You also have to consider that if the human is suspended within the ball on flexible restraints, the G forces will be greatly reduced. Instead of the deceleration being from 80 to 0 mph within the distance of a couple of inches if you hit dirt, the deceleration is 80 to 0 over perhaps a distance of three or four feet. HUGE difference.
Could be a great sport. How about dropping them out of low-flying airplanes over the ocean? They can call it ‘skipping’.
Personally I think you’d need enormous balls to jump out of a plane without a parachute…
…HUGE cojones in fact!
I’ve cleaned a Hamster’s cage. Believe me, there’s plenty of “p” in a Hamster.
This is one of these great SDMB threads that when you visualize what people are talking about it ends up being incredibly hilarious.
I mean, just imagine in it: The army of the future - a huge plane flying overhead dropping giant hamster balls on the ground. They bounce around a bit, and then the soldiers inside take off running en mass towards the terrified enemy. The opposing army has no idea what to do as it sees hundreds of giant hamster balls approaching at alarming speed…
As I understand it, terminal velocity is an irrelevant concept in a vacuum. TV represents the equilibrium point between decceleration due to air resistance (or the equivalent for other mediums) and acceleration due to gravity. Without frictional resistance, there can be no equilibrium and the law of gravitation requires that velocity increase without limit.
No, there is a very definite limit to the velocity of a body in free fall. That limit is equal to the escape velocity at the surface of the parent body. As the distance of the fall increases, the velocity at time of impact asymptotically approaches the escape velocity.
You don’t want something so solid that it will absorb 0 G’s. That’s why cars use airbags instead of concrete blocks.
…assuming that the body starts from rest. You can get an arbitrarily high velocity by giving yourself a downwards initial velocity, but you’re still in free-fall the entire way. </nitpick>
thckhd:
I get the same answer as you, given your assumptions, for the initial radius - 109 km.
However, I was always under the impression that aerodynamic drag was proportional to the square of the velocity, and Stokes’ law doesn’t support that. So I did a bit of Googling and according to here…
http://farside.ph.utexas.edu/teaching/329/lectures/node70.html
Stokes’ law is only valid for low velocity spheres. There is a second drag term proportional to v[sup]2[/sup] that dwarfs it.
I don’t really understant your buoyant force - isn’t it exactly balanced by the weight of the air inside the sphere? Of course, if the sphere is filled with a lighter-than-air gas or a vacuum then you may not hit the ground at all, leaving only the problem of asphyxiation to deal with…
OK people, lets fix these calculations. First, Q.E.D., speed does not asymptotically approach escape velocity. It would asymptotically approach zero if you left the Earth with that speed (and there was no drag).
Second, theckhd, the buoyant force you are using is incorrect. You are making the implicit assumption that there is a vacuum inside the sphere, which there is not. Also, Stokes Law is only good for Reynold’s numbers less than about 1; way lower than we are dealing with.
Now I will make some mistakes myself for everyone else to pick apart:
We really need to combine the equations for drag, to determine the terminal velocity of any size sphere, and the g-forces experienced for that size. No real sphere will match the idealized case, so we will be setting a minimum bound.
The equation we should be using for drag is based on an experimental determination of Cd (coefficient of drag) as it is highly dependent upon velocity and the resulting wake structure. Spheres will have a Cd between 0.5 (very small Re) to 0.07 (higher subsonic Re). We will go with the 0.07 as we are working with very high Reynold’s Numbers.
Cd=2D/(rhoAV^2) where A is the frontal surface area, or pir^2 and D is drag.
For our deceleration equation we have 2Gg*r=V^2, where G is the constant g-force experienced by the jumper. (The force will not really be constant, but we will assume so for this case, not to mention we can’t compress the full radius.)
The maximum survived g-force on record was 179 Gs, but that is a bit higher than we want to go. The human body can withstand very high forces for very short times so long as it is well supported. Fighter pilots train at up to 15 Gs for very short banks. As we are not expecting out of shape people to be taking part in this sport, and the force will be over a VERY short time, 10 Gs should be good.
Finding terminal velocity when D=mg, and using the earlier assumption of the ball having the same mass as it’s occupant we have the constants:
m=136kg
g=9.8m/s^2
rho=1.25kg/m^3
Cd=0.07
G=10
that we put into the equation:
r^3=m/(Cdrhopi*G)
to get:
3.7 m radius sphere.
We have a terminal velocity of 10.5 m/s (25mph) and a deceleration time of 0.7 s.
Damn, the problem with long detailed posts is that you get scooped. Good job, matt.
It’s reversible. You can state it either way, and it means precisely the same thing.
Well, I’ve done MY calc and I’m damned well going to post it, even if flight’s is better :
To continue, the site I linked gave an expression for the additional drag force as:
F=[sup]1[/sup]/[sub]2[/sub] CpAv[sup]2[/sup] where C is the drag coefficient, p is the density, A is the cross sectional area and v is the velocity.
The site below gives the drag coefficient of a smooth sphere as 0.1
http://aerodyn.org/Drag/tables.html
Using air density of 1.29 kg/m[sup]3[/sup] we get:
mg = 0.05 x 1.29 x pi a[sup]2[/sup] x v[sup]2[/sup]
I’m going to put my guy in an inflated thin walled plastic balloon and try and slow him to 11mph, which I vaguely recall is parachutist ground impact speed. That’s 5 m/s and I’ll assume balloon and person together mass 100kg.
1000 = 0.05 x 1.29 x pi a[sup]2[/sup] x 25
Which gives a = 14m, and the guy can stand on the bottom of the ball and do a parachute roll as he hits the ground.
Great thread. On the subject of “extreme sports,” don’t forget that NASA has been using essentially this type of technology to land the most recent round of Mars probes. So it’s hardly an insurmountable engineering challenge.
Nitpick: In Armour of God, Jackie climbs into a Zorb and rolls down a hillside (and then a steep cliff) to escape a horde of natives angry at his theft of an artifact. It isn’t dropped from an airplane or anything; he climbs into it on the ground and frantically hamsters away. Oh, and for the record, Armour of God was a huge hit in Asia; it’s just too early (and too HK-clunky, IYKWIM) to have gotten a big-screen release in the West. Far from “less successful.”
(It’s worth seeing, too, as long as you avoid any of the American-aimed cut versions. Get the HK original. Jam-packed with action, one of Jackie’s better efforts. And it’s the movie that contains the stunt that came closest to killing him; the hole in the skull you sometimes read about was suffered in a relatively straightforward jump from a wall to a branch.)
True, but they also use aerobraking, 2 parachutes, and a set of retrorockets before using the air bags. That would be called “cheating” in our extreme sport.
Perhaps I am misunderstandin, but I don’t think it does. You asymptotically approach zero velocity when leaving the Earth at the right speed, but when falling you ACTUALLY reach escape velocity rather than just approach it. I don’t think that qualifies for the definition.
Oh, and any spher that is going to take the impact is probably going to weigh a lot more than we estimated.
matt: You are indeed correct, there are other terms that i neglected, and stokes’ law is not completely valid at large velocities. I wasn’t about to go through the full problem for a back-of-the-envelope calculation though (I have preliminary exams this week, and have slightly more important problems to be working on).
flight: I am doing no such thing as assuming there is a vacuum inside the sphere. I included the mg term of gravity on the other side of the equation. The buoyant force on the sphere is exactly as i represented it. What i neglected was including the mass of the air inside the sphere in the mg term. I did this for two reasons:
The buoyant force, however, is correct. Buoyant force is ALWAYS equal to the mass of fluid displaced by the object, which is exactly what i have it listed as. Any material inside the object (including trapped air) is considered part of the object’s mass.
You are right that Stokes’ law is only valid for small Reynold’s number. However, my calculation was only an order of magnitude estimate, as i stated in the post. I was just seeing if it was remotely possible for such a sphere to work, and since we all ended up with estimates around 5 to 15 meters, it seems that my simple model managed to do its job. Your calculation (and matt’s) are much more accurate and thorough than mine is, of course.
In respect to the question posed by chrisk, i would agree with flight (assuming you start from a far enough distance that the force of gravity is very weak – you don’t reach escape velocity by dropping from a point close to the surface, after all). I don’t think that it would be fair to call it terminal velocity, since terminal velocity is specifically referencing the velocity reached due to the balance of gravity and drag/buoyant force.
Since i’m an Optics guy, i feel obliged to point out that there is a terminal velocity in a vacuum – 3x10^8 m/s. That has nothing to do with drag forces though, so it’s fairly irrelevant to this question. Just a little nit-pick. 
I don’t think you do. As I understand it, no matter how far an object falls, it will never quite reach escape velocity at the instant of impact. It can get damn close for large distances, but it can’t ever exactly equal it, unless you give it an initial velocity other than zero.
I must agree with QED. A freefalling object that has an initial velocity of 0 should not reach or exceed escape velocity before impacting the surface, assuming it doesn’t fall into some deep chasm or something.
Also, how about calculating the weight of the sphere by the radius of the sphere? I think the only significant mass in one will be the bands, not the plastic, so since the bands would only need to be longer, not thicker or more plentiful, to hold in a person of the same weight with the same G’s (am I mistaken here?), it would be roughly equivelant to the radius minus the radius of the inner sphere. So assuming the inner sphere were 2 meters, the total mass, assuming a 75 kg person, should be roughly the following:
m = 75 + 50 * (r - 2)
I estimated 50 kilos for each meter between the two spheres based on what the website listed.
Using this information for drag, we get into the difficult part. This is going to take a while to figure out if I’m able to, so I’ll post again.
Well I got stuck. Maybe someone with better math skills can finish this off for me. Here’s what I have so far.
First I simplified the force of drag into a formula where only radius and velocity need to be given. Again, the inner sphere has a diameter of 2 meters, the person is 75 meters, and the ball weighs 60 pounds for every meter between the two layers. (That last number is estimated for an actual ball.)
F = 1/2 * C * p * A * v^2
F = 1/2 * 0.1 * ((75 + 60 * (2r - 2)) / (pi * 4/3 * r^3)) * (pi * r^2) * v^2
F = 0.05 * ((pi * (75 + 60 * (2r - 2))) / (pi * 4/3 * r) * v^2
F = 0.05 * (75 + 60 * (2r - 2)) / (4/3 * r) * v^2
F = 0.05 * (15 * (5 + 4 * (2r - 2))) / (4/3 * r) * v^2
F = 0.05 * (20 * (5 + 4 * (2r - 2))) / r * v^2
F = (5 + 4 * (2r - 2)) / r * v^2
F = (5 + 8 * r - 8) / r * v^2
F = (8 * r - 3) / r * v^2
F = v^2 * (8 - 3 / r)
Then I obtained a formula for the gravity exerted on the object with only radius needing to be given. It’s basically mass times acceleration.
F = 9.8 * m
F = 9.8 * (75 + 50 * (r - 2))
F = 735 + 490 * (r - 2)
F = 490 * r - 245
I then came up with a formula for terminal velocity (where the last two equal) based off of radius.
490 * r - 245 = v^2 * (8 - 3 / r)
(490 * r - 245) / (8 - 3 / r) = v^2
245 * (2 * r - 1) / (8 - 3 / r) = v^2
v = sqrt (245 * (2 * r - 1) / (8 - 3 / r))
I came up with a formula that would calculate the velocity at which a given size ball would cause 8 G’s. I assumed the inner sphere decelerated at an even rate to 3/4 of the way to the outer sphere.
a = v / (3/4 * (r - 1))
78.4 = v / (3/4 * (r - 1))
v = 58.8 * (r - 1)
Finally, I was calculating the radius of sphere that would have a terminal velocity which caused 8 G’s.
58.8 * (r - 1) = sqrt (245 * (2 * r - 1) / (8 - 3 / r))
(58.8 * (r - 1))^2 = 245 * (2 * r - 1) / (8 - 3 / r)
86436/25 * (r - 1)^2 = 245 * (2 * r - 1) / (8 - 3 / r)
86436/25 * (r^2 - 2 * r + 1) = 245 * (2 * r - 1) / (8 - 3 / r)
86436/25 * (r^2 - 2 * r + 1) = 245 * r * (2 * r - 1) / (r * (8 - 3 / r))
86436/25 * (r^2 - 2 * r + 1) = 245 * (2 * r^2 - r) / (8 * r - 3)
However, I reached the limit of my math skills. If someone can finish this, I would be very interested in what it yields.