Did we decide if ballasted pullups are higher????????

"Kevin Neave" k 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.

Hi Todd
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??

Todd Pattist wrote:

Robert Ehrlich 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.

But one extreme case is starting at stall speed for the
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.

Todd Pattist wrote:
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 a

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.

"Todd Pattist" wrote in message
news |
| 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

In article ,
"Jim Kelly" wrote:

"Todd Pattist" wrote in message
news |
| 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)?

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

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?
BQ
"Bruce Hoult" a écrit dans le message de
...
In article ,
"Jim Kelly" wrote:

"Todd Pattist" wrote in message
news |
| 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)?

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 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?

Power = Energy/Time = Force x Distance/Time, or Force x velocity. But no
force here, so no power.

I think we all agree on this. I wanted to make sure that no one thinks that
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" a écrit dans le message de
...
"szd41a" 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.
Todd Pattist - "WH" Ventus C
(Remove DONTSPAMME from address to email reply.)

Early in my soaring education I was fortunate enough to share a coffee with
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" wrote in message
...
"szd41a" 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.
Todd Pattist - "WH" Ventus C
(Remove DONTSPAMME from address to email reply.)