Why do balloons hit a wall of air?

[Rosalynn]

This is not a work-related question (at least, I don't think it has any practical application, but I could be wrong!), but it's been bugging me for awhile. Figured HVAC folks would be best equipped to answer it.

Have you ever noticed that, when you hit a balloon with your hand to try to pass it to someone across the room, it goes quite fast, then all of a sudden slows down and drops? It's like it hits a wall.

Balloons don't behave like balls, which gradually slow down due to air resistance. (I can almost remember the math for it from university.) In contrast, the behaviour of balloons in air seems to go through some sort of discontinuity (when they hit that "wall&quot. Why is that?

I know it has something to do with the relative size of the balloon in comparison with its low mass. Also, the properties of air obviously control the situation. But which properties, and how, and why?

I'll be watching eagerly to see what people say. Equations, etc. would be fantastic.

Thanks a lot,
Rosalynn

 

[IRstuff]

Why would you think that it's not air resistance?

Weight to volume of any balloon is probably 50(?) times that of a ball. This is aggravated and additionally masked by the fact that the same applies to the effects of gravity, since a balloon can actually be stopped by air resistance before hitting the ground.

In fact, just consider the relative speeds of a balloon and a baseball dropped at the same time. Same "wall" effect; it's just air resistance

TTFN

 

[wilg]

Ratio of Low mass and large surface area. Will look into equations....get back to you next week.

 

[MintJulep]

I think it may be a perception thing.

Both balloon and ball are acted upon by the same forces, namely drag in the horizontal direction and gravity + drag in the vertical direction.

In the vertical direction, the gravity force is F=ma (sometimes known as the weight).

In both directions, drag is proportional to velocity squared.

Horizontal motion stops when the drag force has eaten up the inital momentum.

Vertical motion stops when the thing hits the ground.

For the balloon, the horizontal drag forces are very much greater than the gravity or vertical drag forces.

For the ball, the gravity force is dominant.

Both must follow a ballistic path through the air, but the shape of the parabolla is different.

The balloon stops moving horizontally before it has lost very much height. The ball stops moving vertically before it has lost all of its horizontal momentum

 

[IRstuff]

Remember that given sufficiently high air resistance, you need to have a force that overcomes that resistance in order to maintain motion.

A balloon in a gravity-free environment would be a neutral buoyancy device, e.g., in the absence of gravity, the balloon would stop and simply hang in the air. In the presence of gravity, the balloon falls, but still does not move forward.

TTFN

 

[wilg]

This equation is from Doug Malewicki's
Altitude Prediction charts
the B is Ballistic Coefficient
which is defined as weight divided by
(drag coefficient x frontal area)
a = (initial force applied to an object)
divided by weight) - 1.
t = time after initial force is applied.
It can be seen that if the ballistic coefficient is
low due to insuffient weight of the object,
projectile or whatever, after a very short
time drag force will be greater than the
initial applied force and the object
will loose velocity very quickly.

 

[imok2]

Rosalynn: See what you started!!

 

[wilg]

Rosalynn:
Most of us enjoyed working on your question.
Apparently IMOK (or "I mock&quot didn't.

 

[Rosalynn]

I have read everybody's answers with interest.

wilg and MintJulep came the closest to answering the question I thought I had asked. Thanks, guys!

To wilg, regarding the Altitude Prediction chart equation:
This looks really good. Are the units SI, metric, or Imperial?
I'm assuming V is velocity. Is it in ft/sec?
Would the frontal area be in sq. in.?
I'm assuming drag coefficient is unitless.
a = (initial force/wt.)-1: is that lb-f? and lb-m?
t is probably in seconds.

So, now all I have to figure out is the drag coefficient for a balloon in air--a function of the Prandtl Number, air density, etc. Maybe it's not as difficult as that?

Thanks again,
Rosalynn

 

[Yeldud]

Actually I preferred postings from IRstuff... No equations necessary...

 

[imok2]

Wlig, That was a cheap shot ...you have no Idea what my intentions were..I was just Joshing..and you owe me an appology!!!

 

[wilg]

Imok,
Sorry about that.

 

[imok2]

Wilg, Thank you.

 

[wilg]

Rosalynn -

The units can be either SI system international (metric)
or IP inch-pound (English,Imperial) as long as all units
remain in the same system.
The equation's derivation is too long to go into it here.
(Doug Malewicki was a great mathematician and aerodynamics
expert).
This equation is only good for an object that is launched,
thrown, or propelled straight up, and the velocity is
a function of time during the boost stage.
A second equation is used to determine the velocity
as a general function of time during coasting phase.
I will post that latter if you are interested.
V = Velocity (feet per second)
B = Ballistic Coefficient is defined as weight (W) in
ounces (in our case, balloon, baseball ect.) divided by
aerodynamic coefficient of drag (CD) of a (balloon)sphere
(not all balloons are spheres) times
the frontal cross sectional area (A) of object or B=W/(CDA)
a = average thrust or push during the powered phase
(in ounces of force) divided by weight of object (in ounces)
minus one, or a = (T/W)-1.
tanh = hyperbolic tangent
Notice the huge difference you would get between
a balloon's Ballistic coefficient if it were filled with air and a baseball's (that is the same size as the balloon)
Ballistic coefficient, hence I would rather get hit
with a 3" diameter balloon filled with air
weighing .1 ounces thrown from 2 feet away than get hit
with a baseball thrown 60 feet away.

 

[wilg]

Frontal area is in square inches in this case.

 

[Rosalynn]

To wilg:

You've given me a lot to think about. I even got out my old text book from school to look up drag coefficients as a result of seeing the Malewicki equation. Thanks!
For now, I think I'll take a pass on the "coast" equation--it probably doesn't have a term for exerting the initial force. Plus the first equation is enough to keep me busy for quite awhile.

It's interesting to see how different engineering fields approach fluid mechanics (and other) problems. Just goes to show that there are many ways of looking at the world. It kind of expands one's mind--makes me happy to be an engineer.

Regards,
Rosalynn

 

[jedwards]

Is this as simple as the balloon has a terminal velocity in air...the same as all other objects falling downward? The direction the balloon is traveling doesn't matter--it will slow to its terminal velocity.

 

[IRstuff]

But it does matter.

Terminal velocity is the point at which the air resistance equals the force applied. When a balloon moves horizontally, there is no acceleration force, only an initial impulse, therefore, the balloon cannot "slow to its terminal velocity", but instead, must stop.

TTFN

 

[wilg]

jedwards and IRstuff
I think your both right, but just visualizing
the problem differently.jedwards is seeing terminal
velocity after the balloon has stopped it's horizontal
progress and begun to fall as an effect of gravity.
IRstuff see's the balloon as it pertains to Rosalynn's
original question of horizontal or force driven movement.

An object which is falling through the atmosphere is subjected to two external forces. One force is the gravitational force, expressed as the weight of the object. The other force is the air resistance, or drag of the object. The motion of a falling object can be described by Newton's second law of motion (Force equals mass times acceleration -- F = m a) which can be solved for the acceleration of the object in terms of the net external force and the mass of the object.

a = F / m

The net external force (F) is equal to the difference between the weight and the drag forces (W - D). When drag is equal to weight, there is no net external force on the object and the object will fall at a constant velocity as described by Newton's first law of motion. The constant velocity is called the terminal velocity .

We can determine the value of the terminal velocity by doing a little algebra and using the drag equation. Drag (D) depends on a drag coefficient, (Cd) the air density, (r) the square of the air velocity (V) and some reference area (A) of the object.

D = Cd * r * V ^2 * A / 2

At terminal velocity, D = W. Solving for the velocity, we obtain the equation

V = sqrt ( (2 * W) / (Cd * r * A) )

The terminal velocity equation tells us that an object with a large cross-sectional area or a high drag coefficient will fall slower than an object with a small area or low drag coefficient. (A large flat plate will fall slower than an a small ball with the same weight.) And if we had two objects with the same area and drag coefficient (two identically sized spheres), the lighter object would fall slower. This seems to contradict the findings of Galileo that all free falling objects would fall at the same rate with equal air resistance. But Galileo's principle only applies in a vacuum, where there is NO air resistance and drag is equal to zero.

 

[235zzzo]

Hi wilg!

I took a fast read on the Rosalynn's thread, and took a look to your post. Forgive me the other ones

Let's see: - Regarding what Rosalynn refers, concerning the moment, you through horizontally the balloon, afterwards the balloon almost stops in its horizontal path, so we have the so-called "hit the wall of air" effect, and then begins to fall down. I think, your analysis is only applied from this precise moment, (not before), the ballooon assumes the vertical path.

In fact we have here forces applied to the balloon, so we should treat those vectorially.

Or, did I missed something?
zzzo