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.
Average height and weight of an American male is 5'9 and 197 lbs.
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.
It's a single solid 8 million ton lump of rock.
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.
It's falling straight down on top of a person from a few feet in the air.
The area that comes into contact with your own body. Think of it like the bumper of the car hitting you in the face.
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?
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.
Let's say it's a dozen feet in the air.
GPE is gravitational potential energy, mass times gravity times height.
Just convert height in feet to meters, convert tons to metric tons and then do this:
This is a hypothetical question.
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
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).
Idk i tried calcing this and failed
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
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/
Not the height of the rock. The height between the ground and the rock when it falls.
The height would be a dozen feet.
Dozen feet huh. 3.6576 meters.
When I said proportional durability of an ironclad beetle I was talking about a 200~ lb man.
Proportional durability of an ironclad beetle scaled up to 200 lbs?
Proportional durability of an ironclad beetle: https://www.cbsnews.com/amp/news/ironclad-beetle-indestructible-survive-run-over-by-car/
That's if the 8 million ton rock fell from 12 feet.
What if said rock moved 12 feet before stopping?
The rock would have to fall first.
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
Are you there dude?
Without the frontal impact area of the 1 million ton rock the calc is simply not something that can be forged.
A dozen feet.
You're just repeating the height. That's not the frontal impact diameter of the rock.
Frontal impact area is the height between the ground and the rock when it falls correct?
No.
I'm curious.
That still doesn't give us the frontal impact area in any shape or form.
