Scooping Ash's Panacea "Hidro" report

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MileHigh

Scooping Ash's Panacea "Hidro" report « on: 2011-04-02, 05:03:08 »

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Quoting Ash:

Quote

(Hidro)
Hidro FREE renewable energy

Guys just a heads up, we were there and evaluated it ourselves , i went on camera for them, we had children with us in front of an over unity machine, a dream i have long sort after, this thing works and ill have full details ASAP, maybe do an interview with Sterling about the report. Wanted to first share some private photos here
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But the worlds only open source free energy university also went there;) , surprise these people are one of us and hate the NWO and are AWARE, 200 grand for 50Kw is the smallest they do, they have cost to cover, and i think that's not bad for now.

15 amps at 24 Volt went in, i was BLINDED by 1Kw of lights out. This thing could build up 20Kw with the fly wheel. I saw a cable that has a 600 kilo tolerance broken like a twig, this thing is POWERFUL.

Will have more details ASAP my friends.

Ash

http://panacea-bocaf.org/hidrofreeenergysystem.htm
http://www.jameskwok.com/tech/hidro.html

So is this too good to be true, or is there something terribly wrong with this picture? I looked into this several months ago and I saw what appeared to be a glaring omission that reduced the whole thing to a collapsing house of cards.

I'm curious to see if anyone else has an opinion.

MileHigh

rizla

Re: Scooping Ash's Panacea "Hidro" report « Reply #1 on: 2011-04-02, 05:37:34 »

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First glance it looks the real deal although if they don`t release detailed enough info on how it`s built there is a 99.99999% chance we the great unwashed will never see this technology. Looks a bit difficult for the average joe to build as well and allmost all goverments around the world will not be in a hurry to use it becouse it can`t be turned into a bomb.

I could really do with one of these in my back garden.

Chet K	Re: Scooping Ash's Panacea "Hidro" report « Reply #2 on: 2011-04-02, 12:40:25 »
Group: Ambassador Hero Member Posts: 4749	MH Well I have been very Busy lately,So I won';t even "Look" at it yet [Distractions] IMO Knowing your mind set, it must be a problem with how to measure power In, Or Power out Accurately? I don't "Bet" but if I did thats where I'd put my Money! Chet

BEP

Re: Scooping Ash's Panacea "Hidro" report « Reply #3 on: 2011-04-02, 13:13:26 »

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Quote from: MileHigh on 2011-04-02, 05:03:08

So is this too good to be true, or is there something terribly wrong with this picture? I looked into this several months ago and I saw what appeared to be a glaring omission that reduced the whole thing to a collapsing house of cards.

I'm curious to see if anyone else has an opinion.

If that glare comes from the statements about adding compressed air to a buoy to make it float and then releasing the air to make it sink, then we should hold off on negative judgment. Yes, adding air to a container increases the weight of that container. This would make it sink, not rise.

If that buoy (container) housed an air bladder, their statements about adding compressed air make sense.

MileHigh

Re: Scooping Ash's Panacea "Hidro" report « Reply #4 on: 2011-04-02, 16:59:53 »

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WaveWatcher:

You are in the ballpark relative to what I am thinking.

The issue is fairly straightforward: The system is being powered by a tank of compressed air. The taller the column of water for traveling buoys, the higher the pressure you need in your air tank to be able to evacuate the water from the buoy so that it can rise to the surface.

Where does the pressurized air come from?

If this system is supposed to be a source of free energy, then that implies that the mechanical power that you can extract from the rising/sinking buoys (or the electrical power derived from a generator setup) can compress enough air to a high enough pressure to keep evacuating the buoys at the base of the water column and at the same time there should be bonus extra power available which you could export to the outside world.

Here is the problem: I looked into this a few months ago and I didn't see any mention of how you maintain a source of compressed air to keep the system powered.

Anybody that is interested in this system and the upcoming "detailed Panacea Hidro report" should keep this issue front and center in their minds.

Will there be any mention about where the power comes from to maintain a tank of high-pressure air at the bottom of the column of water? If you don't see any mention of this in the literature then you are smelling a rat.

MileHigh

e2matrix

Re: Scooping Ash's Panacea "Hidro" report « Reply #5 on: 2011-04-02, 17:17:52 »

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Posts: 393

If it takes more power to run it than it creates how many of these systems do you think they are going to sell? They have put a lot of work into the construction of the device so for now I'm going to assume it works since I looked at the concept and it seemed valid (but what do I know?). They say it works on the same principle as a submarine that can ascend and descend: "Pressure-gradient exists naturally in a water column at 1bar (100kN/m2) per one-meter depth. Nature will always seek to equalise pressure-differences in water and the changes in water-pressure states from static to dynamic-pressure created in Hidro water-tower produces pressure-force that is harnessed as kinetic energy to operate buoyant-floats to drive a linear motor translator, thereby converting the harnessed energy into potential energy in the linear electric generator where the electrical-load is directly generated producing electricity. Many phenomena's regarding the flow of liquids and system mechanics are applied in the Hidro water-tower, including the basic buoyancy [10kN/m3] and water displacement, these phenomena's are utilized in a submarine (capsule) enabling it to rise and descend. The Patented "Reverse Thruster Pressure Equalizer Chamber (RETPEC) and its "operations" in HidroLinear-System is key to the technology commercial utilisation. "

The way I pictured it is not that they need compressed air but just a way to empty water out of the capsules at the bottom so they would only have air in them (not compressed as that would not be any advantage as far as I can see). It seems to me it's a system that is just using gravity (weight of the capsules) plus bouyancy (upward force when empty under water) to generate power. I would think the small amount of energy needed to change the water in and out of the capsules would be a lot less than that generated by the rather strong forces of bouyancy and gravity.

MileHigh

Re: Scooping Ash's Panacea "Hidro" report « Reply #6 on: 2011-04-02, 17:24:31 »

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A few months ago I spent an hour or two looking online at this "Hidro" system and like I said, I saw no mention of how the main air tank remains pressurized to power the system.

Just for fun, let's look at some of the points that make this whole system impossible.

1. The first point is based on pure gut feel: You say to yourself, "It doesn't make sense. It takes air pressure times displacement to evacuate the water from the buoy at the bottom of the column of water. Pressure x displacement = work, and there is no way that you can extract that much work + 25% bonus to cover the overhead to make the system self-sustaining. The taller the column of water, the more work you can extract from the rising/sinking buoys, but at the same time, the more air pressure you need to evacuate the buoys." By pure gut feel you can sense that it simply won't work.

2. You can't extract energy from a gravity field, even if you try to turn things on their head by using buoys in a column of water.

3. When the buoy is at the bottom of the column of water, you have to fill it with pressurized air. When the buoy reaches the top of the column, that pressurized air is released. That represents a nugget of lost energy.

4. Buoys moving through water have to overcome the viscosity/friction of the water. This represents more energy lost as heat.

5. When you compress air you heat it up. That heat dissipates into the environment. This represents more lost energy.

6. There is some considerable overhead to make the system work. You need to power some sort of electro-mechanical system that connects and disconnects the air feeds to the buoys. There are gears and pulleys that have friction, etc.

7. The efficiency of the electrical generator is not 100%.

8. Assuming that you are using an electrical motor to power an air pump to keep the main air tank pressurized, the motor-pump efficiency is not 100%.

There you go, the whole thing is a complete croc of nonsense. The guy that is promoting this system, "engineer" James Kwok is deluded and/or "crazy." Poor Ashtweth gobbles up free energy concepts like jellybeans and has almost no technical background and no qualifying skills.

MileHigh

MileHigh

Re: Scooping Ash's Panacea "Hidro" report « Reply #7 on: 2011-04-02, 17:46:42 »

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E2matrix:

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The way I pictured it is not that they need compressed air but just a way to empty water out of the capsules at the bottom so they would only have air in them (not compressed as that would not be any advantage as far as I can see). It seems to me it's a system that is just using gravity (weight of the capsules) plus bouyancy (upward force when empty under water) to generate power. I would think the small amount of energy needed to change the water in and out of the capsules would be a lot less than that generated by the rather strong forces of bouyancy and gravity.

Do you do any cycling? Have you used a manual air pump to pump up the tire of your 14-speed 'racer' style bike? Using English units, you typically have to pump up the tire to 70 psi. Have you experienced what it is like to do the final two pounds and get the tire from 68 psi to 70 psi? It is a hell of a lot of work. It takes WORK to pump up a tire when it is already at a high pressure.

When the buoy is at the bottom of the column of water and needs to be evacuated with air, the buoy is like the tire where the pressure is a constant 70 psi. It tales a lot of energy to evacuate the buoys of water when they are at the bottom of the column of water. It's possible that the whole scam is perpetuated on the fact that the average person doesn't realize this.

Quoting Ash again:

Quote

Hi Altrez yes they have independent (not just ours and Queensland university) tests with calibrated meters showing 300W in and 1KW out, we saw this also and took a closer look. This was a 20Kw power station, it has a lot of power. you could use 4 of them to run an Aquaponics set up 24/7 and be self sufficiency in ORGANIC food and carbon negative.

300 watts in to do what? I bet you Ash doesn't even know the answer to that question.

I bet you if someone reads all of the literature and follows all of the links, there will be no information at all about how the system drains off power from the rising/sinking buoys to keep the main air tank compressed at high pressure to keep the whole system running. The main air tank that keeps the whole system running is barely mentioned, and I did not see anything about recharging it to maintain its pressure when I read the literature a few months ago.

The whole thing is ridiculous and I challenge the OU and EF lurkers to raise this issue with Ash and James Kwok.

MileHigh

BEP

Re: Scooping Ash's Panacea "Hidro" report « Reply #8 on: 2011-04-02, 19:16:11 »

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Yes, buoyancy and energy for the compressed air - those questions and more.

The things that make me hold back on analysis are related to buoyancy and the air storage/conversion.

Buoyancy is a clear and easy understanding with very well worn rules. Still, I have serious doubts about the precision of those rules.

Compressed air....

This is wide open for me due to experiences with what may be called 'air-batteries'. I have worked with several systems from a few kVA to 250,000 kVA. All using a scroll expander to drive an alternator have the same issue. They always tend to supply more energy out than in, or so it seems, even when you throw all the finest details into the math.

The manufacturers have never claimed anything more than extremely high efficiency. I understand some UK firm recently gained a patent but I know the company 'ActivePower' has been building them for a long time.

My experience with these units is related to troubleshooting anomalies on the load-side. Any personal investigations relating to the above would have been refused.

The only known and obvious symptom is when the stored air is applied to drive the alternator, heat energy is very clearly sucked out of the ambient space. This effect is far more than the usual icing of an A.C. gas line.

As far as I know... If this system is as claimed, the inventor may not realize some of the 'extra' may be due to other parts of the system or a combination of forces, if there is any 'extra'.

I did look into the invention in 2008. I found nothing devious about it or the inventor and shelved it in hopes it was real and could be proven.

MH,

You're always asking about charts and time lines. If I remember correctly this guy posted some very good data for review. Hopefully, you've looked at that.

For now, I don't have a reason or enough info to consider a detailed analysis or post negative comments about this.

MileHigh

Re: Scooping Ash's Panacea "Hidro" report « Reply #9 on: 2011-04-02, 21:58:24 »

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WaveWatcher:

Quote

This is wide open for me due to experiences with what may be called 'air-batteries'. I have worked with several systems from a few kVA to 250,000 kVA. All using a scroll expander to drive an alternator have the same issue. They always tend to supply more energy out than in, or so it seems, even when you throw all the finest details into the math.

I briefly looked this stuff up. Okay, so you have an air capacitor that discharges through a scroll expander. The amount of energy in the air capacitor is akin to an electrical capacitor, = (1/2 x the size of the air tank x the air pressure squared). It's impossible for an air capacitor to supply more energy out than in. When the gas is expanding it draws in thermal energy from the environment. So if you 'overlook' that glaring fact perhaps you can get "more out than you put in."

It harks back to one of my old themes where electrical inductors and capacitors are no different from mechanical springs and masses. An air tank is just an other form of a capacitor and it works with essentially the same differential equations and has the same characteristics. If a compressed air tank fails you get a massive flow of air and a sudden drop in pressure, which is identical to shorting a capacitor out or having a moving mass crash into a concrete wall.

I don't have the drive to search out the data on James Kwok's system but I will read the Panacea report. If anybody thinks it's real and can find some data to back up the claim by all means post some links here.

A very simplistic analysis is as follows: The energy you can extract from the buoyancy is proportional to the height of the column of water. By the same token the pressure at the base of the column of water is proportional to the height of the column of water. Therefore you can say that the buoyancy energy as it floats up and the energy required to evacuate the water from the buoy are both proportional to the height of the column of water. However, for a fixed volume buoy, you are obliged to put higher and higher pressure air into the buoy to evacuate the water from it for a higher and higher column of water. That extra air pressure is released from the buoy when it reaches the surface of the column of water. That released "extra" air pressure represents lost energy. Therefore it's fair to say that the higher the column of water, the less efficient the system becomes. Not forgetting to mention that the system is always under unity.

I shouldn't prejudge but based on past experience I won't be surprised if the Panacea report barely touches on the issue of were the supply of compressed air comes from to keep the device running.

MileHigh

Allcanadian

Re: Scooping Ash's Panacea "Hidro" report « Reply #10 on: 2011-04-02, 22:41:03 »

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@Milehigh

Quote

The first point is based on pure gut feel: You say to yourself, "It doesn't make sense. It takes air pressure times displacement to evacuate the water from the buoy at the bottom of the column of water. Pressure x displacement = work, and there is no way that you can extract that much work + 25% bonus to cover the overhead to make the system self-sustaining. The taller the column of water, the more work you can extract from the rising/sinking buoys, but at the same time, the more air pressure you need to evacuate the buoys." By pure gut feel you can sense that it simply won't work.

I like these devices because they are always a bit of a Rube Goldberg affair which as an engineer I find attractive because of the neat dynamics involved. As well I have found appearances and the standard logic we apply in many cases can mislead us into thinking something is pretty straightforward but seldom is. Here is one example of a device I invented 20 years ago which I really have no interests in but proves an important point about perspective.

Houston we have a problem, at one time I was very interested in a sport called Hookah diving which is where fresh air is pumped under pressure to a diving mask through an attached air hose which leads to a pump at the surface. Now it should be common sense that if the diver is at a depth of 100 feet then the water pressure is 59 psi thus the minimum pressure needed for the supply of fresh air must be greater than 59 psi. It should be obvious that this system is not unlike an air pipe extending 100 feet under the water with a check valve on the end and we have 59 psi water pressure on one side of the valve thus we must need greater than 59 psi air pressure to produce an air flow through the check valve to our diver so we must perform an equivalent amount of work to compress atmospheric air.
That is common sense tells us that there is absolutely no way to get fresh air to the diver at 100 feet under water unless we perform an amount of work equivalent to compressing every cubic foot of fresh air to lets say 70 psi, but in this case common sense has failed us. You see the problem is not pressure that is an illusion the problem is the oxygen required in the fresh air by the diver. My solution to this impossible problem was simple, I use two small lines from a clear plexiglass diving helmet which extend to the surface, one line is fresh air at 70 psi the other exhaled air from the diver at 70 psi. I utilize the exhaled air at 70 psi and use this pressure to accelerate the mass of a piston to velocity X against atmospheric pressure and the piston mass now has momentum. I utilize this same momentum of the piston mass at velocity X and use it to compress fresh air in another air space cyclically, so you see the energy of the outgoing air pressure equals the energy of the incoming air pressure minus system losses through the conservation of momentum in isolated systems. In reality the real work required to provide fresh air to my diver is not the compressing of air from atmospheric pressure to 70 psi it is the system losses due to friction because the work performed by the air leaving the system is almost equal to the work performed on the air entering the system.

As well this process may sound familiar to you and it should, if we take two identical reservoir's of electrical energy in the form of capacitors and close the circuit the capacitors will simply balance their potentials. If however we add an inductance which has the property of electrical inertia and acts like a mass having momentum then we can discharge one capacitor to charge the inductance not unlike setting a mass in motion then discharge this inductive energy to charge the other capacitor. In this case energy is conserved minus losses just as my Hookah diving air delivery system conserves energy as pressure minus the system losses, so you see the notion that we must always perform work equivalent to compressing atmospheric air to 70 psi and then lose all this energy is false, the pressure energy is conserved while a fresh air flow is produced. In theory and practice I found that fresh air could be delivered to a diver under water with 80% less energy than is normally required by simply understanding the problem from the proper perspective.

I am not implying that this Hidro system does anything as I have no proof of what he is doing however I just wanted to point out the fact that our gut feelings and common sense can be very misleading if not completely wrong in many cases as I have shown in my example. I should note that this seemingly impossible task of delivering fresh air to a depth under water is little more than a low flow constant pressure system, one volume of air at pressure is efficiently exchanged with another, nothing more nothing less.
Regards
AC

---------------------------

Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman

Farrah Day	Re: Scooping Ash's Panacea "Hidro" report « Reply #11 on: 2011-04-03, 00:26:07 »
Hero Member Posts: 651	Just the very mention of both 'Ashtweth' and 'Sterling' in the first post is enough to lose my interest. To my mind anything brought to light, promoted or endorsed by two of the most feeble-minded and ridiculously gullible people on the circuit is always a bad sign.

BEP

Re: Scooping Ash's Panacea "Hidro" report « Reply #12 on: 2011-04-03, 02:36:33 »

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Quote from: Farrah Day on 2011-04-03, 00:26:07

Just the very mention of both 'Ashtweth' and 'Sterling' in the first post is enough to lose my interest.

To my mind anything brought to light, promoted or endorsed by two of the most feeble-minded and ridiculously gullible people on the circuit is always a bad sign.

I must agree with you

However, I can't declare an auto accident victim at fault just because of the ambulance chasers(lawyers) showing up

BEP

Re: Scooping Ash's Panacea "Hidro" report « Reply #13 on: 2011-04-03, 06:02:05 »

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Quote from: MileHigh on 2011-04-02, 21:58:24

I briefly looked this stuff up. Okay, so you have an air capacitor that discharges through a scroll expander. The amount of energy in the air capacitor is akin to an electrical capacitor, = (1/2 x the size of the air tank x the air pressure squared).
It's impossible for an air capacitor to supply more energy out than in. When the gas is expanding it draws in thermal energy from the environment. So if you 'overlook' that glaring fact perhaps you can get "more out than you put in."

The amount of energy stored in an air capacitor is akin to that stored in a capacitor. Charging and discharging are also virtually the same. What isn’t similar are the points where current flow sees a sudden change in restriction.

What part of an electric capacitor would become hot where current restriction increases?
I would say the first location noticed would be a part where resistance increases.

What part of an electric capacitor would become cold when current sees a sudden decrease in resistance?
None. A capacitor and an air storage unit are not completely the same. (Cold current explanation?

)

Quote

It harks back to one of my old themes where electrical inductors and capacitors are no different from mechanical springs and masses. An air tank is just an other form of a capacitor and it works with essentially the same differential equations and has the same characteristics. If a compressed air tank fails you get a massive flow of air and a sudden drop in pressure, which is identical to shorting a capacitor out or having a moving mass crash into a concrete wall.

I’ve heard an inductor has a voltage drop across it while carrying current. When you apply pressure to one end of a spring with the other end fixed is there a pressure drop across the spring?

Yes, they are essentially the same but not completely. These analogies only fit at the basic level.

Where I’m burnt is where the physics police hammer ideas with poorly thought out reasons for the impossibilities. Reasons for it not working are easy to announce…. and boring. What ideas could be used to make it work? These are what I need to know.

Never mind. I know the answer….. ‘None. It’ll never work because it is impossible’.

The first useless negativity I saw, for this topic(not on this forum), was the one about air being heavier than vacuum and adding weight to the buoy. Yea, so what? Who cares? Air doesn’t matter. Displacement of the water matters. An easy answer would be to add an air bladder to the buoy. I’m sure the inventor figured that out the first day of the idea.

The next (also not originally seen on this forum) is the thing about having to use high pressure to pump down to the depth of the buoy while at the bottom of the stroke. Why would that method be used? Why not just blow the air into the bottom of the tank so the bubbles fill the buoy from below? No massive pressure needed that way.
Another one the inventor surely figured on the first day. Nothing new about that idea. Remember the mysterious diving and surfacing submarine toy in the Cracker-Jacks box?

Quote

I shouldn't prejudge but based on past experience I won't be surprised if the Panacea report barely touches on the issue of were the supply of compressed air comes from to keep the device running.

Agreed. The chances of answers to pointed questions are slim. Very much so from Panacea.

This project has more believable information than most. When the detail questions are asked, I really hope the inventor provides some answers. Maybe after the patents run out ?

Peterae

Re: Scooping Ash's Panacea "Hidro" report « Reply #14 on: 2011-04-03, 07:07:11 »

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OK i may be missing something here, so fire away.

Imagine this , a tall container of water, in which a buoy has sunk to the bottom because it's heavy and filled with water, there are 2 pipes to this buoy, one goes to the top of the water container and open to fresh air, it has a valve, one pipe goes to the bottom of the column of water through the vessel to the open air under the tall water vessel, it also has a tap.

Now the buoy has water in it, so first thing we want to do is remove it, so surely this is as simple as opening the air valve at the top and opening the valve at the bottom, gravity will drain the water from the buoy and fill it with air from the top valve, close both valves and you have a buoy full of air in a tall water column, in my book this should float to the top.

What did i get wrong, i don't see how the water will not drain on it's own, so no need for compressed air, offcourse we all know this is not possible so what did i get wrong.

Peterae	Re: Scooping Ash's Panacea "Hidro" report « Reply #15 on: 2011-04-03, 07:11:20 »
Group: Administrator Hero Member Posts: 4698	OK i worked it out, the energy in the drained water has been lost to the vessel, so eventually the vessel would empty on the refill cycle

EMdevices

Re: Scooping Ash's Panacea "Hidro" report « Reply #16 on: 2011-04-03, 07:50:07 »

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I'm amazed by the expense that went into this device to bring it to its current state of development, which is nothing more then a perpetual motion idea from the 1800's.

There are dozens of patents and ideas like this from that era. Here's an interesting link. I like the capillary action between the two disks, I almost believed this worked at one time.

http://www.lhup.edu/~dsimanek/museum/unwork.htm

EM

MileHigh

Re: Scooping Ash's Panacea "Hidro" report « Reply #17 on: 2011-04-03, 12:15:56 »

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Quote from: Peterae on 2011-04-03, 07:11:20

OK i worked it out, the energy in the drained water has been lost to the vessel, so eventually the vessel would empty on the refill cycle

We can put that another way: The water that drains out through the pipe at the bottom of the column has to be lifted all the way back to the top. You have to expend Mgh energy to do this. You are back at square-one.

MileHigh

MileHigh

Re: Scooping Ash's Panacea "Hidro" report « Reply #18 on: 2011-04-03, 12:41:57 »

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WaveWatcher:

Quote

I’ve heard an inductor has a voltage drop across it while carrying current. When you apply pressure to one end of a spring with the other end fixed is there a pressure drop across the spring?

In the case of an ideal inductor there is only a voltage drop across it if there is AC current going through it. For DC current the voltage across it is zero. With respect to your question about a spring in this case the spring acts in an ideal fashion. The variable is velocity, and there is no "velocity drop" across the ends of a spring spring when the pressure applied to it is constant.

Quote

Where I’m burnt is where the physics police hammer ideas with poorly thought out reasons for the impossibilities. Reasons for it not working are easy to announce…. and boring. What ideas could be used to make it work? These are what I need to know.

I believe that I've read you state this many times before. It has tinges of hiding your head in the sand. You've got to let people say what they want to say. This is a fun one because Ashtweth is going to try to promote this like it is real. The profile has been raised and before people start throwing heaps of praise on "Panacea University" I think that it's a good exercise to look at the merits or lack thereof for this case. You saw that I extended an open invitation to anybody to argue the merits of this system and put up some links. Plus there is always the possibility that we are preventing somebody from giving a small fortune to James Kwok because they were led astray by Ashtweth. We could also be preventing someone from spending their own money on trying to build one themselves, which would also cost a small fortune in materials and time.

Quote

Why not just blow the air into the bottom of the tank so the bubbles fill the buoy from below? No massive pressure needed that way.

You are wrong, massive pressure is needed. No matter how you look at it you have to overcome the water pressure.

Quote

Remember the mysterious diving and surfacing submarine toy in the Cracker-Jacks box?

Baking soda -> chemical energy

MileHigh

BEP

Re: Scooping Ash's Panacea "Hidro" report « Reply #19 on: 2011-04-03, 14:15:12 »

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Quote from: MileHigh on 2011-04-03, 12:41:57

In the case of an ideal inductor there is only a voltage drop across it if there is AC current going through it. For DC current the voltage across it is zero. With respect to your question about a spring in this case the spring acts in an ideal fashion. The variable is velocity, and there is no "velocity drop" across the ends of a spring spring when the pressure applied to it is constant.

So, a voltage drop only exists when there is a changing electric field. If unchanging then the voltage drop is zero. I'll buy that. I agree 'velocity drop' doesn't work. The variable indicating a possible 'drop' would be acceleration, not velocity.

Quote

I believe that I've read you state this many times before. It has tinges of hiding your head in the sand. You've got to let people say what they want to say.

Back-at-ya

Quote

This is a fun one because Ashtweth is going to try to promote this like it is real. The profile has been raised and before people start throwing heaps of praise on "Panacea University" I think that it's a good exercise to look at the merits or lack thereof for this case. You saw that I extended an open invitation to anybody to argue the merits of this system and put up some links. Plus there is always the possibility that we are preventing somebody from giving a small fortune to James Kwok because they were led astray by Ashtweth. We could also be preventing someone from spending their own money on trying to build one themselves, which would also cost a small fortune in materials and time.

Granted

Quote

You are wrong, massive pressure is needed. No matter how you look at it you have to overcome the water pressure.

I am wrong. All it took was a cup of coffee and I regretted that statement. The pot calling the kettle black? I didn't put much thought into that one.

itzon

Re: Scooping Ash's Panacea "Hidro" report « Reply #20 on: 2011-04-03, 16:22:26 »

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Quote from: WaveWatcher on 2011-04-03, 06:02:05

Where I’m burnt is where the physics police hammer ideas with poorly thought out reasons for the impossibilities. Reasons for it not working are easy to announce…. and boring. What ideas could be used to make it work? These are what I need to know.

My feelings EXACTLY!

There are a lot of self righteous people, not on just this forum, but everywhere, that should take a step back and ask themselves the hard question, "Am I part of the problem, or can I be part of the solution?"

It's easy to say something can't or won't work for reasons that we have been taught, or don't comprehend because of what we haven't been taught. It's the lazy person's answer.

Your imagination and creativity is a terrible thing to waste.

If the world had more problem solvers like allcanadian, it would be a better place.

No need to ramble further...

Itzon

MileHigh

Re: Scooping Ash's Panacea "Hidro" report « Reply #21 on: 2011-04-03, 17:22:16 »

Hero Member

Posts: 1888

WaveWatcher:

Quote

So, a voltage drop only exists when there is a changing electric field. If unchanging then the voltage drop is zero. I'll buy that. I agree 'velocity drop' doesn't work. The variable indicating a possible 'drop' would be acceleration, not velocity.

There is flexibility to choose different analogous variables for your modeling and as long as you are consistent then everything will work. In my case I usually use velocity == voltage and force == current. For a mass == capacitor you have 1/2mv^2 == 1/2cv^2. For a spring == inductor you have 1/2kf^2 == 1/2Li^2, where k is the spring constant and f is the force going through the spring.

Acceleration is d(velocity)/dt. That would translate into dv/dt. For a capacitor it's easy. i = C dv/dt. So if you have a mass in your hand and you accelerate it you feel force between the mass and your hand. In other words i = C dv/dt == f = M dv/dt == f = Ma.

It's a bit more complicated for a spring. If you apply velocity to the end of a spring, the force going through the spring changes, it turns out to be the integral of the velocity with respect to time. (i.e. for an inductor i = 1/L integral v dt) That makes sense, you compress a spring between your arms at a certain velocity and you know the force going through the spring is increasing.

If you integrate the velocity with respect to time you get displacement. So that makes sense because you know that if you compress a spring with a certain displacement you are storing energy. Therefore the integral of the velocity across a spring with respect to time becomes a measurable amount of energy. So that's the analysis going "down" from velocity to displacement.

Velocity translates into changing force going through the spring. You do one step of integration with respect to time and you get displacement translating into energy stored in the spring.

To go from velocity to acceleration you have to go one step "up" and do a differentiation with respect to time.

Therefore accelerating the end of a spring gives you the rate of change of force with respect to time through the spring, df/dt.

A Newton is a kgm/s^2. I am really rusty on my calculus but I think a d(Newton)/dt is = -2kgm/s^3. (??)

I don't think d(Newtons)/dt is a particularly useful unit but there you go.

MileHigh

BEP

Re: Scooping Ash's Panacea "Hidro" report « Reply #22 on: 2011-04-03, 17:30:58 »

Hero Member

Posts: 2205

itzon,

Just remember that 'self-righteous' is usually only a perspective or view of someone. It can also be right or left. Trying to stay in the middle is difficult for anyone.

You can only go by your experiences. Where people foul-up is when they believe completely in what others say without self-experience or sufficient vetting. Many here have plenty of experience in the facts of these matters. We may appear as self-righteous sometimes. Most of the time it is due to previous experience.

I can see both sides. Quite often I'm still on one side or the other. The hard part is seeing clearly through the usual smog accompanying these claims.

MileHigh

Re: Scooping Ash's Panacea "Hidro" report « Reply #23 on: 2011-04-03, 17:32:20 »

Hero Member

Posts: 1888

Itzon:

Quote

There are a lot of self righteous people, not on just this forum, but everywhere, that should take a step back and ask themselves the hard question, "Am I part of the problem, or can I be part of the solution?"

I suppose you can say that there are two parallel approaches to arriving at the solution. One approach is to propose new ideas for what you think will work. Another approach is to propose things that won't work so you don't waste your time searching down blind alleys.

Now anybody that suggests a new idea has to be prepared to debate it. If your idea is debated and found out to not work then you can still consider that progress. In other words, that process is still part of the solution.

How much time, money, and energy do you think was wasted trying to replicate the Mylow scam? I'm convinced that James Kwok's Hidro project is a scam.

Can you propose why you think the "Hidro" project will work? I'm all ears!

MileHigh

Groundloop	Re: Scooping Ash's Panacea "Hidro" report « Reply #24 on: 2011-04-03, 17:50:54 »
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