Force needed to crush an average human into paste?
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Force needed to crush an average human into paste?
I'm writing a story. How would I go about calculating this?
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Re: Force needed to crush an average human into paste?
You can start with looking at best accelerations sustainable by human body.
In general anything over 100 gees of acceleration (or deceleration, the same thing) is fatal.
If you want the force needed to actually make paste out of human body ("raspberry jam on the bulkheads"), imho a good approximation would be around 2 tons per cm2 of pressure, that is the pressure needed to break human bone.
Organs are destroyed with much less pressure, although I don't know if muscles are too.
In general anything over 100 gees of acceleration (or deceleration, the same thing) is fatal.
If you want the force needed to actually make paste out of human body ("raspberry jam on the bulkheads"), imho a good approximation would be around 2 tons per cm2 of pressure, that is the pressure needed to break human bone.
Organs are destroyed with much less pressure, although I don't know if muscles are too.
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--
Stereotypical spacecraft are pressurized.
Less realistic spacecraft are pressurized to hold breathing atmosphere.
Realistic spacecraft are pressurized because they are flying propellant tanks. -Isaac Kuo
--
Good art has function as well as form. I hesitate to spend more than $50 on decorations of any kind unless they can be used to pummel an intruder into submission. -Sriad
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Re: Force needed to crush an average human into paste?
Thank you for the link.
Okay so 1800 kg/cm^2 ~ 177 MPa.
Since pressure = force / area; force = pressure * area.
If we say that a human is a cylinder then area = 2*pi*r(r+h).
Average Human Height (US) = 1.763 meters. But what would the radius be?
Okay so 1800 kg/cm^2 ~ 177 MPa.
Since pressure = force / area; force = pressure * area.
If we say that a human is a cylinder then area = 2*pi*r(r+h).
Average Human Height (US) = 1.763 meters. But what would the radius be?
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Re: Force needed to crush an average human into paste?
You could probably just use the waist size. I'm not sure how practical it would be to measure from shoulder width since they're an extrusion of the body and don't make up the core, so you'd get a skewed number.
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Re: Force needed to crush an average human into paste?
Also, increase your final number by 50%. Just to be on the safe side.
If someone points it out, you can say 'Hey, the wanted to be on the safe side'
If someone points it out, you can say 'Hey, the wanted to be on the safe side'
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It's so when they comment on or approve of something, I know what pages to block/what not to vote for.
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Re: Force needed to crush an average human into paste?
I've always felt humans are better approximated by a rectangular block than a sphere or cylinder, though technically it depends on the human. Lot of round people in America.
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Re: Force needed to crush an average human into paste?
According to the CDC the Average Male Waist Circumference (US) = 39.7 inches = 1.008 meters
Circumference = 2*pi*r; r = C/(2*pi)
Average Male Radius = .1604 meters
Area of Cylinder = 2*pi*r(r+h) = 1.938 m^2
force = pressure * area = (177 MPa)(1.938 m^2) = 343 x 10^6 N
So if a human were a cylinder of bone the minimum force required to crush us would be 343 x 10^6 N.
Now F = ma; m = F/a
So an Earth an object would have to weigh 35 Mg (39 tons) to produce that much force.
Let's say that object is ice.
The density of ice at 0 degrees Celsius is .9998 Mg/m^3.
The volume of the ice would be 35.01 m^3
So if an ice cube, then the side lengths = 3.271 meters.
Or if it were a sphere, then the radius would be = 2.029 meters.
Circumference = 2*pi*r; r = C/(2*pi)
Average Male Radius = .1604 meters
Area of Cylinder = 2*pi*r(r+h) = 1.938 m^2
force = pressure * area = (177 MPa)(1.938 m^2) = 343 x 10^6 N
So if a human were a cylinder of bone the minimum force required to crush us would be 343 x 10^6 N.
Now F = ma; m = F/a
So an Earth an object would have to weigh 35 Mg (39 tons) to produce that much force.
Let's say that object is ice.
The density of ice at 0 degrees Celsius is .9998 Mg/m^3.
The volume of the ice would be 35.01 m^3
So if an ice cube, then the side lengths = 3.271 meters.
Or if it were a sphere, then the radius would be = 2.029 meters.
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Re: Force needed to crush an average human into paste?
This is the cylinder's total area, so you are placing your victim in a pressure chamber, not dropping a rock on it/slamming it with a vehicle/letting it fall from a building.Area of Cylinder = 2*pi*r(r+h) = 1.938 m^2
May sound stupid but I think it is better to point it out.
I would calculate the pressure necessary to crush a 5 cm diameter bone column (a very rough approximation of both backbone and leg bones), and assume it is enough to crush the softer tissues as well.
So, assuming a "rock falling from above" scenario since it is the more likely, I think only the cylinder's upper circular area is relevant for us.
So that's 15.7 cm2 of area, or 0.157 m2.
Since the pressure needed to break bone is around 2 tons per cm2 (as above), I'd say we need an object weighting 31,4 tons (shaped as a cylinder with a radius of 5 cm falling perfectly vertical) to crush this model.
Probably, in realistic conditions you will need more than that (falling not vertical, impact not perfect... and so on). Double or even triple the weight of the "rock" should do.
Feel free to point out gross mistakes, I'm not particularly more skilled than you in this stuff.
I'm nobody. Nobody at all. But the secrets of the universe don't mind. They reveal themselves to nobodies who care.
--
Stereotypical spacecraft are pressurized.
Less realistic spacecraft are pressurized to hold breathing atmosphere.
Realistic spacecraft are pressurized because they are flying propellant tanks. -Isaac Kuo
--
Good art has function as well as form. I hesitate to spend more than $50 on decorations of any kind unless they can be used to pummel an intruder into submission. -Sriad
--
Stereotypical spacecraft are pressurized.
Less realistic spacecraft are pressurized to hold breathing atmosphere.
Realistic spacecraft are pressurized because they are flying propellant tanks. -Isaac Kuo
--
Good art has function as well as form. I hesitate to spend more than $50 on decorations of any kind unless they can be used to pummel an intruder into submission. -Sriad
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Re: Force needed to crush an average human into paste?
I'm not so sure. What about Newton's Third Law? If an object falls on you, wouldn't you exert a force on the ground, which would in turn exert a force on you?someone_else wrote:This is the cylinder's total area, so you are placing your victim in a pressure chamber, not dropping a rock on it/slamming it with a vehicle/letting it fall from a building.Area of Cylinder = 2*pi*r(r+h) = 1.938 m^2
May sound stupid but I think it is better to point it out.
I would calculate the pressure necessary to crush a 5 cm diameter bone column (a very rough approximation of both backbone and leg bones), and assume it is enough to crush the softer tissues as well.
So, assuming a "rock falling from above" scenario since it is the more likely, I think only the cylinder's upper circular area is relevant for us.