Whoops: one additional thing I was going to note that confused me the first time around. The F[sub]D[/sub] coefficient is a function of Reynolds number, which is itself a function of velocity. So the drag force winds up being a function of velocity squared, like it should be.
Your calculations are irrelevant. If you’re only slightly heavier than air, you can jump slowly, keeping your acceleration well below the maximum upward acceleration of the balloon, keeping the string taut the entire time.
Address this point before you claim nobody’s refuting your argument, or you’re just being disingenuous.
zut, I am not even going to attempt to understand your first link. It is attempting to create a model for very specific conditions: tiny bubbles of air (1mm) rising in a narrow tube (r=2cm). They present the whole idea as an hypothesis (which I think is not applicable to general cases) and in the conclusions page they say
That model is unnecessarily complicated for the simple problem we are studying here and furthermore, the authors recognise the results are invalid. On top of that you are probably interpreting the formula wrongly. You cannot just use formulas without understanding their validity, limits and meaning.
Well, first of all I graphed the equation from my earlier post on a graphing calculator varying the radius of the balloon from 1m to 20m, and using densities of air and helium of .0755 and .011. The following are the velocities the balloon reaches before the acceleration drops below 1g.
R = 1m………. v = 3.0 m/s
R = 5m…………v = 6.5 m/s
R = 10m……… v = 9.5 m/s
R = 20m……… v = 14.0m/s
Plus I don’t think the air has to drop from the top of the balloon to the bottom. My guess is that at the bottom the air pressure propelling the balloon automatically takes care of this.
Yeah, but where do you think that air pressure comes from? The gravitational force felt by the air above it.
Yes, but in the page I linked to, they are using the theoretical particle force balance for a general particle. Scale effects are tucked into the Reynolds number, and as far as I can see there’s no particle size restriction on the equations. The wall effects of the tube are added to the model later (at the bottom of the page, in fact). The force balance equation should hold for any particle; just because the authors apply it to small bubbles in a tube doesn’t mean we can’t apply it to larger objects in the unbounded atmosphere.
Their bubble-in-a-tube results are suspect because of wall interactions and questionable computational modeling. That doesn’t invalidate the theoretical model. Neither is the model “unnecessarily complicated,” particularly when you use it to calculate max acceleration as I did above. I’m surprised you would suggest that.
Would you like to point out specifics where I’m “probably interpreting the formula wrongly”? Or would you prefer to devolve into playground “Yes it is”/“No it isn’t” taunts? If the latter, I’m not particularly interested in playing.
Perhaps you could consider the following argument: It seems intuitive to me that a long thin body should accelerate faster than a spherical body for the same reason that rowing sculls are long and thin. So surely the underlying model should have some dependency on shape, correct?
Here is the same model (pdf) applied to a balloon (a balloon on Venus, actually, so it’s doubly cool). The equation of motion used is:
(M[sub]tot[/sub] + C[sub]m[/sub][symbol]r[/symbol][sub]a[/sub]V)(d[sup]2[/sup]z/dt[sup]2[/sup]) = g([symbol]r[/symbol][sub]a[/sub]V - M[sub]tot[/sub]) - (1/2)C[sub]D[/sub]A[symbol]r[/symbol][sub]a/sub[sup]2[/sup]
Where
d[sup]2[/sup]z/dt[sup]2[/sup] = balloon acceleration
M[sub]tot[/sub] = total balloon mass
V = balloon volume
C[sub]m[/sub] = virtual mass coefficient for the balloon (0.5 for a sphere)
[symbol]r[/symbol][sub]a[/sub] = atmospheric density
g = gravitational acceleration
and the last term on the right is the drag term.
If you ignore drag and solve for acceleration, you get the same equation I posted above (only on a mass basis rather than density). Comments?
Yes, but at a specific height that pressure is homogenous throughout the volume. Iow it doesn’t just depend on the cross section of the air directly above the balloon.
I wonder why I’m even participating in this thread, I don’t know diddly about this stuff.
I plugged Ring’s equation:
with a 20’ diameter balloon (isn’t that about what parabounce was using?) and plotted acceleration and time. The balloon starts out w/ a max acceleration of about 57 m/s^2, and slows to zero acceleration after about 1.5 sec and a terminal velocity of about 11m/s.
If I change the equation to add the mass of the air displaced, the acceleration goes from about 7.25m/s^2 to zero in about 12 seconds. And velocity becomes terminal about 11 m/s.
I’m still on the fence as to which is correct. I think the “can’t go faster than g” group is not realizing how quickly the acceleration falls below zero due to the drag (see above). I think when some people are saying, “well, you have to include the weight of the air that’s being displaced”, that that force you’re talking about is the same as the drag force, which increases as speed squared. Doesn’t it make sense that the faster you push something through a liquid (or gas), the faster you push it, the greater the drag is, or looking at it from the other perspective, the faster that air (or liquid) that’s being displaced has to move around it? And form definitely plays a big part. A square balloon will not rise as quickly as a round one, which won’t rise as quickly as a torpedo shaped one.
Hopefully the folks sailor wrote to will respond. WHERE’S CECIL WHEN WE NEED HIM???
In any case, it doesn’t change the answer to the OP, which is that the balloon will make the weight rise above it’s orginal trajectory (if the buoyancy of the system is fairly close to zero, but still negative, i.e., it will stay on the ground, taps notwithstanding, which would make it rise, but whereafter it would then fall back to ground)
I plugged Ring’s equation:
with a 20’ diameter balloon (isn’t that about what parabounce was using?) and plotted acceleration and time. The balloon starts out w/ a max acceleration of about 57 m/s^2, and slows to zero acceleration after about 1.5 sec and a terminal velocity of about 11m/s.
If I change the equation to add the mass of the air displaced, the acceleration goes from about 7.25m/s^2 to zero in about 12 seconds. And velocity becomes terminal about 11 m/s.
I’m still on the fence as to which is correct. I think the “can’t go faster than g” group is not realizing how quickly the acceleration falls below zero due to the drag (see above). I think when some people are saying, “well, you have to include the weight of the air that’s being displaced”, that that force you’re talking about is the same as the drag force, which increases as speed squared. Doesn’t it make sense that the faster you push something through a liquid (or gas), the faster you push it, the greater the drag is, or looking at it from the other perspective, the faster that air (or liquid) that’s being displaced has to move around it? And form definitely plays a big part. A square balloon will not rise as quickly as a round one, which won’t rise as quickly as a torpedo shaped one.
Hopefully the folks sailor wrote to will respond. WHERE’S CECIL WHEN WE NEED HIM???
In any case, it doesn’t change the answer to the OP, which is that the balloon will make the weight rise above it’s orginal trajectory (if the buoyancy of the system is fairly close to zero, but still negative, i.e., it will stay on the ground, taps notwithstanding, which would make it rise, but whereafter it would then fall back to ground)
I SWEAR I only “Submit” once!!! It’s not my fault!!!
I talked with my co-worker with the bubble background today and here’s what he said:
The proper formulation of the buoyancy force is Vg([symbol]r[/symbol][sub]f[/sub] - [symbol]r[/symbol][sub]p[/sub])/[symbol]r[/symbol][sub]p[/sub], where f and p are fluid and particle, respectively. This force is balanced by drag in the steady state. For accelerating bodies, you need to include the effect of the fluid acceleration; this effect is modeled using “virtual mass.”
The virtual mass is slipped into the equation as I’ve shown above; it’s modeled as a fraction of the displaced mass of the bubble. The exact fraction (1/2 in the standard case of a sphere in an infinite field) is determined by integrating the acceleration field in the fluid around the body. Different shapes have different coefficients; additionally, virtual mass coefficients are changed by the presence of walls.
Although his experience with this formulation is applied to bubbles, there’s no assumed scale, so it’s equally well applied to larger bodies (although he said very large bubbles have odd surface tension effects that must be included; I think we can ignore these).
Finally, the argument that buoyant bodies rise at a max acceleration of g because surrounding fluid falls at that acceleration is incorrect. The rising body slips through the surrounding fluid. The fluid is accelerated, in all directions, but there’s no bulk fluid acceleration downward at g. The virtual mass term takes care of the acceleration.
So let me recap: Using a buoyancy force of Vg([symbol]r[/symbol][sub]f[/sub] - [symbol]r[/symbol][sub]p[/sub])/([symbol]r[/symbol][sub]p[/sub] + [symbol]r[/symbol][sub]f[/sub]) is incorrect. Using a buoyancy force of Vg([symbol]r[/symbol][sub]f[/sub] - [symbol]r[/symbol][sub]p[/sub])/[symbol]r[/symbol][sub]p[/sub] to calculate maximum acceleration without including virtual mass is also incorrect. Maximum acceleration of a buoyant sphere is 2g.
zut! That’s great! But…unless I’m screwing something up here, the units don’t work out for what you’re calling “buoyant force”? (m^3m/s^2kg/m^3)/(kg/m^3) = m^4/s^2 which is not a measure of force? I imagine you’ve just mixed up your terminology or something, but can you make sure you’ve got the words right, or the formula right? (Also, I’m assuming the V in your equation = volume?)
Whoops! Right you are. Lesson: always check units. Let’s try this:
The proper formulation of the buoyancy force is Vg([symbol]r[/symbol][sub]f[/sub] - [symbol]r[/symbol][sub]p[/sub]), where f and p are fluid and particle, respectively; g is the gravitational acceleration and V is volume. Or, if you prefer, the acceleration component due to buoyancy is g([symbol]r[/symbol][sub]f[/sub] - [symbol]r[/symbol][sub]p[/sub])/[symbol]r[/symbol][sub]p[/sub]. Substitute the correct formulation into the rest of the above post as needed.
Also, this gives me a chance to note that the text I cite in Exhibit B above gives a table of virtal mass coefficients for ellipsoids, so you can get a feel for dependence of virtual mass on shape:
length/diameter virtual mass coeff.
0.1 0.960
0.2 0.895
0.5 0.702
1.0 0.500
2.0 0.209
5.0 0.059
10.0 0.021
The upshot being that, since the max acceleration of a body = g([symbol]r[/symbol][sub]f[/sub] - [symbol]r[/symbol][sub]p[/sub])/([symbol]r[/symbol][sub]p[/sub] + C[sub]m[/sub][symbol]r[/symbol][sub]f[/sub]), the max acceleration for very light bodies ([symbol]r[/symbol][sub]p[/sub]->0) is near g for squatty bodies, is 2g for a sphere, and gets huge for long thin bodies (is near 50g for a 10:1 ellipsoid!).
Excellent! Thanks! It only took us 4 pages to get to the bottom of this!