, prompting my friend to wonder, "what is the probability of two submarines colliding in the ocean?" Ignoring that we now know the true probability is 1, we can estimate the probability of collision by calculating the ratio of the volume of water through which the world's fleet of submarines travels in, say, 30 years to the total volume of the earth's oceans.
Fortunately, the news article, Wikipedia, and the rest of the internet have most of the values we need:
- (the smaller of the two crashed submarines)
- Displacement ≈ 15,000 tons
- Speed ≈ 46 km/h
- Length ≈ 140 m
In 30 years (2.6×105 hours), one submarine will travel about 46 km/h × 2.6×105 h ≈ 1.2×107 km. That is about 8.5×107 times its own length. One submarine displaces 15,000 tons (1.36×10-5 km3) of water, so during that time it will travel through a volume of about 1.36×10-5 × 8.5×107 = 1,178 km3. With 250 such submarines, the total volume traveled comes to 2.94×105. The ratio of that to the volume of the world's oceans is 2.94×105 / 1.3×10^9 = 2.3×10-4.
Therefore, the probability that a fleet of 250 randomly-traveling Le Triomphant-sized submarines will have at least one collision in 30 years is around 1 in 5000.
This assumes that the average displacement of all subs deployed during those 30 years is roughly equivalent to 250 Le Triomphant class-sized subs. I have no idea if that is a valid assumption because I was unable to find a definite number for how many submarines are currently active, much less the year-to-year count or average size. Does anyone want to scrape to aggregate the average displacement and deployment numbers by year? Also, the number of submarines has decreased since the end of the cold war.
Check my math with the following Python script.
print "Q. What is the probability of two subs accidentally colliding in the ocean?"
print " http://www.dailymail.co.uk/news/worldnews/article-1146124/"
def years_to_hours(years):
days_per_year = 365.24
hours_per_day = 24.0
return years * days_per_year * hours_per_day
def tons_to_cubic_km(tons):
lbs_per_ton = 2000.0
liters_per_lb = 1.0 / 2.2 # of water
cubic_km_per_liter = 1.0e-12
return tons * lbs_per_ton * liters_per_lb * cubic_km_per_liter
ocean_volume = 1.3e9 #km^3 http://hypertextbook.com/facts/2001/SyedQadri.shtml
number_of_subs = 250.0 # http://en.wikipedia.org/wiki/Nuclear_navy
travel_time = years_to_hours(30.0)
# Submarine Characteristics
# http://en.wikipedia.org/wiki/Le_Triomphant_(S_616)
# http://en.wikipedia.org/wiki/HMS_Vanguard_(S28)
sub_displacement = tons_to_cubic_km(15000.0)
sub_speed = 46.0 #km/h
sub_length = 0.14 #km
# calculation
travel_distance = sub_speed * travel_time
volume_traversed = (travel_distance / sub_length) * sub_displacement
fleet_volume = number_of_subs * volume_traversed
ratio = fleet_volume / ocean_volume
print "A. %g" % ratio
Update
says the following on the mailing list on which I originally posted the calculation:
nice! I don't know if the analysis is right, but it gives us an upper bound at least!
Brett's calculation is equivalent to the following: set out 250 subs that randomly teleport to cover the right volume. None of the subs ever visits space that it or any other sub has been before. Now, what's the probability of choosing a spot that a sub has been if I were to randomly teleport a new sub into the ocean after 30 years?
I imagine this will over estimate a bit.
Since the total volume of the subs is much much smaller than the volume of the ocean, I think we can safely ignore "overlapping" yet non-colliding paths. Other than that, I think it is a good characterization.