Do you have a solution to the Deepwater Horizon Oil Spill?

Do you have a solution to the Deepwater Horizon Oil Spill?

Yes. My solution is to fill the well hole with objects that are dense enough and hydrodynamically streamlined enough to descend by gravity alone through the upward flow of oil. As they accumulate in the 3+ mile deep well hole, those objects will impede the flow until it becomes a trickle. Large steel balls (e.g., cannonballs) should do the trick. If they are large enough, they will have a downward terminal velocity, even as they move through the upward flowing oil. Because they descend, they will eventually accumulate at the bottom of the well hole and form a coarse “packed powder.” That powder will use its enormous weight and its resistance to flow to stop the leak. Most importantly, building the powder doesn’t require any seals or pressurization at the top of the well hole, so it should be easy to do.

The packed powder will exert downward drag forces on the upward flow of oil and gas, slowing its progress and decreasing its pressure faster than gravity alone. With 3+ miles of hole to fill, the dense steel objects should impede the flow severely. As the flow rate diminishes, the diameters of the metal spheres can be reduced until they are eventually only inches or even centimeters in diameter. The oil and gas will then be forced to flow through fine channels in the “powder,” allowing viscous drag and pressure drag to extract all of the pressure potential energy from the flow and convert that energy into thermal energy. The flow will, in effect, be attempting to lift thousands of pounds of metal particles and it will fail. It will ooze though the “packed powder” at an insignificant rate.

Another way to think about my technique is that it gradually increases the average density of the fluid in the well hole until that column of fluid is so heavy that the high pressure at the bottom of the hole is unable to lift it. The liquid starts out as a light mixture of oil and gas, but it gradually transforms into a dense mixture of oil, gas, and iron. Viscous forces and drag forces effectively couple the materials phases together to form a single fluid. Once that fluid is about 50% iron by volume, its average density will be so high (4 times the density of water) that it will stop flowing upward. If iron isn’t dense enough (7.8 times water), you could use silver cannonballs (10.5 times water). Then you could say that “silver bullets” stopped the leak! The failed “top kill” concept also intended to fill the well hole with a dense fluid: heavy mud. But it required pushing the oil and gas down the well hole to make room for the mud. That displacement process proved to be impossible because it required unobtainable pressures and pumping power. My approach takes no pressurization or pumping at all because it doesn’t actively displace the oil and gas.

Including deformable lead spheres in the mixture will further plug the upward flow. The lead will deform under the weight of metal overhead and will fill voids and narrow channels. Another refinement of this dense-fill concept would be to drop bead chains down the well hole. The first large ball in such a chain would be a “tug boat” that is capable of descending against the upward flow all by itself. It would be followed by progressively smaller balls that need to draft (travel in the wake of) the balls ahead of them in order to descend into the well. Held together by steel or Kevlar cord, those bead chains would accumulate at the bottom of the well and impede the flow more effectively than individual large balls. Especially streamlined (non-spherical) objects such as steel javelins, darts, rods, and rebar could also be dropped into the well at the start of the filling process. In fact, sturdy sacks filled with junk steel objects—nuts and bolts—might even work. Anything that descends into the well hole is good and smaller particles are better. The point is not to form a seal, since the enormous pressure that will develop beneath any seal will blow it upward. The point is always to form narrow channels through which the oil and gas will struggle to flow.

A video of this idea appears at: and a manuscript detailing this idea appears on the Physics ArXiv: I’m trying to find a home for it in the scientific literature, but so far Applied Physics Letters, Physic Review E (which includes the physics of fluids), and PLoS (Public Library of Science) One have turned it down—they want articles with new physics in them, not articles applying old physics to new contexts, no matter how important those contexts. It’s no wonder that the public views science as arcane and irrelevant.

Why are high heeled shoes so damaging to floors?

Why do high heels worn by a lady walking on a wooden floor leave impressions on the floor? — AK, Abbottabad, Pakistan

High heeled shoes can produce enormous pressures on a wooden floor and dent it permanently. To understand why that happens, let’s start with a pair of flat-heeled shoes and consider the forces and pressures in that situation.

When a women stands on the floor, the floor must support her weight. Specifically, she isn’t accelerating so the net force on her must equal zero. That implies that the floor must exert an upward force on her that exactly cancels her downward weight. She is motionless and stays motionless because there is no overall force on her.

Because the floor is pushing upward on her shoes, her shoes must be pushing downward on the floor. It’s an example of the famous “action and reaction” principle known as Newton’s third law: if you push on something, it pushes back equally hard in the opposite direction. Anyway, her shoes are pushing down hard on the floor.

Now for the pressure part of the story. Because she is wearing flats, her shoes are pushing against a large area of the floor and the pressure—the force per area—she produces on the floor is relatively small. For example, if she weighs 130 pounds (580 newtons) and her shoes have a contact area of 10 square inches (65 square centimeters), then the pressure she exerts on the floor is about 13 pounds-per-square-inch (9 newtons-per-square-centimeter or 90,000 pascals). That’s a gentle pressure that won’t permanently dent most woods. It might dent cork or balsa, but that’s about it.

But when she wears high heels, most of her weight is supported by a very small area of flooring. If the heels are narrow spikes with a contact area of 0.1 square inches (0.65 square centimeters) and she puts all of her weight briefly on one of the heels, she may exert a pressure of 1300 pounds-per-square-inch (9000 newtons-per-square-centimeter or 90 million pascals) on the floor. That’s an enormous pressure that will permanently dent most wooden floors.

You can experiment with these ideas simply by supporting the weight of your right hand with the open palm of your left hand. If you lay your right fist on your left palm, you won’t feel any discomfort in your left hand. The pressure on your left palm is very small. But if you instead point right index finger into your left palm and use that finger to support the entire weight of your right hand, it won’t feel so comfortable. If you shift all of the weight to your fingernail, it’ll start to hurt your left palm. What you’re doing is reducing the area of your left palm that is supporting your right hand and as that area gets smaller, the pressure on your left palm increases. Beyond a certain pressure, it feels uncomfortable. Long before your palm dents permanently, you’ll decide to stop the experimenting.

Why do deep wells require pumps at the bottom?

Why do deep water wells need a pump at the bottom rather than one at the top? — LG, Vancouver

While it’s easy to push on water, it’s hard to pull on water. When you drink soda through a straw, you may feel like you’re pulling on the water, but you’re not. What you are actually doing is removing some air from the space inside the straw and above the water, so that the air pressure in that space drops below atmospheric pressure. The water column near the bottom of the straw then experiences a pressure imbalance: the usual atmospheric pressure below it and less-than-atmospheric pressure above it. That imbalance provides a modest upward force on the water column and pushes it up into your mouth.

So far, so good. But if you make that straw longer, you’ll need to suck harder. That’s because as the column of water gets taller, it gets heavier. It needs a more severe pressure imbalance to push it upward and support it. By the time the straw and water column get to be about 40 feet tall, you’ll need to suck every bit of air out from inside the straw because the pressure imbalance needed to support a 40-foot column of water is approximately one atmosphere of pressure. If the straw is taller than 40 feet, you’re simply out of luck. Even if you remove all the air from within the straw, the atmospheric pressure of the water below the straw won’t be able to push the water up the straw higher than about 40 feet.

To get the water to rise higher in the straw, you’ll need to install a pump at the bottom. The pump increases the water pressure there to more than 1 atmosphere, so that there is a bigger pressure imbalance available and therefore the possibility of supporting a taller column of water.

OK, so returning to your question: once a well is more than about 40 feet deep, getting the water to the surface requires a pump at the bottom. That pump can boost the water pressure well above atmospheric and thereby push the water to the surface despite the great height and weight of the water column. Suction surface pumps are really only practical for water that’s a few feet below the surface; after that, deep pressure pumps are a much better idea.

Breathing Underwater through a Hose

How deep under water can I go while breathing from a hose that rises above the surface of the water? — DF, Downers Grove, IL

You can only go a few feet under water before you’ll no longer be able to draw air into your lungs through that hose. It’s a pressure problem. The water pressure outside your chest increases rapidly as you go deeper, but the air pressure inside the hose and your mouth barely changes at all. Pretty soon, you’ll have so much more pressure outside your lungs than inside them that you won’t be able to draw in any more air. Your muscles just won’t be strong enough.

The water pressure increases quickly with depth because each layer of water must support the weight of all the water layers above it. Since water is dense, heavy stuff, the weight piles on quickly and it takes only 10 meters (34 feet) of descent to increase the water pressure from atmospheric to twice atmospheric. In contrast, the air in the hose is light, fluffy stuff, so its pressure increases rather slowly with depth. Even though each layer of air has to support the weight of all the layers of air above it, the rise in pressure is extremely gradual. It takes miles of atmosphere above the earth for the air pressure to build up to atmospheric pressure near the ground. The air pressure in your hose is therefore approximately unchanged by your descent into the water.

With the water pressure outside rising quickly as you go deeper and the air pressure in your mouth rising incredibly slowly as you go deeper, you quickly find it hard to breathe. Your muscles can push your chest outward against a modest pressure imbalance between outside and inside. But by the time you’re a few feet below the surface, you just can’t draw air into your lungs through that hose anymore. You need pressurized air, such as that provided by a scuba outfit or a deep-sea diver’s compressor system.

Why do faster moving fluids have lower pressure?

Why do faster moving fluids have lower pressure? — JH

Actually, faster moving fluids don’t necessarily have lower pressure. For example, a bottle of compressed air in the back of a pickup truck is still high-pressure air, even though it’s moving fast. The real issue here is that when fluid speeds up in passing through stationary obstacles, its pressure drops. For example, when air rushes into the open but stationary mouth of a vacuum cleaner, that air experiences not only a rise in speed, it also experiences a drop in pressure. Similarly, when water rushes out of the nozzle of a hose, its speed increases and its pressure drops. This is simply conservation of energy: as the fluid gains kinetic energy, it must lose pressure energy. However, if there are sources of energy around—fans, pumps, or moving surfaces—then these exchanges of pressure for speed may no longer be present. That’s why I put in the qualifier of there being only stationary obstacles.

How do the automatic soda dispensers at fast food joints know when the cup is fu…

How do the automatic soda dispensers at fast food joints know when the cup is full? — MB

They measure the volume of liquid they deliver and shut off when they have dispensed enough soda to fill the cup. Accurate volumetric flowmeters, such as those used in the dispensers, typically have a sophisticated paddlewheel assembly inside that turns as the liquid goes through a channel. When the paddlewheel has gone around the right number of times, an electronic valve closes to stop the flow of liquid.

How come if I stand on the balcony of my third story apartment and drop a hose t…

How come if I stand on the balcony of my third story apartment and drop a hose to the swimming pool down below, I can’t suck any water up through the hose into my mouth?

While it may seem that you are somehow attracting the water to your mouth when you suck, you are really just making it possible for air pressure to push the water up toward you. By removing much of the air from within the hose, you are lowering the air pressure in the hose. There is then a pressure imbalance at the bottom end of the hose: the pressure outside the hose is higher than the pressure inside it. It’s this pressure imbalance that pushes water into the hose and upward toward your mouth.

But air pressure can’t push the water upward forever. As the column of water in the hose rises, its weight increases. Atmospheric pressure can only lift the column of water so high before the upward force on the water is balanced by the water’s downward weight. Even if you remove all of the air inside the hose, atmospheric pressure can only support a column of water about 30 feet tall inside the hose. If you’re higher than that on your balcony, the water won’t reach you no matter how hard you try. The only way to send the water higher is to put a pump at the bottom end of the hose. This pump can push upward harder than atmospheric pressure can and it can support a taller column of water. That’s why deep home wells have submersible pumps at their bottoms—they must pump the water upward because it’s impossible to suck it upward more than 30 feet from above.

How does a standard water pump work?

How does a standard water pump work? — ML, Wilmington, NC

The water pumps in most cars are centrifugal pumps. These pumps work by spinning water around in a circle inside a cylindrical pump housing. The pump makes the water spin by pushing it with an impeller. The blades of this impeller project outward from an axle like the arms of turnstile and, as the impeller spins, the water spins with it. As the water spins, the pressure near the outer edge of the pump housing becomes much higher than near the center of the impeller. There are many ways to understand this rise in pressure, and here are two:

First, you can view the water between the impeller blades as an object traveling in a circle. Objects don’t naturally travel in a circle—they need an inward force to cause them to accelerate inward as they spin. Without such an inward force, an object will travel in a straight line and won’t complete the circle. In a centrifugal pump, that inward force is provided by high-pressure water near the outer edge of the pump housing. The water at the edge of the pump pushes inward on the water between the impeller blades and makes it possible for that water to travel in a circle. The water pressure at the edge of the turning impeller rises until it’s able to keep water circling with the impeller blades.

You can also view the water as an incompressible fluid, one that obeys Bernoulli’s equation in the appropriate contexts. As water drifts outward between the impeller blades of the pump, it must move faster and faster because its circular path is getting larger and larger. The impeller blades do work on the water so it moves faster and faster. By the time the water has reached the outer edge of the impeller, it’s moving quite fast. But when the water leaves the impeller and arrives at the outer edge of the cylindrical pump housing, it slows down. Here is where Bernoulli’s equation figures in. As the water slows down and its kinetic energy decreases, that water’s pressure potential energy increases (to conserve energy). Thus the slowing is accompanied by a pressure rise. That’s why the water pressure at the outer edge of the pump housing is higher than the water pressure near the center of the impeller.

When water is actively flowing through the pump, arriving through a hole near the center of the impeller and leaving through a hole near the outer edge of the pump housing, the pressure rise between center and edge of the pump isn’t as large. However, this pressure rise never completely disappears and it’s what propels the water through the car’s cooling system.

How does a toilet work?

How does a toilet work? — JJ, Stafford VA

A toilet is actually a very clever device that makes use of a siphon to extract the water from its bowl. A siphon is an inverted U-shaped pipe that can transfers water from a higher reservoir to a lower reservoir by lifting that water upward from the higher reservoir and then lowering it into the lower reservoir. In fact, the water is simply seeking its level, just as it would if you connected the two reservoirs with a pipe at their bottoms. In that case, the water in the higher reservoir would flow out of it and into the lower reservoir, propelled by the higher water pressure at the bottom of the higher reservoir. In the case of a siphon, it’s still the higher water pressure in the higher reservoir that causes the water to flow toward the lower reservoir, but in the siphon the water must temporarily flow above the water levels in either reservoir on its way to the lower reservoir. The water is able to rise upward a short distance with the help of air pressure, which provides the temporary push needed to lift the water up and over to the lower reservoir. At the top of the siphon, there is a partial vacuum—a region of space with a pressure that’s less than atmospheric pressure. The same kind of partial vacuum exists in a drinking straw when you suck on it and is what allows atmospheric pressure to push the beverage up toward your mouth.

In the toilet, the bowl is the higher reservoir and the sewer is the lower reservoir. The pipe that connects the bowl to the sewer rises once it leaves your view and then descends toward the sewer. Normally, that rising portion of the pipe isn’t filled water—water only fills enough of the pipe to prevent sewer gases from flowing out into the room. As a result of this incomplete filling, the siphon doesn’t transfer any water. But when you flush the toilet, a deluge of water from a storage tank rapidly fills the bowl and floods the siphon tube. The siphon then begins to function. It transfers water from the higher reservoir (the toilet bowl) to the lower reservoir (the sewer) and it doesn’t stop until the bowl is basically empty. At that point, the siphon stops working because air enters the U-shaped tube with a familiar sound and water again accumulates in the bowl. When the storage tank has refilled with water, the toilet is ready for action again.

How does one calculate the pressure of air flowing in a tube? My specific applic…

How does one calculate the pressure of air flowing in a tube? My specific application is air traveling in a 1/2-inch tube at a velocity of 14 inches/second. I know that Bernoulli would have the answer, but I cannot find it myself. — NT, Cambridge, MA

Without more information about the air in your tube, it’s not possible to determine its pressure. Bernoulli’s equation is frequently misunderstood to say that high-speed air is low-pressure air and that low speed air is high-pressure air—two observations that aren’t necessarily true. Just because air is moving rapidly doesn’t mean that its pressure is low. For example, the air in an airplane cabin is moving quickly but its pressure is higher than that of the air outside the cabin. Similarly, if you were to throw a tank of compressed air across the room, its pressure would remain high despite its increase in speed.

What Bernoulli’s equation really says is that air has three forms for its energy and that as long as that air flows smoothly and without significant friction through a system of stationary obstacles, the sum of those three energies can’t change. The three energies are kinetic energy (the energy of motion), gravitational potential energy, and an energy associated with pressure that I call pressure potential energy. The obstacles must remain stationary so that they can’t do work on the air and thus change its total energy. Since the sum of those three energies doesn’t change as air flows through a stationary environment, its pressure typically falls whenever its speed rises and vice versa. If the air also changes altitude significantly, then gravitational potential energy must be included in these energy exchanges.

So the reason why I can’t answer your question about air in a pipe is that I don’t know what the air’s total energy was before it flowed through the pipe. While I can calculate the air’s kinetic energy from its speed and we can neglect gravitational potential energy because the air isn’t changing altitudes much in the pipe, I need to know what the air’s total energy is in order to determine its pressure potential energy and thus its pressure.