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 time.) Here are the detailed calculations that I did to get these figures: __________ 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 This is due to boiloff from these dewars even when nothing is withdrawn.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. ____________________