Given that there are 350 liters of helium at Pole on June
30th, 2000, we could:

(a) run both SPARO and AST/RO for 4.6 weeks.

(b) warm SPARO now, and run AST/RO for 8.3 weeks.

(Warming AST/RO does not help SPARO much, but each day we
run SPARO costs AST/RO approximately one day of observing

Here are the detailed calculations that I did to get these


We assume SPARO uses 5.5 liters/day.  This is based on
actual experience at Pole this season.  This loss rate
includes the transfer losses in filling SPARO but does not
include the boiloff from the storage dewars sitting on the
roof of MAPO.  The boiloff from these dewars is a cost of
storing Helium, not specifically a cost of running SPARO.
(SPARO can run off of either 100's or 250's.)

We assume AST/RO uses 2 liters/day.  This comes from Tony
Stark.  I assume that this includes transfer losses, but,
again, not losses from the 250 liter boiloff.

We assume that a 250 liter loses 3 liters/day, and a 100
liter loses 2 liters/day.  This is due to boiloff from these
dewars even when nothing is withdrawn.  These estimates are
from Tony Stark and they look about right to me.  The 100
liter boiloff cannot be much higher or Greg would not get 2
weeks of SPARO operation out of one 100 liter, which is what
he has been doing.

This is the calculation:

let t = 0 be the time when 3000 gallon runs dry.
let t_h = time when station runs out of Helium
let t_s = time when SPARO is warmed
let t_a = time when AST/RO is warmed

let H(0) = amount of Helium on station when 3000 gallon runs dry.
let H(t) = amount of Helium on station as a function of time
then H(t_h) = 0

let B_h(t) = bioloff rate of storage dewars 
let B_s(t) = boiloff rate of SPARO (= 5.5 for t < t_s; 0 for t > t_s)
let B_a(t) = boiloff rate of AST/RO (= 2.0 for t < t_a; 0 for t > t_s)

Here is a key assumption:  as you lose helium, the loss rate
of the storage dewars goes down.  I will assume it goes down
linearly with time, though this is an approximation.  This
is due to some storage dewars running dry:

B_h(t) = B_h(0) [ 1 - (t/t_h) ]

OK, now do the calculation:

(Note we will assume that B_h(0) is 8, because two 250's
and one 100 have Helium, and we are assuming boiloff from
250 is 3 liters/day and boiloff from 100 is 2 liters/day.)

dH/dt = -B_h(t) - B_s(t) - B_a(t)

integrate from 0 to t_h to get

0 = H(0)  -  B_h(0) x t_h x (1/2)  -  B_s(0) x t_s  -  B_a(0) x t_a

solve for t_h for three cases:

case a: Operate both to the end: t_s = t_a = t_h

t_h =  H(0) / { 0.5 x B_h(0) + B_a(0) + B_s(0) } 
    =  350  / {4 + 2 + 5.5}  = 32 days 

case b: Operate only AST/RO: t_a = t_h; t_s = 0

t_h =  H(0) / { 0.5 x B_h(0) + B_a(0) }
    =  350  / {4 + 2}  =  58 days

case c: Operate only SPARO: t_s = t_h; t_a = 0

t_h =  H(0) / { 0.5 x B_h(0) + B_a(0) }
    =  350  / {4 + 5.5}  =  37 days

So warming AST/RO does not help SPARO much.  But warming
SPARO will help AST/RO a lot.  One way to state this
trade-off is to estimate the loss of AST/RO observing days
for each day that we keep SPARO cold.

There are two methods Tony and I have used to estimate this.
Both give about 1 day of AST/RO observations lost for each
day of SPARO observations gained.  The details are:

(method 1) Go back to 

0 = H(0)  -  B_h(0) x t_h x (1/2)  -  B_s(0) x t_s  -  B_a(0) x t_a

Assume t_a = t_h, solve for t_h, and differentiate, to get:

d(t_h)/d(t_s) = -  B_s(0) / { (1/2) B_h(0)  +  B_a(0) }

d(t_h)/d(t_s) = - (5.5) / {4 +2} = 0.9

(method 2)

Imagine a situation where there is one 250 liter dewar full,
and both AST/RO and SPARO running.  Then, if SPARO runs to
the end, the duration is

250 liters/(3 liters/day + 2 liters/day + 5.5 liters/day) = 24 days

If SPARO is not run, the duration is

250 liters/(3 liters/day + 2 liters/day) = 50 days

So the cost is:  26 days of AST/RO lost

And the gain is: 24 days of SPARO gained.

Thus each day of SPARO observations costs AST/RO 1.1 (=26/24) 
days of data.

So, combining the methods: d(t_h)/d(t_s) = 0.9-1.1

Thus each day of SPARO operation from here on out costs
AST/RO about one day of observation.