The title pretty well says it.
Assuming you had a very sturdy sumo suit (one of those air-filled suits) that wouldn’t be punctured by rocks and twigs and whatnot, and you managed to not land on your head, would it save you if you fell from a large height? 3m? 15m? 100m?
I have no idea why I was wondering about this, but there you go.
I would guess not. AFAICT, the suit might save your ribcage, but would do little to keep the blood inside your veins if the suit didn’t slow you down slowly enough.
Actually, I don’t think you want a really sturdy one unless it was also very flexible. The basic idea is to stretch your deceleration out over as long a period of time as possible. For example, I believe that the air bags that stuntmen use deflate in a controlled manner when they land on them. And automotive air bags also deflate rapidly. So you want a suit that won’t burst like a balloon, but also won’t maintain air pressure, thus squeezing the occupant in a very unpleasant manner.
But given that those two examples will successfully cushion someone moving in the area of 60+ mph, then yeah, I can see a sumo suit protecting you to a certain degree. If nothing else, the added surface area will slow your terminal velocity.
I was sort of wondering if you would bounce. You know - like a balloon that’s dropped bounces.
It’d help.
'Round these parts, there’s an annual Halloween competition, conducted at the College of Engineering and Mineral Resources building - of which I am a former student - called the Pumpkin Drop. Many educational institutions have similar events, oftne involving eggs instead of pumpkins.
The object, of course, is to get the payload to the ground intact (and additionally, close to the target). If you pad it well enough, it just might survive.
Of course, human beings are seldom pumpkins, and most peoples’ grey matter is more sensitive to sudden deceleration trauma than pumpkin goo - which, in all honesty, is already goo.
Whether it helps you enough to survive? Too many variables to say. After all, we hear reports of the occasional sky diver surviviing tremendous falls with only mild injury…
Well, bouncing * per se * isn’t necessarily a good thing. Instead of decelerating from, say, 100 mph to zero, you’re instead decelerating to zero and then accelerating all over again. That works with bungie jumps because your deceleration/acceleration is gradual over many feet. A sumo suit could only deform a foot or two before the contents (you) contacted the ground. So you’re not spreading the impact out over a particularly long period of time. I really don’t think you want to bounce – that would result in whiplash and, for lack of a better term, sloshing about.
To apply some real-world numbers to this:
Back in the 1980s I watched film of Dar Robinson setting the (at the time) high-fall record, 300 feet from a helicopter into an airbag. The airbag was enormous, bigger than a house and pretty tall, IIRC it looked like about 20-30 feet thick.
If we ignore air resistance you’d be falling about 132 fps (90mph) after a 270 foot fall (300 feet less 30 feet of airbag). Slowing down to zero, assuming constant decelleration over 30 feet of airbag, gives about 9G over .45 seconds. An athletic person taking this right on their back (as he did) would probably be OK - stunned for a few seconds but probably not much worse. Fighter pilots can pull about 9G sitting down before they start to black out (with G-suits and training and etc). We’re in the right ballpark.
If you run the above calculations but only give yourself 2 feet to slow down…the decelleration varies directly with the stopping distance, so 2 feet of sumo suit stopping you from 90mph gives about 135G. Admittedly it’s for a much shorter period of time (a thirtieth of a second) but I sure wouldn’t want to try it, I think you’d get pretty badly hurt.
Henri La Mothe did a 28 foot high dive into 12 inches of water on TV (I saw that one) and got up right away. That’s about 28G for 0.05 seconds. He was 75 at the time.
So to answer your original question, 3m sure thing, 15m probably, 100m no way.
What about if you’re in a Zorb?
Well, the thing is we can’t really ignore air resistance. If you are wearing a 30 foot thick airbag or a Zorb, it will slow you down significantly. Significantly enough to make the Zorb anything more than a convenient container to haul of your remains is another matter.
I am certain we did this about two months ago. I will look for that and come up with an answer if I can’t find it. In short, though, the answer is yes. The only question is how big and how strong the suit must be.
OK, this site says that the limit for a properly oriented and supported person is 80g for an impact (as opposed to longer term effect). Say we limit the sport to people in good health and put a 1.5 safety factor in there we get about 53g. For this there would have to be strong neck support and I think the person would have to land on their back.
Lets assume we design our suit well enough that there will be an approximately constant deceleration over the entire thickness of the suit. Once you have stopped the suit will then try to recover its deformation and you will bounce, but the gs there will be equal or less than the initial acceleration.
So, a free body diagram shows us that:
F[sub]i[/sub] - F[sub]g[/sub] = ma
Where F[sub]i[/sub] is the force from the ground and F[sub]g[/sub] is the force of gravity. F[sub]i[/sub] is what your body feels, and that is what we are concerned with, or more specifically F[sub]i[/sub]/m where m is your mass.
So,
F[sub]i[/sub] - mg = ma
a = F[sub]i[/sub]/m - g
v = (F[sub]i[/sub]/m - g)*t - v[sub]term[/sub]
x = (F[sub]i[/sub]/m - g)*t[sup]2[/sup]/2 - v[sub]term[/sub]*t + x[sub]thick[/sub]
where x[sub]thick[/sub] is how much suit we are compressing (the distance over which the acceleration occurs) and v[sb]term[/sub] is the terminal velocity.
Solving for t[sub]final[/sub], the time when velocity equals zero we get:
t[sub]final[/sub] = v[sub]term[/sub]/(F[sub]i[/sub]/m - g)
0 = v[sub]term[/sub][sup]2[/sup]/(2*(F[sub]i[/sub]/m - g)) - v[sub]term[/sub][sup]2[/sup]/(F[sub]i[/sub]/m - g) + x[sub]thick[/sub]
0 = -v[sub]term[/sub][sup]2[/sup]/2 + (F[sub]i[/sub]/m - g)x[sub]thick[/sub]
F[sub]i[/sub]/m = v[sub]term[/sub][sup]2[/sup]/(2x[sub]thick[/sub]) + g
Lets divide that by g so the left side becomes the acceleration in g’s that the body can take that we found from the website.
53 = v[sub]term[/sub][sup]2[/sup]/(2gx[sub]thick[/sub]) + 1
52 = v[sub]term[/sub][sup]2[/sup]/(2gx[sub]thick[/sub])
x[sub]thick[/sub] = v[sub]term[/sub][sup]2[/sup]/(104*g)
Hmm, now we need a way to find v[sub]term[/sub], which we have already determined will be a function of the size of the doomed faller, or x[sub]thick[/sub].
So, this is really difficult for a human shaped object, but a good bit easier for a sphere. So lets assume that we are going with the Zorb idea. The terminal velocity formula is nicely derived for us here, so no need to rederive it.
v[sub]term[/sub] = sqrt( (2mg) / (C[sub]d[/sub]rhoA))
where C[sub]d[/sub] is the drag coefficient, rho is the density of air, and A is the frontal area of the object. Since we are going with a sphere
A = pi*R[sup]2[/sup]
and as a simplification we will assumbe the ball is a good bit bigger than the thickness of the subject, so:
R = x[sub]thick[/sub]
That gives us:
R = (2mg) / (104gC[sub]d[/sub]rhopiR[sup]2[/sup])
R[sup]3[/sup] = (mg) / (52grhopi*C[sub]d[/sub])
Next task, find a good approximation of C[sub]d[/sub].
If we use those numbers we can figure the whole thing out, since they specify that 80g is for 0.1s.
Initial velocity (speed when decelleration began) would be at, where a = 5332.2 (using your factor of safety) and t = 0.1sec, comes out to about 170fps which is about 120mph - same as terminal velocity for a skydiver in a normal arched position in a normal jumpsuit.
Distance it takes to stop is 0.5a(t^2) which works out to about 8.5 feet.
Somebody with 8.5 feet of Special Straight Dope Padding wrapped around their body will not have the same terminal velocity as a skydiver in coveralls (unless you have weights strapped to their body) so assuming that we could build this thing and ensure a proper impact orientation, you could “safely” (this is up to 53G after all!) fall from any height.
Keep in mind that being in the center of an 8.5 foot radius ball means that you’ve got 225 square feet of wind resistance helping you, significant parachute effect.
Tell me when you want to try it, I’ll bring a lawn chair, some beer and a camera.
How about the system of airbags that the Mars Rover used to land with? If you were inside that little pyramid would you survive?
I’ll take it from here. The coefficient of drag for a sphere is .5, and the density of air is about 1 kg/m[sup]3[/sup]. The mass of the person + padding is, let’s say, 100 kg, and g is of course 10 m/s[sup]2[/sup]. So grinding that all together, we get just over a meter for R, quite attainable.
Fun eh? To know the drag coefficient we need to know Reynolds number (Re). Reynolds number depends upon the density of the the air, the viscosity of the air (miu), and the diameter of the sphere and the velocity. But wait, we are trying to find the diameter and velocity. Well, can’t we just plug the equation in and solve simultaneously? No, because there is no explicit equation, just look up tables since it is all solved determined experimentally. This is something you see a lot in aerodynamics, iterative solution.
So, lets guess. I guess that the radius of the Zorb will be 1 meter and the final velocity will be 45 m/s (since 56 m/s is a good estimate for someone without a Zorb).
Re = v2R*rho / miu
Re = 45 m/s * 2 * 1 m * 1.25 kg/m[sup]3[/sup] / 1.8E-5 Ns/m[sup]2[/sup]
Thus Reynolds number is about 6.2 million (it is unitless).
Look at this, I found a neet little drag calculator. When I enter in my data I get about the same Re and a C[sub]d[/sub] of .255. A good engineer always checks his assumptions, so I saw what they used, and they just used a constant of .255 for anything over 4 million, but since I can’t find a better value anywhere else (including textbooks) I will just have to take it. So, using:
R[sup]3[/sup] = (mg) / (52grhopiCd)
R[sup]3[/sup] = m / (52rhopi*Cd)
and my weight of 200 lb (90 kg)
R[sup]3[/sup] = 90 kg / (52 * 1.25 kg/m[sup]3[/sup] * 3.14 * .255)
R = 1.2 m
This says that our terminal velocity will be:
v[sub]term[/sub] = 35 m/s
or about 78 mph. Not too fast. Now we reiterate. I get a recalculated Re of 5.8 million, so I can’t get a better C[sub]d[/sub].
So, in conclusion, you need a 8’ diameter sphere that will not pop, yet will be able to compress half that much, or at least have you supported my some sort of elastic structure inside that will. It also must have you facing upward (lying on your back) and have your head and neck rigidly supported with the rest of your back. Definitely possible. And, the bigger you make it, the more room you have to slow down and slower you will be falling. Eight feet is just the minimum.
Oh, sure, if you do it the easy way. 
Interesting that the different approximations you made from me cancelled out to give about the same answer. Heh.
Maybe but it is hard to say. The thing was designed to take 40g’s, but it only ever saw about two on impact. Remember, they had retro rockets which would be cheating.
Ah - excellent.
So, assuming an 8 foot suit (or larger - 12 feet maybe) what would be the maximum height you could be droped from? Infinity?
(assuming that the wind didn’t blow you into any unfortunate obsticles.)
Oh, interesting side note. Once you reached terminal velocity (which would occur rather quickly), you would no longer be weightless. In fact, if the thing fell in a stable manner (unlikely) and was opaque the only hint that you were falling would be vibration and noise.