Can you give us an example?
Also, drag is generally used to describe friction forces and not forces from ther inertia of an accelerating mass but I am not going to argue about that as long as we all know what we are talking about. As long as we understand each other I’m fine with any notation.
All you have to do is plug numbers into the formula. Use a large density difference and a large radius. I’ll do it when I get some time but that isn’t right now.
**
Even in a completely frictionless medium you have drag; as an object forces it’s way through the material it displaces particles of the medium and you get a reaction force to the F = ma of the displaced particles.
Oops, sorry Sailor I just realized that that’s basically what you said, but drag definitely includes these retarding forces.
Jeez, I am so stupid. The fomula I gave above is for velocity not acceleration… let me start another search. Sorry.
Okay, I believe it then. Thank you.
Note that at terminal velocity a=0. There is no acceleration and the only “drag” is caused by friction.
OK, from the first site listed below you can derive the equation for the upward acceleration of a sphere in a fluid.
a = g [ (p[sub]fluid[/sub] / p[sub]sphere[/sub]) –1] – (1/2)(p[sub]fluid[/sub] / p[sub]sphere[/sub])C[sub]d[/sub](A[sub]s[/sub]/V[sub]s[/sub])*U[sup]2[/sup]
p = density
C = coefficient of drag
U = velocity
A = cross section of the sphere
V = volume of the sphere
From the second listed site if the density of water is 1 then the density of a helium filled balloon is approximately .00018. (does this sound reasonable?)
It appears to me that the upward acceleration would initially be much greater than g but would then decrease until the buoyancy force is equal to the drag force. Please correct me if I am seeing this incorrectly.
http://www.me.umn.edu/courses/me3031/S03/manual/lab1A2.pdf
http://www.elmhurst.edu/~chm/vchembook/123densitygas.html
No that’ not right. As long as the balloon’s moving it’s accelerating the particles of the medium out of it’s way.
Dammit I just can’t seem to get this right. The above should read "until the first term in the equation equals the second term. My only excuse is that I know practically nothing about fluid dynamics so this is all just winging it.
Yes you are seeing incorrectly but you still do not post any concrete example so I cannottell you where you are wrong. It is difficult to argue against generalities.
a = g [ (pfluid / psphere) –1] – (1/2)(pfluid / psphere)Cd(As/Vs)*U2
On the second side of the equation we have two terms. The first one represents the acceleration due to gravity and is exactly what I said. It cannot be greater than g. The second term represents friction and can only decrease the first term so, there is no way in the world a can be greater than g.
I have already posted the density of air and of helium and calculated the acceleration neglecting friction which is the first term of the equation. Consider now friction and you get a lower a.
I just don’t know where you get a>g. can you do the actual calculations step by step so I can follow them?
Before blindly applying a formula you need to understand what it represents and how it works. This is how that page explains the formula we are using:
The maximum acceleration achieved at the moent of release is represented by the first term of the second side and can never be greater than g no matter how you slice it. And then a decreases until it is zero at terminal velocity.
Sailor please keep in mind that you made a general statement that buoyancy could never accelerate anything faster than g. This general statement of yours should apply equally well to a helium filled sphere in a water medium.
The density of water = 1
The density of a helium filled sphere = .00018
Instantaneously after release the velocity equals zero and the acceleration equals:
a[sub]inst[/sub] = 9.8 ( 1 / .00018 – 1) = 54434 m/sec[sup]2[/sup]
This is slightly greater than 9.8m/sec[sup]2[/sup]
Or from the below site the density of helium is .011 pounds per cubic foot and the density of air is 0.0755 pounds per cubic foot
a[sub]inst[/sub] = 9.8 ( .0755 / .011 – 1) = 57 m/sec[sup]2[/sup]
Once again I’m not sure if I’ve made any errors please correct me if I’m wrong.
Also, in most advanced physics texts you never see any examples worked out using numbers. Instead you derive and study abstract equations, so your statement *” but you still do not post any concrete example,” * doesn’t hold any water.
Ring, I do apologize as I had looked at the formula quickly and missed something but now that I have looked at it again I believe it to be in error and I will explain why. As I cannot do greek letter and such, let us call df the density of the fluid and ds the density of the sphere.
In that formula that first term is the same a (df-ds)/ds . In other words, it considers the force actuating as the difference of the weights, which is correct. But it considers the mass to be accelerated as only the mass of the sphere which is incorrect. I believe the correct formula is (df-ds)/(df+df) which I posted above and is supported by the webpage mentioned above.
The formula as given in that page produces absurd results. A bubble of air would have an a=806g which is patently ridiculous. A bubble of helium would have a=6097g which is even more preposterous. Do you want to calculate the a for a bubble of H? It defies our evidence and it defies common sense because it implies water is accelerating faster, much faster, in its fall when it is replacing the void left by a bubble than when it is falling in a vacuum where it does not have to replace anything. It contradicts common sense. And it defies the observation of anyone who has blown bibbles underwater. They do not have an acceleration faster than g. My formula (supported in the other web page, predicts a bubble of air will rise with a=0.997 *g which makes sense. The fall of the water is slowed a bit by the bubble it is displacing. The author of that document blew it. I am emailing the authors to let them know my view and see what they say.
I’m still working on the air that the balloon displaces having to fall. It seems to me that the balloon will create a slightly high pressure in front and a slightly low pressure behind and simply push the air sideways as it passes and then the air will fill in the low pressure area behind without much verticle motion of the displaced air.
“verticle” = “vertical” in this case.
No the equation is correct. What you’re forgetting is that these high acceleration values are rapidly damped down by the drag force which is proportional to v[sup]2[/sup]. I gotta run but I’ll do some figuring tomorrow. In the meantime I’m sure you can get an idea of what’s happening by going back and playing with the equation for the terminal velocity.
I don’t think you can jump higher, but you can probably broad jump further. The balloon will not accelerate at 1 g. As it rises most of the buoyant force is used up in accelerating the air out of its way. And as Ring said that drag force rises as the v[sup]2[/sup] because higher velocity means more acceleration to the surrounding air.
So when you jump the line goes slack. As has been said during the time the line is slack the balloon rises a little so that when you start down your descent is slowed. You didn’t jump any higher though. So if you jumped straight up you wouldn’t notice any difference. However if you jumped with some forward speed your slowed descent would give that forward speed a longer time to act and you would travel further forward than in the no-ballon case.
I think.
So imagine the bubble in the water. If it starts accelerating with an acceleration greater than g you have two possibilities:
a) Water is accelerating down with a=g or less in which case it is not taking the space of the bubble fat enough and the bubble is leaving a vacuum behind. What exactly is pushing the bubble upwards in this case?
b) Water is accelerating downwards with an a equal to the a of the bubble. How can water fall many thousand times faster when it has to displace a bubble than if it were falling in a vacuum and there is no bubble. It is the falling fluid which makes the bubble rise.
Anyay, I have emailed the author of that paper and I will let you know what happens.
Just to reiterate: I believe the correct formula is a= (df-ds)/(df+ds) as it appears in other places.
I defer to those with a greater knowledge of physics (which seems to be everybody else in this thread), but would the follwing method work?
Assume a 200lb man. Rig him up with 20’ of cord (as inelastic as you want) and balloon so that if Man + Rig steps on a scale, scale reads 20lb.
Man holds cord just below where the balloon is attached and squats down. Man releases balloon which rises at whatever rate it rises. Just before the cord goes taut, Man jumps with all his strength.
In my mind, this would produce a Hulk-like leap by the Man. The only question I have is the Man’s ability to hold the balloon. Would this be equivalent to trying to lift a 180lb weight? In other words, not impossible but very taxing?
Oh man, am I gonna throw a wrench in the works. I started looking up bubble accelerations, and… it looks like everyone’s wrong. Apparently, rising particles can accelerate faster than g, depending not only on the difference in density between the particle and surrounding fluid, but also the shape of the rising particle.
Here’s Exhibit A, Particle force balance equation used in FLUENT (commercial fluid dynamics analysis software): The equation of motion used in FLUENT is
du[sub]p[/sub]/dt = F[sub]D[/sub](u - u[sub]p[/sub]) + g([symbol]r[/symbol] - [symbol]r[/symbol][sub]p[/sub])/[symbol]r[/symbol][sub]p[/sub] + F[sub]x[/sub]
where the subscript “p” refers to the particle in question, so:
u[sub]p[/sub] = velocity of the particle,
u = velocity of the surrounding fluid,
du[sub]p[/sub]/dt = acceleration of the particle,
F[sub]D[/sub] = drag force coefficient,
g = gravity,
[symbol]r[/symbol] = fluid density,
[symbol]r[/symbol][sub]p[/sub] = particle density, and
F[sub]x[/sub] = a virtual force due to the “virtual mass” of the fluid; for a sphere,
F[sub]x[/sub] = 1/2([symbol]r[/symbol]/[symbol]r[/symbol][sub]p[/sub])*d/dt(u - u[sub]p[/sub])
What does this all mean? Well, in still fluid (u = 0), and for low velocities, like on takeoff (u[sub]p[/sub] = 0), the EOM simplifies to:
du[sub]p[/sub]/dt = g([symbol]r[/symbol] - [symbol]r[/symbol][sub]p[/sub])/[symbol]r[/symbol][sub]p[/sub] - 1/2([symbol]r[/symbol]/[symbol]r[/symbol][sub]p[/sub])*du[sub]p[/sub]/dt
or
du[sub]p[/sub]/dt = g([symbol]r[/symbol] - [symbol]r[/symbol][sub]p[/sub])/([symbol]r[/symbol][sub]p[/sub] + 1/2[symbol]r[/symbol])
And that means, for very light spherical particles ([symbol]r[/symbol][sub]p[/sub] ~= 0), that the maximum acceleration = 2g. Moreover, this max acceleration depends on the shape of the particle (or balloon, if you will). This wasn’t what I was expecting, but it does make some intuitive sense to me: a long slender balloon will be “moving” less mass around at a given time than a lumpy-shaped balloon.
And, for those of you who want even more background, I present Exhibit B, Textbook deriving equations of motion for bubbles. If you scroll down to section 5.10 (equations of motion), you’ll see the following:
Noting the change in nomenclature, this agrees with the FLUENT page quoted above. This conclusion is supported by extensive derivation and citation, so I tend to think it’s pretty definitive.
However, I must admit that the subject matter in the derivation is more than a little out of my area of expertise, so I can’t comment on it’s validity directly. I do, though, work with a guy who got his PhD working on cavition bubbles, so I’ll ask him for input on Monday.