Recovery

Let's assume a disaster strikes.  You have an incredibly efficient logistics team, who have access to large ships, air freight, and helicopters capable of dropping supplies into any location needed.

Assume you pack everything in a 20' shipping container.  This gives you 33.1 m^3 of volume, and a net load of 28,200kg.  What do we need to pack into this?

1. Water.  1 m^3 of water weighs 1000 kg, and contains 1000 liters.  Woo metric system.  ($6.20 / 12L)
2. Food. Let's go with MREs, and this page, so a case is 12 MRE/case, which weighs 20lbs and is 16"x11.5"x10".  Metricize, and we find that 12MRE = 9.07185 kg and 0.0301522 m^3. ($60/case)
3. Blankets. Let's get some of those space blankets.  A case of these contains 200, 0.2lb each, case size 17.5"x9"x10". ($1.35ea)  Let's do regular blankets too.  6"x6"x15", 3.6lb, $17.
4. Other supplies.  Let's say we want to reserve 25% of the container for medical supplies/generators/communication equipment/etc.

Ok, now solve.  One person needs about 2 liters of water a day under regular circumstances.  Let's bump this to 10L/day because you're rebuilding civilizationStaten Island.  One person also requires 3 meals a day, and should probably have one space blanket and two regular blankets.  Let's bump that to four regular blankets, and assume the excess represents extra clothing, etc.  This gives us the equations:

Usage = [W, M, S, B] = [10 * D * N, 3 * D * N, 1 * N, 4 * N] 

Weight (kg)  /unit   = [1        , 0.75599  , 0.090718, 1.6329]

Volume (m^3) /unit   = [0.001    , 0.0025127, 1.2905e-4,0.008849]

W = k1 * W_w + k2 * W_m + k3 * W_s + k4 * W_b

V = k1 * V_w + k2 * V_m + k3 * V_s + k4 * V_b

N_people = k4 / 4 = k3 => 4 * k3 = k4

         = k2 / (3D) = k1 / (10D) => k2 = (3/10) k1

W = D * N * (10 * W_w + 3 * W_m) + N * (W_s + 4 * W_b)

V = [...]

[12.267970 6.622318;0.017538 0.035525] * [N * D ; N]

Ok, I come up with plausible solutions:

N  D  Weight     Volume   W / Wc W / Wm   V / Vc V / Vm

58 30 21382.3622 32.5766 [1.0110 0.7582] [1.3122 0.9842]

60 29 21383.6069 32.6476 [1.0110 0.7583] [1.3151 0.9863]

62 28 21335.7796 32.6485 [1.0088 0.7566] [1.3151 0.9864]

64 27 21238.8805 32.5793 [1.0042 0.7532] [1.3124 0.9843]

65 27 21570.7380 33.0883 [1.0199 0.7649] [1.3329 0.9996]

67 26 21412.4990 32.9314 [1.0124 0.7593] [1.3265 0.9949]

69 25 21205.1882 32.7043 [1.0026 0.7520] [1.3174 0.9880]

72 24 21243.8591 32.8635 [1.0044 0.7533] [1.3238 0.9929]

75 23 21208.9221 32.9174 [1.0028 0.7521] [1.3260 0.9945]

Where Wc and Vc are the 75% critical limits, and Wm and Vm are the absolute container limits.

How much is this going to cost?  Let's take the maximal people solution.  This means we need:

75 * 23 * 10 L of water      @ $6.20 / 12 ea = $8912.5

75 * 23 * 3  MRE             @ $60 / 12 ea   = $25875

75           space blanket   @ $1.35 / 1 ea  = $101.25

75 * 4       regular blanket @ $17   / 1 ea  = $5100

1            shipping container @ $3000 / 1ea= $3000

============================================== $42988.75

So, for something in the $50000 range, you can support 75 people for three weeks.  Ok, so we can't solve all of Staten Island's problems with these, as supporting its population of 470467, you'll need 6300 containers, at a cost of about a third of a billion dollars.  Still, this seems like the kind of thing the government could just stockpile in a few places, put them on a train towards where it looks like disaster might be happening, and drop them into place as needed.  Having a dedicated recovery unit that is ready to go anywhere at anytime, the initial recovery impact is likely to be more dramatic, providing a more efficient (and PR popular) recovery.