5 ms·

Hillwalkers should love the Comte de Buffon (2020)

89 points by ColinWright 5y ago

IanCal 5y ago
Really nice post, hits a few sweet spots * Practically useful * Neat bit of approachable maths * Surprising result
Luc 5y ago
For kilometer people: Length of path in km= 0.8 * number of gridlines crossed (Because pi/4 = 0.785 = ~0.8)
deleted 5y ago
[deleted]
skywal_l 5y ago
As the formula depends on pi, it means you can estimate pi by droping needles on the floor. That's called monte-carlo.
EdwardDiego 5y ago
Heavily terrain dependent, this. You need the low rolling hills, at most, for length in miles to be useful. As "length of a tramp" based on a 2D grid doesn't account for ascent/descent if needed. I've met hills I can ascend in 5 hours, and hills that took 9 hours, and both were the same gain in height. In my country, backcountry travel is best described by time, not kilometres.
- KineticLensman 5y ago
  Yes. When I used to hike in the UK we used Naismith's rule [0] to figure out the additional time due to ascents / descents. We didn't generally climb or scramble and IIRC the rule gave us a likely minimum time, not accounting of course for breaks, really steep bits, goofing around, etc. [0] https://en.wikipedia.org/wiki/Naismith%27s_rule https://en.wikipedia.org/wiki/Naismith%27s_rule
  - ghaff 5y ago
    There certainly are situations where the terrain is unusually challenging which throws off any formula. (Also conditions like snow.) But, yes, in general a combination of distance and elevation gain usually gives you a reasonable approximation.
  - vincebowdren 5y ago
    Naismith must have been a hard mountaineer; I can't keep up with his timings, certainly not in Scottish hill country.
  - EdwardDiego 5y ago
    We have similar rules of thumbs in New Zealand, ours are a bit more generous though, given how much of our ascent is very steep. On a track, Igenerally estimate 1 hour for 300 meters ascent, 1 hour for 600 meters descent. But in the bush, with no tracks but animal tracks, in the North Island or on the West Coast of the South Island, where the forest is denser, a good rule of thumb is 1 hour to ascend 100m, 1 hour to descend 200m. In the eastern beech forests, which are typically more open, 1 hour for 200m up, 1 hour for 300m down is realistic. The big issue of course, is that descending in untracked bush is a great way to find impassable waterfalls or bluff lines that require sidling or ascending to a ridge to bypass.
georgecmu 5y ago
It's the problem #28 in this set: https://news.ycombinator.com/item?id=27884973 https://news.ycombinator.com/item?id=27884973
murbard2 5y ago
This is missing an interesting argument regarding "Buffon's noodle". A straight needle crosses a gridline with probability (4 / pi) (l / t) where l is its length. Surprisingly, a flexible noodle of length l also crosses an expected (4 / pi) (l / t) gridlines. To see why consider that the noodle can be approximated by a set of N straight segments of length (l / N). While the events of different segments crossing a gridline are correlated, the expected number of crossing for each segment is (4 / pi) (l / (N t)) and since, expectation is linear, the expect number of crossings for the noodle is still (4 / pi) (l / t).
dougSF70 5y ago
This is quality.
- dougSF70 5y ago
  Also, pi is an amazing construct. It is evidence of so much magic in the universe.

89 points by ColinWright 5y ago

IanCal 5y ago

Really nice post, hits a few sweet spots * Practically useful * Neat bit of approachable maths * Surprising result

Luc 5y ago

For kilometer people: Length of path in km= 0.8 * number of gridlines crossed (Because pi/4 = 0.785 = ~0.8)

deleted 5y ago

[deleted]

skywal_l 5y ago

As the formula depends on pi, it means you can estimate pi by droping needles on the floor. That's called monte-carlo.

EdwardDiego 5y ago

Heavily terrain dependent, this. You need the low rolling hills, at most, for length in miles to be useful. As "length of a tramp" based on a 2D grid doesn't account for ascent/descent if needed. I've met hills I can ascend in 5 hours, and hills that took 9 hours, and both were the same gain in height. In my country, backcountry travel is best described by time, not kilometres.

KineticLensman 5y ago
Yes. When I used to hike in the UK we used Naismith's rule [0] to figure out the additional time due to ascents / descents. We didn't generally climb or scramble and IIRC the rule gave us a likely minimum time, not accounting of course for breaks, really steep bits, goofing around, etc. [0] https://en.wikipedia.org/wiki/Naismith%27s_rule https://en.wikipedia.org/wiki/Naismith%27s_rule
- ghaff 5y ago
  There certainly are situations where the terrain is unusually challenging which throws off any formula. (Also conditions like snow.) But, yes, in general a combination of distance and elevation gain usually gives you a reasonable approximation.
- vincebowdren 5y ago
  Naismith must have been a hard mountaineer; I can't keep up with his timings, certainly not in Scottish hill country.
- EdwardDiego 5y ago
  We have similar rules of thumbs in New Zealand, ours are a bit more generous though, given how much of our ascent is very steep. On a track, Igenerally estimate 1 hour for 300 meters ascent, 1 hour for 600 meters descent. But in the bush, with no tracks but animal tracks, in the North Island or on the West Coast of the South Island, where the forest is denser, a good rule of thumb is 1 hour to ascend 100m, 1 hour to descend 200m. In the eastern beech forests, which are typically more open, 1 hour for 200m up, 1 hour for 300m down is realistic. The big issue of course, is that descending in untracked bush is a great way to find impassable waterfalls or bluff lines that require sidling or ascending to a ridge to bypass.

georgecmu 5y ago

It's the problem #28 in this set: https://news.ycombinator.com/item?id=27884973 https://news.ycombinator.com/item?id=27884973

murbard2 5y ago

This is missing an interesting argument regarding "Buffon's noodle". A straight needle crosses a gridline with probability (4 / pi) (l / t) where l is its length. Surprisingly, a flexible noodle of length l also crosses an expected (4 / pi) (l / t) gridlines. To see why consider that the noodle can be approximated by a set of N straight segments of length (l / N). While the events of different segments crossing a gridline are correlated, the expected number of crossing for each segment is (4 / pi) (l / (N t)) and since, expectation is linear, the expect number of crossings for the noodle is still (4 / pi) (l / t).

dougSF70 5y ago

This is quality.

dougSF70 5y ago
Also, pi is an amazing construct. It is evidence of so much magic in the universe.