However, I am not convinced that there are incredible g-forces experienced upon impact with water. In fact, when I roughly calculate deceleration (neglecting surface tension and drag) I find that the jumper experiences less than 1g no matter how fast they hit the water. My opinion is that surface tension is probably insignificant in all but the calmest water, therefore this means that drag is really important to this type of calculation. The right body position and a teflon coated wetsuit could probably reduce drag enough to make any fall survivable.
Lance, I’m not sure I agree with your numbers. Granted, I don’t know what kind of calculation you did, but I’ll toss my back-of-the-napkin numbers out.
More than once, an impact velocity of roughly 80 mph has been mentioned for GGB jumpers - I’ll go with that. 80 mph ~= 36 m/s.
Decelerating uniformly from that speed to a halt at 1 G would require (36 m/s)/(9.81 m/s[sup]2[/sup]) = 3.7 s. This sounds like an awfully long slowdown time to me, even if you’re streamlined and going in feet-first. I get that you’d be 66 meters under the water by the time you stopped;
x = v[sub]0[/sub]t + 0.5at[sup]2[/sup];
v[sub]0[/sub] = 36 m/s; t = 3.7 s; a = -9.81 m/s[sup]2[/sup]
x(t=3.7s) = 66.1 m.
That sounds unrealistically far to me - 1 g just doesn’t sound like nearly enough, IMO.
If you smack into the water in an uncontrolled fashion, I’d think that in well under 1 second your velocity would be down pretty much to zero. Even if we go with that 1-second figure, that’s 3.7 g’s. That’s hardly unsurvivable, but like I said I think that’s a pretty conservative estimate.
I’d also suspect that the deceleration would be “front-loaded” - initially, it would be well above the time-average value, and would rapidly drop off. Just how high it would be, and how quickly it would drop off (both of which seem important to the question of just how badly they’d screw the jumper up) would require a considerably fancier analysis to determine.
I didn’t do all that much of a calculation, but I throw it out here for all to see and disect.
Neglecting drag and surface tension, there are only two forces acting on the body in the water. Gravity and buoyancy. I don’t happen to know the average density of a human, so I solved for what density would result in a deceleration of 1g. That density is half the density of water. (I may be wrong about the whole concept of buoyancy and density) I know that a human is less dense than water, but only slightly. In fact, when I’m in the pool and I forcefully expell as much air as I can from my lungs, I am more dense than water. So I made the assumption that the density of the average human is more than half the density of water.
All of this neglects drag. Without drag, the speed a body enters the water is unimportant. Anyone who has ever gone off a diving board knows that this isn’t true. Drag increases with velocity. This would also lead to the front-loaded deceleration predicted by brad_d. Drag makes all the difference, and I’m pretty sure that body position and material science could reduce drag enough to make any fall survivable.
Oh, OK. I see what you are doing, Lance. It makes sense, and I think I agree with your ultimate conclusion: that any impact speed should be survivable. You’ll just go down awfully deep before you stop - I suppose there might be some practical limit there.
My suspicion is that drag is far larger than buoyancy, at least until you’ve slowed down pretty seriously. I don’t know where I’d look for a Re/C[sub]D[/sub] curve for a streamlined human 
so I can’t really dig up numbers.
WEll…think of it this way-remember how much it hurts to do a belly flop into the water?
Imagine that times 20.
Crafter_Man, The surface of water is hard. It’s not soft.
Its almost like hitting cement from that height.
BTW some guy on tv said its best to land on your back.
I’m convinced. I retract all my previous statements on this topic, and heartily agree with handy. I mean, if some guy on tv said it, how can I argue?
Wait, I’ve changed my mind again. I’m back to believing that it is suvivable to hit water from a great height.
In addition to this shocking double 180 of my view on the topic I have something of value to add. There is a possibility (I haven’t thought it all the way through) that you could increase you chance of survival by holding onto something big and dense. Like an anvil, a canonball, or a solid cubic meter of platinum. The way this works is that you’d experience less deceleration due to buoyancy, and all you’d have to worry about is drag. Once you and your anvil slow down to terminal velocity (due to drag), you let go of the anvil and then decelerate due to buoyancy mostly with very little drag. There are a few problems with this. The first being that you’d probably hit the water a lot faster, and the second being that you’d go pretty deep (possibly too deep).
I am not an anatomist, but it seems to be that the spinal column might keep the heart from moving such a distance that it would rip off of the aorta. Just a guess, of course, but it seems many animals expose their backs to predators, rather than their soft underbelly.
Or you could place yourself in a strong missile-shaped tube and work out a way that it would plunge straight in (as opposed to hitting on its side). People have gone over Niagra Falls in “barrels”, but most of them snuff it. But a sufficiently padded structure will help.
I think that depends on the terminal velocity of the falling object. A draggy object will have a slower terminal velocity. (For example, a parachute has much more drag than a lawn dart, so it falls slower.) AFAIK, an unpropelled object will only fall at 32 feet per second per second, no matter how heavy it is. So in a fall from a height that is not great enough for it to reach a higher terminal velocity than the terminal velocity of another object, the heavier object will not fal faster than the lighter one (keeping aerodynamic drag in mind).
Being an adrenaline junky, I’ve personally jumped off of quite a few bridges and other assorted high places and I report in with first hand data.
First, no, I’ve never done the GGB. The highest bridge I’ve jumped from is the one that goes over the intercoastal canal at Freeport, Tx. I’m guessing it’s about 80+ft. A long way from 220, but still pretty darn high.
Through trial and error (ouch!) I’ve developed this method.
- Feet must be CROSSED! I can’t stress this enough (see"trial and error" above). Toes pointed down.
2.Hands tightly over your face, elbows together and held tightly against your stomach. This prevents the’uppercut’ effect and protects the solar plexus. Also keeps water out of your nose. - Nothing short of rubber pants can completely stop the ‘enema’ effect, however a good pucker will reduce it almost nothing.
- Yes, you must enter vertically. But I have never had a problem with this.
5.Be patient. This is my WAG as to why so many people get hurt/die. When you are standing on an 80+ precipice, your mind makes a mental calculation of about how long it will take to hit the water. And it probably does a pretty good job. HOWEVER, as you’re falling it SEEMS to take forever and the suspense is overwhelming. So you look down to see whats taking so damn long. I jumped off of a 64 ft bridge (at low tide) with a friend once (video taped by a third) who did just that, as the video clearly shows. I suffered no more trauma than I would of off of a high diving board, my friend ended up in the hospital with displaced clavicle, fractured collar bone, concussion and shock trauma. He had to stay in that hospital 2 weeks before they could move him to one closer to his home. To date I have made 12 jumps off of that same bridge with no injury.
My highest measured jump is 102 ft at Lake Powell. other than the enema (I was wearing only a ‘Speedo’ type bathing suit) I came through unharmed. And LP is damn cold!
And regardless of what calculations ‘prove’, hitting the water from these heights is nothing like hitting cement. the “shock” of hitting is negligible. There is no sudden stop.
I won’t run the math, but I’m guessing the speed at which you hit the water from 102 isn’t that much less than 220. Even so, I’ve still never been hurt since refining this method, so I believe I could go 220 o.k.
Disclaimer: I never jump in high wind.
Finally, there is a case of a military man ejecting over the sea, chute didn’t open, fell 2 MILES (!) and lived (Reader’s Digest).
warmgun, something in your post reminded something posted by darkcool once.
Feel free to read the post here and let me know if you pick up on the subtle similarity between these posts.
You two should hang out.
You’re right - seems like a cool guy. But I only hang with the 100+ club. For those of you who haven’t tried Lake Powell, you gotta try it!
Johnny L.A. Said:
I don’t think that’s correct. The way to model a falling object is to take the force due to gravity and mass, and subtract the force from air friction. The resultant force vector defined both the terminal velocity and the acceleration rate of the object.
As I said, I’m not an expert. What I was thinking of at the time was the famous experiment where Galileo purportedly dropped two balls of different weights from the Tower of Pisa. Although their weights were different, they hit the ground at virtually the same time: http://www.endex.com/gf/buildings/ltpisa/ltpnews/physnews1.htm
Predict the next Skyway Bridge jumper. Win a t-shirt!
Yes, it’s in poor taste, but I still found it kind of funny and the information within is neat. On the other hand, sites with only lower case letters make me want to jump off a bridge.
“BTW some guy on tv said its best to land on your back.”
His reasoning was simple. He simply stated that if you did that any part of the rest of your body could be fixed.
Also, if you noticed stunt men always land on their backs when then jump from high buildings onto a pillow or whatever they use to land on.
I’m allegedly a fluid dynamicist, so I decided to try hacking together some kind of more detailed analysis of this problem. Partially for my own amusement, but here goes:
I submit that any ill effects caused by smacking the water are not caused water’s incompressibility. Air is effectively incompressible, too, at speeds below Mach 0.3 (225 mph at sea level). True, water “doesn’t want to get out of your way,” but that’s not an incompressibility effect, that’s what drag is. The density difference (factor of 1000) plays a big role in that - a little packet of water in your way has 1000 times more mass and, hence, inertial resistance to getting the hell out of the way as the same volume of air.
I think it’s simple hydrodynamic drag causing you to slow down far faster than your body can really stand.
Assumption #1: That the drag coefficient of a diver/object/whatever will remain essentially constant
Water and air, in these speed ranges, are both well-approximated as incompressible Newtonian fluids, so the only relevant differences between their behaviors will be their different densities and viscosities. The beauty of this is that a Reynolds number (Re) vs. drag coefficient (C[sub]D[/sub]) relation for a given object is applicable in any incompressible Newtonian fluid - property differences between the fluids are wrapped up in the definitions of Re and C[sub]D[/sub]. Both Re and C[sub]D[/sub] depend on the properties of the object, the speed at which it’s moving, and the properties of the fluid.
For most objects, C[sub]D[/sub] changes with Re. However, the changes are generally not too drastic. When an object moving in air at some speed abruptly enters water, the property differences will cause Re to jump by a factor of about 10. Looking at a chart I have handy, this should not cause a drastic change in C[sub]D[/sub]. Perhaps by a factor of 3 or 4, but not 10 or 100. I’m going to assume that C[sub]D[/sub] is constant for a human diver.
This is useful. C[sub]D[/sub] is defined as (drag force)/(0.5rhoU[sup]2[/sup]*A), where rho is the fluid density, U is the velocity, and A is (usually) the frontal area of the object. In other words, if C[sub]D[/sub] is constant, drag force is exactly proportional to U[sup]2[/sup] - in the same fluid. When we suddenly switch from air to water, drag force will change as the density changes - which is a factor of 1000 between air and water. Drag force will suddenly multiply by 1000!
What Assumption #1 Means: drag force on an object is proportional to density[sub]fluid[/sub]U[sup]2[/sup] in any fluid. This has been demonstrated in experiments to be more or less true for many objects, and I’ll assume it’s true for a human.
OK, moving on to a human, diving in a streamlined position. This thread mentions 200 mph (83 m/s) as the terminal velocity of vertically oriented human, so I’ll use that for now.
If he’s at terminal velocity, than the drag force is exactly equal to his weight. Assume he weighs 170 lb, or 77 kg. I don’t know his frontal area, and I don’t want to try estimating it, so I’ll just leave it as unknown, and I’ll look for his C[sub]D[/sub]A, which will also be constant from air to water.
C[sub]D[/sub]A for this guy is (drag force)/(.5rho[sub]A[/sub]U[sup]2[/sup]) = (77 kg)(9.81 m/s[sup]2[/sup])/(0.5(1 kg/m[sup]3[/sup])*(83 m/s)[sup]2[/sup]) = 0.191 m[sup]2[/sup].
This figure, we expect to be the same in water. Take into account the different density of water (rho[sub]W[/sub]) and we get this relation for drag in the water:
D[sub]W[/sub] = 0.5*(rho[sub]W[/sub])C[sub]D[/sub]AU[sup]2[/sup] = 0.5*(1000 kg/m[sup]3[/sup])*(0.191 m[sup]2[/sup]) = (95.5 kg/m)U[sup]2[/sup]. We’re assuming that this will apply for all speeds as he slows down.
Newton’s 2nd Law has F=ma, so his deceleration rate will be that figure divided by his mass: a = (1.24 m[sup]-1[/sup])U[sup]2[/sup]. This is assumed to apply at any U because of constant C[sub]D[/sub], so it’s valid even if he hits the water at less than that, and it’s valid as he slows down.
This creates a nice differential equation: x’’(t) = -(1.24 m[sup]-1[/sup])(x’(t))[sup]2[/sup] for his travels beneath the water (note that it’s nonlinear). I’ll spare the details, but it’s solvable - plugging in initial conditions (x(0) = 0; x’(0) = v[sub]0[/sub], the impact velocity), here’s what I get. Call K = 1.24 m[sup]-1[/sup] for brevity:
x(t) = (1/K) * log[sub]e[/sub](Kv[sub]0[/sub]t + 1)
v(t) = v[sub]0[/sub] / (Kv[sub]0[/sub]t + 1)
a(t) = -Kv[sub]0[/sub][sup]2[/sup] / (Kv[sub]0[/sub]t + 1)[sup]2[/sup]
Suppose he hits the water at 80 mph, a figure we’ve been using for the GGB. 80 mph = 36 m/s. I’m having a MATLAB routine plot those three curves, and I see that he will slow down damn quick. After 0.2 seconds, he’s down to about 4 m/s, is down to about 2 g’s of deceleration, and is about 1.8 meters deep.
The problem is time before that. Initial deceleration is about 160 g’s, even though it drops off very fast. Time-average deceleration over that first 0.2 seconds is 16.3 G’s, which is a lot, although it might be survivable.
Now, I’ve ignored buoyancy, but I think that’s justified until drag deceleration gets down to around 1 G, since buoyancy will surely be less than that - it looks like it’ll happen within the first half-second of so, but you may be dead by then.
It’s also true that a falling human is almost 2 meters long, so what happens within the first 2 meters will certainly be a bit distorted. I think that that may help, but my suspicious is that it won’t be much. At 36 m/s, somebody goes 2 meters in 0.05 seconds, so it’ll only delay things a tad. Well before 0.2 seconds is up, he’ll be completely in the water and exposed to this wicked deceleration.
Even if my main assumption is wrong, and that C[sub]D[/sub] does change somewhat with Re, I think the point is this: suddenly getting exposed to a fluid that’s 1000 times more dense than air, and thus drag forces 1000 times higher, results in some nasty G-forces on your body. Since they’re proportional to U[sup]2[/sup], what isn’t so bad at low speeds becomes lethal at higher speeds - much more quickly than our intuition might lead us to expect.
God this was long. I’d be interested in hearing from anybody who finds mistakes or has issues with the assumptions here - they are significant, but (I think) justifiable.
And you voluntarily wrote all that for your own amusement?
Wow :o, no shit it’s long. Now, do you mind translating that into layman’s terms for those of us who aren’t fluid dynamicists?
This could also be because the ocean side of the bridge is closed to pedestrians. It’s open to bicyclists on the weekends.
The last time I walked across the bridge, over on the bay side and back on the ocean side, in 1963, there was a suspended walkway under the bridge on the Marin County end. I guess they’ve not only closed that, but the ocean side walkway as well. That day I was stopped by a black and white on my way back asking me what I was doing. Someone had forgotten to lock the walkway gate at 6pm. In any case, that still leaves at least 26 years for jumpers to go over on the ocean side. Maybe the jumpers just want to be sure their body is found and not washed out to sea.