OK Folks here we go with the maths, including that
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 !!

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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

:-))