# Exploring physical concepts (buoyancy, air drag) but afraid of the math.

1. Aug 27, 2011

### pgiustino

Hello! I have just rediscovered this place since a few years back when I first signed up. This will be my first post, so forgive me if it is a bit long. I have been thinking about this for a while.

So I have been thinking about buoyancy. I have learned that the buoyancy force is caused by a change in pressure per change in depth, and that for incompressible fluids, the buoyancy force remains constant at any depth. So this got me thinking about compressible fluids and that the buoyancy force changes at different depths. My first question is does the change in pressure per depth increase or decrease as you go deeper in a compressible fluid? Then this got me thinking about different situations, like can I tell at what depth an object is neutrally buoyant in a compressible fluid?

My mind then started to draw parallels from buoyancy force in compressible liquids to air drag force, because air drag force is changing as an object's falling velocity increases. Then I started to think about how I could derive at what falling distance does it take for an object to reach terminal velocity.

This stuff is all interesting to me now, but I've never been able to put any of the math I learned to good use. No one really showed me how to derive things, especially when dealing with changing rates. Is it too late to ask?

2. Aug 27, 2011

### Lobezno

Welcome to PhysicsForums!

No offence, but you'll have a hard time doing physics if you're afraid of maths. :D

The air drag equation is fairly simply though:
F(resistive)=c1*v*r + c2*v^2*r^2

As you can see with the inclusion of r (radius), this expression is for spheres. Generally a decent approximation. c1 and c2 are constants for any substance (fluid of any kind, liquid or gas). In plain air, the second part is negligible at common velocity and for most sizes, so air drag is generally given by: F=c1*v*r.

3. Aug 27, 2011

### MrNerd

Looking at your first question about how the buoyancy force changes in relation to depth, I;m going to guess that it would the change would increase with depth. As you go deeper, the liquid is compressed more. This, in turn, allows more liquid to be in a volume at a deeper depth than a higher depth. Because the density changes with depth, I would think that would cause the pressure's rate of increase to increase(basically positive acceleration of pressure).

4. Aug 27, 2011

### olivermsun

This is incorrect. The buoyancy force is due to a difference in relative density difference between a given parcel of fluid and the surrounding fluid. For example, differences in temperature and composition could cause some bit of fluid to be a bit light for its surroundings (think hot air balloon). However, there are certainly depth-varying buoyancy forces in (mostly) incompressible fluids, such as the ocean, where temperature and salinity are the major factors.

It is indeed possible for the change in pressure to vary for fluids in different states. It is not guaranteed that two parcels of fluid which are currently equally buoyant at some given pressure to be equally buoyant at some other pressure.

To find the static stability height for an incompressible object then of course you just compare it to the density of fluid at various heights. If you have a compressible object then you would need to find its density at various pressures and then find where it becomes neutrally buoyant with respect to the surroundings at that pressure.

5. Aug 28, 2011

### pgiustino

I am fine with algebra and formula manipulation, but when I start deriving formulas using changes in rates, I get lost with the calculus. I think what I need is a tutorial on how the kinematics equations are derived to give me a better idea on how to set up solutions to my own questions. Btw, what do you mean by common velocity?

Ok, so I need to think more in terms of density and the variables like temperature that affect it as depth increases. But in a perfectly incompressible fluid, the density would not change as depth increases, right?

Ok, that makes sense. Say I knew there was a constant change in the density going deeper in a fluid, what other givens would I need to know about the fluid to find an object's neutrally buoyant depth (given the object's density)? Basically I become lost when I try to set up an equation.

6. Aug 28, 2011

### Lobezno

At common velocities, sorry. That is, at speeds you'd find in every day life. And when I say common sizes, I don't mean it doesn't work on a quantum mechanical scale, that much is a given. I mean for very large and very small things.

And the equation for buoyancy would be, I think, simply equating the density of the object to it's surroundings. Someone should confirm this, but it makes intuitive sense, right?

7. Aug 29, 2011

Not it isn't. Any typical shape at a typical velocity will have a drag in air that is almost perfectly proportional to $v^2$ At very low velocities or in very viscous liquid (Stokes flow), you can find instances of drag being proportional to simply $v$, but that generally isn't the case. Of course, in general the viscous drag is much more complicated than this.

Not only is this not the general equation for air drag, but it is also not the correct form for air drag under any but the most specific of circumstances. The value of what you refer to as $C_1$ is not necessarily constant for changing fluid properties or even changing velocities either. Going back to the original question, the simplest form of an equation for drag is:
$$F_d = \frac{1}{2} C_d \rho v^2 A$$
Where:
$F_d$ is the drag force
$C_d$ is the drag coefficient, which is typically found empirically
$\rho$ is the fluid density
$v$ is the velocity
$A$ is a reference area that depends on the shape of your object and the type of drag

Of course, this still isn't the general formula. The general formula would be much more complicated and would include several types of drag. For example, the general formula for purely viscous drag would be:

$$F_{d,v} = \int_{z_1}^{z_2} \int_{x_1}^{x_2} \mu \frac{d V(x,y)}{dy}\; dx \; dz$$

Where:
$F_{d,v}$ is viscous drag
$\mu$ is viscosity
$y$ is normal to the surface in question
$x,z$ are along the surface

For pressure drag, you would have:

$$F_{d,p} = \int_{z_1}^{z_2} \int_{x_1}^{x_2} P(x,y=0,z) \; dx \; dz$$

For what it is worth, liquids are essentially incompressible, though your guess is true for gases.

This isn't true. The buoyancy force alone is a function of the pressure difference on the top and bottom of an object. The overall net force boils down to a density difference assuming the fluid is incompressible and the object is fully submerged. True, in the ocean, the buoyancy force would vary due to salinity, but this is going to be slight and still ultimately boils down to a pressure difference between the upper and lower portions of an object. It isn't just density.

This is akin to asking at what point in the atmosphere is a helium balloon neutrally buoyant, and it can be done. You simply account for compressibility when finding the pressure distribution from the top to the bottom of the fluid the object is immersed in. For the atmosphere, this has been done many times. If you have an object that doesn't change size, you just find the point in the fluid where the pressure distribution integrated over the surface of the object exactly cancels out the weight of the object and you have neutral buoyancy. This is a fairly easy thing to do unless you want to go back to the case of the helium balloon, in which case the balloon will be getting larger as it moves upward and eventually pops before it can find a neutral point. That is a much more complicated problem, but still theoretically doable.

For pressure drag, the mechanism of action is very similar to buoyancy, only you have an added dynamic pressure term in addition to the static pressure. Still, if you just find the pressure distribution around an object you can get a reasonable approximation of the drag by integrating over the surface for many shapes. The idea breaks down when you have a shape that experiences a large amount of viscous drag or goes supersonic and therefore experiences wave drag, but for relatively slow-moving objects in air, you can find the drag quite readily if you know how to compute the flow around the object. This can sometimes be done fairly simply with potential flow theory, though it will never take into account viscosity if you use potential flow.

It is never too late to learn something new.

8. Aug 29, 2011

### olivermsun

Perhaps you could clarify what you mean. My understanding of "buoyancy" works like this: A "neutrally buoyant" object would experience a pressure difference between its upper and lower surfaces, but this would be exactly canceled by the weight of the object. On the other hand, an object with the same density as its surroundings might experience a pressure difference and therefore an acceleration due to, say, an upward flow -- yet we wouldn't necessarily call it "positively buoyant." At least, that's how I interpret the "buoyancy force" term as it is often written,
$$b = -\rho^\prime g.$$

Last edited: Aug 29, 2011
9. Aug 29, 2011

### cjl

I agree with bonh3ad on this one - a neutrally buoyant object still has a buoyancy force - it just is exactly balanced by the object's weight. Buoyancy comes from the hydrostatic pressure gradient in the fluid.

(Adding an upward flow complicates the matter, since that is by definition a drag force and not a buoyancy force. That is why I specified "hydrostatic" in my statement above)

10. Aug 29, 2011

### olivermsun

I accept that this is a perfectly reasonable definition for buoyancy. Is this the usual way the term is used in aero engineering (since I noticed that both of you are from this field)?

Perhaps a difference is that in my field we generally look at perturbation effects to a parcel that is near neutrally stable, while you are accustomed to treating the forces on an object that is not (approximately) a part of the background?

11. Aug 29, 2011

That is just what buoyancy is in the most basic sense. Density is how it is often expressed, but that isn't the actual nature of buoyancy.

12. Aug 29, 2011

### olivermsun

Hmm... I suppose what I should ask is how you would describe the "nature" of buoyancy (in its "most basic sense").

It can't be just pressure differentials, since those can happen even in the complete absence of buoyancy. Is it the delta in the total force due to the purely hydrostatic part of pressure as the previous poster seemed to state? As I said, that seems a perfectly fine definition, although it does involve density...

Last edited: Aug 29, 2011
13. Aug 29, 2011

Yes. That is probably a decen definition, but at it's most basic, buoyancy amounts to a fluid force on and object manifested through a pressure differential. The governing equation can often be reduced to a simple density relation, but that isn't the underlying principle. A ball with the same density as water would still have a buoyant force on it despite the lack of density difference. That force would just be canceled by weight.

14. Aug 29, 2011

### olivermsun

So, for example, an object could be buoyant sideways due to a horizontal pressure gradient?

15. Aug 29, 2011

### cjl

I believe that the point olivermsun is trying to express is that buoyancy is not from an arbitrary pressure gradient, but rather from a hydrostatic one. All non-viscous fluid forces on an object are from a pressure differential, but they certainly are not all buoyancy forces.

16. Aug 29, 2011