Cadman’s Hydrostatic Displacement Engine © 2019 Cadman Weyland

[Grumage]

Hi Gambeir.

A quick question.

Do you know if by making the power piston hemispherical ( convex ) effectively adding to its surface area, if we would see some additional force applied to the piston rod?

The other obvious effect would be a small reduction in the volume of fluid being needed to be transferred during the next stroke.

Cheers Graham.

[Gambeir]

Hi Gambeir.

A quick question.

Do you know if by making the power piston hemispherical ( convex ) effectively adding to its surface area, if we would see some additional force applied to the piston rod?

The other obvious effect would be a small reduction in the volume of fluid being needed to be transferred during the next stroke.

Cheers Graham.

I appologize Graham, yes I did miss your post, and no I can't claim to be that advanced with motor design, but of course there's a reason for doomed and convex/concave pistons. Hopefully this advice is taken in and adopted to the design. Any other suggestions I'm sure would be appreciated.

[HuntingRoss]

A fun idea

"what is the weight of a liter of water in a vacuum at sea level on Earth "

What an intriguing question.

The phrase 'sea level' is a shorthand to allow the presumption of atmospheric pressure being 14.5psi.

The 'weight of water' presumes containment (how else would you weigh it?)

As we are presuming the presence of atmosphere, 'in a vacuum' implies the contained litre of water is within a vacuum container.

The implied presence of a vacuum container defeats 'sea level'. The weight of a litre of water would be the same in the vacuum container irrespective of its altitude. Unless the vacuum container is so distant from the effect of gravity to discount its effect.

A further twist to this notion is, a vacuum is the absence of matter, which is demonstrably incorrect in the question which includes the presence of one litre of water.

A fun idea though.

[Grumage]

Thanks.

I appologize Graham, yes I did miss your post, and no I can't claim to be that advanced with motor design, but of course there's a reason for doomed and convex/concave pistons. Hopefully this advice is taken in and adopted to the design. Any other suggestions I'm sure would be appreciated.

Hi Gambeir.

The title says it all.

The various search engines cite many reasons but sadly all are referring to the ICE.

Cadman's design is quite unique insofar as we are working with a low pressure application where, in this case, increasing surface area would seem to be advantageous. In the arena of hydraulics with very large pressures little to no advantage would be noticed IMO.

I've made a start today on ordering a 10" by 6" automotive Rubber tire inner tube. Those that have been following Cadman's progress over at OU.Com will know that I presented an alternative design. Rather than a piston and cylinder I chose to use a " bellows " approach in the hopes of reducing some of the friction and sealing issues.

If anyone is interested I can attach a drawing and explanation of how I see its operation.

Cheers Graham.

[RAMSET]

yes hurts my brain too

HuntingRoss
Grum and I have been struggling to grasp this ...your thoughts
do give more focus ...and are greatly appreciated .

but I have to confess I am still pondering the reasons for the question too
EA is not a sadist...he does and has shared way more than I or Grum can absorb .

There is something about the "annular gap" buoyancy and the gravity field
as it applies to mass manipulation ?
and Vacuum or establishing a nullification of atmospheric pressure by
utilizing this vacuum which manifests at times in these systems.
to cycle when it should not cycle...[the holy grail ?]

Grum
IMO you should post your idea ? the nice thing is the ITP [original thread
Poster Cadman] can ask us to remove or whatever ...

nice feature of this Venue IMO

also will try to post a quick link To EA open source work here or get a PDF ?

thx
Chet

[Grumage]

Ok Chet.

Here goes.

My single drawing shows the device at rest with water from the upper cistern piped down to the underside of the bellows/concertina, admission valve closed.

Upon opening of the admission valve the water fills the bellows causing it to rise ( transfer valve is shut ) we have allowed it to rise by 150mm or 6" lifting the displacement cylinder vertically through a sliding seal in the centre of the cistern 2 meters above.

We close the admission valve and open the transfer valve. The combined weight of the " bias " and cylinder now causes the water to flow into the displacement cylinder and travel upward round the sides of the solid Steel displacer piston. The bellows have now returned to rest as in my picture.

The next upward stroke see's the bellows fill and rise again, as we have 75Lbs of force available acting against just the volume of water plus that of the bias and cylinder. The displacer which is rigidly fixed from above the device, displaces the water through a cup washer at the top.

We now open the transfer valve and the bellows and cylinder collapse once again. As the cylinder collapses any water above the displacer piston falls over the sides to refill the cistern.

Back at rest again and ready for the next cycle.

For an automatic reciprocating engine we shall have to design some mechanism to operate the two valves.

But....

This all seems far too easy, what has been missed, am I dreaming?

Cheers Graham.

Post script.

I felt that perhaps my description of operation was lacking one detail, I'd forgotten to mention that the system had been previously purged of any air.

Sorry for any inconvenience.

[HuntingRoss]

My tuppence worth

Ok Chet.

Here goes.

My single drawing shows the device at rest with water from the upper cistern piped down to the underside of the bellows/concertina, admission valve closed.

Upon opening of the admission valve the water fills the bellows causing it to rise ( transfer valve is shut ) we have allowed it to rise by 150mm or 6" lifting the displacement cylinder vertically through a sliding seal in the centre of the cistern 2 meters above.

We close the admission valve and open the transfer valve. The combined weight of the " bias " and cylinder now causes the water to flow into the displacement cylinder and travel upward round the sides of the solid Steel displacer piston. The bellows have now returned to rest as in my picture.

The next upward stroke see's the bellows fill and rise again, as we have 75Lbs of force available acting against just the volume of water plus that of the bias and cylinder. The displacer which is rigidly fixed from above the device, displaces the water through a cup washer at the top.

We now open the transfer valve and the bellows and cylinder collapse once again. As the cylinder collapses any water above the displacer piston falls over the sides to refill the cistern.

Back at rest again and ready for the next cycle.

For an automatic reciprocating engine we shall have to design some mechanism to operate the two valves.

But....

This all seems far too easy, what has been missed, am I dreaming?

Cheers Graham.

Post script.

I felt that perhaps my description of operation was lacking one detail, I'd forgotten to mention that the system had been previously purged of any air.

Sorry for any inconvenience.

Stage 1.
The displacement cylinder, bellows and displacement cylinder above the non-return valve (DP2) contain only air. The transfer valve and inlet valve are closed.

Stage 2
The inlet valve opens and water flows from the cistern through the supply column to the bellows.

The head of water in the cistern is sufficient to displace the bellows, bias weight, weight of the displacement cylinder and some random friction from the sealing ring to raise it 6" (150mm). Pressure in the bellows and cistern equalise.

The inlet valve closes.

Stage 3.
The transfer valve opens allowing the pressurised water to flow into the displacement cylinder and annular gap around the displacement piston. The bellows, bias weight and displacement cylinder move down to 0" (0mm).

The transfer valve closes.

The bellows, displacement cylinder and DP2 contain only water.

Stage 4.
Repeats as Stage 1 with the exception that the displacement cylinder, bellows and DP2 are flooded.

Thoughts.
The open end of the displacement cylinder at Stage 1 must be level or higher than the water level in the cistern.

The head of water at Stage 4 will equal the head of water in the cistern and no water will flow.

The presence of air or not doesn't effect the outcome as the conditions in the cistern is shared with DP2.

[Grumage]

Thanks for your two penneth.

Hi HuntingRoss.

I can get to stage 3 without a problem in your description.

At stage 4 we have a collapsed bellows that is flooded. With the transfer valve shut we can now refill with a fresh charge of water. This action causes the fluid above to be displaced through the annular gap and through the cup washer
( non return valve ) at the top.

Some of the displaced mass is now held above the displacer piston leaving a small amount below.

I wonder if by adding some dimensions we can get the figures needed to see if the design is plausible?

Cheers Graham.

[HuntingRoss]

Dims may help

Evening Grumage

Dimensions would help but I think the assessment is accurate based on current data. Assumptions include the level of the cistern water and open ended displacement cylinder per your drawing AND the cistern is the same diameter as the lower section containing the bellows.

On that basis the bellows would draw down a considerable volume to raise 6" (150mm) reducing the head pressure significantly.

Good hunting

[RAMSET]

seems odd for a reason

The question and I suppose this is the point !

"what is the weight of a liter of water in a vacuum at sea level on Earth "

just updating here .

this question has roots in our measurement standards and "their beginnings or roots" and has relevance toward a specific observation and claim.
EDIT
Apparent weight
https://en.wikipedia.org/wiki/Apparent_weight

International System of Units
https://en.wikipedia.org/wiki/Intern...ystem_of_Units

Kilogram
https://en.wikipedia.org/wiki/Kilogram

Snip
This information forms the basis of what we work from, not our own personal opinion on the matter which can differ. If we do differ then we have to calmly and rationally explain why.

Pay particular attention to the following statement and try and identify what information it does not state clearly:

“The kilogram was originally defined in 1795 as the mass of a litre of water. “

Ask yourself the question... Was the kg defined as it's actual weight in 1795 or was it defined as it's apparent weight ?

I can find no reference anywhere, ever, of the measurement being done inside a vacuum chamber. Therefore I must draw the conclusion that it was not..

What this means is that if we assume that the scientists at the time were breathing air, and they were immersed in air as was the 1 litre of water and measuring scales, then they measured the apparent weight of 1 litre of water and not it's actual weight.

Therefore it follows that the weight they measured was the actual weight of 1 litre at sea level (1 atmospheric pressure) with a prime moving stress force of 1g. The actual weight of that 1 litre of water should be 1kg of water + 1 kg of air because the buoyancy force acting on that 1 litre of water is equal to the amount of air displaced by that 1 litre of water, and this must be taken into account in the measurement.

So, 1 litre of water = 1kg is the apparent weight of that mass and not the actual weight.

When I ask “what does 1 litre of water at sea level weigh in a vacuum?” I am asking you to do an apparent weight measurement.

Do not get sidetracked by “sea level”, I ask this to provide location where g = 1, and as we know the column of air that covers 1 square inch of surface area starts at sea level and stretches upward to the upper atmosphere and weighs 14.7 pounds... therefore we have a pressure at the bottom of that column that is equal to 14.7 pounds per square inch (psi).

To weigh 1 litre of water at sea level in a vacuum the mass and scales must be inside a vacuum chamber. The air pressure at sea level acts on the vacuum chamber walls and not the 1 litre of water inside the vacuum chamber. The vacuum chamber walls are strong enough to withstand 14.7psi of pressure and therefore do not collapse inward.

So atmospheric pressure is absent inside the vacuum chamber and therefore there can be no buoyancy force because the 1 litre of water is immersed in a vacuum, nothing! Therefore the measured weight at sea level is the actual weight of 1 litre of water under 1g of stress from Earth's gravity, and not the apparent weight of water, which is the actual weight minus the buoyancy force from the air that is acting upon it in opposition to gravity.

Phew... got that ?

Ok, so what does 1 litre of water weigh at sea level immersed in a container of Mercury ?

https://www.youtube.com/watch?v=Rm5D47nG9k4
end snip
just a partial snip of the query..
more info to come


HuntingRoss ...I see your contributions here and elsewhere recently

I am grateful you take the time to analyze,critique and comment here.
this is the only way forward...brutal honesty [the scientific method}
and 100% transparency .
hopefully you will not be disappointed.


respectfully
Chet K

[HuntingRoss]

Having fun with units

Hello Ramset

"Ask yourself the question... Was the kg defined as it's actual weight in 1795 or was it defined as it's apparent weight ?"

Unless I am missing a nuance, the litre (a measurement of volume) defined the Kg. A litre does not vary according to altitude or proximity to a planet.

Gravity is attraction of mass and is measured as force (SI units - Newtons). Measuring gravity = 1 (atmosphere) at sea level is to quantify pressure.

To discuss weight, one is always considering it relative to its surroundings. The weight of 1kg in space maybe 9.8N (for sake of discussion) but in orbit the same Kg weighs 0N because it is in a permanent state 'of falling' and is therefore absent the effect of gravity.

Gravity acts at the centre of Earth, so 1kg at the centre of the Earth weighs 0N (there is no acceleration at the centre of the Earth).

"So, 1 litre of water = 1kg is the apparent weight of that mass and not the actual weight."

1kg is not the apparent weight - only it's mass.

"When I ask “what does 1 litre of water at sea level weigh in a vacuum?” I am asking you to do an apparent weight measurement."

At sea level, the mass of 1 litre of air is 1.225kg/m^3 divided by 1000 (litres/m^3) = 0.001225kg

1kg of water displaces 0.001225kg of air = 9.787995 N (assuming 9.8m/s^2)...A metric ton of water 1000kg would displace 1.225kg of air.

Absent the buoyancy effect of the air, 1kg in a vacuum container at sea level would weigh 9.8N and would probably weigh that for some considerable distance from the surface.

Hope this makes sense

Happy hunting

[RAMSET]

a reply ...and a build is moving forward.

“Unless I am missing a nuance, the litre (a measurement of volume) defined the Kg. A litre does not vary according to altitude or proximity to a planet. “

A litre is a measure of volume and does not vary, correct. The litre defined the Kg, correct. Was this litre's weight measured with the buoyancy force of air acting on it at 1g of acceleration, or without the buoyancy force of air acting on it, is the question. In the absence of air's buoyancy force a litre would not weigh 1 Kg if it was originally defined immersed in air.

https://en.wikipedia.org/wiki/Apparent_weight

“The apparent weight can also differ from weight when an object is "partially or completely immersed in a fluid", where there is an "upthrust" from the fluid that is working against the force of gravity.[2] “

“To discuss weight, one is always considering it relative to its surroundings. The weight of 1kg in space maybe 9.8N (for sake of discussion) but in orbit the same Kg weighs 0N because it is in a permanent state 'of falling' and is therefore absent the effect of gravity. “

This statement is incorrect. w=mg therefore an astronaut weighs the same as the same as he would on Earth because his mass and his acceleration have remained the same. What the astronaut experiences as the feeling of weightlessness is the absence of stress forces acting in opposition from a normal plane. What you can say is that the 1Kg has a weight of 9.8N but exerts a force of 0N on it's surroundings.

https://en.wikipedia.org/wiki/Weightlessness

Weightlessness in Newtonian mechanics

In Newtonian mechanics the term "weight" is given two distinct interpretations by engineers.

Weight1: Under this interpretation, the "weight" of a body is the gravitational force exerted on the body and this is the notion of weight that prevails in engineering. Near the surface of the earth, a body whose mass is 1 kg has a weight of approximately 9.81 N, independent of its state of motion, free fall, or not. Weightlessness in this sense can be achieved by removing the body far away from the source of gravity. It can also be attained by placing the body at a neutral point between two gravitating masses.

Weight2: Weight can also be interpreted as that quantity which is measured when one uses scales. What is being measured there is the force exerted by the body on the scales. In a standard weighing operation, the body being weighed is in a state of equilibrium as a result of a force exerted on it by the weighing machine cancelling the gravitational field. By Newton's 3rd law, there is an equal and opposite force exerted by the body on the machine. This force is called weight2. The force is not gravitational. Typically, it is a contact force and not uniform across the mass of the body. If the body is placed on the scales in a lift (an elevator) in free fall in pure uniform gravity, the scale would read zero, and the body said to be weightless i.e. its weight2 = 0. This describes the condition in which the body is stress free and undeformed. This is the weightlessness in free fall in a uniform gravitational field. (The situation is more complicated when the gravitational field is not uniform, or, when a body is subject to multiple forces which may, for instance, cancel each other and produce a state of stress albeit weight2 being zero. See below.)

To sum up, we have two notions of weight of which weight1 is dominant. Yet 'weightlessness' is typically exemplified not by absence of weight1 but by the absence of stress associated with weight2. This is the intended sense of weightlessness in what follows below.

A body is stress free, exerts zero weight2, when the only force acting on it is weight1 as when in free fall in a uniform gravitational field. Without subscripts, one ends up with the odd-sounding conclusion that a body is weightless when the only force acting on it is its weight.

The apocryphal apple that fell on Newton's head can be used to illustrate the issues involved. An apple weighs approximately 1 newton. This is the weight1 of the apple and is considered to be a constant even while it is falling. During that fall, its weight2 however is zero: ignoring air resistance, the apple is stress free. When it hits Newton, the sensation felt by Newton would depend upon the height from which the apple falls and weight2 of the apple at the moment of impact may be many times greater than 1 N. It was great enough—in the story—to make the great man invent the theory of gravity. It is this weight2 which distorts the apple. On its way down, the apple in its free fall does not suffer any distortion as the gravitational field is uniform.

A complicated subject indeed...

https://www.livescience.com/46560-ne...econd-law.html

Force, Mass & Acceleration: Newton's Second Law of Motion

Acceleration and velocity

Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force. However, if the object is already in motion, or if this situation is viewed from a moving inertial reference frame, that body might appear to speed up, slow down, or change direction depending on the direction of the force and the directions that the object and reference frame are moving relative to each other.

The bold letters F and a in the equation indicate that force and acceleration are vector quantities, which means they have both magnitude and direction. The force can be a single force or it can be the combination of more than one force. In this case, we would write the equation as ∑F = ma

The large Σ (the Greek letter sigma) represents the vector sum of all the forces, or the net force, acting on a body.

It is rather difficult to imagine applying a constant force to a body for an indefinite length of time. In most cases, forces can only be applied for a limited time, producing what is called impulse. For a massive body moving in an inertial reference frame without any other forces such as friction acting on it, a certain impulse will cause a certain change in its velocity. The body might speed up, slow down or change direction, after which, the body will continue moving at a new constant velocity (unless, of course, the impulse causes the body to stop).

There is one situation, however, in which we do encounter a constant force — the force due to gravitational acceleration, which causes massive bodies to exert a downward force on the Earth. In this case, the constant acceleration due to gravity is written as g, and Newton's Second Law becomes F = mg. Notice that in this case, F and g are not conventionally written as vectors, because they are always pointing in the same direction, down.
The product of mass times gravitational acceleration, mg, is known as weight, which is just another kind of force. Without gravity, a massive body has no weight, and without a massive body, gravity cannot produce a force. In order to overcome gravity and lift a massive body, you must produce an upward force ma that is greater than the downward gravitational force mg.

The key part here to note is the last sentence...

An upward force ma that is greater than the downward gravitational force mg will lift a massive body, as demonstrated in the real world by the Mythbusters team:

Mythbusters - Ping pong salvage

https://www.youtube.com/playlist?lis...C02E9ABC7C1144

In conclusion it is possible to switch a massive bodies apparent weight by manipulating the buoyancy force acting upon it. The ratio between water and air density is:

1000/1.225 = 816.3265306122449

https://www.youtube.com/watch?v=t-0iJ25zbcs

https://www.youtube.com/watch?v=DDfI2piICvM

https://www.youtube.com/watch?v=qDfXC7GKHZw

[HuntingRoss]

Moving forward to where ?

Hello Ramset

I'm not certain where all of this is going and I'm not seeing any questions that need addressing.

My only comment / observation :

A 1kg mass sitting in a plane flying in the same direction around the earth as a 1kg mass in orbit do not weigh the same. The plane is slow and the wing lift exerts an equal and opposite force on the mass. The orbiting mass is fast to the point where it's not slow enough to drop to Earth and not fast enough to be thrown out to space, it is therefore considered to be constantly falling. The centrifugal / centripetal forces cancel. The same mass in space would have weight unless it is so far distant from the attractive mass that it is considered weightless. I think in your comment "What you can say is that the 1Kg has a weight of 9.8N but exerts a force of 0N on it's surroundings" you are resolving the two definitions of weight. Fair enough.

"In the absence of air's buoyancy force a litre would not weigh 1 Kg if it was originally defined immersed in air". A litre of water does not weigh 1kg and I haven't seen a definition for the weight of water and the conditions under which it was measured where the buoyancy effect should/would have been considered.

I'm assuming this path of enlightenment is leading to a bigger question or point from your title "a build is moving forward" ?

Good hunting

[RAMSET]

HuntingRoss
Quote
I'm not certain where all of this is going and I'm not seeing any questions that need addressing.
end quote

Thanks for your reply , energy harvesting and repurposing ,a proof of
concept empirical test bed.

Open sourced .

should not be possible !

respectfully
Chet K

[HuntingRoss]

Was this litre's weight measured with the buoyancy force of air acting on it at 1g of acceleration, or without the buoyancy force of air acting on it, is the question. In the absence of air's buoyancy force a litre would not weigh 1 Kg if it was originally defined immersed in air.

In conclusion it is possible to switch a massive bodies apparent weight by manipulating the buoyancy force acting upon it.

Hello Ramset

I've been reflecting on your reply. Was your question relating to the weight of water rhetorical or do you know the answer ? I am unaware of a defined weight of 1kg of water because it is dependent on so many other factors...to the point that I doubt it has ever been defined.

In my previous post I gave a quick estimation for the weight of 1kg of water at sea level by accounting for the bouyancy effect of air but you have returned to the question with the misconception that "a litre would not weigh 1kg"...was this a typo error ?

It seems elementary to comment on the buoyancy force manipulating the apparent weight of a body, hot air balloons for example, which still makes me think you are alluding to something 'bigger' but your response seems quite cryptic. Are you at an early stage of something ?

Happy hunting