
having fun,no? Me I have no proof to offer, and I feel a little obligated,
being the one who set the fire. So we are desperatly begging anyone who
actually did some testing to come forward and put an end this nightmare
;-), before we start throwing punches at each other!!!!We have alrerady
witnessed people using capital letters in their message!!!!!
Again, as someone mentionned earlier, please stop throwing ping-pong and
golf balls at us. The Cx of a sphere is .42, which makes it the worst
aerodynamic shape, an it quickly build a force that will oppose accelaration
at a low speed, which is not the case of our beautiful gliders, at he
operating speed, ther is not much difference in form (parasite drag),
neighter for induced drag for which the heavy will suffer more during the
pull-up
The debate is not over yet!!!!We need a proof!!
BQ
"Derrick Steed" <REMOVE_TO_REPL...@eds.com> a écrit dans le
message de news:bl1pqc$73o8v$1...@ID-49798.news.uni-berlin.de...
Stefan
>
> I forgot, how did you arrive at this? The light glider has
> the advantage of being able to slow to a lower stall,
Stall speed isn't the limit. If you go ballistic at the top, i.e. if you
push to zero G, then you can slow down to zero km/h if you wish (or to
zero knots, if you live in the USA).
Stefan
Kevin Neave
glider having the advantage in the 2G case.
In normal flight at high speed we have relatively little
induced drag, and the major component of our total
drag is profile.
With it's higher reserve of energy the heavy glider
gains the benefit 'cos even though the induced drag
is higher it's a small proportion of the total.
If we now start to pull G the induced drag for both
gliders goes up, it now becomes a more significant
proportion of the total for each glider so surely the
advantage of the heavy glider is reduced?
(And I notice that you have admitted you're in the
'Heavy Glider Wins' camp) :-)
At 21:06 26 September 2003, Todd Pattist wrote:
>
>The advantage of the heavy glider (in terms of lost
>altitude) remains throughout the increased G-load portion
>of
>the pullup. Think of the two gliders suddenly doubling
>their
>weight (2G pullup). The ballasted glider would have
>a lower
>sink rate in the 2G case for the same reason that it
>has a
>lower sink rate in the 1G case.>
szd41a
aircrafts, complete with some graph and theory and all.
http://www.av8n.com/how/htm/4forces.html and
http://www.go.ednet.ns.ca/~larry/bsc/winch/winch.html
I hope this help
BQ
"Todd Pattist" <tpat...@DONTSPAMME.snet.net> a écrit dans le message de
news:nja9nvo1mj36n6nvr...@4ax.com...
> "szd41a" <girar...@videotron.ca> wrote:
>
> >checked the drag, the only
> >force that could make a difference and came to conclusion the the light
> >glider has an edge, contrary to popular belief.
>
>
> far as drag goes, the heavy glider has the advantage (in
> terms of altitude lost) at all times.
> Todd Pattist - "WH" Ventus C
> (Remove DONTSPAMME from address to email reply.)
szd41a
So no advantage if final speed is 0 mph. But, my intuition led me into the
following, to accelerate vertically, you will need a bigger force for the
ballasted, the old F=ma m being the mass. What force pulls the glider
upwards???l Lift. You will need more lift on the ballasted to change
direction upwards, and more induced drag thus a little advantage to the
light. (difference in angle for the same speed is minute) After we go
uniformely accelerated , induced drag and form drag being of the same
magnitude (every equation for solving drag includes Cx, density of air,
speed and area,load factor for induced drag) nothing about mass!!!! So same
results(roughly...). Notice that the more speed, the more drag!!!!. Now
let's have the ping pong ball again, why is it that for the same initial
speed, the golf ball will travel wayyyyy further???? I thing the answer is
in the very high Cx of a sphere (.42) which is not the case of our glider
( .00something.) The ping pong ball will be stopped cold at very low speed
because of the high force oppsed to direction of travel (drag).Of course the
ping pong has less kinetic energy to oppose and the equilibrium point will
be attained at very lower speed than the heavy golf ball. But for our
gliders, drag will never be relatively high enough to equilibrate the
kinetic energy of both ballasted or unballasted gliders for the range of
speed we are travelling at.The only results according to that are those from
Udo, whose sims trials gave a litlle(very little) advantage to the
unbalasted (probably because of the induced drag during pull-up). I gave an
adress that shows how to claculate the force produced by drag on an
aircraft. Using theese equations, I was never able to equilibrate a 100 feet
difference of altitude for the ballasted, respecting the conservation of
energy law. I could easily solve the ping pong ball problem (intuitively).
BQ
"Stefan" <stefan@mus._INVALID_.ch> a écrit dans le message de
news:3F74B2E8.A6A2C301@mus._INVALID_.ch...
Mike Borgelt
Might have been better to miss English and History class. Physics and
maths is useful for pilots.
Mike
Bruce Hoult
Kevin Neave <REMOVE_TO_REP...@kneave.freeserve.co.uk>
wrote:
> I'm not sure I follow your reasoning about the heavy
> glider having the advantage in the 2G case.
>
> In normal flight at high speed we have relatively little
> induced drag, and the major component of our total
> drag is profile.
> With it's higher reserve of energy the heavy glider
> gains the benefit 'cos even though the induced drag
> is higher it's a small proportion of the total.
>
> If we now start to pull G the induced drag for both
> gliders goes up, it now becomes a more significant
> proportion of the total for each glider so surely the
> advantage of the heavy glider is reduced?
But how hard are you pulling?
Minimum drag (min sink) for the dry glider is probably at around 50
knots, so to get the same AOA at 110 knots you have to pull well over
4G. Min sink for the wet glider might be -- what -- 55 knots? So
you're still talking 4G at 110 knots to get that AOA.
If you're only pulling 2 or 3 G then you'll be above best L/D speed as
well.
The heavy glider still clearly has an advantage.
-- Bruce
Bruce Hoult
Stefan <stefan@mus._INVALID_.ch> wrote:
> Stall speed isn't the limit. If you go ballistic at the top, i.e. if you
> push to zero G, then you can slow down to zero km/h if you wish (or to
> zero knots, if you live in the USA).
Sure, but that doesn't help you fly your circuit.
-- Bruce
Stefan
>
> > Stall speed isn't the limit. If you go ballistic at the top,
>
This whole thread is only marginally relevant to real flying. (Besides:
It does help you when flying dolphin.)
Stefan
Jose M. Alvarez
My ASW24, being lighter than, say, a ASK-21, will get 200mts (600ft) from a
high speed pull up. The ASK will get 100 to 130mts.
A Blanik will get much less.
A ballasted ASW-24 gets more from the pull up than I do unballasted.
In our kind of flying drag is everything! Why an ASK-13 flyes less than
a -21, a -24, a -25 and so on? How can we explain such large differences in
performance? Drag is reduced for the newer gliders.
You are all right with the math, and in the total energy equations mass is
nonrelevant as it is a constant. But you're all disregarding the effect of
drag and you cannot do that! That is the reason why all the fancy math
you're doing does not match with our real life experience. And if a
theorical reasoning does not match with reality, then it's obvious that the
theory is somewhere wrong.
So please if you want to get into math please do account drag and fluid
mechanics into it. If you simplify it so much, it will be inaccurate enought
to be false.
Ballasted gliders will go higer because of the increased mass, more
penetration and more energy for the same aerodinamic drag.
"Bruce Hoult" <br...@hoult.org> escribió en el mensaje
news:bruce-3025AC....@copper.ipg.tsnz.net...
> In article <bl2daj$7h47g$1...@ID-49798.news.uni-berlin.de>,
> Kevin Neave <REMOVE_TO_REP...@kneave.freeserve.co.uk>
> wrote:
>
Lots of maths snipped...
Kevin Neave
& 100kgs ballast.
Why will none of the 'Heavy Glider Wins' contingent
actually give any numbers as to what they think the
advantage here is?
At 09:24 29 September 2003, Jose M. Alvarez wrote:
>Drag is the thing here...
>My ASW24, being lighter than, say, a ASK-21, will get
>200mts (600ft) from a
>high speed pull up.
And how much EXACTLY will it get from 100kts?
>A ballasted ASW-24 gets more from the pull up than
>I do unballasted.
How much ballast EXACTLY are we talking about here?
And EXACTLY how much extra height do you think you'd
gain in a pull up from 100kts.
>You are all right with the math, and in the total energy
>equations mass is
>nonrelevant as it is a constant. But you're all disregarding
>the effect of
>drag and you cannot do that!
And you're disregarding TIME, there's simply not enough
time in which the slight difference in performance
of the ballasted/unballasted glider has to operate.
That is the reason why all the fancy math
>you're doing does not match with our real life experience.
>And if a theorical reasoning does not match with reality,
then it's obvious that the theory is somewhere wrong.
I think you're confusing 'perception' with reality
here
>So please if you want to get into math please do account
>drag and fluid mechanics into it. If you simplify it
so much, it will be inaccurate enought to be false.
My analysis, including drag, but with some approximations
(All in favour of the heavy glider) is on the way
>Ballasted gliders will go higer because of the increased
>mass, more penetration and more energy for the same
aerodinamic drag.
>
Ballasted Gliders average higher cross country speeds,
finish faster, and with the pilot feeling better about
life! That's why the perception is that ballasted gliders
gain more in the pull-up!!!
:-)
>
Kevin Neave
old chesnut drag, and before any of you start to pick
nits I'm going to use kilograms as a unit of 'Force'
to avoid confusing our American friends who seem to
use pounds as both a unit of mass and force.
But I'm assuming that the analysis I'm doing here is
for our own planet earth where G is about 9.8m/s/s
and the difference in Gravitational field between the
start and end of the pull up is negligible.
Let's start with a Glider weighing 300kgs with a stall
speed of 38kts (19m/s), and a best L/D of 40:1 at 60kts.
If we now add 100kgs of ballast all the numbers are
multiplied by sqrt(4/3) so we have a stall speed of
44kts (22m/s) and a best L/D of 40:1 at 69kts
Now I believe that at Best L/D the Profile and Induced
drag are equal (And I'm sure someone will correct me
if I'm wrong).
So for our light glider at 60kts we have a total 'drag'
of 7.5kgs (300/40) which means that we have Induced
= Profile = 3.5kgs
I also believe that Induced Drag goes down with the
square of the speed (i.e 1/Vsquared) (Again I'm sure
I'll be corrected!) and that profile goes up with the
square of the speed.
Induced drag is also proportional to wing loading so
our Heavy Glider will always have 4/3 the induced drag
of the light one.
So going back to our light glider...
At 100kts our induced 'drag' is now (60/100) * (60/100)
* 3.75 = 1.35kgs
And our profile is (100/60) * (100/60) * 3.75 = 10.41kgs.
Total 'drag' = 11.76kgs
Assuming that adding the ballast doesn't alter the
shape of our glider too much then for the heavy glider
Induced = 1.35*4/3 = 1.8kgs
Profile = 10.41kgs
Total = 12.21kgs.
Now we're ready to pull up & I'll make a few assumptions
here.
1) That we're not going to pull up to below the stall
speed of either glider. This is a reasonable assumption,
pulling up to an airspeed of 0kts followed by a spin
recovery and return to normal flight almost certainly
hands the 'advantage' to the light glider (It'll hit
the ground less hard!!)
2) We're going to pull up into a 45deg climb & maintain
a straight line up to our recovery speed. This is not
a ballistic trajectory 'cos in order to maintain this
straight course the wings will have to generate some
lift and so will upset my next assumption, which is...
3) We can ignore changes drag!! This is a pretty big
assumption but here goes... At the same level of 'G'
the induced drag is propotional to the wing loading
i.e the heavy glider will have 4/3 times the induced
drag. However the 'Energy Fuel Tank' of the heavy glider
also has 4/3 as much as the Light one so I think the
two effects cancel out
(And once again I'm sure someone out there will correct
me!). Secondly, as the speed drops off our profile
drag will also be reduced with the square of our speed,
in some ways this makes up for ignoring the increasing
induced drag required to maintain our straight 45deg
climb.
Thirdly, making these assumptions about drag gives
a slight advantage to the heavy glider. And Lastly
it makes the maths simpler!!
So here we go...
Both gliders start to pull 2G. The induced drag for
each is doubled (i.e goes to 2.7kgs for the light,
3.6 for the heavy) but since the change is not significant
compared to the total I'm ignoring it!
And we start to climb (I'm assuming that the time taken
to transition from level flight to 45deg climb is small
so any transient effects in our 2G pull won't be significant)
The retarding force for the light glider is now 300kgs
* 1/sqrt(2) due to gravity plus the 11.76kgs due to
drag = 212.13 + 11.76 = 223.89kgs
So our deceleration will be 223.89/300 * 9.8 = 7.31
m/s/s
For the heavy glider we have 400 * 1/sqrt(2) for gravity
plus 12.21kgs due to drag = 295.05kgs
So our deceleration will be 295.05/400 * 9.8 = 7.23
m/s/s
Now finally we're going to continue up our 45deg climb
to our respective stall speeds, all of which is done
at Newtonian rather than Einsteinian speeds
So V*V = U*U - 2*a*s :- where V is final velocity,
U is initial, A is acceleration and s is distance travelled.
So ((U*U) - (V*V)) / (2*a) = s
For the light glider
V=19m/s, U=50m/s, a=7.31m/s/s
((50 * 50) - (19 * 19)) / (2 * 7.31) = 146.31 metres.
So Our height gained = 1/sqrt(2) * 146.31 = 103.46m
For the heavy glider
V=22m/s, U=50m/s, a=7.23m/s/s
((50 * 50) - (22 * 22)) / (2 * 7.23) = 139.24 metres
!! Height gained = 98.46 metres !!
------------------------------------------------------------------
---------------------------
OK, So there's some assumptions in the above, but I
think all of them were made in favour of the heavy
glider.
But I say once again, for a pull up from 100kts with
100kgs of ballast, 'It's too close to call'...
Over to you Todd
:-))
Scott Correa
"Kevin Neave" <REMOVE_TO_REP...@kneave.freeserve.co.uk> wrote in
message > So ((U*U) - (V*V)) / (2*a) = s
>
> For the light glider
>
> V=19m/s, U=50m/s, a=7.31m/s/s
>
> ((50 * 50) - (19 * 19)) / (2 * 7.31) = 146.31 metres.
> So Our height gained = 1/sqrt(2) * 146.31 = 103.46m
>
> For the heavy glider
>
> V=22m/s, U=50m/s, a=7.23m/s/s
>
> ((50 * 50) - (22 * 22)) / (2 * 7.23) = 139.24 metres
> !! Height gained = 98.46 metres !!
>
> ------------------------------------------------------------------
> ---------------------------
>
> OK, So there's some assumptions in the above, but I
> think all of them were made in favour of the heavy
> glider.
>
> But I say once again, for a pull up from 100kts with
> 100kgs of ballast, 'It's too close to call'...
>
> Over to you Todd
>
> :-))
>
Kevin.
I don't see anywhere in your in your math the higher initial rate of sink
for
the lighter ship. If you look at polars for ballasted vs empty, for any
given
speed, the rate of sink is higher unloaded. The assumption that both
gliders are at the same height when they reach the the 0 fps up point
of the pullup looks flawed in my mind. You must wait longer in the
unloaded glider to establish a climb. In fact I think the softer pullup,
the greater the difference in starting heights of the newtonian decayed
climb.
Scott.
szd41a
This debate is pointless, some say. By looking at the numbers of visit, it
is of a certain interest to many people!!! I set fire on a european forum
too, the debate was hot, but quickly ended with a tie, or nothing measurable
from the cockpit. This debate allowed me to review some flight theory, and I
think it is beneficial to anybody who participate to this. Drag is something
that is anything but intuitive, and has to be looked at closely, actually
kinetic energy is far from intuitive, too(but easy to calculate), and many
had to review their opinion when they did the simple maths.
I read your comment here;
"Not true. The heavy glider pays for the higher drag with
less altitude lost, just like it pays for the higher drag in
1G flight with less altitude loss. That's why we carry
ballast. This is an advantage for the heavy glider here"
and I would like your comment. The reason we carry ballast is not is not
quite that drag-altitude trade, but because when we are ballasted, the
component of weight parralell to the direction of flight is bigger (Bigger
engine) For a given AOA, you will go faster ballasted. Drag is not running
the show, it will be a consequence of the flight attitude. Weight is the
motor. The polar curve reflect the situation as long as you are gliding,
that is going towards Mother Earth, weight helping you along. Your wings
will create enough lift to equilibrate weight. It is so well designed that
it will generate lift at the same best L/D even with a bigger loading ( and
perhaps an even better L/D..but not much!!!!) Now, generaly speaking in
aircrafts, lift does that and only that. If you want to gain height, you
throttle up or you crank that thermal, not counting on lift per say. Maybe
we are all saying the same thing, it is just what comes first the egg or the
hen????
And then, following your reasonning...
>
> Your assumptions are unconvincing to me. We can show by
> simple analysis that the heavy glider is always losing less
> altitude per second, at least until it gets down to its best
> L/D at the current load factor. The largest rate of
> altitude loss will be at the beginning- high speed and high
> load factor.
Now, I think that as soon as your nose is pointing up, the polar does not
reflect the flight characteristics anymore. Why?? Maybe because weight is no
more the motor . It has no component in the direction of flight. It has
become a load that you have to carry. I am not too sure we can say we are
loosing less altitude for a given speed, and conclude that it will go
higher. The polar does not apply to this part flight. What is the motor
then?? Kinetic energy of course and we are going ballistic. Lift can even be
a nuisance pulling you away from optimal path. If we are uniformely
accelarated, speed decrease will be 9.8m/s/s for both glider.
Does this sound right??
Robert Ehrlich
are possible, no need to do the experience, just think of
it at 2 extreme cases.
As the first extreme case, imagine a very light glider,
even if no technological possibility exists that such a glider
could be build, say 10 or 100 times lighter than current ones,
but with the same shape. At the speed for a nice pullup for
a ballasted glider, such a glider would be almost immediately
stopped by its drag, using up almost all the very little kinetic
energy it has, without a chance even to start the pullup. So
in this case the heavy glider is clearly the winner.
As the second extreme case, imagine 2 normal gliders, one ballasted
and the other not, but starting their pullups at a speed just above
the stalling speed of the ballasted one. The ballasted glider is
unable to do any pullup, while the light one can do it, so the
light glider is the winner.
Robert Ehrlich
>
> Robert Ehrlich <Robert....@inria.fr> wrote:
>
> >It seems obvious, with a little thinking, that both cases
> >are possible, no need to do the experience, just think of
> >it at 2 extreme cases.
>
> heavy glider and depends on the stall speed differential. I
> think everyone agrees that's a possibility, but it's not an
> answer to the legitimate question of what happens in a real
> pullup.
> Todd Pattist - "WH" Ventus C
> (Remove DONTSPAMME from address to email reply.)
From the extreme cases it seems obvious that there somewhere
in the speed range where both are equal. Below that speed the
lighter glider wins, above that speed the heavier wins. My
conviction is that the speed usually used for low passes
and pullups is in the upper range, but some calculation and/or
experimentation is needed to confirm this.
nafod40
> I suppose this is true, but either glider can go as fast as
> they want - just point the nose down. The ballast isn't
> carried for speed. It's carried because the power we get at
> any sink rate is weight times sink rate. The heavy glider
> gets more power, and has more fuel.
I don't get this. given that the glide angle for a ballasted and
unballasted glider is approximately the same, i.e., in a no-wind zero
lift day the two gliders will touch down at the same spot, what other
advantage is their other than speed? In fact, the heavier glider will
land sooner.
If both gliders H and L start their maneuvers at the same speed v and
same altitude (assume they start at zero altitude), and fly the same
trajectory and pull into the vertical to climb to height h, then the
energy equations for each are:
1/2 mv^2 - Fd = mgh
Where Fd is the work done against the glider by drag, over the distance
flown. F is a complex function of speed and lift, which is changing
throughout the profile.
to solve for the height reached, you get
(1/2)v^2/g - Fd/mg = h
The leftmost term is the same for heavy and light glider. The rightmost
differs between the two obviously. for a heavy glider, the term is
smaller, and if you only consider parasitic drag, which would be the
same for both gliders, the heavier would clearly go higher. When you add
in induced drag due to lift, it might get a little muddy. But I doubt
the increase in induced drag due to the pullup is that much more, and it
certainly does not occur for long. so the heavier should go higher. So
sayeth Sir Isaac Newton.
Kevin Neave
in velocity is the same for both gliders.
If you limit the two to the same speed at the end of
the pull-up of course the heavy glider wins. But due
to the lower stall speed the lighter glider has a few
more m/s change in v, therefore more change in h
:-)
At 14:12 02 October 2003, Nafod40 wrote:
>
>to solve for the height reached, you get
>
>(1/2)v^2/g - Fd/mg = h
>
>The leftmost term is the same for heavy and light glider.
>The rightmost
>differs between the two obviously. for a heavy glider,
>the term is
>smaller, and if you only consider parasitic drag, which
>would be the
>same for both gliders, the heavier would clearly go
>higher. >sayeth Sir Isaac Newton.
>
>
Jim Kelly
"Todd Pattist" <tpat...@DONTSPAMME.snet.net> wrote in message
news:ppnjnv8vdf6t9d4ke...@4ax.com...
|
| You add to that aerodynamic power any power required
| to climb. That power is climb rate times aircraft weight.
| The total of those two is power required to fly.
Hi Todd,
Would you mind extending this thought to declare a useful quote to
use in "Air Experience Flights" as to what "horsepower" we are
getting from a thermal showing 5 knots lift in a typical club
twin-seater (say a Janus B)?
Thanks,
Jim Kelly
Bruce Hoult
"Jim Kelly" <kell...@compuREMOVE-TO-EMAIL-MEserve.com> wrote:
That's pretty easy.
Call it 500 kg total mass, moving at 90 km/h (50 knots, 25 m/s), with an
L/D of 40, climbing at 2.5 m/s (5 knots).
Drag = 500 / 40 = 12.5 kgf = 125 N.
glide power = 125 N * 25 m/s = 6250 W = 4 HP
climb power = 5000 N (500 kgf) * 2.5 m/s = 12500 W = 16.75 HP
So total HP is about 20.
Change the numbers to suit your reality :-)
-- Bruce
szd41a
2.5 m/s, so is the boat. How many H.P has the boat?
BQ
"Bruce Hoult" <br...@hoult.org> a écrit dans le message de
news:bruce-CDB56F....@copper.ipg.tsnz.net...
nafod40
> An empty motorless 500 kg boat is drifting on a river. Current is moving at
> 2.5 m/s, so is the boat. How many H.P has the boat?
Power = Energy/Time = Force x Distance/Time, or Force x velocity. But no
force here, so no power.
szd41a
a heavier glider has more "climbing power" as stated in "climb power = 5000
N (500 kgf) * 2.5 m/s = 12500 W = 16.75 HP". Our problem was stated in still
air, and we should keep it this way. We all know that we are going down when
themalling, it's just that the air around is climbing faster, and we are
pumping fuel in the glider (PE). May we add that the energy spent in a
rising bubble of air is tremendous!!.(air density is around 2.5 #/ cubic
yard or 1.2kg/m3).
At this point in our debate, I suggest that both parties agree on the
following:
-ignoring drag, and for equal initial and final speed, it is a tie
-we deduct that drag is the force holding the" lower" glider (wow!!!!how
smart!! ;-)
-any results less than 15 m (50 feet) of difference declares a tie (what we
can read on the alti.)......less than that is not worth debating!!!
-from now on, we set entry speed at 55m/s (110 knts)and final speed at 25
m/s(50knts) (more room to set proof, pro-ballasted say around 100 feet
(30m) higher)
So far, Kevin gave us the best "math proof" declaring a tie (perhaps a hair
to the light glider). Theese results are very close to simulations done by
Udo Rumpf (what sofware???).
Last week-end, David and Chris (we salute them) flew their Ventus wings
abreast to run a reality check (one wet and one dry) for ten pull-ups.
Apparently, they were dolphining in cloud street (positive varios) and one
pull-up was estimated at 50 feet in favour of the wet. we have no mention of
average height achieved ( delta V: 100 knts-60knts).
So here we are, still debating and learning from it, I believe.
Todd's theory is based on the fact that, while coasting nose-up, gliders
will fly accordingly to their gliding polar, i.e. less form drag up to best
L/D speed =>more height.
We have difficulty resolving induced drag in the vertically eccelarated
pull-up phase. I am sure that the ballasted will suffer more there (F=Ma)
BQ
"Todd Pattist" <tpat...@DONTSPAMME.snet.net> a écrit dans le message de
news:6i6rnvgn638h826p9...@4ax.com...
> "szd41a" <girar...@videotron.ca> wrote:
>
> >An empty motorless 500 kg boat is drifting on a river. Current is moving
at
> >2.5 m/s, so is the boat. How many H.P has the boat?
>
> I think you want to ask how much HP it takes to move the
> boat. The answer would be "None." The boat is not rising
> against gravity, and not storing potential energy. It is
> not moving relative to the water and not dissipating energy
> in the form of fluid drag. It's the same as a drifting
> balloon at constant altitude.
Kevin Neave
>If the difference is 1 foot or 200', I'm
>interested in the answer. I've pulled up from 145
>knots
>ballasted and from that speed without ballast. I had
>the
>distinct impression, like the ballasted aerobatic pilot
>who
>posted, that the ballasted pullup was longer and higher.
How much higher?
How steep was the pull-up?
Did you have a straight line trajectory after the pull,
or a ballistic parabola?
What was your exit speed, ballasted & un-ballasted?
I'm pretty sure that at these speeds the heavy glider
wins, but I still don't think the margin is all that
great.
>I've looked at it enough now, to be reasonably confident
>that the heavy glider has an advantage everywhere except
>at
>the low speed stall regime. I've also concluded that
>getting numbers mathematically will be hard without
>a
>defined flight profile and a lot of work.
>
>
So how about defining a flight profile for us, the
original post way back when simply defined 100kts entry
& 100kgs ballast. I've suggested starting at above
100kts flying straight & level 'til the speed drops
back to 100 then pulling up to a constant 45deg climb.
Chris's flight tests were done from a normal glide
at 100kts then pulling up to 30deg
:-)
Fat Albert
Helmut Reichman, who kindly signed my copy of Cross Country Soaring
In the conversation I mentioned that the benefit from his advice in CCS to
'always cross the start line with your sailplane at max speed and max
gross*, pull up to your anticipated/estimated climb derived McCready speed
for that wing loading, glide to your first climb where you dumped ballast to
your anticipated ideal wing loading for the day'
was practically all derived from the cruise to the first thermal
After some discussion we agreed that in the real world of 1G pull ups the
only benefit was that obtained by a heavier sailplane due to its lower sink
rate at higher speeds during the time it took to decelerate from initial
cruise to exit cruise (say 130 kts to 80kts) and then its lower sink rate
during the time it took to reach the first climb at the estimated McCready
speed.
A quick calculation gives a deceleration time < 5 seconds and an average
lower sink rate < 2ft/sec giving a benefit of < 10ft
Wow, what a lot of hot air expended to find that, in a pull up the heavier
glider will always end up higher but by such a small amount as to make this
debate an intellectual exercise
*at the time with ground based observed start lines this meant crossing a
3000ft start line as close to VNE as you dared, definately not PC today, hey
JJ?
"Todd Pattist" <tpat...@DONTSPAMME.snet.net> wrote in message
news:6i6rnvgn638h826p9...@4ax.com...
> "szd41a" <girar...@videotron.ca> wrote:
>
> >An empty motorless 500 kg boat is drifting on a river. Current is moving
at
> >2.5 m/s, so is the boat. How many H.P has the boat?
>
Ewald Bombelka
> OK people, what was the verdict.
> I'm sure some logger equipped pullups were made.
> Who wins?? Wet or dry.
> I still think wet pullups go higher, but I can't prove it.
>
> Scott.
Let's continue a bit with physics and math:
I solved the equation of motion for an ascending flight path of a
constant angle with the following parameters:
mass of the glider = 325 and 525kg (= 750 and 1160 lbs)
drag coefficient = 0.015 (constant with velocity)
wing area = 10.5 m^2
start velocity = 185 km/h (100kt)
final velocity = 110 km/h (60kt)
density of air = 1 kg/m^3
gravit.acceleration= 10 m/s^2
I did not account for the initial rotation from normal glide path into
the ascending flight path and I did not account for the subsequent
rotation back into normal glide.
Results for an ascending constant glide path angle of 10°
for the 325kg glider compared to the 525kg glider:
The heavier glider can go up with an angle of 10.8° in order
deaccelerate in the same time (9s).
Both gliders reach the final velocity after 380m flight path.
The 325kg glider climbs to 65m,
however the heavier 525kg glider reaches 70m.
The reason for this is, that the second deaccelarating term in the
equation of motion is: drag/mass, when resolved to acceleration; hence
deaccelaration due to drag is less for the heavier glider.
Robert has already given a vivid explanation for it.
This result (altitude difference=5m=16ft) is less than the observed
altitude difference for the two Venti
(where is the discussion thread ?).
Ewald
Cliff Hilty
wrong constant. All of these base the constant on airspeed.
I am not a mathematition (I can't even spell it) but
if we base it on sink rate it should be alot more
obvious. After all thats exactly why we carry water
ballast in the first place, to increase speed for the
same amount of altitude loss right? So if your going
10 knots faster in the ballasted glider and at the
same sink rate as the unballasted glider. You will
gain more in pullup, right?
Cliff
At 21:00 07 October 2003, Ewald Bombelka wrote:
szd41a
">
> http://www.tux.org/~milgram/papers/alds_auvsi2003.pdf
>
> This paper was referred to me by an aerodynamics
> professional/glider pilot. I hope he won't mind if I post
> this limited excerpt from his email to me:
>
> "I notice you carrying the torch for good physics on r.a.s.,
> more power to you. But don't fret over it (especially not on
> r.a.s.), since as you've noted you have to look at the
> energy losses to see any difference, and no analysis can
> answer this without incorporating some kind of trajectory
> analysis. That would have to include some kind of
> representation of the glider's long-period longitudinal
> dynamics (a.k.a. phugoid dynamics), and a way to optimize
> the pullup for each given weight. Plus you have to decide
> what the rules are (pullup from same airspeed, pullup from
> weight-appropriate MacCready speed, etc.).
>
> Todd Pattist - "WH" Ventus C
> (Remove DONTSPAMME from address to email reply.)
We can appreciate that very intelligent view on the question. In no ways, it
denies the fact that in our system, no energy is created, so ultimaly we
will have to balance the conservation of energy theorem. This is simple:
kinetic energy, potential energy and drag. Not matter what trajectory you
will choose, you will have to balance the equations even if you use phugoid
dynamics. There are no rule here for trajectory, just go for max height!!!
We did the maths, and we look at our total energy probe (i.e vario), we
notice we're trashing energy in a pull-up (if the probe is well calibrated),
otherwise in still air, nothing is gained.
This is going way further than I expected!!! We had all kinds of comments,
this one from a mathematician CFI who conclude that ballasted will go twice
as high, another from an outsider power pilot who candidly state that he
noticed that pull-ups with a load (passengers) were more demanding (we all
know that power pilots are no physicians ;-)... and that soaring is an art
that defy laws of physics) and all kind of rewritten equations that give
a substantial advantage to the ballasted. Don,t get me wrong, we are talking
one hundred feet difference on an avergage 300'+ pull-up!!!!! Now, that is a
lot of drag!!!!
We sill expect a probant mathematical proof, or a reality check in still
air.
BQ