Question about "Sorbothane"

[elementcollector1]

I've recently been considering buying a material titled "sorbothane", a very soft and energy-absorbent derivative of polyurethane. The product list can be found here: http://www.sorbothane.com/blog/wp-content/uploads/2012/12/11-26-12-Sorbothane_SPG_11.2012_v4.pdf

My question is, when one examines the 4"-by-4"-by-1/2" square pad of durometer 70(product 0204025-70-10), it claims to be able to withstand a load of up to 7400 pounds, or about 33 kN. Now, this already seems kind of ridiculous, but maybe that's just how good it is. Maybe. But how does that translate to impact force?

I've been trying to calculate, as a rough example, how much force a person would experience if they dropped from freefall at terminal velocity and hit the ground with one of these pads on each of their shoes. However, calculating impulse seems to be useless, as the amount of time the person takes to hit the ground is essentially zero. Attempting to derive from I = delta(p) where p is momentum, I arrived at something like 1600 kg*m/s impulse given a person who weighs approximately 600 N, but I don't know if that's correct, or even what to do with it if it is.

Not only that, wouldn't the person experience some sort of force based on all the potential energy - turned - kinetic energy from the distance they fell?

[swansont]

I've only used sorbothane for nominally static geometries where it's there for vibration isolation. I'm not sure it's supposed to be used for impacts.

[elementcollector1]

It's mentioned on the webpage (http://www.sorbothane.com/) that it can be used for both "shock" and vibration applications.

[swansont]

It's mentioned on the webpage (http://www.sorbothane.com/) that it can be used for both "shock" and vibration applications.

 

I think shock assumes there is already contact such as protecting an object during shipping, rather than dropping it onto the pad. One issue with impact is the force depends on the contact surface area, so how an object hits can drastically change what the impact force is.

[elementcollector1]

Okay, so assuming two pads, one on each of the person's feet, the contact area for the body would be about 20 cm by 10 cm, or 200 cm². How would the force exerted on this area be calculated, then?

[John Cuthber]

 

My question is, when one examines the 4"-by-4"-by-1/2" square pad of durometer 70(product 0204025-70-10), it claims to be able to withstand a load of up to 7400 pounds, or about 33 kN. Now, this already seems kind of ridiculous,

Why is that ridiculous?

 

Anyway, if a rigid body falls for a distance x and is brought to a halt over a distance y then the average impact force is x/y times the body's weight. (neglecting air resistance of course).

 

"I've been trying to calculate, as a rough example, how much force a person would experience if they dropped from freefall at terminal velocity and hit the ground with one of these pads on each of their shoes. "

Well, I'm not sure how far you have to fall to get to terminal velocity (strictly speaking, you never quite do) but it must be something of the order of hundreds of metres. Call it 1200 metres- just to have a number to work with.

If you are then brought to a halt over a distance of 12 mm (the thickness of the pad) your feet would be subject to an acceleration of about 100,000g

That's going to be messy.

 

Consider your head. It will be slowed down and stopped over the course of about 1.8 metre (your height).

1200/1.8 = an acceleration of about 700 g

 

A human head weighs about 5Kgf so it would be subject to about 3.3 tonnes force

 

Still messy.

Of course, it would also suffer some damage as it ploughed through the rest of your body.

It wouldn't matter much since it would trash the lungs heart etc.

Edited October 2, 2014 by John Cuthber

[elementcollector1]

Huh. So energy and shock absorption do nothing in this case, because the deceleration itself is enough to damage the human body?

[John Cuthber]

Not exactly, They would do something, but not enough.

You would do better using a bigger thickness of something much softer.

That way the deceleration would be smaller and the forced would be less.

[elementcollector1]

What would be an appropriate "soft" substance? Foam, or maybe a type of rubber?

[John Cuthber]

Foam, feathers, some such thing.

One important criterion is that the density must be low.

[imatfaal]

What would be an appropriate "soft" substance? Foam, or maybe a type of rubber?

 

Empty cardboard boxes or bags of air - these seem to be what the stunt industry use to break falls; but in both cases you need lots. As explained above you need to slow down at a slow rate - so you must travel a fair distance whilst slowing

[elementcollector1]

So why is an airbag able to decelerate a person safely? After all, it's not all that big compared to a person, yet it stops collisions very well and allows the driver to survive mostly intact.

[Enthalpy]

The force given for the PU part amounts to 3MPa, which is a reasonable compression load for PU. Not ridiculous at all. Anyway, 3MPa is a moderate load for a polymer, and a tiny one for a metal, which are commonly in the 200-1000MPa range. The advantage of PU is that is deforms much at 3MPa without damage, and this means much energy per volume unit.

 

I confirm PU is the standard choice to absorb shocks. Not the materials that rebounds most nor least, but the elastomer that absorbs the biggest energy without destruction, over a wide range of impact durations.

 

-----

 

An airbag saves a person's head only, thanks to a deceleration distance of 0.1m, from a speed like 40km/h or hopefully 50km/h. I've seen once a car crash-test at 70km/h and the obvious and intuitive conclusion is that you better have braked the car to a smaller speed before the impact.

[John Cuthber]

The airbag slows the person's head to a stop over a distance of something like 10 cm. The steering wheel (or dashboard or whatever)would do so over about 1 cm or less.

So the acceleration and hence the force is 10 times less.

Also, and probably more significantly, that force is distributed over the whole of the persons face rather than just the bit that hits the wheel.

So there's a lot less pressure acting on the person and it does a lot less damage.

[elementcollector1]

So if the acceleration is directly proportional to the stopping distance, what is the upper limit on deceleration for a human?

Also, would it be possible to 'transfer' the acceleration somewhere else - as in, direct the shock somewhere it can be less harmful? Not sure on how shock 'moves', at any rate.

[John Cuthber]

According to this

http://en.wikipedia.org/wiki/G-force#Human_tolerance_of_g-force

You might be lucky and survive 100 g

 

And the acceleration is inversely proportional to the stopping distance, not directly proportional.

 

There is a way to survive much higher accelerations but it's neither morally acceptable not practical.

A foetus will survive accelerations that kill the mother - mainly because it has no airspaces and it's well "padded".

[elementcollector1]

Well, unless time travel is involved, I don't think I'll be applying the second method.

After some thought, I wondered why we could survive airplanes landing (after all, that's quite some momentum!), and looked it up - apparently something called an 'oleo strut' absorbs most of the shock, apparently by damping with compressed air or nitrogen and oil.

 

 

An oleo strut consists of an inner metal tube or piston, which is attached to the wheel axle, and which moves up and down in an outer (or upper) metal tube, or cylinder, that is attached to the airframe. The cavity within the strut and piston is filled with gas (usually nitrogen, sometimes air – especially on light aircraft) and oil(usually hydraulic fluid), and is divided into two chambers that communicate through a small orifice. When the aircraft is stationary on the ground, its weight is supported by the compressed gas in the cylinder.^[1] During landing, or when the aircraft taxis over bumps, the piston slides up and down. It compresses the gas, which acts as a spring, and forces oil through the orifice, which acts as a damper. A tapered rod may be used to change the size of the orifice as the piston moves, and a check valve may be used to uncover additional orifices so that damping during compression is less than during rebound.

-Wikipedia

 

Hypothetically, what if a human pogo-sticked one of these off of a 1200-meter cliff (or some similar setup where the strut would take the brunt of the impact), in keeping with the earlier numbers - would they still be injured from deceleration? I would think they wouldn't if a few of these things could stop the vertical momentum of a passenger plane, but maybe I'm missing something more subtle.

[John Cuthber]

It's still the same problem. The strut would have to be absurdly long- tens of metres- so you wouldn't be able to pack one per passenger into a plane.

 

It's considered a better idea to stop planes falling out of the sky in the first place.

For the weight of all those struts, you could have a spare engine and that's probably a more sensible idea.

On the other hand the flights would cost a little more, and the industry is lethally competitive.

[elementcollector1]

I don't get it - if the strut can stop an airplane, why wouldn't it stop a human (ignoring practicality)?

[John Cuthber]

Mostly the wings already stopped the plane falling.

 

A plane might be doing a couple of hundred miles an hour when it lands, but nearly all of that speed is along the ground, rather than down. The wheels etc only have to absorb the impact from the vertical component of the speed. They then slow the plane down over many metres of runway.

 

If you could fly, it would work for you too.

[elementcollector1]

Hmm. But shouldn't the plane be heavy enough that the reduced vertical momentum (from the action of the wings) is still greater than the vertical momentum of a person at terminal velocity?

 

 

 

The latter figure comes from FAA certification requirements, which require that a landing gear be designed to handle a hard landing with a sink rate of 10 feet/second (or 600 FPM). A simple calculation shows that for a main landing gear strut with a typical maximum throw length of 12 inches, the required deceleration from such a landing is about 3g. Thus, the strut needs to be able to dissipate a worst-case landing force equivalent to about three times the aircraft's normal static weight.

 

Source: http://www.avweb.com/news/savvyaviator/192153-1.html?redirected=1

 

3g might not be much acceleration, but considering the plane that's a whole lot of force.

Then again, if deceleration's the issue, wouldn't there still be a problem even if all the force of landing was absorbed?

Back to your earlier calculations on deceleration, the minimum distance to get the deceleration within "safe" (relatively speaking) limits would be 12 m. Is there any way to make this length less - as in, could it be coiled up or otherwise rendered shorter while still acting as if it were the full length?

[John Cuthber]

The calculation I gave earlier in terms of distance fallen and stopping distance is derived directly from the conservation of energy.

There's not a tot of opportunity to get round it.

 

3 g on a plane is quite a lot of force, but it's a big lump of metal so it's very strong.

[Enthalpy]

Why not use maths? They have proven useful again and again at physics problems.

[elementcollector1]

Not sure what you mean, Enthalpy - I've been doing calculations this whole time.

Anyway, if there's no way to "cheat" the work equation (W = Fd) into thinking there's more distance than there really is, I think I'll have to settle for a system with a lower tolerance for falls - better than what a human could manage, but not perfect.