World War II by country Puns

Wikipedia:

> The U.S. dropped 2,756,941 tons of ordnance on 113,716 Laotian sites in 230,516 sorties between 1965 and 1973 alone. By September 1969, the Plain of Jars was largely deserted.

> U.S. aircraft dropped more ordnance on Laos than on all countries during World War II, leaving Laos with about 78 million pieces of unexploded ordnance (UXO) by the end of the war. Casualties continue to mount from UXO dropped by the U.S. and Laotian Air Forces from 1964 to 1973. It has been reported that, between 1964 and 1973, areas controlled by the invading communist North Vietnamese and Pathet Lao were hit by an average of one B‑52 bomb-load every eight minutes, 24 hours a day. Xiangkhouang Province was the most heavily bombed province. Thirty percent of bombs failed to explode immediately.

The citation for the 78 million is to this PDF: Khamvongsa, Channapha; Russell, Elaine (2009). "Legacies of War: Cluster Bombs in Laos"

This is jaw-dropping, and raises many questions.

1. Was a 30% failure-to-explode rate considered normal? How did this rate compare to rates in previous decades (e.g., WW2) or later ones (e.g., Persian Gulf War)?

2. Is the figure of 78 million unexploded bombs a widely accepted estimate?

3. Given some calculation (see below), and assuming the numbers in the quoted passage are accurate, we can infer either (a) a low number of planes per mission along with a high number of bombs per plane (like 5 & 226), (b) a high number of planes per missions along with a low number of bombs per plane (like 200 & 6), or (c) somewhere between the two. Which is closest to the truth?

Calculations

To bastardize the Drake equation, to achieve u = 78 million unexploded bombs we can multiply together:

• Number of missions m: 230,516 (over nine years this implies 70/day on average)

• Average number of planes per mission p: ?

• Average number of bombs per plane b: ?

• Probability that a bomb will fail to explode f: 30%

The two parameters to be estimated bear the relationship b = u/(m*p*f), or, equivalently, p = u/(m*b*f). Plugging in the known values of u, m, and f, we have:

1. b = 78,000,000 / ( 230,516 * p * 30%)

2. p = 78,000,000 / ( 230,516 * b * 30%)

In other words, allowable pairs