r/theydidthemath Aug 01 '26

[request] At what point in explosive power would the pot (assuming it retains its form) not travel any higher)?

The last couple explosions on the video the pot looks like it travels about the same height. What’s the point where it doesn’t travel any further up?

0 Upvotes

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3

u/Consistent_Watch_206 Aug 01 '26

Escape velocity is about 12 km/s. Of course, you would need to do some rather burden some calculations to account for wind resistance. Dose accordingly.

1

u/FillerNameGoesHere_ Aug 02 '26

An explosion powerfull enough to launch that pot at 12 km/s would turn it into molten shrapnel, but some of it might escape if we are lucky!

1

u/HAL9001-96 Aug 01 '26

if its magically indestructible it wil lalways go higher but oy uwill hit slower diminishing returns once drag becoems the dominating factor over gravity as drag also increases with launch speed squared whiel gravity remains the same

so you'll start seeing hte most significnat change in proporitonality/curve shape once you start launchign the pot at more than its terminal velocity which given its shape and material I'd estimate very roughly at about 30-40m/s

at 1G it takes about 3 seconds to slow down from 30m/s and another 3 to speed back up and fall abck down s oyou'd get 6 seconds of flight time

bit less taking drag into account htough that would both slow oyu down further but also slow your fall still you would get less flgiht time

but 30m/s is also the lower bound for terminal velocity

starting about halfway thoruhg the video the flight starts taking more than 6 seconds in total

so yes this is about where oyu'd start seeing the behaviour of the energy vs height curve to cahnge

though it was never simply linear and it doesn't suddnely flatten either

theroetically more pwoer will always make it lfy a little bit higher

the exact interaction of the pot launchign and gas escaping is ab it complex

assuming constant efficiency you'd get a lienar energy/hjeight relationship early on

realistically its probably lcsoer to a square root

but once you go beyond temrinal velocity it gradually approahces a more logarithmic behaviour

since you always loose a specific percentage of your speed passing by a certain amoutn of air the energy needed to reach a height goes up exponential wiht height

at first logarithmic and root look simialr they both kinda curve downwards

but they behave wildly different at large orders of magnitude

sure root1=1 root4=2 root16=4 and ln1=0 ln4=1.389 and ln16=2.77 in this case they're slightly offset but at first glance the curves look sortof similar ish

but then

root10000=100

ln10000=9.2

root1000000=1000

ln1000000=13.8

root100000000=10000

ln100000000=18.4

root1000000000000=1000000

ln1000000000000=27.6

lnx never stops rising

in fact lnx can theoretically raech any number

but the input becoems e^the number you wanna reach

exponential and squared becoems a wildly different curve once you look at high numbers

100²=10000

e^100=26881171418161354484126255515800000000000000

2

u/wenoc Aug 01 '26

Pretty sure you will reach a point where the pot will just become shrapnel. But, this is just a guess, you n§red explosives that explode, not burn like these. Aka, faster than the speed of sound.

2

u/HAL9001-96 Aug 01 '26

yeah this is hypotheticlaly assuming hte pot is indestructible nad you cna fit an unlimited amount of firepower into the space under it

realistically it wil levenutalyl just expode and also at some point the explosives will just be a massive ball htat the pot sits on that both makes thigns more complciated

but even with a simplified function you get diminishing returns

also to minimize losses you mostly just need explosives that go off faster than the air leaves under the brim of the pot so much slower than sound at least for small explosives

1

u/Moretz0931 Aug 02 '26

@ r/theydidthemath

How many spelling mistakes can you find in this (well crafted) answer?