Question of the Day

Since mass is a constant factor on both sides of the equation, it cancels out. Therefore there should theoretically be negligible difference in the pullup altitude gained between the ballasted and unballasted cases.True except for two things:The ballasted glider has more induced dragwhile at the same airspeed as the unballasted oneThe ballasted glider also has a higher stall speedSo the unballasted glider will go higherMark Boyd

Mark.
What about the L/D polar being skewed to the right to the benefit of the
higher
wing loaded vehicle. At VNE the heavier a/p is cleaner and will glide
farther.
If both a/p's pull up together, the cleaner a/p runs out of energy last. It
looks to me like
heavier climbs further.....

Scott


"M B" wrote in message
...
Since mass is a constant factor on both sides of the equation, it
cancels out. Therefore there should theoretically be negligible difference
in the pullup altitude gained between the ballasted and unballasted
cases.True except for two things:The ballasted glider has more induced
dragwhile at the same airspeed as the unballasted oneThe ballasted glider
also has a higher stall speedSo the unballasted glider will go higherMark
Boyd

Indeed you are correct that at high speed the ballasted
Glider 'bleeds' height (not energy) more slowly. That's
why we fly with ballast when we're cruising across
country, flying at higher speeds for longer periods

However if we look at that nice man Mr Johnson's test
flight of the Discus we find that carrying 183lbs of
ballast reduces the sink rate at 100kts from 3.3 m/s
to 2.24m/s.

If we pull up into a 45deg climb our velocity will
reduce at about .7g, i.e about 7m/s/s

So... if we're slowing from 100kts (50 m/s) to 40kts
(20 m/s)this will take about 30/7 seconds (i.e about
4)

Even if our ballasted glider could maintain it's sinkrate
advantage for the whole period we'd gain less than
5 metres.

At 12:36 09 September 2003, Scott Correa wrote:
Shouting is unbecoming a gentleman..................

Somehow I don't think you understood what I said.
Every test I have seen published shows the max L/D
point moving to the right (ie occuring at a higher
speed)
with an increase in wing loading. The sink rate curves
do the same thing. So again I ask, doesn't the heavier
airplane bleed energy more slowly..................

This has nothing to do with starting the engine......

Oh Yeah I also forgot to mention that although you
cannot
create energy, you can add it to the glider by flying
in air
going up faster than you are sinking thru it......................
...

Last time I looked at total energy systems, it read
airspeed
(kinetic energy) and barometric pressure (potential
energy)

Scott



'szd41a' wrote in message
.. .
YOU CANNOT CREATE ENERGY UNLESS YOU FIRE YOUR ENGINE!!!!!!!
'Scott Correa' a écrit dans le message de
...
Mark.
What about the L/D polar being skewed to the right
to the benefit of the
higher
wing loaded vehicle. At VNE the heavier a/p is cleaner
and will glide
farther.
If both a/p's pull up together, the cleaner a/p runs
out of energy last.
It
looks to me like
heavier climbs further.....

Scott


'M B' wrote in message
...
Since mass is a constant factor on both sides of
the equation, it
cancels out. Therefore there should theoretically
be negligible
difference
in the pullup altitude gained between the ballasted
and unballasted
cases.True except for two things:The ballasted glider
has more induced
dragwhile at the same airspeed as the unballasted
oneThe ballasted
glider
also has a higher stall speedSo the unballasted glider
will go
higherMark
Boyd

The heavier glider will get more altitude.But not very much...If the ballasted glider stalls at exactly 100 knots,it cannot gain any altitude, while the unballastedglider will gain some altitude. Thereforethere is at least one case where the unballastedglider will outclimb the ballasted one.The proposed equations I have seen do not account for this case and must therefore be insufficient.

Yeah, I was wrong, the mass does cancel out.

My mommy and daddy.



Jim Vincent
CFIG
N483SZ

It will gain more height with ballast. The
kinetic energy is defined as 1/2*m*v squared.
[...]
The potential energy is m*g*h,
[...]
So for example, if a gldier weighs twice as much, it will
gain twice the height, or at least I think so!

Again, take out the constants. Both aircraft have the same velocity at the
beginning, 100Kts. Assume, for the sake of argument, that they have the same
velocity at the end, say 30 kts (I know the heavier one will stall first, but
in a vertical pull up, the wing loading is zero, so the stall speed would be
very close).

SO at the beginning, the delta in kinetic energy for two ships travelling the
same speed is only proportional to the mass. Since the heavier one weighs more,
it has more kinetic energy. At the end of the pull up, when all the kinetic
is converted to potential, take out the constants again (g), and the only
remaining variable is h. h is proportionally more for the heavier ship. And,
as I said before, this is not accounting for drag.

P.S. I f'in hate calcusus. R dR d theta double dot!
Jim Vincent
CFIG
N483SZ

"Jim Vincent" wrote in message
...
It will gain more height with ballast. The
kinetic energy is defined as 1/2*m*v squared.
[...]
The potential energy is m*g*h,
[...]
So for example, if a gldier weighs twice as much, it will
gain twice the height, or at least I think so!

Again, take out the constants. Both aircraft have the same velocity at
the
beginning, 100Kts. Assume, for the sake of argument, that they have the
same
velocity at the end, say 30 kts (I know the heavier one will stall first,
but
in a vertical pull up, the wing loading is zero, so the stall speed would
be
very close).

SO at the beginning, the delta in kinetic energy for two ships travelling
the
same speed is only proportional to the mass. Since the heavier one weighs
more,
it has more kinetic energy. At the end of the pull up, when all the
kinetic
is converted to potential, take out the constants again (g), and the only
remaining variable is h. h is proportionally more for the heavier ship.
And,
as I said before, this is not accounting for drag.

P.S. I f'in hate calcusus. R dR d theta double dot!
Jim Vincent
CFIG
N483SZ


Wrong again. If you can't do the math right, at least stop doing it in
public.
"The only remaining variable is" not h, it is mh. mh is "proportionally
more for the heavier ship"; h is the same.

Wrong again. If you can't do the math right, at least stop doing it in
public.

Yo' daddy
Jim Vincent
CFIG
N483SZ

Since mass is a constant factor on both sides of the equation, it
cancels out.

You need to compare one equation to the other. The masses are different. Yes,
for one glider, the masses cancel out, but not when comparing two different
masses, cetarus parabus.
Jim Vincent
CFIG
N483SZ

"Jim Vincent" wrote in message
...
Since mass is a constant factor on both sides of the equation, it
cancels out.

You need to compare one equation to the other. The masses are different.
Yes,
for one glider, the masses cancel out, but not when comparing two
different
masses, cetarus parabus.
Jim Vincent
CFIG
N483SZ



You don't know what you are talking about.