Atmospheric Pressure Energy

[RainmanTime]

Hi Einstein,

I figured I would pick up our discussions here in a new thread. Let’s start out by talking to each other about the parameters of this design concept. You have begun by pointing out a very important aspect that is of utmost importance to model correctly:

I don’t quite know how to figure out the head pressure. Although at 14.7 lbs/in^2 at sea level, I suspect that a large diameter inlet will definitely increase head pressure.

Head pressure in any fluid is a pretty easy concept to model, based on the hydrostatic relationships. The head pressure for any fluid can be expressed in terms of the height of a column of that fluid, independent of the shape of the column at either end. For any height of a fluid “h” we can compute its head pressure over that height as:

P-head = (Fluid Density)g(fluid height) P = rhogh

So whatever the height of the device you wish to build, we can model the head pressure difference between the bottom and the top with respect to the height (h) between them.

The energy equation written in terms of fluid pressure could be represented as:

P-static + P-head + P-dynamic = Constant. OR

P-static1 + Phead1 + P-dynamic1 = P-static2 + Phead2 + P-dynamic2

The surface area at the mouth of the funnel is just like a large pipe conducting a fluid. And just like in a pipe as the diameter grows smaller the fluid pressure and velocity goes up.

OK, but this generalization is usually applied to a LEVEL flow (no head pressure difference). What you say is very true and is a basic model for aerodynamic flows that do NOT go through a change in height. Head pressure differences in non-level flow system changes this generalization.

Now I know that higher air temperature will result in lower air pressure.

Maybe I can convince you to think about this differently… You will note that as you go UP in altitude away from the earth’s sea level that the temperature goes DOWN. For example, when it is 85 Deg F at Huntington Beach, CA, it would be quite a bit cooler up in the mountains of SoCal near Big Bear. It would probably only be about 65-75 Deg F at altitude. This is consistent with how the atmosphere is actually structured. From Sea Level to about 36000 feet the temperature goes down linearly with height above sea level. And the other thing we notice as we go up in altitude and down in temperature: The air pressure goes down. This is because the density of air molecules is being lowered. This is why it is very handy to have a simple, Excel spreadsheet that models the ICAO standard atmosphere, as it can help you study atmospheric pressure questions.

Also some type of data chart with the rate at which atmospheric pressure falls off with height would definitely figure in.

Yes, this is the ICAO standard atmosphere model you are talking about. It is an exercise I give to my freshman AERO engineering students.

And then I do believe there would be some loss due to the small angle or slope of the pipe.

Any pressure differential due to height is handled by the head pressure equation I have given above. The only other force field that needs to be accounted for is the frictional fluid forces inherent in the boundary layer near the inner surfaces of the conduit. That would be called the frictional pressure losses and they would be added to the head pressure difference.

But how big a structure would be necessary to produce enough head pressure to say run a one megawatt generator? Such a simple question.

It is kind of simple, you know. That’s because head pressure only depends upon height within a fluid column. Look at the head pressure differential equation: Delta-P = densitygdelta-height. It does not depend at all upon how large of a vessel you have at the surface of the earth. Head pressure is only about height, not the shape of the vessel.

Figuring out the answer will be fun. Like what would the optimum pipe angle slope be? There would be structural limits that would put a cap on how much atmospheric pressure could safely be contained within a large structure at the base. Also I do want to point out that I do suspect a structure as big as a mountain might be what we calculate out to be.

You’re absolutely right, it is fun! Before we think about the optimum pipe slope angle, let’s finalize some basics about what sort of altitude differences we could reasonably accomodate. There is a mountain in SoCal called Mt. Baldy, and it is roughly (give or take 200 feet) 10,000 feet above sea level. So if we ran a pipe up Mt Baldy from the ocean we could get a “head pressure” of atmosphere of about 10,000 feet. Using our equation and assuming that air density is constant at the sea level value (.002377 slugs/cubic feet), we could calculate a total head pressure of:

P-head = 764.78 Pounds per square foot. That would be the approximate head pressure differential between the bottom and the top of the pipe. That is the pressure exerted by that column of air.

But like I said before, it is free energy waiting to be tapped. Of course the projected yields can be scaled down too. But somewhere in the calculated yields an optimum power generating solution can be configured.

OK, if we agree on a 10,000 foot head pressure differential, then we only need to talk about static pressures and dynamic pressures at each end in order to model this system. The ICAO standard atmosphere will tell us what the typical static pressures are for the atmosphere at 10,000 feet above sea level and at sea level. So the only thing that is left to answer is what are the velocities you wish to achieve at both the top and the bottom? That will establish the dynamic pressure requirement for the system.

RMT

[Einstein]

RMT

I have been giving this some thought. For instance the 14.7lbs/in^2 at sea level. That square inch of air probably has 14.7lbs directly above that square inch and it extends outward into space. Now if we define that square inch within a circle (area of circle is one square inch) then draw lines through the circle so that all lines intersect at the center of the earth. The volume described is a conical slice. As the height increases away from the surface of the earth, The cross sectional area going through the conical slice increases. The mouth of the conical slice somewhere out in space is much bigger than a square inch. If the conical slice defines the walls of a pipe then the air pressure inside the pipe and outside the pipe is in equilibrium. But as you can see the pipe is not your typical pipe with parallel walls. As the cross sectional area at sea level increases, the angle of walls of the pipe increases. This angle would be the equilibrium angle where inside and outside air pressure are the same. This equilibrium angle will be the zero point where all additional angles are based from. Now just for one square inch, the angle is very close to zero. But for a pipe with say an inlet of 200 hundred feet across, the angle becomes probably measureable. But this is something that can be calculated out too.

Now there is something else that I thought of that could offset this equilibrium angle. The gravity field of the earth is curving the space too. But I don’t know how to calculate the curvature into our calculations. I’m not even sure if this curvature would affect the final outcome or not.

For now I’m going to just pick say a 200 foot diameter inlet and try and calculate the equilibrium angle. Then change it by say one degree at the base and try and figure out how long the pipe would have to be if I want a ten to one increase in air pressure at the base. The base of the pipe will be sealed off for now. So it wont equalize air pressure with the outside air pressure. I’ll just work at it with one variable at a time.

Also, how do I calculate flow rate? Like what percentage of the full head pressure is allowable to use?

[RainmanTime]

Hi Einstein,

I am going to ask you to be patient with me here, and slow down a bit. You are moving quickly from one topic/assumption to another, and I am not sure they are all valid jumps. Let’s start with this:

For instance the 14.7lbs/in^2 at sea level. That square inch of air probably has 14.7lbs directly above that square inch and it extends outward into space.

Yes and no. 14.69 PSI is what is defined as the “standard day, sea-level” total pressure for temperate zones of the atmosphere. That pressure is a result of the column of atmospheric air above it, but it is not constant. This is where the hydrostatic equation comes into play along with head pressure. Atmospheric total pressure (where velocity is zero) decreases as you go up in altitude, and it is directly dependent upon the temperature decrease. So say for a standard day, if your atmospheric pressure is 14.69 PSI at sea level, then it would be about 3.20 PSI by the time you reached the tropopause altitude (about 36000 feet). Pressure keeps going down until you reach the ionosphere where it is MUCH lower than even 1 PSI.

I do not understand the rest of your description with respect to the angle of the walls of a cone. But suffice it to say that is not how you can accurately model the atmosphere. And I’d also like us to limit our discussions to altitude changes of less than 10,000 feet. If we work a few problem at this low level, then we can advance to the higher levels (where atmospherics change their governing equations… like at the tropopause where temperature becomes constant as you move upward in the atmosphere through the stratosphere).

Can we describe a simple vertical pipe from 10,000 feet down to sea level, and then see what happens with regard to head pressures, temperatures, and flow rates? That will help us decide how (or if) we can modify the cross sectional areas to induce airflows.

RMT

[Einstein]

RMT

Let me clarify something first. A conical slice is a three dimensional slice off of a sphere. It is analogous to a slice of pie. The bigger piece of pie you take will result in a larger angle. But I was doing some quick calculations to actually see just how much this conical slice angle would be for a pipe with a 200 foot diameter mouth. The circumferance of the earth is around 25000 miles. Divide that by 360 and that results in 69.4 miles per degree. Convert that into feet: 366667 ft. So if I divide my 200 feet by 366667 I get the percentage of one degree. It calculates out to be .000545454 percent of one degree. This does appear to be small enough to ignore. But this is the angle to be used if you wanted a pipe with equal air pressure on inside and out.

Yesterday I spent some time calculating out the length of the pipe with a one degree slope. The length is only slightly longer than the height by a few inches. But the height is very close to 3/4 of a mile. I took a picture of my scribble sheet:

Basically you can see I used just a little trigonometry to do the calculations. I chose a surface area at the base of the pipe that is one tenth the area of the upper inlet of the pipe. From that area I calculated the base radius of the pipe (31.6227766 ft).

Now I want you to know that this idea might not come to fruition. Mainly because I am making an assumption that gravitational head pressure is equivalent to mechanically produced head pressure. And more often than not when making assumptions about nature, quite often I am wrong. But if I am wrong I will have gained some more knowledge about the behavior of gravity. But this idea is very interesting to explore.

Off to the head pressure. I used 14.7 lbs/in^2 and just multiplied by the area of the inlet. I’m not sure if it is legal to do it this way. But I get 33,250.616 tons of air pressure spread across the mouth of the pipe. At the base of the pipe the area is 1/10, so that pressure per area goes up by 10. 332.506.16 tons of available pressure at the base. The pressure differential at the base inside the pipe should be 10 times what is outside the pipe. 147 lbs/in^2 is ideally what I expect to be there but I don’t know how to calculate what kind of available flow rate that could be tapped off to drive an electrical generator.

Now lets pick this apart. I know the air pressure at 3/4 miles up is not 14.7 lbs/in^2. But if the pipe were a straight pipe. the pressure at its base would be 14.7 lbs/in^2. By reducing the area at the base the pressure should go up. I still have the same amount of available head pressure at the inlet no matter what the area at the base is. So I did just multiply by ten because I reduced the area by 10. But is this correct. Could there be something in the behavior of gasses that is different than the behavior of liquids? Or is there something I’m not seeing?

Also the one degree deflection angle of the pipe should support some of the weight of the head pressure. But I am thinking I could just use a trig function to calculate the loss. It would be small but something that should be calculated in.

[creedo299]

Yeah’ Im thinking a venturi farm, where there is nothing but funnel stacks fro miles, that go hundreds of feet up into the air.

Since Bernoulli said that an increase in velocity of air flows over a partial recess reduces pressure within that recess, then of course, you hook up, up-draft turbines to the bottom of the funnels.

Free power.

But what is the principle of paying in this society?

People will be good?

Who determines good?

I mean this is like Rainman sitting in his swivel office chair, only in a T-shirt and undies&socks and the room being full of blonds in bathing suits.

Yeah, he will be good. we all will!!??

[Einstein]

Creedo

Yeah’ Im thinking a venturi farm, where there is nothing but funnel stacks fro miles, that go hundreds of feet up into the air.

Since Bernoulli said that an increase in velocity of air flows over a partial recess reduces pressure within that recess, then of course, you hook up, up-draft turbines to the bottom of the funnels.

I know this principle works in reverse. In fact I wouldn’t be surprised if you found something in your home that works in the reverse mode. A chimney, an exhaust over the kitchen area. Any pipe open to air at the top of a building will allow upward flow of air. And the cooling towers at nuclear facilities. But I don’t see anywhere where we use a pipe to allow the weight of the atmosphere to compress the air in a pipe. Probably because it has to be on a large scale to get any appreciable power output. I think you misunderstood the priciple behind the pressure increase. I should get air flowing out of the bottom of the funnel pipe. If I inverted the funnel pipe so the large mouth was at the bottom, then it would work as you describe. The air pressure inside would be probably 1/10 that outside. But vaccuum pressure only allows up to one atmosphere of pressure. By placing the mouth of the funnel pipe at the top I sould get a multiplication of air pressure at the base. I hope I didn’t confuse you. But using the funnel pipe with the mouth on top should get power increases at least ten times greater.

There is something else too. The question comes to mind: What would be the optimum size? Should we make lots of little funnel pipe generators? Or just a few really big ones? Economics comes into play here. It may be more feasable to produce lots of little ones. What if an economoical size could be made for home use?

I still don’t know what kind of power capacity can be generated yet. It’s just an explorative idea at present. But RMT and I should be able to examine in closer detail if it is at all possible to create a power generator this way.

[creedo299]

I’ve been thinking of the venti ure updraft principle for two weeks, now.

They also had a tower like this, in the old forbidden city, in antartica.

I’ve been there, but other humans are expressly forbidden to go there.

[RainmanTime]

Hi Einstein,

Let me clarify something first. A conical slice is a three dimensional slice off of a sphere. It is analogous to a slice of pie.

OK, thanks. I did understand this, but I am not sure why you are racing off in this direction already. Your math is good, and it may well come in handy once we better understand the nature of fluid pressure fields. Hang onto this thought for awhile and see if you can follow me in some of the BASICS, because I guarantee you that building a model with these basics will help us.

Now I want you to know that this idea might not come to fruition.

I know exactly where this idea will end up, and you might be surprised as it will support something else I have been saying in this forum. But I would like to lead you through an engineering analysis and construction of a mathematical model to demonstrate to you how this process can help you avoid going down paths that are either based on incorrect assumptions, or paths that cannot be realized. Are you willing to stick with me on this?

Mainly because I am making an assumption that gravitational head pressure is equivalent to mechanically produced head pressure. And more often than not when making assumptions about nature, quite often I am wrong.

Exactly. And this is why the most successful and fruitful development efforts begin with earnest analysis and the construction of computational models. It’s worked well in the past and has taken man to the moon and back. I assure you it will help in this case.

Off to the head pressure. I used 14.7 lbs/in^2 and just multiplied by the area of the inlet. I’m not sure if it is legal to do it this way. But I get 33,250.616 tons of air pressure spread across the mouth of the pipe. At the base of the pipe the area is 1/10, so that pressure per area goes up by 10. 332.506.16 tons of available pressure at the base. The pressure differential at the base inside the pipe should be 10 times what is outside the pipe. 147 lbs/in^2 is ideally what I expect to be there but I don’t know how to calculate what kind of available flow rate that could be tapped off to drive an electrical generator.

In this you are attempting to attack a specific piece of the pressure energy situation. This will lead to dead ends and incorrect assumptions. I can tell you right now that you cannot assume that head pressure is constant, as air density changes (lowers) as you rise in altitude. To compute this piece of the pie correctly, you will need to use an integral and integrate density as a function of altitude over the lower and higher altitude limits. I know you don’t like integrals, so let’s see if we can do this a different way. We have to look at the entire energy balance situation, and so we use Bernoulli’s equation in its full-form that includes head pressure terms. Again let’s not worry about a container (cylindrical or conical) as yet. Let’s just model the energy balance at two distinct altitudes: 0 feet MSL and 10,000 feet MSL. If we call 0 feet point 1 and 10Kft point 2, we can write the Bernoulli pressure energy balance equation as:

P-static1 + P-dynamic1 + P-head1 = P-static2 + P-dynamic2 + P-head2 (assumes no frictional energy losses)

Now, things we already know are:
P-static1 and P-static2 are the atmospheric pressures at the two altitudes. This is the “zero velocity” pressure if we assume the atmosphere is a giant reservoir at these altitudes. The ICAO standard atmosphere model is an empircal model for these pressures. And using this model we can pick values for these static pressures on a standard day in a temperate region:
P-static1 = 14.69 lb/in^2 (2116.2 lb/ft^2)
P-static2 = 10.10 lb/in^s (1455.2 lb/ft^2)

P-dynamic1 = 0.5density1Velocity1^2
P-dynamic2 = 0.5density2Velocity2^2
P-head1 = density1gh1 (since we define h1= 0 feet, this term goes away. All head pressures are then referenced to Mean Sea Level)
P-head2 = density2gh2

We can assume “g” is constant since it changes by a VERY small amount over the 10,000 feet we are interested in. And we can also use the ICAO standard atmosphere to tell us what the air densities are at out two altitudes on a standard, temperate day:

density1 = .002377 slugs/ft^3
density2 = .001755 slugs/ft^3

So now we are down to two unknowns (Velocity1 and Velocity2), but only one equation. The other equation we can use is that for steady flow the mass flow rate must be constant. The equation for mass flow rate for a compressible fluid is:

m-dot = density1Area1Velocity1 = density2Area2Velocity2

I’d encourage you to solve these equations for one of the velocities (either the one at the top or bottom, doesn’t much matter) and more importantly to program the solution into a tool like MS-Excel so you can study how things change as you change the pressure and velocity at one end.

What you should find is that the “standard atmosphere” as we define it cannot generate sufficient pressure differentials to spin a turbine. That is because, as we have defined the ICAO standard atmosphere, we have modeled the atmosphere as STABLE in how temperature changes with respect to altitude. In order to induce significant pressure differentials to spin a turbine, what we find is that we REQUIRE the atmosphere to be UNSTABLE.

If you’d like to continue this, next time we can begin to dig deeper into atmospheric stability and how temperature profiles within the atmosphere will be a major factor in determining the static pressures at each end of our duct.

RMT

[Einstein]

RMT

I know exactly where this idea will end up, and you might be surprised as it will support something else I have been saying in this forum. But I would like to lead you through an engineering analysis and construction of a mathematical model to demonstrate to you how this process can help you avoid going down paths that are either based on incorrect assumptions, or paths that cannot be realized. Are you willing to stick with me on this?

Definitely. But don’t be so sure on where this idea will lead to. I see it as opening the doors to many possibilities. I do suspect you are referring to the upward and downward spirals. But what I am working with here is more basic than that.

Now I did spend the better part of yesterday afternoon reaquainting myself with Bernoulli’s equation. Although I like it and it even seems to work. I doesn’t seem to account for the situation we are presently discussing. So I delved a little deeper and this morning woke up with a realization. Bernoulli’s equation doesn’t make a distinction between gravitationally induced pressure and mechanically induced pressure. Well I asked myself are there any differences? Yes. There is. A large container with a fluid or gas in it will experience a pressure that increases with a decrease in altitude. A mechanically induced pressure will produce equal pressure throughout the container. In fact this is directly analogous to another phenomena in physics that I am very familiar with. Electrical phenomena. Mechanically induced pressure is analogous to voltage in an open circuit. Voltage in an open circuit is the same everywhere in the circuit. Gravitationally induced pressure is directly analogous to magnetically induced voltage. The voltage will increase in a wire under the influence of a moving magnetic field. I can think of three ways to make that voltage increase proportionally within the wire. Increase the speed that the magnetic field cuts across the path of a conductor, increase the magnitude of the magnetic field, and finally increase the length of the path in the conductor for the electrons to flow. So as you can see I do believe there is a distinction between gravitational pressure and mechanical pressure. It appears to me that we are mixing the two types of pressure in an equation without regard for the inherent differences each type of pressure has. That’s fine if it defines situations with predictable outcomes. But here I have a model that falls outside of Bernoulli’s equation. So apparently another way has to be used to predict the outcome. You have made me aware that I will probably need a longer pipe to get the necessary gravitational head pressure. But I still need some type equation to predict atmospheric pressure at different altitudes to get some idea of how long my funnel pipe will have to be. But lets move back to basics just a bit.

I have been assumming all along that you understood how I was creating the pressure differential within the funnel pipe. But you made a statement that leads me to suspect I need more clarification.

I did understand this, but I am not sure why you are racing off in this direction already

Your statement was in reference to the conical slice angle that I described earlier. But it leads me to suspect that you need to see this through my eyes. So lets switch over to a solid object for clarification. Lets say that I have an object that weighs one ton. If the base of the object is configured so that only one square inch comes in contact with the ground, I have a one ton per square inch pressure. But I could configure the base so that the weight was spread over one square foot instead. That would calculate out to only 13.88 lbs/in^2. As you can see I can change the contact pressure that the object exerts on the ground without changing the objects weight just by changing the area, that the weight of the object rests upon. That seems pretty easy to comprehend. But I had to ask myself, could the same concept be applied to air? If I could get the air to weigh more over a given area, I would have a pressure differential. That pressure differential would be all I need to run an electrical generator.

Now the idea of using a funnel to concentrate the weight of the air above it just might be feasable on a large scale. By keeping the angle of the pipe walls very small to the direction of the gravitational pressure, should result in a deflection of the gravitational pressure toward the center of the funnel pipe. That translates into more air mass weight over the center section. Because the deflection angle is so small, only a small portion of the gravitational head pressure is absorbed by the walls of the pipe. The rest of the weight is directed toward the center section. That means that the pipe walls are not supporting the air mass weight directly overhead. Now you might disagree because Bernoulli’s equation doesn’t predict this behavior. But you could derive this from other areas in physics that do behave this way. Ballistics comes to mind. A bullet will deflect off a surface when hitting it at an angle. Just like air molecules do. And I do have to point out that gravitational head pressure is directional. Mechanical head pressure is not directional. This is interesting because there is some predicted new type of pressure phenomena going on in the center of the air mass in the pipe. The mathematics for this just could be worth a Nobel prize. But for now it is still in the hypothetical stage.

[RainmanTime]

Einstein,

But don’t be so sure on where this idea will lead to.

I am quite sure where it will lead to, because they are forms of vortex energy that are exhibited by nature all the time. [quote]
I do suspect you are referring to the upward and downward spirals.
[/quote]

You suspect correctly, even though we only see one half of the opposing spirals in the high-energy formations of atmospheric weather phenomenon. [quote]
Bernoulli’s equation doesn’t make a distinction between gravitationally induced pressure and mechanically induced pressure.
[/quote]

Yes, it does. Each has a separate term in Bernoulli’s equation. Static pressure is, by definition, mechanically induced pressure (due to a specific concentration of a fluid within a given volume). Head pressure is the gravitationally-induced pressure term. The ONLY thing that Bernoulli’s equation, as I have stated it, does not account for is frictional losses, which would need to be accounted for once your vision of a conduit is introduced. [quote]
Mechanically induced pressure is analogous to voltage in an open circuit.
[/quote]

Yes, static pressure is directly analogous to EMF. [quote]
So as you can see I do believe there is a distinction between gravitational pressure and mechanical pressure.
[/quote]

Yes, and the head pressure term is directly analogous to magnetically induced voltage, because it is a result of the gravitational forces upon a fluid of a given density and a given column height. Just as magnetically-induced voltage is due to a magnetic field, so head pressure is due to a gravitational field. [quote]
But here I have a model that falls outside of Bernoulli’s equation.
[/quote]

No. Bernoulli’s accounts for your analogy. [quote]
But I still need some type equation to predict atmospheric pressure at different altitudes to get some idea of how long my funnel pipe will have to be.
[/quote]

That equation is the ICAO standard atmosphere model. Here are the equations, and they start with an assumed linearly-decreasing temperature gradient:

T-alt = T-sealevel + a*h (h=altitude above mean seal level, and a=lapse rate, which for the standard atmosphere is defined as -.00356 Deg. Rankine/foot. T-sealevel is the standard day sea level temperature which is defined as 518.69 Deg. Rankine)

To compute the pressure at any altitude, the ICAO model defines a pressure ratio equation which is a function of the temperature ratio (T-alt/T-sealevel) as follows:

P-alt/P-sealevel = (T-alt/T-sealevel)^(-g/R/a)

In this equation, the “a” is still the same lapse rate defined above, and “R” is the specific gas constant for air (defined as 1716 in units which are consistent with Deg. Rankine, feet, and seconds). Now, this set of equations is ONLY valid for linearly-decreasing segments of the atmosphere. The troposphere (from sea level up to about 36000 feet) is such a level. However, at that altitude, the temperature becomes constant for the next layer of the atmosphere called the stratosphere. There is another set of relations that will be used for that layer. From here on outwards, the atmosphere is made up of alternating linear gradient layers, and constant temperature layers. [quote]
That seems pretty easy to comprehend. But I had to ask myself, could the same concept be applied to air? If I could get the air to weigh more over a given area, I would have a pressure differential. That pressure differential would be all I need to run an electrical generator.
[/quote]

Yes it is easy to understand, and I did understand it the first time you presented it. However, I need to point out that there is only ONE way to achieve what you wish to achieve here, and it requires vorticity. That was what I was hoping to lead you towards by modeling the equations, rather than simply relying on your visualizations.[quote]
Now the idea of using a funnel to concentrate the weight of the air above it just might be feasable on a large scale. By keeping the angle of the pipe walls very small to the direction of the gravitational pressure, should result in a deflection of the gravitational pressure toward the center of the funnel pipe. That translates into more air mass weight over the center section. Because the deflection angle is so small, only a small portion of the gravitational head pressure is absorbed by the walls of the pipe. The rest of the weight is directed toward the center section. That means that the pipe walls are not supporting the air mass weight directly overhead. Now you might disagree because Bernoulli’s equation doesn’t predict this behavior.
[/quote]

No, I wouldn’t disagree for that reason. I would disagree because you must now consider shear forces of the fluid’s boundary layer over the surface of your funnel. These shear forces are where friction is created, heat is dissipated, and therefore energy is lost. BTW, you are correct that the “simple” form called the Bernoulli equation would not be useful for your situation. However, once you come to the conclusion (by doing the math) that you MUST have vorticity to induce the kind of pressure differentials you would need to run a turbine, then you would be forced to use the full-blown Navier-Stokes, non-linear differential equations to model this phenomenon. The simple closed-form solution represented by the Bernoulli equation will no longer work because the required vorticity will cause the entire pressure field to vary with more than just altitude. It will now vary with position with respect to the center of the vortex. [quote]
And I do have to point out that gravitational head pressure is directional. Mechanical head pressure is not directional.
[/quote]

Well…yes…kind of. Head pressure is directional because it aligns with the direction of the gravity vector. But mechanical pressure (static pressure) is still directional… it is simply OMNI-directional. The concept of pressure equalization in a container is what causes this phenomenon. [quote]
This is interesting because there is some predicted new type of pressure phenomena going on in the center of the air mass in the pipe. The mathematics for this just could be worth a Nobel prize. But for now it is still in the hypothetical stage.
[/quote]

Yes, it is interesting, and yes the center of the airmass inside a cone (with the required vortex) IS the interesting phenomenon. But the mathematics for this are well-known, and fully described in the Navier-Stokes differential fluid dynamics equations, if you were to derive them in a cylindrical coordinate system. And BTW, this is precisely what Computational Fluid Dynamic (CFD) computer programs solve, numerically. The CFD aerodynamic computer simulations simply allow you to create a static flowfield definition, one or more “forcing functions” (in this case it would be the vorticity required), and then the program will numerically integrate the Navier-Stokes equations to provide complete flowfield parameters (velocities, accelerations, static pressures, head pressures, and dynamic pressures). This is why such programs are so valuable to aerodynamicists who design aircraft wings such that vorticity is MINIMIZED at the wing tips (called a wake vortex).

I hope you understand that I have been trying to lead you towards the solution, and at the same time show you that the math IS required. What you are searching for with your “visualization” process is the exact same thing I have commented on in this forum. Namely, you are seeking a system that will tap-into nature’s vortex phenomenons in order to turn an electrical generator.

You need VORTICITY. That is the key. And nature creates such vortex fields as a result of two regions of atmospheric instabilities combining with each other to form the vortex. If you’d like to continue with even more explanations of the math and derivations, we can do so.

RMT

[Einstein]

RMT

Yes it is easy to understand, and I did understand it the first time you presented it. However, I need to point out that there is only ONE way to achieve what you wish to achieve here, and it requires vorticity. That was what I was hoping to lead you towards by modeling the equations, rather than simply relying on your visualizations.

Yes. I agree. We are going to get to that. But first a math model, to describe this static predicted behavior, before moving on to a vortex model, is needed. Here is a drawing I made to indicate the suspected high pressure zone.

Now idealy I would like to run a drain pipe right up the middle of the high pressure zone and bleed off pressure to run a generator. As you can see Bernoulli’s equations don’t predict this high pressure zone. So in reality I need a static math model to show the pressure zone. Now there is something I noticed that has got my curiousity up just a bit. Remember I said that gravitational head pressure is directional and varies with altitude making it different from static pressure. Well now we have a new type of pressure that varies radially away from the center of the pipe. The gravitational pressure just pushes in one direction or one dimension. This radial pressure is two dimensional but shares the same varying intensity with distance as gravity pressure. Now being two dimensional in nature makes me wonder if this configuration is stable. Because two dimensional objects can rotate.

So now we can move into the vortex model you were talking about. But you can now see the importance of looking at the whole picture and not just one segment. As far as I know rotating air masses have a low pressure zone in the center. So if the air mass in the funnel pipe does start to rotate, The high pressure zone in the center will equalize with the pressure on the pipe walls. But the air mass would have to be rotating for equalization to take place.

However, once you come to the conclusion (by doing the math) that you MUST have vorticity to induce the kind of pressure differentials you would need to run a turbine, then you would be forced to use the full-blown Navier-Stokes, non-linear differential equations to model this phenomenon. The simple closed-form solution represented by the Bernoulli equation will no longer work because the required vorticity will cause the entire pressure field to vary with more than just altitude. It will now vary with position with respect to the center of the vortex.

We’re getting there. I’m using the vortex model as an equal but opposite force to balance the static pressure zone initially produced. Once the pressure balances we now have a new problem. A spinning vortex that probably will generate an atmospheric disturbance. A turbine shaft could be run up the middle of the pipe to suck off some of the rotational energy. But the well would seem to never run dry. But now the atmospheric problem wouldn’t go over to well with the public. Hurricane force winds as a byproduct of free energy probably would get a lot of negative reaction. So I suspect that if we do make a vortex power generator, we will have to make them in counter rotating pairs.

Did any of this ring any bells? It did for me. I think we are looking at the force dynamics of electron structure. The way the rotatating pressure field changes in intensity from positive pressure to negative pressure at the center is identical to my rotating gravity field concept. Remember I said a rotating gravity field becomes an antigravity field with increased rotational speed. And I do believe my theory on electron behavior changinng from an attractive state to a repulsive state with the addition of voltage parallels this pipe pressure concept exactly.

So now for the math. It’s not as easy as it looks. We have a pressure gradient in existance before we even start to rotate the air mass. Rotatating the air mass doesn’t completely nullify the initial pressure gradient at all radiuses. I think defining the parameters of the initial pressure gradient is where we should spend some time. Also a math model to show just how much air density is being dumped into the center section. There probably is an optimum size to search for. Like pipe slope angle to radius values to determine the most efficient pressure gradient producer.

[RainmanTime]

Einstein,

Here is a drawing I made to indicate the suspected high pressure zone.

OK, so here is where we are going to have to have some more discussions, and I’d like to ask you to move slowly.

Now idealy I would like to run a drain pipe right up the middle of the high pressure zone and bleed off pressure to run a generator. As you can see Bernoulli’s equations don’t predict this high pressure zone.

Perhaps Bernoulli’s does not predict the high pressure zone because it does not really exist (absent vorticity in the flowfield)? I would like you to spend some more time explaining the physical basis of your picture and why you think a static airmass works this way, because current measurements and math models do not support this. The arrows/lines you draw bouncing off the pipe walls… I can tell you that an airmass with little to no velocity does not act in this manner (remember, the static pressure at the bottom is HIGHER than the pressure at the top).

And when you write “predominantly downward collisions”, what are you referring to? If you are referring to intermolecular collisions, this is where the gas dynamic theories and predictions related to TEMPERATURE will come into play. Certainly, temperature gradients will have an effect on pressures and molecular motion. I don’t want you to think I am ignoring your other points, but based on my experience and knowledge I think we need to focus in on some topics, as what you are discussing does not reflect what we already know.

We have a pressure gradient in existance before we even start to rotate the air mass.

How so? This is precisely why I introduced the Bernoulli equation. There is a STATIC pressure gradient between the top and bottom, as predicted by the ICAO atmosphere model. But in the absence of any fluid velocity, that static pressure gradient is offset by the head pressure difference between the two altitudes.

Again, we need to talk specifics, use math, and start looking at example values. Words alone are not going to result in an adequate and/or accurate analysis.

RMT

[Einstein]

RMT

Perhaps Bernoulli’s does not predict the high pressure zone because it does not really exist (absent vorticity in the flowfield)? I would like you to spend some more time explaining the physical basis of your picture and why you think a static airmass works this way, because current measurements and math models do not support this. The arrows/lines you draw bouncing off the pipe walls… I can tell you that an airmass with little to no velocity does not act in this manner (remember, the static pressure at the bottom is HIGHER than the pressure at the top).

And when you write “predominantly downward collisions”, what are you referring to? If you are referring to intermolecular collisions, this is where the gas dynamic theories and predictions related to TEMPERATURE will come into play. Certainly, temperature gradients will have an effect on pressures and molecular motion. I don’t want you to think I am ignoring your other points, but based on my experience and knowledge I think we need to focus in on some topics, as what you are discussing does not reflect what we already know.

This is the heart of the whole concept. I believe I recall Brownian motion as being applicable. All the air molecules ARE in a state of constant motion. It’s part of the property of matter in the gas state. So in reality the air mass is in a state of constant motion. Air has weight. All the air molecules moving in the downward direction toward the pipe walls will be deflected with a deflection angle equal in incidence to the impact angle. Technically all OTHER collision angles should average out to zero when added up. So we are just left with the downward collisions to deal with. An air molecule moving in a downward direction toward a pipe wall will deflect toward the center of the pipe. Since the pipe is round, all force vectors converge at the center. Mathematically the air has to be denser in the center because of the converging force vectors. I know my drawing over exaggerates the angle of impact and deflection. With only a one degree deflection angle the area of higher pressure would be much further down inside the pipe.

I haven’t really told you anything that you didn’t already know. I merely pointed out a suspected behavior of a gas in a pipe. I don’t see an equal but opposite force to prevent the pressure buildup in the center section. It is very possible that temperature and existing air pressure can either enhance or retard the predicted behavior. I would tend to think that higher temperature should result in higher collisions producing a higher pressure in the middle section. But higher temperature air tends to prefer a smaller density configuration. So temperature and pressure may balance each other out maintaining a constant pressure zone.

Now as you can probably see it would be very easy to construct a math model to parallel the suspected pressure buildup. But would we be wasting our time? That’s what we’re doing all the time anyway. But if this does parallel reality, then this math model could give us some insight into just how tornadoes and hurricanes originate. Playing around with the math could actually give us some new tools to help predict when this phenomena occurs in our atmosphere.

So you are right in going slow. If there is a radial air pressure field here, then this phenomena would be something new to investigate. Changing the angle of the pipe to a larger one actually increases the height of the suspected pressure zone. That would probably reduce the pressure within the zone. So there may be an optimum pipe angle to try and find. I have yet to think of anything that would disallow the pressure buildup. If you can think of something, throw it up for discussion.

Using my one degree pipe angle I can calculate just how high the pressure zone extends within the pipe. Just a little more trigonometry for that. That will be interesting since my base pipe area is one tenth my pipe mouth area. Then I will extend a cone from the base area to the predicted height. I will be kind of curious to compare the actual volume of space within the suspected pressure zone to the actual volume of the entire pipe volume. Will it be the ten to one area ratio or something else? Have to do that later…

[RainmanTime]

Einstein,

I believe I recall Brownian motion as being applicable.

No. Brownian motion is referenced to a particle suspended in a fluid. It is not applicable to a homogeneous fluid. I think the more appropriate phenomenon would be diffusion.

All the air molecules ARE in a state of constant motion. It’s part of the property of matter in the gas state. So in reality the air mass is in a state of constant motion.

Yes, but there is no preferred direction that is evident in this motion, which is precisely why we can ignore the individual molecular effects and treat the airmass as a single, static fluid reservoir. In other words, the RMS position of any single molecule of air is zero.

All the air molecules moving in the downward direction toward the pipe walls will be deflected with a deflection angle equal in incidence to the impact angle. Technically all OTHER collision angles should average out to zero when added up. So we are just left with the downward collisions to deal with.

No. This is a good example of how trying to think this through without math can lead one astray. In this case, if you wish to analyze molecular motion and see its effects on the overall airmass, you would need to write the equations for a single molecule with a generalized velocity vector, and then integrate over all air molecules with a Markovian distribution of velocity vectors. Your assumption is again based upon a “perferred” direction in which air molecules are moving more in the downward direction than in any other direction, and this is simply not supported by empirical evidence.

But there is a more important effect that you are not including in your analysis, and it is this effect which serves to damp-out the effect of any molecules colliding with the surface of your funnel. The Boundary Layer effect is well known, well studied, and well documented in fluid dynamic models. It is a result of the viscous properties of a fluid and how they come into play with respect to fluid motion when there is a fixed surface present. This viscous action of the fluid in the boundary layer region of a fixed surface act as a damper on fluid velocities. If you look at THIS web page you can see the damping effect of the boundary layer. In the situation we are discussing, the laminar boundary layer is the appropriate picture to look at, since the velocities you are discussing when you reference Brownian motion (or diffusion) are certainly slow enough so at to not induce turbulent flow. Note that the velocities become slower as you approach the fixed surface. These are the viscuous damping forces of the fluid. These same viscuous forces will essentically cancel-out any accumulation of molecular motion due to impacts with the fixed surface. But even more important than that is the fact that pressure will always equalize when given the chance. Thus…

Since the pipe is round, all force vectors converge at the center. Mathematically the air has to be denser in the center because of the converging force vectors.

No. This ignores the property of diffusion which leads to pressure equalization. Any TENDENCY for air to increase its density in the center will be countered by diffusion which seeks to equalize pressure energy when there is no physical entity present that prevents diffusion and pressure equalization.

I merely pointed out a suspected behavior of a gas in a pipe. I don’t see an equal but opposite force to prevent the pressure buildup in the center section.

Diffusion and the pressure equalization that results from it. That is the part you are missing.

So temperature and pressure may balance each other out maintaining a constant pressure zone.

Yes, but this is true of ANY temperature. In point of fact, if you investigate the concepts behind atmospheric stability you will see that temperature, and specifically the temperature lapse rate (the “a” in the equations I gave above) is the factor that is most directly responsible for whether a section of the atmosphere is stable or unstable. You no doubt know what an “inversion layer” is. This is a segment of the atmosphere where temperature does not descrease linearly with altitude, but in fact either remains constant or increases with altitude. In this state (inversion layer) we call the atmosphere “overdamped”. This is what causes pollution to get “locked in” and held down because the temperature inversion damps out any tendency for the higher-pressure air at the surface that contains the pollution to diffuse upward and outward. The value of “a” that I gave you above is the value at which the atmosphere is neutrally stable. In other words, under this value of “a”, if I displaced a volume of air upwards in the atmosphere (say about 100 feet) that volume of air would stay where it is. It would not move further upward nor would it move back down. It would simply diffuse at the point it was displaced to. As the negative magnitude of “a” gets larger (becomes more negative), the atmosphere becomes less and less stable. This is typically what we observe in “tornado alley” when the conditions are right to spawn tornados. We see the gradient at which temperatures decrease with altitude become much larger (more negative) in a warm, high pressure region. When this mixes with a cooler, low pressure region at a weather front this is what gives rise to large airmass fluxes, and this is what can lead to buildup of vorticity.

I have yet to think of anything that would disallow the pressure buildup. If you can think of something, throw it up for discussion.

Diffusion and pressure equalization. I do not believe that empirical measurements support your current theory. Ergo, I do not think there is something here that is “new” or that we do not understand. Again I must point out, vorticity is the key. Your idea will only become useful when we introduce a vortex field, and at THAT point you will indeed see an accumulation of LOW PRESSURE air within the vortex. Remember that the center of a tornado is a low pressure region.

RMT

[Einstein]

RMT

Yes, but there is no preferred direction that is evident in this motion, which is precisely why we can ignore the individual molecular effects and treat the airmass as a single, static fluid reservoir. In other words, the RMS position of any single molecule of air is zero.

I disagree completely and I think this just might be because the equations we work with don’t treat gravitational air pressure as being equal but opposite in magnitude to the induced mechanical air pressure. The gravitational air pressure does have a preferred direction. We describe this as the weight of the air. Each molecule of air is in constant motion. Any motion would be influenced by the gravity vector. Air has a property that does distinguishes it from liquids and solids. It is compressable.

No. This is a good example of how trying to think this through without math can lead one astray.

Your’e not fooling me. All math comes from visualizations. Without the visualizations, math would have never been developed.

In this case, if you wish to analyze molecular motion and see its effects on the overall airmass, you would need to write the equations for a single molecule with a generalized velocity vector, and then integrate over all air molecules with a Markovian distribution of velocity vectors. Your assumption is again based upon a “perferred” direction in which air molecules are moving more in the downward direction than in any other direction, and this is simply not supported by empirical evidence.

What you say can actually be true. But it just means that we have not bothered to look to see if this type of evidence exists. My assumptioon that the air molecules have a preferred direction downward, isn’t really an assumption. It’s a fact. It’s just another way to say air has weight. An undeniable fact.

No. This ignores the property of diffusion which leads to pressure equalization. Any TENDENCY for air to increase its density in the center will be countered by diffusion which seeks to equalize pressure energy when there is no physical entity present that prevents diffusion and pressure equalization.

But there is a physical entity present that prevents the pressure equalization. Remember all this extra pressure is coming from the diverted weight of the airmass above.

Diffusion and pressure equalization. I do not believe that empirical measurements support your current theory. Ergo, I do not think there is something here that is “new” or that we do not understand. Again I must point out, vorticity is the key. Your idea will only become useful when we introduce a vortex field, and at THAT point you will indeed see an accumulation of LOW PRESSURE air within the vortex. Remember that the center of a tornado is a low pressure region.

I do think the diffusion and pressure equalization would be the equal but opposite force to my proposed high pressure zone. But the equalization just takes place in a different manner than you are accustomed to. If the air starts to rotate around the pressure buildup, centrifugal force plays a role in pulling air out of the pressure zone. With enough rotational speed, the center high pressure zone could actually be lowered into a low pressure zone. I actually kind of like this model because it directly links rotating phenomena in our atmosphere to a gravitational influence. But then one has to develop the math from the visualizations. Are you up to this? I find it fascinating to come up with understandable mathematical scenarios that could lead to free energy.

[RainmanTime]

Einstein,

I disagree completely and I think this just might be because the equations we work with don’t treat gravitational air pressure as being equal but opposite in magnitude to the induced mechanical air pressure.

I’m sorry you feel that way, but what I am describing to you are well-understood phenomena of hydrostatics. And you have not yet described anything that Bernoulli’s equation for pressure energy does not account for. [quote]
The gravitational air pressure does have a preferred direction. We describe this as the weight of the air. Each molecule of air is in constant motion. Any motion would be influenced by the gravity vector.
[/quote]

Yes, indeed this is all correct. This is the head pressure term in Bernoulli’s. And this term is actually what causes the static pressure term to rise as you get closer to the earth. The weight of the column of air above it is the forcing function for the static pressure at any altitude. Diffusion is the means by which head pressure in the downward direction is converted to NORMAL (as in perpendicular to a surface) static pressure.

So what I must report to you is that your drawing is actually backwards. Once you place a funnel-shaped object in the airmass, you will observe that the static pressure at any altitude in the funnel exerts itself normal to the funnel’s inner surface. See HERE. Moreover, the diffusion of the head pressure as static pressure will actually serve as another damper to Brownian movement, in addition to the boundary layer effect I have described above. In a static airmass with a funnel in it, I can assure you that the principle of diffusion still works as it always does, and there is no tendency for the core of the funnel to develop a hi pressure differential. Now once things become dynamic (air velocities) that is when things change, and Bernoulli’s tell us how via the dynamic pressure term.

Air has a property that does distinguishes it from liquids and solids. It is compressable.

Yes, a very good point. And in fact when we look at compressibility of air, this will prove my point even further about diffusion in a static airmass. You see, while air is a compressible fluid, it only shows compressibility effects at higher velocities (Mach 0.3 and above is the rule of thumb in aerodynamics). So for static airmasses, and low velocity flows, air can be treated as incompressible, and incompressible flow relations work well. It is not until you exceed 3/10th the speed of sound that you see air compressibility coming into play.

Your’e not fooling me. All math comes from visualizations. Without the visualizations, math would have never been developed.

And I am not trying to fool you, nor do I disagree with your statement. My point was only that you cannot rely on visualization solely without cross-checking with both math and empirical observations. But we’ve had that discussion all along.

My assumptioon that the air molecules have a preferred direction downward, isn’t really an assumption. It’s a fact. It’s just another way to say air has weight. An undeniable fact.

Again I am not disagreeing with this, only the assumption of yours that leads to the high pressure zone in the center, and your claim that Bernoulli’s is missing a term that describes what is going on. For Hydrostatic situations, which is what we are talking about, pressure energy balance is dictated by the head pressure and static pressure terms of Bernoulli’s. [quote]
I do think the diffusion and pressure equalization would be the equal but opposite force to my proposed high pressure zone. But the equalization just takes place in a different manner than you are accustomed to.
[/quote]

How so? Please explain, as I’ve worked this stuff my whole career, and I’d like to know if I’ve been looking at things wrong all this time.

If the air starts to rotate around the pressure buildup, centrifugal force plays a role in pulling air out of the pressure zone. If the air starts to rotate around the pressure buildup, centrifugal force plays a role in pulling air out of the pressure zone. With enough rotational speed, the center high pressure zone could actually be lowered into a low pressure zone. I actually kind of like this model because it directly links rotating phenomena in our atmosphere to a gravitational influence. But then one has to develop the math from the visualizations. Are you up to this? I find it fascinating to come up with understandable mathematical scenarios that could lead to free energy.

Ahhhh, OK so now we are going to move on to the vortex (dynamic) situation. Sure I am up for it, but I want to make sure that we have reached understanding and agreement about the hydrostatic situation described by Bernoulli’s. If we don’t have a common understanding of the hydrostatic condition, then we will have trouble agreeing on what happens when we introduce velocity. Such as when you say:

With enough rotational speed, the center high pressure zone could actually be lowered into a low pressure zone.

But there is no high pressure zone in the hydrostatic condition, before you introduce rotational velocity. This is what I think we need to agree upon.

RMT

[Einstein]

RMT

Time seems to be in short supply for me lately. So I wont be addressing every point you brought up.

You have seen my approach on how I want to mathematically construct the high pressure zone. So I think I have really said all I can behind the mathematics for the suspected high pressure zone. Now what is amusing to me is that you are familiar with turning the funnel upside down and creating a low pressure zone. This would create an upward air flow. But when I examined this situation it was exactly in reverse with the mathematical approach I showed you I used initially. The weight of the air over the surface area of the smaller end produces less gravity induced air pressure over the surface area of the larger base area. Technically you have created a low pressure zone. The funnel pipe in this manner actually shields the air beneath it from what the pressure would be if the funnel pipe wasn’t there.

Now today I was thinking about the gravitational head pressure and it occurred to me that a pressure sensor could be made that is directional only, just for the purpose of indicating gravity head pressure. Just combine a regular pressure sensor with a sealed vessel that has a sliding weight inside it. Pointed in the up direction causes the weight to increase internal air pressure. But pointed downward indicates negative internal air pressure. At right angles no gravity pressure exists. Yes there is a method to my madness. This kind of sensor could be used inside a funnel pipe to indicate either an increase or decrease in the gravity induced head pressure. Then the thought occurred to me that maybe this sensor has already been invented and may already be in use. You might have access to information like that.

A digital sensor could be constructed that is very sensitive to minute changes in air pressure. The funnel pipe idea could be tested on a much smaller scale in a laboratory setting to monitor minute changes in air pressure by using different pipe configurations. Now my question is: Does such a sensor already exist? Or is this going to be my next experimental project?

[RainmanTime]

Hi Einstein,

Now what is amusing to me is that you are familiar with turning the funnel upside down and creating a low pressure zone. This would create an upward air flow.

No, it wouldn’t. Not unless there was some induced velocity, and again I am trying to reach common ground on the static situation before we move on to dynamic.

It has finally occurred to me what you are thinking may be the case, and I can again assure you that this is well-known phenomenon that I teach in my ARO 101 course when we discuss pressure. That principle is stated as: the shape of a container has no bearing on the pressure at any given depth. HERE is another web link that explains this fact of hydrostatic theory.

“Note: Only the density of the fluid and the difference in depth affects the pressure. The shape and size of the container are irrelevant. Thus the water pressure 6 inches below the surface of the ocean is the same as it is 6 inches below the the surface of a glass of salt water.”

This same principle applies to the static atmosphere, and as I have already pointed out the compressibility of air is not a factor at low (or no) fluid velocities.

Now today I was thinking about the gravitational head pressure and it occurred to me that a pressure sensor could be made that is directional only, just for the purpose of indicating gravity head pressure.

Yes indeed, and it does not even need to be complex. The simple water manometer, or even a barometer, measures head pressure. The difference in pressure between atmospheric and an evacuated (vacuum) chamber creates a fluid column that is measured to determine the head pressure with respect to vaccuum.

A digital sensor could be constructed that is very sensitive to minute changes in air pressure. The funnel pipe idea could be tested on a much smaller scale in a laboratory setting to monitor minute changes in air pressure by using different pipe configurations. Now my question is: Does such a sensor already exist? Or is this going to be my next experimental project?

Precision pressure sensors are available, but not to the levels of micro-precision that you would need for this experiment. And when you get down to such precise measurements, you cannot ignore other effects that introduce errors. Thermal noise is a big problem in aerospace pressure sensors, and we don’t need near the accuracy you are talking about. Furthermore, pressure is usually measured with a vibrating diaphragm to improve accuracy. Yet you would also have to wonder if some of the miniscule measurements you might see are attributable to vibration frequency error.

Good luck in finding that high pressure zone, but don’t be upset if I don’t follow along. I don’t really see the point since this is well established by many experiments and many scientists over the years.

RMT