Question of the day

You have just destroyed your own conclusion. The light one will damp out
faster and eventually stop sooner.

The weight acts as thrust, not just mass.

Weight acts like thrust when you and your load are travelling towards the
center of the earth. When you try to carry you load away from Mother Earth,
I am nt too sure it is helping you along!! Does that sound right?????Maybe I
put my foot in my mouth here. HELP ME SOMEBODY I AM GOING NUTS

.......but i am having fun. Maybe we are setting a record for the longest and
wittiest sring on rec.avistion ;-))))))

Kevin Neave wrote:

One trivial point for Jere's post is that the stalling
speed for the unballasted glider will be lower, so
this guy can pull up to a slower speed & therefore
regain more altitude.

That's right. I left that thing out just to make the calculation simpler.

Lots of discussion about different initial and final speeds and their
effects. Having two gliders with different masses, different initial
speeds and different final speeds makes things complicated. Adding drag
to that makes it even worse. We would need the polar curves for
different loads. Taking the dynamic behavior (1g pull-up and possibly
0g flight path) into account would require the polar curve (or surface)
from 0g to Ng. It would also take some simulating as well I guess.

I'm not going that far. But leaving the drag out of the question we can
get some simple results. Starting with the good ol' conservation of energy:
1/2*m*v1^2 + m*g*h1 = 1/2*m*v2^2 + m*g*h2
This time, dividing by m cancels it out completely, as we've seen
multiple times. So:
1/2*v1^2 + g*h1 = 1/2*v2^2 + g*h2
For the two gliders that is:
(1) 1/2*v01^2 + g*h0 = 1/2*vf1^2 + g*hf1
(2) 1/2*v02^2 + g*h0 = 1/2*vf2^2 + g*hf2
v is speed, where subscript 0 means initial, f final and number is the
glider.
substracting (1)-(2):
1/2(v01^2 - v02^2) = 1/2(vf1^2 - vf2^2) + g(hf1 - hf2)
that is
hf1 - hf2 = 1/2g * (v01^2 - v02^2 - vf1^2 + vf2^2)
But it's kind of hard to see what's happening there. So say the initial
speed of glider 2 is a times the speed of glider 1:
v02 = a*v01
and similarly for final speeds:
vf2 = b*vf1
So we get:
hf1 - hf2 = 1/2g * ((1 - a^2)*v01^2 + (b^2 - 1)*vf1^2)
Then, say the speed change for glider 1 from initial to final is c:
vf1 = c*v01
So:
hf1 - hf2 = 1/2g * ((1 - a^2) + (b^2 - 1)*c^2)*v01^2
that is:
hf1 - hf2 = v01^2/2g * (c^2*b^2 - a^2 - c^2 + 1)
Reality check:
If c=1 (glider 1 has constant speed) increasing a (glider 2 faster in
the beginning) makes the altitude difference negative (glider 1 is
lower). Increasing b (glider 2 faster in the end) makes the difference
positive (glider 1 is higher).
Seems OK.
When c 1 (so at least glider 1 pulls up to a slower speed) a^2 has a
bigger factor (1) than b^2 (1).

What that means:
20% more speed initially has more effect than 20% in the end.
So if glider 1, starting with speed 150 km/h, pulling up to 70 km/h and
glider 2, starting with 180 km/h (150 km/h +20%), pulling up to 84 km/h
(70 km/h +20%) are competing, number 2 gets higher.

Some people have been comparing two gliders flying at Vne. In that case
the lighter wins since it's able to pull up to a slower speed. Plus, I'm
not sure about what different glider manufacturers say about flying at
Vne with full ballast. Might be OK, since the weight and the lift are
both in the wings...

Jere
jere at iki.fi

Eric Greenwell wrote:
The heavier glider is flying at a higher L/D than the lighter one, so

I don't think that can be right.
The heavier glider will achieve the same max L/D ratio than the lighter
one, but at a higher speed.

Jere
jere at iki.fi

In article , says...
Eric Greenwell wrote:
The heavier glider is flying at a higher L/D than the lighter one, so

I don't think that can be right.
The heavier glider will achieve the same max L/D ratio than the lighter
one, but at a higher speed.

It's true they will have the same max L/D ratio; however, at any speed
above the ballasted glider's max L/D, the ballasted glider will have
an L/D better than the unballasted glider (same speed, lower sink
rate). That L/D will be lower than the max L/D, of course, since it is
flying faster the max L/D speed.

Take a look at a polar plot for a glider that shows the both the
unballasted and ballasted polars, compare them for the same speed, and
you will see what I mean.
--
!Replace DECIMAL.POINT in my e-mail address with just a . to reply
directly

Eric Greenwell
Richland, WA (USA)

Eric Greenwell wrote:
It's true they will have the same max L/D ratio; however, at any speed
above the ballasted glider's max L/D, the ballasted glider will have
an L/D better than the unballasted glider (same speed, lower sink

Oh yes, of course. On the other hand, at a speed slower than max L/D
speed for the lighter, the sink rate will be smaller for the lighter
glider. And the gliders will have the same sink rate at a speed
somewhere between their max L/D speeds.

So, during a manouver where two gliders pull up from a speed greater
than max L/D speed for the heavier one to a speed slower than max L/D
speed for the lighter, the altitude difference is a function of the
speed profile during the manouver. It's quite hard to compare the
altitude gains without heavy calculations, but I wouldn't say the effect
would clearly benefit either glider. What it _does_ say, in my opinion,
is that the pilot should probably only pull up to the max L/D speed or
minimum sink speed (depending on the situation) for _his_ glider.

Jere

jere at iki.fi