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What's the durability required to withstand being crushed by 8 million tons of rock unscathed?
Edit: It's actually 8 million lbs instead of tons.
Edit: It's actually 8 million lbs instead of tons.
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Crushing Durability Question
- Thread starter Continuum19
- Start date
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Depends on the frontal impact area of that 8 million tons of rock and the displacement (i.e. how far those 8 million tons of rock went before they stopped moving), and then there's your body to take into account.
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- #3
Average height and weight of an American male is 5'9 and 197 lbs.
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Now you need the frontal impact area of that 8 million tons of rocks. Is it a single solid lump of rock or is it many countless chunks? If it's the latter, then it might be incalculable.
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It's a single solid 8 million ton lump of rock.
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Well then, better get to finding the frontal impact area of the rock then. And possibly the GPE if it was tossed from a height.
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- #7
It's falling straight down on top of a person from a few feet in the air.
What is frontal impact area by the way?
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The area that comes into contact with your own body. Think of it like the bumper of the car hitting you in the face.
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- #9
So if the person is laying flat on his back while the 8 million tons of rock falls on top of him from a few feet in the air?
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Well if it hits him dead center he'd prolly scale to the full value. But you'd still need to find the rock's GPE and how many feet it was in the air before collision happened.
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- #11
Let's say it's a dozen feet in the air.
What is GPE by the way?
Apologies for my ignorance.
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GPE is gravitational potential energy, mass times gravity times height.
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- #13
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Just convert height in feet to meters, convert tons to metric tons and then do this:
mass in kg * 9.81 * height in meters
It's not that hard.
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- #15
This is a hypothetical question.
How do I find the height of an 8 million ton lump of rock?
Eight million US tons is 7257477.92 metric tons so that's easy enough.
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depends on its shape. If you have an image you can use pixel scaling but if it doesn't then just assume its a circle and use density which would depend on material
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- #17
So let's say it's a circle and we're using the density of granite (which is apparently between 2.65 and 2.75 g/cm).
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Idk i tried calcing this and failed
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- #19
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yeah i got it to be like the size of the sun or sth and it made absolutely no sense. You could probably use inverse square law or sth idk. Good luck
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- #21
I was trying to figure out the proportional durability of an ironclad beetle: https://www.cbsnews.com/amp/news/ironclad-beetle-indestructible-survive-run-over-by-car/
I guess I'll have to figure it out some other way.
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Not the height of the rock. The height between the ground and the rock when it falls.
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- #23
The height would be a dozen feet.
7257477.92 metric tons is 7257477920 kilograms.
What do I do next?
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Dozen feet huh. 3.6576 meters.
Gravity is 9.81 m/s^2
GPE= 7257477920 * 9.81 * 3.6576= 2.60405971666e+11 J or 62.23852095267686 tons of TNT (8-B+, City Block level+)
Now, you need to find the frontal impact area of the rock and the cross-sectional area of the ironclad beetle that would tank it. We will need to find the ratio of the two to find out how much the beetle would take, then we'd multiply that ratio with the above energy value.
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- #25
When I said proportional durability of an ironclad beetle I was talking about a 200~ lb man.
200 lbs times 39,000 is nearly 8 million tons.
An ironclad beetle can withstand 39,000 times their own body weight unscathed.
I'm taking about a 200 lb man with the proportional durability of an ironclad beetle not the actual beetle itself.
Do you understand?
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Proportional durability of an ironclad beetle scaled up to 200 lbs?
What is the durability of this ironclad beetle? Because I'm pretty sure if we scale up this durability to a 200 lb man that'd be what you call calc-stacking.
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- #27
Proportional durability of an ironclad beetle: https://www.cbsnews.com/amp/news/ironclad-beetle-indestructible-survive-run-over-by-car/
If a 200 lb man had the proportional durability of an ironclad beetle they would be able to withstand being crushed by nearly 8 million tons unscathed.
That's what this entire discussion has been about from the start.
You calced earlier that this would be in the 8-B range (62+ tons) unless I'm missing something?
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That's if the 8 million ton rock fell from 12 feet.
But here I'm not seeing that. Here it's just pressure.
You can't calculate energy from pressure without finding out the displacement (AKA the distance it moved before stopping) of said rock.
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- #29
What if said rock moved 12 feet before stopping?
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The rock would have to fall first.
Then you'd need the frontal impact area of the rock and the cross-sectional area of the 200 lb 5 ft 9 in man (Cross-sectional area is half that of total surface area, there are surface area calculators online).
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- #31
This link states that someone with the proportional durability of an ironclad beetle would be able to withstand being crushed by 2 space shuttles (4 million tons each) so the earlier calc you did should work just fine: https://www.businessinsider.in/scie...figured-out-how-/amp_articleshow/78819035.cms
Which means a man with the proportional durability of an ironclad beetle would have 8-B durability (62+ tons of tnt).
Thanks by the way.
Can you do one more calc?
It would be a 200 lb man being crushed by 1 million tons.
So you'd have the exact same calculation that you did earlier but 1 million tons instead of 8 million.
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- #32
Are you there dude?
If you aren't busy can you do another calc?
It would be a 200 lb man being crushed by 1 million tons.
So you'd have the exact same calculation that you did earlier but 1 million tons instead of 8 million.
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Without the frontal impact area of the 1 million ton rock the calc is simply not something that can be forged.
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- #34
A dozen feet.
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You're just repeating the height. That's not the frontal impact diameter of the rock.
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- #36
Frontal impact area is the height between the ground and the rock when it falls correct?
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No.
Think of it like the face on a cylindrical crusher. But you're measuring the face. Basically, it would be the diameter of said face from which you derive an area using ye olde pi * r^2 (R is radius, half of a diameter).
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- #38
I'm curious.
How did you do this calc earlier?
By the way,.one million tons is 907184.74 metric tons and 907184.74 metric tons is 907184740 kilograms.
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- #39
One million tons is 907184.74 metric tons and 907184.74 metric tons is 907184740 kilograms.
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That still doesn't give us the frontal impact area in any shape or form.
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