Being validated as we speak, no human can "Not" do this at home with a small glass jar in a bucket of water. It is over. and this is the third machine all different physics, number 4 is one its way with another physics application.
NO ONE ON EARTH CANNOT DO THIS AT HOME
as per usual will be no comments read or written by the author except the validation note from the engineers not that this needs it
as usual, extraordinary claims requires extraordinary evidence...
Quote from: The Eskimo Quinn on November 26, 2014, 02:46:58 PM
Being validated as we speak, no human can "Not" do this at home with a small glass jar in a bucket of water. It is over. and this is the third machine all different physics, number 4 is one its way with another physics application.
NO ONE ON EARTH CANNOT DO THIS AT HOME
as per usual will be no comments read or written by the author except the validation note from the engineers not that this needs it
NB: Engineer Validated 28
th Nov 2014 – "Looks good"
Notes: this machine is more viable than a combustion engine that can run out of fuel or fail in spark – for Buoyant objects will always rise to the surface no matter the depth, falling Objects will always create energy – Nothing can alter these facts :
At less than 10 percent lift cost on the described height, there is a minimum 90 percent output for free – estimated fall time 60 second driving the generator @10,000KG (not including bonus encasement weight) 10,000 falling 1 Metre x 9.81 = 98100 joules x 100m = 9,810,000 joules divided by the 60 second fall time = 163,500 watts x 60 mins per hour =9,810,000 watts per hour, less lift cost of 10 percent, = 8,829,000 watts. However the cycle time for rise is not accommodated even with multiple blocks in play, so it may add up to 5 minutes to the cycle. This one such unit would be then calculated at producing =8,829,000 divided by 6 =1,471,500 watts per hour clean free energy per tower unit. Or 1471.5 kilowatts per hour= 1.4 megawatts or min 75 houses assuming massive 20kw usage per day.
upload it in another form. it is impossible for me to open this odt
ok i opened it through this site zamzar.com
Buoyancy is something i know very well,having owned my own boat building firm,and i was also one of the builders ;).
Because of this,I too have an OU buoyancy device that i designed many years ago -but where to get the money to fund such a device or project.
No broken laws of physics,and the complete math to boot to back up said OU device. The opperation is so simple--.increasing and decreasing mass density/surface area. This is where the !buoyant force! comes into play.If the object's weight is less than the buoyant force, the object will float. If the object's weight is greater than the buoyant force, the object will sink.The buoyant force is the force that pushes against the bottom and top surface areas of the object in the body of water,as the side ways forces cancel each other out.Because the bottom of the object has a greater depth than that of the top of the object(as the object will have hight/thickness),the object will feel slightly lighter under water,as the water pressure under the object will be slightly higher than it is on top of the object-and this depends on many factors in regards to the object. So all you need to do is increase the objects buoyant surface area without increasing it's weight,and the object will float. We then decrease the objects buoyant surface area,and the object will sink.
So here is the kicker. Once the object begins to sink,it will keep sinking with the same downward force until it hits the bottom-ocean floor. And when it begins to float,it will rise with the same force until it reaches the surface. So if it takes say 10000 joules of energy to get the object to sink say 10 feet,it takes the same amount of energy to get the object to sink 1 000 000 feet-->and this is done by decreasing the objects buoyant surface area,while maintaining the object weight.
They say that it takes x amount of energy to lift y amount of weight a set distance. Well using buoyancy as a tool,this is where that law dose not apply.This is one case it takes x amount of energy to lift y amount of weight to any hight,and return it back to it's starting point. 8)
Quote from: tinman on November 27, 2014, 09:04:26 AM
Buoyancy is something i know very well,having owned my own boat building firm,and i was also one of the builders ;).
Because of this,I too have an OU buoyancy device that i designed many years ago -but where to get the money to fund such a device or project.
No broken laws of physics,and the complete math to boot to back up said OU device. The opperation is so simple--.increasing and decreasing mass density/surface area. This is where the !buoyant force! comes into play.If the object's weight is less than the buoyant force, the object will float. If the object's weight is greater than the buoyant force, the object will sink.The buoyant force is the force that pushes against the bottom and top surface areas of the object in the body of water,as the side ways forces cancel each other out.Because the bottom of the object has a greater depth than that of the top of the object(as the object will have hight/thickness),the object will feel slightly lighter under water,as the water pressure under the object will be slightly higher than it is on top of the object-and this depends on many factors in regards to the object. So all you need to do is increase the objects buoyant surface area without increasing it's weight,and the object will float. We then decrease the objects buoyant surface area,and the object will sink.
So here is the kicker. Once the object begins to sink,it will keep sinking with the same downward force until it hits the bottom-ocean floor. And when it begins to float,it will rise with the same force until it reaches the surface. So if it takes say 10000 joules of energy to get the object to sink say 10 feet,it takes the same amount of energy to get the object to sink 1 000 000 feet-->and this is done by decreasing the objects buoyant surface area,while maintaining the object weight.
They say that it takes x amount of energy to lift y amount of weight a set distance. Well using buoyancy as a tool,this is where that law dose not apply.This is one case it takes x amount of energy to lift y amount of weight to any hight,and return it back to it's starting point. 8)
The net downward force on a submerged object is simply the difference between the dry weight of the object and the equivalent weight of fluid displaced by the object's volume. The energy involved sinking or raising the object is offset by the complementary movement of the displaced fluid. Altering the object's volume while submerged requires work to increase the volume of displaced fluid, or absorbs work by shrinking. Rheology aside, there is really no difference in the mechanics of a rock falling through the atmosphere to the land below or a rock sinking in the ocean. the net energy per unit height is simply reduced in the case of falling through the denser fluid.
Hi guys,
there was a French patent where the guy had it with the weight pushing out pistons on one side to
increase the byouncy. And on another side to pull inward to decrease it.
Regards
This is a PDF that has the original doc, the validation and notes, and full mechanical design instructions as well now on the end. all in one doc.
if you are building one as a model you can reverse the side drives, so the chain is simply a bicycle chain and the protruding post are on the box. clear perspex is ideal or glass and silicone for models.
Quote from: MarkE on November 27, 2014, 10:23:57 AM
The net downward force on a submerged object is simply the difference between the dry weight of the object and the equivalent weight of fluid displaced by the object's volume. The energy involved sinking or raising the object is offset by the complementary movement of the displaced fluid. Altering the object's volume while submerged requires work to increase the volume of displaced fluid, or absorbs work by shrinking. Rheology aside, there is really no difference in the mechanics of a rock falling through the atmosphere to the land below or a rock sinking in the ocean. the net energy per unit height is simply reduced in the case of falling through the denser fluid.
There is a very large difference between the mechanics of air and water. A heavier than air object will not float in the atmosphere,but a heavier than water object will float in water. You can hollow out a rock all you like,and it will not float in air,but it will in water.This means the mechanics are different between fluid displacement and atmospheric displacement. No work is required to alter the objects volume when submerged,as the work is done by way of pressure equilization between the internal and external pressures of the submerged object. Work is only required to reduce surface area of the object at the begining of the cycle,which is while the object is afloat.Once that work is done,the object will sink an infinite depth that is limited only to the deepest water body.If we take a balloon and fill it with a lighter than air gas,the balloon will loose lifting force as it gains altitude,while the same balloon would maintain the same lifting force in water regardless of depth(we are asumeing the balloon is strong enough to withstand crushing forces in this example)
Quote from: tinman on November 27, 2014, 04:50:43 PM
There is a very large difference between the mechanics of air and water. A heavier than air object will not float in the atmosphere,but a heavier than water object will float in water.
Sorry, you are wrong. An object that is heavier than the water it displaces will not float in water, and an object that is heavier than the air it displaces will not float in air. An object that is lighter than the air it displaces, like a big blimp, will float in air, and an object that is lighter than the water it it displaces, like a huge container ship, will float in water. Buoyancy is buoyancy, no matter the surrounding material.
QuoteYou can hollow out a rock all you like,and it will not float in air,but it will in water.This means the mechanics are different between fluid displacement and atmospheric displacement.
No, you are comparing apples and oranges. If your hollow rock displaces enough air mass, it will float, sure enough. Our air isn't dense enough for any real hollow rock to float, but it will float just fine in a much denser gas environment.
Quote
No work is required to alter the objects volume when submerged,as the work is done by way of pressure equilization between the internal and external pressures of the submerged object. Work is only required to reduce surface area of the object at the begining of the cycle,which is while the object is afloat.Once that work is done,the object will sink an infinite depth that is limited only to the deepest water body.If we take a balloon and fill it with a lighter than air gas,the balloon will loose lifting force as it gains altitude,while the same balloon would maintain the same lifting force in water regardless of depth(we are asumeing the balloon is strong enough to withstand crushing forces in this example)
Again, you are confounding your facts with false comparisons. Take a look at how high-altitude research balloons work. They start at the ground only "partially" inflated, just enough to be buoyant (they displace slightly more airmass than they weigh) and as they rise, the gas inside expands, so they displace more and more volume, of less and less dense air, so they continue to rise until they finally burst from being completely overfull. If you sink a balloon in water by deflating it at the surface, the only way to get it to rise up again is to pump gas into it, and you will have to pump harder the deeper the balloon is. A volume of water must be displaced that has a mass that is greater than the balloon's mass for it to rise up, and the deeper the balloon the more gas must be pumped into it, because the gas is compressible and the pressure inside the balloon must be slightly greater than the pressure of the water, for the balloon to expand and displace the necessary volume of water. When the balloon is expanded enough to displace more mass of water than the balloon weighs, the balloon will begin to rise _and expand even more_ as the gas inside expands to match the water pressure at whatever depth it is at.
When you sink a closed, constant volume thing like your hollow rock, the water level rises. Take a bucket and mark the water level on the side of the bucket, then sink your rock to just below the surface. You will note that the water level is now above your mark; an equal volume of water has been _lifted up_ as your rock has been sunk. This lifting of water takes work. Now let your rock sink more, say one-rock-diameter further. Now you have lifted up another volume of water, but you don't see the water level increase because you aren't changing the overall volume of the system any more, but you still have displaced one rock's worth of water from underneath the rock and moved it to above the rock. This takes work. If you have a rope attached to the rock, the work available by the rock pulling on the rope is _less than_ the work available from dropping the rock/rope in air, less by the amount of work it takes to raise up that water, continuously all the way down. When you pull the rock up, it's easier than in air because the water is flowing back underneath the rock as you lift it. This is buoyancy. There is no free lunch, even underwater.
Quote from: The Eskimo Quinn on November 27, 2014, 04:33:40 PM
This is a PDF that has the original doc, the validation and notes, and full mechanical design instructions as well now on the end. all in one doc.
if you are building one as a model you can reverse the side drives, so the chain is simply a bicycle chain and the protruding post are on the box. clear perspex is ideal or glass and silicone for models.
Feel free to show us your working model.
Quote from: The Eskimo Quinn on November 27, 2014, 04:33:40 PM
This is a PDF that has the original doc, the validation and notes, and full mechanical design instructions as well now on the end. all in one doc.
if you are building one as a model you can reverse the side drives, so the chain is simply a bicycle chain and the protruding post are on the box. clear perspex is ideal or glass and silicone for models.
This is just silly, your pdf document is a bunch of nonsense. Are you Archer Quinn, of "sword of god" fame?
"Full mechanical design instructions"... bull hockey. Not even a sketch. Where are the dimensioned blueprints, but more importantly... where is your working model? Why don't you "teach" us how to build it, like you did your "sword of god".
Quote from: The Eskimo Quinn on November 26, 2014, 02:46:58 PM
Being validated as we speak, no human can "Not" do this at home with a small glass jar in a bucket of water. It is over. and this is the third machine all different physics, number 4 is one its way with another physics application.
NO ONE ON EARTH CANNOT DO THIS AT HOME
as per usual will be no comments read or written by the author except the validation note from the engineers not that this needs it
Yep... my "gandmother" (sic) could make it, but for some reason YOU can't.
Quote from: TinselKoala on November 27, 2014, 11:27:54 PM
Sorry, you are wrong. An object that is heavier than the water it displaces will not float in water, and an object that is heavier than the air it displaces will not float in air. An object that is lighter than the air it displaces, like a big blimp, will float in air, and an object that is lighter than the water it it displaces, like a huge container ship, will float in water. Buoyancy is buoyancy, no matter the surrounding material. No, you are comparing apples and oranges. If your hollow rock displaces enough air mass, it will float, sure enough. Our air isn't dense enough for any real hollow rock to float, but it will float just fine in a much denser gas environment. Again, you are confounding your facts with false comparisons. Take a look at how high-altitude research balloons work. They start at the ground only "partially" inflated, just enough to be buoyant (they displace slightly more airmass than they weigh) and as they rise, the gas inside expands, so they displace more and more volume, of less and less dense air, so they continue to rise until they finally burst from being completely overfull. If you sink a balloon in water by deflating it at the surface, the only way to get it to rise up again is to pump gas into it, and you will have to pump harder the deeper the balloon is. A volume of water must be displaced that has a mass that is greater than the balloon's mass for it to rise up, and the deeper the balloon the more gas must be pumped into it, because the gas is compressible and the pressure inside the balloon must be slightly greater than the pressure of the water, for the balloon to expand and displace the necessary volume of water. When the balloon is expanded enough to displace more mass of water than the balloon weighs, the balloon will begin to rise _and expand even more_ as the gas inside expands to match the water pressure at whatever depth it is at.
When you sink a closed, constant volume thing like your hollow rock, the water level rises. Take a bucket and mark the water level on the side of the bucket, then sink your rock to just below the surface. You will note that the water level is now above your mark; an equal volume of water has been _lifted up_ as your rock has been sunk. This lifting of water takes work. Now let your rock sink more, say one-rock-diameter further. Now you have lifted up another volume of water, but you don't see the water level increase because you aren't changing the overall volume of the system any more, but you still have displaced one rock's worth of water from underneath the rock and moved it to above the rock. This takes work. If you have a rope attached to the rock, the work available by the rock pulling on the rope is _less than_ the work available from dropping the rock/rope in air, less by the amount of work it takes to raise up that water, continuously all the way down. When you pull the rock up, it's easier than in air because the water is flowing back underneath the rock as you lift it. This is buoyancy. There is no free lunch, even underwater.
I am not wrong TK,you misunderstood what i said-Quote: a heavier than water object will float in water. My statement is correct,and i said nothing about displacement. I was refering to net weight of the bouyant object-EG,steel is heavier than water per net volume,but can be made(shaped) to float in water-made to be bouyant.
Quote TK: No, you are comparing apples and oranges. If your hollow rock displaces enough air mass, it will float, sure enough. Our air isn't dense enough for any real hollow rock to float, but it will float just fine in a much denser gas environment.
No ,im not the one comparing apples to oranges. Im the one saying that the apples and oranges are differnt-Quote: the mechanics between water and air are very different. My statements are also based on the makeup of our planet,not Helion Prime,thus my statement is correct-hollow out a rock all you like,but it will not float in air(planet earths air/atmosphere)
Quote TK: Again, you are confounding your facts with false comparisons. Take a look at how high-altitude research balloons work. They start at the ground only "partially" inflated, just enough to be buoyant (they displace slightly more airmass than they weigh) and as they rise, the gas inside expands, so they displace more and more volume, of less and less dense air, so they continue to rise until they finally burst from being completely overfull. If you sink a balloon in water by deflating it at the surface, the only way to get it to rise up again is to pump gas into it, and you will have to pump harder the deeper the balloon is. A volume of water must be displaced that has a mass that is greater than the balloon's mass for it to rise up, and the deeper the balloon the more gas must be pumped into it, because the gas is compressible and the pressure inside the balloon must be slightly greater than the pressure of the water, for the balloon to expand and displace the necessary volume of water. When the balloon is expanded enough to displace more mass of water than the balloon weighs, the balloon will begin to rise _and expand even more_ as the gas inside expands to match the water pressure at whatever depth it is at
Although i quoted:-->we are asumeing the balloon is strong enough to withstand crushing forces in this example.lets look at this a different way. Let's use a steel sphere insted of our crushless balloon. I will take my steel sphere(lets say with a volume of 1 cubic meter),and pump air into it to a pressure of 100psi-->my sphere will float in water. Now do the same with your sphere of the same size,and see if it floats in air.
Second test. I will now pull an absolute vacume in my sphere,and my sphere will still float just as well as it did with a pressure of 100psi-->now do the same with your sphere,dose it float in air yet?.
Quote TK: When you sink a closed, constant volume thing like your hollow rock, the water level rises. Take a bucket and mark the water level on the side of the bucket, then sink your rock to just below the surface. You will note that the water level is now above your mark; an equal volume of water has been _lifted up_ as your rock has been sunk. This lifting of water takes work. Now let your rock sink more, say one-rock-diameter further. Now you have lifted up another volume of water, but you don't see the water level increase because you aren't changing the overall volume of the system any more, but you still have displaced one rock's worth of water from underneath the rock and moved it to above the rock.
This we know.
Quote TK: If you have a rope attached to the rock, the work available by the rock pulling on the rope is _less than_ the work available from dropping the rock/rope in air, less by the amount of work it takes to raise up that water, continuously all the way down. When you pull the rock up, it's easier than in air because the water is flowing back underneath the rock as you lift it. This is buoyancy. There is no free lunch, even underwater.
This is not correct as far as my device is designed to work,and another reason that the mechanics/dynamics between water and air are very different. A bouyant vessle correctly engineered can be made to sink and then rise again in water without the addition of any air(or gas of any type),as bouyancy in water is not reliant on a gas volume of any kind. As i stated above,the internal area of a sealed bouyant vessle can be under an absolute vacume,and it will have no effect on the bouyancy of the vessle-->not the case with air im afraid. Now you take what ever vessle you want,and make it float in air without the use of a gas,but by simply increasing the size of that vessle. ;)
You must also take into account the weight of the vessle it self when calculating bouyancy,not just the volume of water it displaces.
This is a French patent I was referring too:
http://vitanar.narod.ru/files/02830575A1.pdf
I would like to hear what you, guys, think about its viability.
Regards
Quote from: tinman on November 28, 2014, 09:20:18 AM
I am not wrong TK,you misunderstood what i said-Quote: a heavier than water object will float in water. My statement is correct,and i said nothing about displacement. I was refering to net weight of the bouyant object-EG,steel is heavier than water per net volume,but can be made(shaped) to float in water-made to be bouyant.
Yes, you are indeed wrong. The only reason a heavier-than-water object floats is because the water it displaces weighs more than the dry object itself. Look at how ships are labeled: Their "weight" are given in _displacement_, that is, how many tonnes, etc. of water they will displace when placed into the water. And the same is true for air: if your object displaces more air mass than it "weighs" it will float in the air. Do you think a packed-up blimp will float in air? Of course it won't. But when it is expanded to full size by filling it with some gas that is lighter than air, so that it _displaces_ a volume of air that weighs more than its "packed" weight, it will float in air. There is no difference in the dynamics, except that water is virtually incompressible and gases aren't. You can fill a balloon with kerosene and it will float in water, you can make a "balloon" out of carbon aerogel (as close as you can get to hollowing out a rock), fill its gaps with helium and it will float in air, because it displaces as much (neutrally buoyant) or more (positively buoyant) weight of air than it weighs itself.
Quote
Quote TK: No, you are comparing apples and oranges. If your hollow rock displaces enough air mass, it will float, sure enough. Our air isn't dense enough for any real hollow rock to float, but it will float just fine in a much denser gas environment.
No ,im not the one comparing apples to oranges. Im the one saying that the apples and oranges are differnt-Quote: the mechanics between water and air are very different. My statements are also based on the makeup of our planet,not Helion Prime,thus my statement is correct-hollow out a rock all you like,but it will not float in air(planet earths air/atmosphere)
Quote TK: Again, you are confounding your facts with false comparisons. Take a look at how high-altitude research balloons work. They start at the ground only "partially" inflated, just enough to be buoyant (they displace slightly more airmass than they weigh) and as they rise, the gas inside expands, so they displace more and more volume, of less and less dense air, so they continue to rise until they finally burst from being completely overfull. If you sink a balloon in water by deflating it at the surface, the only way to get it to rise up again is to pump gas into it, and you will have to pump harder the deeper the balloon is. A volume of water must be displaced that has a mass that is greater than the balloon's mass for it to rise up, and the deeper the balloon the more gas must be pumped into it, because the gas is compressible and the pressure inside the balloon must be slightly greater than the pressure of the water, for the balloon to expand and displace the necessary volume of water. When the balloon is expanded enough to displace more mass of water than the balloon weighs, the balloon will begin to rise _and expand even more_ as the gas inside expands to match the water pressure at whatever depth it is at
Although i quoted:-->we are asumeing the balloon is strong enough to withstand crushing forces in this example.lets look at this a different way. Let's use a steel sphere insted of our crushless balloon. I will take my steel sphere(lets say with a volume of 1 cubic meter),and pump air into it to a pressure of 100psi-->my sphere will float in water. Now do the same with your sphere of the same size,and see if it floats in air.
Second test. I will now pull an absolute vacume in my sphere,and my sphere will still float just as well as it did with a pressure of 100psi-->now do the same with your sphere,dose it float in air yet?.
Nice strawman argument. Anything will float in any fluid if it displaces as much or more _weight_ of the fluid than it itself weighs. It doesn't matter what the pressure is inside of a _rigid_ object, only the volume of the object matters. Don't you know the story of Archimedes, the gold crown and "Eureka"? You put any object in water, the level of the water rises because the object displaces the same volume of water as its own volume, no matter the shape or composition of the object. If the object is heavier (dry weight) than the water it displaces it will sink, if it is lighter than the water it displaces it will float... and guess what, a floating object on the surface displaces exactly as much water by weight as it weighs itself. You can dunk your scale into the water and prove this to yourself easily enough. Measure the volume of water that a thing displaces, this will of course give you the weight of that displaced water since the density of water is 1 gram per cubic centimeter. The weight of your object fully submerged and sitting on the bottom, on the scale, is exactly the (object's dry weight outside the water) - (water displaced weight). Do the experiment yourself and see. This is buoyancy.
Quote
Quote TK: When you sink a closed, constant volume thing like your hollow rock, the water level rises. Take a bucket and mark the water level on the side of the bucket, then sink your rock to just below the surface. You will note that the water level is now above your mark; an equal volume of water has been _lifted up_ as your rock has been sunk. This lifting of water takes work. Now let your rock sink more, say one-rock-diameter further. Now you have lifted up another volume of water, but you don't see the water level increase because you aren't changing the overall volume of the system any more, but you still have displaced one rock's worth of water from underneath the rock and moved it to above the rock.
This we know.
Quote TK: If you have a rope attached to the rock, the work available by the rock pulling on the rope is _less than_ the work available from dropping the rock/rope in air, less by the amount of work it takes to raise up that water, continuously all the way down. When you pull the rock up, it's easier than in air because the water is flowing back underneath the rock as you lift it. This is buoyancy. There is no free lunch, even underwater.
This is not correct as far as my device is designed to work,and another reason that the mechanics/dynamics between water and air are very different. A bouyant vessle correctly engineered can be made to sink and then rise again in water without the addition of any air(or gas of any type),as bouyancy in water is not reliant on a gas volume of any kind. As i stated above,the internal area of a sealed bouyant vessle can be under an absolute vacume,and it will have no effect on the bouyancy of the vessle-->not the case with air im afraid. Now you take what ever vessle you want,and make it float in air without the use of a gas,but by simply increasing the size of that vessle. ;)
You must also take into account the weight of the vessle it self when calculating bouyancy,not just the volume of water it displaces.
Your last statement is of course correct. But you cannot make an object that is heavier than the water it displaces, to float, unless you can increase its volume somehow without also increasing its dry weight. A balloon's volume is increased by putting some gas into it. We just don't have the magic materials (except for aerogels) that are strong enough to float in air unless they contain some lighter than air gas that can balance the internal and external pressures. If we had such a material that could be strong enough to hold outside air pressure against a vacuum inside, it certainly would float in air as long as it displaces more airmass than it weighs itself.
How is your rigid sealed object to increase its volume, against the pressure of the water surrounding it, so that it will float? The only way is to do work against the water pressure somehow or another. Expanding nested cylinders, for example? You have to push the cylinder outward against the water pressure trying to push it in. And watch the water level in your container rise as you are doing this: you are lifting water, which requires work.
ETA: You might enjoy watching my various buoyancy and "wayne travis" demonstrations on YouTube. And I'd love to see anything you can come up with in support of your arguments.
Here are a couple of mine selected more or less at random (it's been a long time since I've had this discussion):
http://www.youtube.com/watch?v=1iijUjtkV-E (http://www.youtube.com/watch?v=1iijUjtkV-E)
http://www.youtube.com/watch?v=rPEOPWG_gh8 (http://www.youtube.com/watch?v=rPEOPWG_gh8)
Quote from: tinman on November 27, 2014, 04:50:43 PM
There is a very large difference between the mechanics of air and water. A heavier than air object will not float in the atmosphere,but a heavier than water object will float in water. You can hollow out a rock all you like,and it will not float in air,but it will in water.This means the mechanics are different between fluid displacement and atmospheric displacement. No work is required to alter the objects volume when submerged,as the work is done by way of pressure equilization between the internal and external pressures of the submerged object. Work is only required to reduce surface area of the object at the begining of the cycle,which is while the object is afloat.Once that work is done,the object will sink an infinite depth that is limited only to the deepest water body.If we take a balloon and fill it with a lighter than air gas,the balloon will loose lifting force as it gains altitude,while the same balloon would maintain the same lifting force in water regardless of depth(we are asumeing the balloon is strong enough to withstand crushing forces in this example)
Tinman there is no difference. Something that is denser than air falls through air. Something that is denser than water falls through water. The reason that hollowing a rock out doesn't work is that the net density is still greater than air because the hollowed out volume is filled with air and the rest of the rock is still denser than air. The analogy with water would be to fill the rock with water. We call those things water logged ships. And they do sink.
It absolutely takes work in one direction or another to change the volume of a submerged object. The pressure on the hull is the surrounding water pressure. Increasing the volume does the incremental work in each direction: dWx = p * dx, dWy = p * dy, dWz = p * dz.
The reason that the balloon loses lifting force is that air is compressible. As you go up the air density changes. The basic mechanics in air and water remain the same.
Quote from: TinselKoala on November 28, 2014, 01:38:07 PM
Yes, you are indeed wrong. The only reason a heavier-than-water object floats is because the water it displaces weighs more than the dry object itself
Look at how ships are labeled: Their "weight" are given in _displacement_, that is, how many tonnes, etc. of water they will displace when placed into the water. And the same is true for air: if your object displaces more air mass than it "weighs" it will float in the air. Do you think a packed-up blimp will float in air? Of course it won't. But when it is expanded to full size by filling it with some gas that is lighter than air, so that it _displaces_ a volume of air that weighs more than its "packed" weight, it will float in air. There is no difference in the dynamics, except that water is virtually incompressible and gases aren't. You can fill a balloon with kerosene and it will float in water, you can make a "balloon" out of carbon aerogel (as close as you can get to hollowing out a rock), fill its gaps with helium and it will float in air, because it displaces as much (neutrally buoyant) or more (positively buoyant) weight of air than it weighs itself. Nice strawman argument. Anything will float in any fluid if it displaces as much or more _weight_ of the fluid than it itself weighs. It doesn't matter what the pressure is inside of a _rigid_ object, only the volume of the object matters. Don't you know the story of Archimedes, the gold crown and "Eureka"? You put any object in water, the level of the water rises because the object displaces the same volume of water as its own volume, no matter the shape or composition of the object. If the object is heavier (dry weight) than the water it displaces it will sink, if it is lighter than the water it displaces it will float... and guess what, a floating object on the surface displaces exactly as much water by weight as it weighs itself. You can dunk your scale into the water and prove this to yourself easily enough. Measure the volume of water that a thing displaces, this will of course give you the weight of that displaced water since the density of water is 1 gram per cubic centimeter. The weight of your object fully submerged and sitting on the bottom, on the scale, is exactly the (object's dry weight outside the water) - (water displaced weight). Do the experiment yourself and see. This is buoyancy.Your last statement is of course correct. But you cannot make an object that is heavier than the water it displaces, to float, unless you can increase its volume somehow without also increasing its dry weight. A balloon's volume is increased by putting some gas into it. We just don't have the magic materials (except for aerogels) that are strong enough to float in air unless they contain some lighter than air gas that can balance the internal and external pressures. If we had such a material that could be strong enough to hold outside air pressure against a vacuum inside, it certainly would float in air as long as it displaces more airmass than it weighs itself.
How is your rigid sealed object to increase its volume, against the pressure of the water surrounding it, so that it will float? The only way is to do work against the water pressure somehow or another. Expanding nested cylinders, for example? You have to push the cylinder outward against the water pressure trying to push it in. And watch the water level in your container rise as you are doing this: you are lifting water, which requires work.
ETA: You might enjoy watching my various buoyancy and "wayne travis" demonstrations on YouTube. And I'd love to see anything you can come up with in support of your arguments.
Here are a couple of mine selected more or less at random (it's been a long time since I've had this discussion):
http://www.youtube.com/watch?v=1iijUjtkV-E (http://www.youtube.com/watch?v=1iijUjtkV-E)
http://www.youtube.com/watch?v=rPEOPWG_gh8 (http://www.youtube.com/watch?v=rPEOPWG_gh8)
OK-how do i split quote's so as i can sepperate the quotes in my reply. This copy paste crap is driving me. :-\
Quote from: tinman on November 28, 2014, 06:10:09 PM
OK-how do i split quote's so as i can sepperate the quotes in my reply. This copy paste crap is driving me. :-\
Hi Brad,
You can quote text as follows:
start the text to be quoted with this symbol and word:
[quote and now use this symbol
] right after the word quote like this quote] and then include the text to be quoted.
And at the end of the text include the following symbols and word:
[/quote and again use this closing symbol
] right after the word quote without a space. Of course, no need for using bold characters.
For the next text to be quoted, apply the same symbols and words, i.e. embed the text to be quoted between the two symbols and words.
Hope this may be useful.
Gyula
yes
and I hope you learn so you can teach Me
:o
@ TK,MarkE
You are telling me thing's i already know.
What you are not doing is showing me that what can be done with water can also be done with air in regards to bouyancy.
Quote: We just don't have the magic materials (except for aerogels) that are strong enough to float in air unless they contain some lighter than air gas that can balance the internal and external pressures. If we had such a material that could be strong enough to hold outside air pressure against a vacuum inside, it certainly would float in air as long as it displaces more airmass than it weighs itself.
So as we are dealing with devices we can build, and do have the material's for,then the pratical mechanics between air and water are different.They only become the same once the materials become avaliable to put together a device that act's like a bouyant vessle dose in water. Your statements remain theoretical until such time as you can prove otherwise-->is this not what you demand of anyone that claims things are different to what you think they are. My claims are in the here and now. The facts i state remain until such time as you can show me a device that contains a vacume,but will rise in air when the device is expanded or increased in outer surface area. I can do this right now in water with the avaliable materials we have here on earth-can you do the same right now in air? If's and but's are something we cannot work with. Lets deal with builds we can actually build,and have the material for. I think a strawman argument would be more of one of a device made from materials that dont yet exist ;)
Quote TK: you are lifting water, which requires work.
This i know,what you dont understand is that the work has already been done before the vessle hits the bottom.
Quote MarkE-It absolutely takes work in one direction or another to change the volume of a submerged object
Once again,this i know-please see reply 8- Quote: Work is only required to reduce surface area of the object at the begining of the cycle.
Guy's
I can build my device right now with materials at hand-can you do the same-->can you build a device with materials we have right now,so as it performs the same in air as mine will in water?.
Also,please read what i have said,as you are making statements that seem to imply that i disagree with things that i have already agreed with,or have posted before you actually made your statement.
@TK
I think you may remember this video. Now,why wouldnt this work?,and how would it be different if we used the same type of theory in water?.
https://www.youtube.com/watch?v=4nKltbQ8PBQ
Quote from: gyulasun on November 28, 2014, 06:53:01 PM
Hi Brad,
QuoteYou can quote text as follows:
start the text to be quoted with this symbol and word: [quote and now use this symbol ] right after the word quote like this quote] and then include the text to be quoted.
And at the end of the text include the following symbols and word:
[/quote and again use this closing symbol ] right after the word quote without a space. Of course, no need for using bold characters.
For the next text to be quoted, apply the same symbols and words, i.e. embed the text to be quoted between the two symbols and words.
Hope this may be useful.
Gyula
Test
Quote from: gyulasun on November 28, 2014, 06:53:01 PM
Hi Brad,
QuoteYou can quote text as follows:
second test
Quotestart the text to be quoted with this symbol and word:
third test
Gyula
Still screwing this up some how ???
Now my replies are in blue too lol.
Hi Folks,
I have made this explanation as attached, I cannot explain it simpler or better... 8)
Gyula
Quote from: gyulasun on November 29, 2014, 04:46:45 PM
Hi Folks,
I have made this explanation as attached, I cannot explain it simpler or better... 8)
Gyula
Thanks Gyula
Even a dumbass like me should be able to get this right now :D
Gangmother or Grandmother ?
Funny, my grandmother invented a wave generator (in the sea) in the 70:tis
pity she did not patented it....
Gyula
that makes it very clear.
@Archer
I sent you a message ,please let me know if your interested.
Thanks
Chet
Doesn't anyone else see the rather obvious (to me) and glaring flaw in the "boot" device described in this document? I've been waiting for someone to post their analysis that shows the flaw, but I'm getting tired of waiting. I sent my analysis to a couple of people a little while ago but perhaps they are sleeping.
Archer, if that's you, you are all wet, and that's no joke.
You start with a thing that looks like a "boot", with a "foot" chamber sticking out to the side at the bottom, and a tall skinny "leg" portion that rises up much higher. There is a water- and pressure-tight door separating the foot from the leg, and there is some kind of water- and pressure-tight lid you can put on the top of the foot to seal it tight. This lid is open to start, and the door between foot and leg is closed, and both foot and leg are filled with water.
Now you have a box that is heavy but is of such a volume that it is just barely negatively buoyant, that is, it just barely sinks in water. So when it is in water, it doesn't take much force to lift it up, as long as it is completely submerged.
Now you lift this box up from the ground outside the foot using a heavy crane, lift it over the wall of the foot, and place it into the foot. It sinks in the water that is in the foot. But the water level in the foot rises! You have added a volume of box to the fixed volume of water in the foot, and now the level of the water in the foot is _higher_ than it was before you added the box. Right? Or perhaps it just runs out onto the ground over the edge of the foot wall, but that's pretty wasteful, don't you think? Also messy.
OK, so now you put the watertight lid on the foot, even with the top of the water so there is no air in there. Now you can open the connecting door between the foot and the leg. Slide the box over into the leg and close the door again. Remember that the level of water in the foot is still higher than it was before you placed the box in there, or you have lost a volume of water over the edge, either one. Right?
Now you can lift up the slightly negatively buoyant box all the way up to the top surface of the water in the tall leg of the boot. This doesn't take much work, you can even do it with a weak 12-volt winch. But then... you have to lift the box up out of the water to place it on the receiving platform, and this takes a bit more work... since the box is actually quite heavy out of water. AND--- the water level in the leg of the boot DROPS, because you have removed the volume of the box from the leg. Let's recap: you have lifted a box from the ground to the top platform, and the water level in the FOOT is HIGHER than it was before you started, and the water level in the LEG is LOWER than it was before you started. You have not only lifted a box, but you have LOWERED an equal volume of water from the top of the leg, down into the foot of the boot.
Go ahead and do it again with another box. Now you've transferred another volume of water from the leg to the foot. The water level in the leg is now lower by another increment of volume equal to the volume of the box and the water level in the foot is now higher by the same increment. Repeat ... until the water level in foot and leg are equal and you are lifting your box way up off the ground to get it into the foot, and way up out of the water in the leg with a crane to get it to the upper platform.
The only way to get around this is to have the water in the leg constantly replaced, as you move your box from the foot to the leg, which transfers an equal volume of water _out_ of the leg and _into_ the foot. Got a convenient river at the top of your boot leg to replace this water? At the top of the pyramid? No? Then your device will eventually run out of the _stored energy_ represented by the head of water in the leg of the boot. It is this stored energy, released as power of water falling, that is helping you to raise up that seemingly light box to the top of the _water level_ , which is falling one box volume at a time, in the leg of the boot.
Looks like Newton isn't "crushed like a bug" after all. No, it is Archer Quinn who is crushed like a bug, on the windscreen of Conservation of Energy and the conservative field of gravitational force. There is no free lunch to be had from buoyancy, which is just gravity acting in a way that is difficult for some people to grasp.
Hi TK,
Nice explanation of the worst case interpretation. Can you do another one where the "heavy box" is actually empty and the weight is the in system water? Empty box lowers into the boot, fills with water except for enough air to remain buoyant, enters the leg, box rises up/pulled up, water drained/emptied to fall through a turbine back into the boot, rinse and repeat?
I have no prediction on whether the losses will be offset ....
tak
Quote from: tak22 on December 01, 2014, 01:05:36 PM
Hi TK,
Nice explanation of the worst case interpretation. Can you do another one where the "heavy box" is actually empty and the weight is the in system water? Empty box lowers into the boot, fills with water except for enough air to remain buoyant, enters the leg, box rises up/pulled up, water drained/emptied to fall through a turbine back into the boot, rinse and repeat?
I have no prediction on whether the losses will be offset ....
tak
If you have enough air in the box for it to "remain buoyant" it will still displace a volume of water when you put it in the foot to begin with. So you are in the same trouble as before. By the way, I am not describing a "worst case" interpretation, I am describing exactly what Quinn put forth in the document, except I am including what he (deliberately?) left out: the displaced volume that winds up transferring water (stored energy of position) from the top of the leg down into the foot, doing work as it goes and requiring work to replace.
So why not just use a bucket? Lower the bucket down the leg to the bottom then hoist it up. If the bucket is heavier than water it will sink by itself. Then you can bring up a bucket full of water from the bottom of the leg, to the top level, with only the cost of lifting the weight of the bucket itself (minus the weight of the water the bucket shell itself displaces, which presumably can be made very small with magic bucket-materials), since water is exactly neutrally buoyant in water... duh. Now you can pour your water from the bucket through your turbine... and then where does it go? The water level in the leg has now dropped by one bucket-volume. If you pour the water through the turbine into the top of the leg... you have to have your turbine above the surface of the water and the only work you can get is the work from the height above the _surface_ and back to the surface, through the turbine. In other words the whole affair of lifting the bucket up through the column of water is wasted and you are just pouring water from the bottom of the turbine, lifting it up to the top of the turbine, and your turbine is only returning part of the work you put in to lift that water thru the air.
If you pour the water thru the turbine at the bottom of the leg, hence using the entire head of water pressure to drive the turbine... how do you get the water back into the leg? You can pump it back up to the top, with the usual losses, or you can pump it into the bottom; either way you still have to pump it against the full head of the water pressure. Or you can let the water run out of the bottom and forget about it... and then you are running on the stored energy of the water column which is getting lower all the time. You might as well just dispense with the bucket or box and just punch a hole in the bottom of the leg and put your turbine in the flow from the hole.... and when the water runs out without being replaced from the _top_, you are once again out of the stored energy represented by the water head in the leg of the boot.
No, the losses will not be offset, and this is not a "prediction" it is a verifiable fact. Unless you have an outside source of energy replacing the water you will quickly run out of the stored energy that you put into the water column to fill up the leg of the boot in the first place.
"worst case" was just a poor turn of phrase, I just appreciated you doing a full walk through. Also, I used too few words so you didn't quite get all of what I meant last post so I'll add a few more thoughts:
- the turbine is positioned just above the boot so it gets maximum head pressure.
- the box is exactly the size of the boot minus enough space to allow enter/exit
- the box lowers into the boot with a bottom opening so it fills with water already in the boot
- when the box reaches the top it is drained directly into the turbine or a head pond, then moved empty to be lowered back to the boot
still no prediction from me, just tossing out refinements :)
tak
Quote from: tak22 on December 01, 2014, 02:30:47 PM
"worst case" was just a poor turn of phrase, I just appreciated you doing a full walk through. Also, I used too few words so you didn't quite get all of what I meant last post so I'll add a few more thoughts:
- the turbine is positioned just above the boot so it gets maximum head pressure.
- the box is exactly the size of the boot minus enough space to allow enter/exit
- the box lowers into the boot with a bottom opening so it fills with water already in the boot
- when the box reaches the top it is drained directly into the turbine or a head pond, then moved empty to be lowered back to the boot
still no prediction from me, just tossing out refinements :)
tak
This is equivalent to the bucket I described above, then, and the box itself is just the same as the bucket.
If the turbine is at the top, then your available pressure head is just small. If you drain the water out of the system thru the turbine into a pond at the top, you have only the small head and a short fall of water, and the level in the leg of the boot drops. If the turbine is at the bottom so you get the full head pressure, then you still have to move the water back to the top, and you can't use the counterweight of the now-empty, light, box to do it because it's not nearly heavy enough to offset the weight of the water you have to move. Again, you will either run out of the stored energy of the water column in the leg, or you have to use an external "river" to replace the water in the leg that you are pouring out with every boxful you lift up. Or you have to lift the boxful of water _out of the water_ to pour it thru a turbine at the top so that it falls back into the top of the bootleg and thus you only have a small head, equivalent to the lift _out of the water_ that you did to get the box full of water up to where you can empty it into the turbine.
Still not a prediction from me either, just a simple statement of fact: There is no way this device will work by anything except the stored energy of the water column, which will either run out one bucket (box) volume at a time, or will need to be replaced by an outside source of energy greater than you are recovering from your turbine and lowering mechanism.
Thanks TK for staying on this thread. Here's the first two steps in a process and I think it's the transition that's the 'sticky spot'. Assuming the use of magically good low loss doors and valves, the energy to move the box from the boot into the leg (which raises the water level), is greater than the possible energy through the turbine?
Well, now you have changed the problem again.
A couple of things should be clear from your diagram. First, you really do need "magic doors" for this version, which are not needed for Quinn's original device or my analysis of it. You cannot push your box into the tall column through any real door without losing a _lot_ of water.
Second, it should be clear, given the magic doors, that when you do push your box into the tall column you are raising the entire column of water by one box-height. In other words you are pushing the box in against the full pressure head of the column of water, raising it up in the process. Of course you cannot recover this work from just draining a box of water from the top (again pushing through a magic door at the top) down thru the turbine. Or rather, assuming fully magic doors and a 100 percent efficient turbine and no drag in the plumbing, or other losses, etc. you will get back exactly as much work as you put in in the first place.
Magic doors and other impossibilities like 100 percent efficient turbines are not allowed in the real world. After all, _every_ non-working perpetual motion machine will work if you are allowed to use magic. For example I have a handful of permanent magnet motor/generators that will run forever, generating excess power, if only one could find some bearings with a negative coefficient of friction. The "magic door" problem, and buoyancy drives in general, are discussed quite well at Simanek's Museum.
https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm
https://www.lhup.edu/~dsimanek/museum/unwork.htm#buoy4 (scroll up and down for much more)
Well it's a good thing I didn't have anything invested in this other than a willingness to toss out ideas and discuss them. ;)
tak
What you think about this
http://www.youtube.com/watch?v=KhUCesdIyOY
yg_ 34
from the comments on your Vid
"" For more visit:[/font][/size]http://isparktube.com/ (http://isparktube.com/)[/color][/font]
Finally An energy system that all can understand. for the common folk, what is happening here, is air is being used to float containers that are being held under water. This produce a rotation, which turns a generator that produce energy. It's is essentially a gravitational system. it also uses the buoyancy principle. The idea is to understand that the power output produces far exceeds the power input. Therefore once started the system runs on itself. This does not go against the second law of thermodynamic, which uses a localize close system. This also uses a close system, but a much bigger one, the gravitation of the earth. You can look at it this way, this system uses the gravity of the earth as fuel[/font][/size] ""
If you have a contact or Phone number for this Fellow could you forward to
ChetKremens@Gmail.com
?
not sure this video was posted by the inventor ??
Thx
Chet
No Chet, I don't have any info.
I couldn't find either any info about this video on their website . Did you ?
No
I think He's just using the vid without the owners permission?
to much unrelated stuff at the web site he links to.
thx
Chet
http://www.rosch.ag/index.php/en/rosch-innovations
I very much doubt they can produce an excess of the energy taking in account a very low efficiency of the air compressor.
But this idea is very similar, but works completely from the gravity - no air compressor is involved.
http://vitanar.narod.ru/files/02830575A1.pdf
Quote from: telecom on December 02, 2014, 12:15:01 PM
But this idea is very similar, but works completely from the gravity - no air compressor is involved.
http://vitanar.narod.ru/files/02830575A1.pdf (http://vitanar.narod.ru/files/02830575A1.pdf)
You mean the "inventor" THINKS it should work, but of course it does not.
Come on people, these buoyancy drives are nothing new, they have been completely and rigorously analyzed for years, and yet every once in a while somebody thinks he's invented a new one, when he really hasn't. No working model of any such drive has ever been produced.
Here it is again. You will note that the devices shown and analyzed in this link are functionally _identical_ to what is described in that patent application link.
https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm (https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm)
Or rather... non-functionally identical, since they do not function, and of course neither does the one in the patent app. link.
Seriously... why are you wasting your time with this stuff? The arrogant Quinn's device, as I have hopefully proven to you, will not function except on the stored energy of the water column which is soon depleted. And it is just as easy to refute the rest of the "buoyancy drive" and "gravity drive" devices once you get down to the actual mathematics involved.
If you want to do some experiments that show otherwise, please go ahead, but be sure to report your _negative results_ honestly as well as any positive results you get. The reason you don't hear of all the failures is because people don't want to report them... hence the perpetuation of the belief that someday one of the same-old-same-old ideas will suddenly start working. But they won't.
Quote from: TinselKoala on December 02, 2014, 02:31:59 PM
You mean the "inventor" THINKS it should work, but of course it does not.
Come on people, these buoyancy drives are nothing new, they have been completely and rigorously analyzed for years, and yet every once in a while somebody thinks he's invented a new one, when he really hasn't. No working model of any such drive has ever been produced.
Here it is again. You will note that the devices shown and analyzed in this link are functionally _identical_ to what is described in that patent application link.
https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm (https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm)
Or rather... non-functionally identical, since they do not function, and of course neither does the one in the patent app. link.
If you want to do some experiments that show otherwise, please go ahead, but be sure to report your _negative results_ honestly as well as any positive results you get. The reason you don't hear of all the failures is because people don't want to report them... hence the perpetuation of the belief that someday one of the same-old-same-old ideas will suddenly start working. But they won't.
Hi,
I really enjoy reading your informative responses, but this case is different from what have been described.
He is using the weights to create or limit the buoyuncy, quite different from the described examples,
which are short of this application of the weights, or at least, not pointed out in the design.
They write:
Now reconsider the full version with piston chambers on a belt over two pulleys. Each pair of pistons gains energy moving on the straight portions of the belt, but loses the same amount of energy going around the pulleys to the other side of the apparatus.
This is the critical mistake they are making by pointing out at the pairs of the cylinders during the transition,
at the same time not mentioning that there can be many more cylinders at the vertical stage of the machine,
producing the work ( for free).
BTW, this is not a patent application, but the actual patent.
Regards.
Quote from: telecom on December 02, 2014, 06:57:20 PM
Hi,
I really enjoy reading your informative responses, but this case is different from what have been described.
He is using the weights to create or limit the buoyuncy, quite different from the described examples,
which are short of this application of the weights, or at least, not pointed out in the design.
They write:
Now reconsider the full version with piston chambers on a belt over two pulleys. Each pair of pistons gains energy moving on the straight portions of the belt, but loses the same amount of energy going around the pulleys to the other side of the apparatus.
This is the critical mistake they are making by pointing out at the pairs of the cylinders during the transition,
at the same time not mentioning that there can be many more cylinders at the vertical stage of the machine,
producing the work ( for free).
BTW, this is not a patent application, but the actual patent.
Regards.
Note that the pdf in your link says, in great big letters on the first page, DEMANDE DE BREVET, in all caps.
Please put "Demande de brevet" into your Google French translator and tell me what you get back.
If you think it will "produce work for free" then why, since the document was published in 2003, are we not seeing working buoyancy drives based on this _really simple_ plan all over the place by now?
And you are wrong that it is not fully covered by the designs and analyses in the Simanek museum. There is nothing new in your "patent" and furthermore it will not work. Go ahead and build it yourself and see. Be sure to report your costs and the test results you get, even, or especially, if they are negative. If they are positive, that is, if you can make it work, then we will all benefit from your investment and hard work. When will you be starting?
Quote from: TinselKoala on December 03, 2014, 12:25:09 AM
Note that the pdf in your link says, in great big letters on the first page, DEMANDE DE BREVET, in all caps.
Please put "Demande de brevet" into your Google French translator and tell me what you get back.
If you think it will "produce work for free" then why, since the document was published in 2003, are we not seeing working buoyancy drives based on this _really simple_ plan all over the place by now?
And you are wrong that it is not fully covered by the designs and analyses in the Simanek museum. There is nothing new in your "patent" and furthermore it will not work. Go ahead and build it yourself and see. Be sure to report your costs and the test results you get, even, or especially, if they are negative. If they are positive, that is, if you can make it work, then we will all benefit from your investment and hard work. When will you be starting?
Dear TK,
first of all, the patent was actually granted:
http://www.patfr.com/200304/FR2830575.html
secondly, I really don't know why the inventor haven't continued his quest and tried building his machine,
but my best guess is that its because of the lack of funds...
lastly, rather then saying that I was wrong, lets take an intelligent look at this design, w/o relying
on the opinion of others, and try to figure it out. I already found at least one serious flow in Semanek's
explanation.
Best regards.
https://www.youtube.com/watch?v=423HiKP5JGk
explanation video for archurian rocket for pressure, still unbeaten in any mechanics or math by 300 physics professor it was sent to. One tried and after more than 30 emails back and forward and me beating him at every turn and answering every question, he crumbled back to the great netwoinian answer of well it can't work "coz physics says so",(pity none of that physics had the mechanic failure or math proof to go with it, oohh it's like the bible, how can there be different families descended from adam and eve if god wiped them all out in Noahs flood? oohh i get it, his name was Noah Newton)
The FACT that you do not actually DEMONSTRATE anything in that waste of time video, except your silly imitation of an adult (we know the child's voice is your real voice) sure means a lot more than any words you might emit. And it is clear that you are simply lying about the "300 professors" just as you lied baldly back in the old days about your "Sword of God" magnet-gravity wheel. How is that going, by the way? Did you apologise to all the people you duped back then, return the money and magnets they sent you? Of course you did not.
Why don't you give us a list of a few of the "professors" who could not figure out what was wrong with your system. You won't, I know, and I know why, too.
Here, once again, is the solid refutation of the system you described in the "Crushing Newton like a bug" pdf document, which I believe describes the same system you fail to demonstrate in that video. And, by the way, even the ancient Greeks and Phoenicians knew that the Earth was spherical, not flat, and anyone with half a brain and the eyes to observe natural phenomena like solar and lunar eclipses can figure that much out for themselves easily enough.
--------------------------
So you have Quinn's "boot" shaped system of water locks. The foot of the boot is where you insert your "just barely sunk" box with the heavy block in it. It is sealed off from the tall "leg" of the boot by a water and pressure tight door. The "foot" is, say, 1 meter deep and has a pressure and watertight lid somehow. You start with ten meters of water in the "leg" of the boot and one meter of water in the foot. You put your block in the foot part and the water level rises in that part, and the rise in water level indicates a volume increase, equal to the volume of the box you just put in. Now you put the watertight lid on the foot part so that there is no air in there. Remember that the water level is higher than it was before you put the block in there. Now you open the sealed door into the leg of the boot. Boom, the pressure in the foot is now the pressure resulting from the full 10 meter head in the leg of the boot. You slide your barely sinking box over into the bottom of the leg, and close the watertight and pressuretight door between the foot and the leg. But the pressure in the foot is still the same 10 meter head as before. You raise up your barely sinking box with little work, with your "12 volt" automobile winch. You raise it up to the top and remove it from the leg -- and the water level in the leg _drops_ indicating that you have now removed a volume (the box volume) from the leg. The water level in the leg is now lower, by the same volume that the water level in the foot rose up when you put the box in the foot. When you open the lid on the foot, you find that that pressure now reduces, perhaps some water splashes out or not, but the level of the water is still higher than it was for the first block, and the water level in the leg is lower, by the same amount. You now put another box in the foot. The water level rises _again_ by that same volume amount. You now seal off the roof of the foot with your pressuretight seal, and open the door into the leg of the boot. Boom, the pressure in the foot is now equal to the _somewhat less than_ 10 meters head in the leg. You slide your barely sunk box over into the leg and close the water and pressuretight door between foot and leg. You raise up the box with the winch and remove the box at the top... and the water level in the leg _drops again_ by the same volume as the volume of the box. And the water level in the foot is higher by that same volume amount. Lather rinse repeat. Do you see what is happening? You are LOWERING THE WATER LEVEL IN THE LEG by a volume amount equal to the volume of the box, every time. This volume transfers from the leg, all the way down to the foot. You are working with the _stored energy_ of the lifted water in the leg of the boot. By the time you've lifted a dozen, or however many, blocks, the water level in the leg is waaaay down and the water level in the foot is waaaay up, by the same amount. It is this falling water, a volume equal to the volume of the box, that is adding to the power of the winch in order to lift up your blocks. Without some way to replace the water in the leg of the boot, you will run out of stored energy in short order. So for this scheme to work you need a river flowing water into the top of the boot, replacing the water that falls into the foot with each block transfer. There was no river at the top of the pyramid, just one at the bottom (the Nile).
There is no free lunch, Newton is resting happily in his grave, not crushed like a bug, and Quinn is an idiot, still. I would not be calling him an idiot, except for the fact that he is so disrespectful and arrogant in his own statements in the PDF, in addition to being utterly and totally WRONG.
--------------------
And that is why you can only wave things around and make silly voices, rather than actually DEMONSTRATING anything.
We remember you Archer Quinn and we expect just the same nonsense from you now, as you delivered all those years ago. And you are showing us that our expectations will be fully met.
Quote from: telecom on December 03, 2014, 12:54:45 AM
Dear TK,
first of all, the patent was actually granted:
http://www.patfr.com/200304/FR2830575.html (http://www.patfr.com/200304/FR2830575.html)
Fine, you set me up by presenting the APPLICATION first. And there are many non-functional patents in every country's patent database as I am sure you must know.
Quote
secondly, I really don't know why the inventor haven't continued his quest and tried building his machine,
but my best guess is that its because of the lack of funds...
Really? Someone who has invented a _working_ free energy device as simple as that can't develop it because of lack of funds? Come on, pull the other one why don't you. That is a garbage excuse. Are you telling me that you can't build a model of that device for a few tens or hundreds of dollars, to show that it works? That is a load of bull hockey. The reason it hasn't gone any farther is because it _does not work_ and every person who has tried to build a model of it knows that.
Quote
lastly, rather then saying that I was wrong, lets take an intelligent look at this design, w/o relying
on the opinion of others, and try to figure it out. I already found at least one serious flow in Semanek's
explanation.
Best regards.
If you think you have found a serious "flow" in Simanek's explanations you really should email him and let him and his graduate students and the rest of the engineering world know about it. Be sure to let us know the results of your correspondence with Simanek. Or, just go ahead and build the design that you think he has explained incorrectly and show it working.
Take an intelligent look at this design? Are you now questioning my intelligence, or perhaps you know more information than you are telling us, like when you presented the _application_ instead of the granted patent?
Quote from: telecom on December 03, 2014, 12:54:45 AM
Dear TK,
first of all, the patent was actually granted:
http://www.patfr.com/200304/FR2830575.html
secondly, I really don't know why the inventor haven't continued his quest and tried building his machine,
but my best guess is that its because of the lack of funds...
lastly, rather then saying that I was wrong, lets take an intelligent look at this design, w/o relying
on the opinion of others, and try to figure it out. I already found at least one serious flow in Semanek's
explanation.
Best regards.
I have some really bad news for you: Buoyancy: the weight of displaced fluid exerted upward on the displacing object doesn't provide free energy anymore than the weight of a sack of rocks does on one side of a teeter-totter or a pulley.
Doing the calculations or building and measureing.
Either way it's still trying out ideas that are new to the explorer.
Making claims that can't be backed up is the only problem.
This is what I would Try/calculate/measure in a bouancy/O.U. experiment.
Cheers
floor
Quote from: Floor on December 08, 2014, 03:14:48 AM
Doing the calculations or building and measureing.
Either way it's still trying out ideas that are new to the explorer.
Making claims that can't be backed up is the only problem.
This is what I would Try/calculate/measure in a bouancy/O.U. experiment.
Cheers
floor
As has been shown time and time again whenever these types of machines have been proposed:
In the idealized case that can never be realized, the work performed pushing the buoyant object into the bottom of the column can exactly be recovered as the displaced fluid falls back down and the buoyant volume rises. In the real world, the energy cannot be fully recovered for several reasons. There is never any energy gain possible.
Quote from: The Eskimo Quinn on December 06, 2014, 09:47:38 PM
https://www.youtube.com/watch?v=423HiKP5JGk
explanation video for archurian rocket for pressure, still unbeaten in any mechanics or math by 300 physics professor it was sent to. One tried and after more than 30 emails back and forward and me beating him at every turn and answering every question, he crumbled back to the great netwoinian answer of well it can't work "coz physics says so",(pity none of that physics had the mechanic failure or math proof to go with it, oohh it's like the bible, how can there be different families descended from adam and eve if god wiped them all out in Noahs flood? oohh i get it, his name was Noah Newton)
It's a silly presentation that is full of fallacies. An object is buoyant when its SG is less than the surrounding fluid. Under the premise that you have filled up a uniform cross-section tower with water, you performed work filling that tower with m
fluidGh
tower/2. That is a sunk energy cost. Now, using any mechanism that you like and as perfect as you like you introduce a float of some smaller height and lower SG into the tower. Water from the tower surrounds the float and it rises. What you ignore is the absolute fact that water from the tower must flood the airlock you placed your float in, dropping the level of water in the tower, which will either have to be drained from the airlock or pushed back up into the tower during the next cycle. If you go the drain route you lose energy that you expended filling the tower in the first place and the machine runs down. If you push the water back up into the tower then "Jimmy" is right and you are wrong. Your scheme does not work.
QUOTE "In the idealized case that can never be realized, the work performed pushing the buoyant object into the bottom of the column can exactly be recovered as the displaced fluid falls back down and the buoyant volume rises. In the real world, the energy cannot be fully recovered for several reasons. " END QUOTE
Agreement. The work performed in EITHER sinking or inserting the cube is equal to the work of the cube riseing
BEFORE LOSSES.......There will always be losses in either and or both of these actions.
These losses may or may not be equal to each other.
These losses are caused by friction / turbulence / inertia versus acceleration.
Let's call these losses "LOSS SET 1"
Lets calll the force of PUSHING the cube into the column (sinking the cube) BEFORE LOSSES " EXPENDITURE 1"
Let's call the force in the riseing of the cube through the fluid within the column (floating up) BEFORE LOSSES "RETURN 1 "
Before the losses in "LOSSS SET 1", "EXPENDITURE 1" and "RETURN 1" are equal.
The falling of the qube while exterior to the fluid column, will have air friction losses ?
Let's call these losses "LOSS SET 2"
Let's call the Force in the falling of the cube while exterior to the fluid column,
BEFORE FRICTION LOSSES "RETURN 2 "
Lets calll the following losses combined " LOSS SET 3"
Overcoming:
1 the inertia of the cube during lateral repositioning,
2 friction of the cube against the seal in the column wall,
3 friction and inertia of the door movements
EXPENDITURE 1 and RETUN 1 are equal before losses.
Do we still have a gain in energy after subtracting LOSS SETS
1, 2, and 3 from RETURN 2 ?
Consider this, the material from which the walls of the hollow cube are composed
is all so buoyant in the fluid. The energy in it's buoyancy is RETUN and it's weigh
is not not subtracted from the energy of the cube's fall while exterior to the column.
Quote from: MarkE on December 06, 2014, 11:16:53 PM
I have some really bad news for you: Buoyancy: the weight of displaced fluid exerted upward on the displacing object doesn't provide free energy anymore than the weight of a sack of rocks does on one side of a teeter-totter or a pulley.
You probably didn't bother to read the actual document.
why not to read it first before breaking the " really bad news"?
Quote from: TinselKoala on December 06, 2014, 10:59:59 PM
Fine, you set me up by presenting the APPLICATION first.
That wasn't my intent, I didn't know the difference until you pointed it out.
And there are many non-functional patents in every country's patent database as I am sure you must know. Really? Someone who has invented a _working_ free energy device as simple as that can't develop it because of lack of funds? Come on, pull the other one why don't you. That is a garbage excuse. Are you telling me that you can't build a model of that device for a few tens or hundreds of dollars, to show that it works? That is a load of bull hockey. The reason it hasn't gone any farther is because it _does not work_ and every person who has tried to build a model of it knows that.
There can me miriads reasons why something isn't built - this is not the issue, but the issue is the merit of the invention.
If you think you have found a serious "flow" in Simanek's explanations you really should email him and let him and his graduate students and the rest of the engineering world know about it. Be sure to let us know the results of your correspondence with Simanek.
I have no idea who this guy is, he can write anything he wants on his web page.
Or, just go ahead and build the design that you think he has explained incorrectly and show it working.
Lets examine it first!
Take an intelligent look at this design?
This is why I ask you to look at it because I value your intelligent opinion.
Regards.
Quote from: telecom on December 08, 2014, 02:12:09 PM
You probably didn't bother to read the actual document.
why not to read it first before breaking the " really bad news"?
I read the document and the bad news still applies.
Quote from: The Eskimo Quinn on November 26, 2014, 02:46:58 PM
Being validated as we speak, no human can "Not" do this at home with a small glass jar in a bucket of water. It is over. and this is the third machine all different physics, number 4 is one its way with another physics application.
NO ONE ON EARTH CANNOT DO THIS AT HOME
as per usual will be no comments read or written by the author except the validation note from the engineers not that this needs it
Did you right this?
"funny because I watch a ton of trapped air, lift hot air balloons every day, so it does not fall at all, in fact it costs energy to get hot air down, no matter how many ton there is. There is always an exception if not many exceptions to the "rules"."
---- Did this air become hot all on its own?-----
"Ok so take your rock, place it in an air filled box and seal it, submerse it in 2 metres of water, and it will sit on the bottom, but with the right amount of air you can lift it up and down all day long."
---- Go see what your actually capable acceleration curve is once hydrodynamic drag is calculated. Plus you have severely underestimated the amount of work that must be done to create your circumstances. Oh and you seem to have forgotten here what you point out at the end,("you lift ten ton ten metres it falls you get the same output")that what you have pointed out is zero sum. Why wouldn't you just set up a pulley system arguably using less energy and wasting fewer precious resources? Remember water is not infinite and it actually takes a lot of energy to move it due to how fucking dense it is! In the ocean you would be forced to use a displacement system to change depth just like a submarine or a scuba diver which both require excess energy because you will loose energy every time you try to manipulate a fluid with human means. So once you have wrapped your head around that adda load against this oscillating system and see how much real usable work you can exert. Remember you mind is abstract reality is concrete.. ----
"So the blocks are all placed in boxes sealed like the rock for semi buoyancy, they are dumped into a canal where a cables pulls them forward into a loch,(think roller coaster car drive from underneath ) the loch is a cube. A gate closes, and the block moves forward into a cubic loch, Remember the entire cube loch is underwater, the gate closes. The next lock is a cube attached to a vertical rectangular cube, so it looks like a boot, the door between the two opens and the vertical water cannot fall because the lochs are sealed, the block is moved forward into the elevator loch, and the gate is closed, a car winch lifts the semi buoyant giant block up to the next loch, the gate opens and it moves across, the gate closes and it is raised again. Unlike an open boat loch where water needs to be pumped or flow downstream to fill the lochs and lift the boats, the sealed loch has no such issues as the water remains where it is always. And yet giant stone blocks with little weight are now climbing Mount Everest. Showing buoyancy does have an energy gain if you know what to do with it. So instead of megawatts of power, it is simply kilowatts. What your are really looking at is a single giant pipe filled with water that has an abject that water displacement took place in the first loch, the same as a submarine only displaces water when it is first submerged after that it is simply an object passing through water just like watching the old retro bubble lamps run, the water displacement never alters, and the top is simply open to a canal where it is removed and then taken away. The boxes simply slid back down at no energy cost. The pressure on each gate is no more than the water weight of two lochs at any one time."
---- Great idea, build the system that can withstand the water weight plus displacement pressure without bursting that takes less energy to build then moving the blocks conventionally. Theory and practice are two very different things i suggest you hit the bench and do some work son. Not to mention show me your methodology for encasing these rocks. You need to keep in mind many things can be abstracted without keeping in mind all pertinent variables, you have kicked your own ass and probably ruined your reputation with people who do not spend their time mentally masturbating. ----
"I can always get more energy out than in, in every possible use of physics"
---- With your methods of abstraction i am sure you can! ----
May god or someone have mercy on your soul because you are treacherous to those trying to learn something.
- David
Quote from: TinselKoala on December 02, 2014, 02:31:59 PM
You mean the "inventor" THINKS it should work, but of course it does not.
Come on people, these buoyancy drives are nothing new, they have been completely and rigorously analyzed for years, and yet every once in a while somebody thinks he's invented a new one, when he really hasn't. No working model of any such drive has ever been produced.
Here it is again. You will note that the devices shown and analyzed in this link are functionally _identical_ to what is described in that patent application link.
https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm (https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm)
Or rather... non-functionally identical, since they do not function, and of course neither does the one in the patent app. link.
Seriously... why are you wasting your time with this stuff? The arrogant Quinn's device, as I have hopefully proven to you, will not function except on the stored energy of the water column which is soon depleted. And it is just as easy to refute the rest of the "buoyancy drive" and "gravity drive" devices once you get down to the actual mathematics involved.
If you want to do some experiments that show otherwise, please go ahead, but be sure to report your _negative results_ honestly as well as any positive results you get. The reason you don't hear of all the failures is because people don't want to report them... hence the perpetuation of the belief that someday one of the same-old-same-old ideas will suddenly start working. But they won't.
For beginners , I understand i did not show displacement, and many housewives and children were confuse also, so a slight variation was drawn up to show it in simple form, I did not think i had to describe every nut and bolt, but this should suffice, save anyone who thinks periscopes lift oceans or submarine do not equalize pressure for divers, or that falling objects do not create energy. I think I am done on this now. here is free energy for beginners
Quote from: The Eskimo Quinn on December 08, 2014, 07:21:17 PM
For beginners , I understand i did not show displacement, and many housewives and children were confuse also, so a slight variation was drawn up to show it in simple form, I did not think i had to describe every nut and bolt, but this should suffice, save anyone who thinks periscopes lift oceans or submarine do not equalize pressure for divers, or that falling objects do not create energy. I think I am done on this now. here is free energy for beginners
Your machine does not work as you describe it for reasons that at least three people here have already explained. Feel free to address these correct explanations either in theory or practice. Alternatively, you can as you have in the past simply proclaim that your unworkable scheme works. That's Wayne Travis' approach. Ask him how his buoyancy drives: The ZED, the TAZ, the rotary TAZ have managed to fare against objective proof of concept tests.
The facts:
Your float cannot rise until you first fill your tower with water.
Each time your float rises it is because an equal volume of water falls by the same distance.
The water is more dense than the float.
Hence the energy loss in the water exceeds the energy gain in the float.
The energy that can be reclaimed by dropping the float is no more than that gained by lifting the float, which is less than the energy lost by the water column each time the float is lifted.
Ergo the system loses energy each cycle.
Ergo your scheme is a bust.
QED.
And all these magic rams moving water against a pressure head, and doors that open without letting water pass, and valves that work when you want them to, they all operate without anyone doing any work.... Just like Archer "Give me money" Quinn. How much work did you put in to fill up your 120 meter tall water tower?
Answer: none at all, since it all happens by magic, in your dreams.
You're an utter failure, Archer, and that is why you cannot demonstrate anything except your squeaky pretend-voice and ten minutes of waving your magic wand around.
And you never answered my question: What happened to the Sword of God gravity-magnet wheel that you utterly FAILED to demonstrate, after taking money and magnets from people all those years ago, before you faded away into the outback?
Quote from: The Eskimo Quinn on November 26, 2014, 02:46:58 PM
Being validated as we speak, no human can "Not" do this at home with a small glass jar in a bucket of water. It is over. and this is the third machine all different physics, number 4 is one its way with another physics application.
NO ONE ON EARTH CANNOT DO THIS AT HOME
as per usual will be no comments read or written by the author except the validation note from the engineers not that this needs it
Being put into 3d cad simulation as we speak, this should make it easier.
This can be solved on the back of a napkin. No complex CAD calculations are necessary to establish without any doubt the already proven complete futility of your scheme:
What a joke! Quinn states in his image above that a "periscope" extended up from a submarine displaces the water around it, but doesn't "lift" any water. How silly can you get? Anyone who has ever taken a bath can tell you what's wrong with that claim. But of course that leaves Archer Quinn out, doesn't it. If your ocean is big enough you can just ignore the fact that the periscope is lifting water, can't you? If you are Archer Quinn, you can.
MarkE, your diagram doesn't include the magic "rams" that can force water out of a submerged chamber into another chamber full of water without doing any work against the pressure, nor does it include the magic doors which let solid objects through without letting any water pass, nor does it include the "disappearing water" that goes through the little hole to equalize pressure and then magically vanishes, only to reappear at the top of the tall water column, also without requiring work. And your floater displaces water! While Quinn's obviously does not, it simply floats up through the water without affecting it at all. When your floater is removed from the top of the tall chamber, your water level is now one floater's height lower, because you have not waved your magic wand, like Archer Quinn did in that silly video!
Quote from: TinselKoala on December 09, 2014, 05:37:52 AM
What a joke! Quinn states in his image above that a "periscope" extended up from a submarine displaces the water around it, but doesn't "lift" any water. How silly can you get? Anyone who has ever taken a bath can tell you what's wrong with that claim. But of course that leaves Archer Quinn out, doesn't it. If your ocean is big enough you can just ignore the fact that the periscope is lifting water, can't you? If you are Archer Quinn, you can.
MarkE, your diagram doesn't include the magic "rams" that can force water out of a submerged chamber into another chamber full of water without doing any work against the pressure, nor does it include the magic doors which let solid objects through without letting any water pass, nor does it include the "disappearing water" that goes through the little hole to equalize pressure and then magically vanishes, only to reappear at the top of the tall water column, also without requiring work. And your floater displaces water! While Quinn's obviously does not, it simply floats up through the water without affecting it at all. When your floater is removed from the top of the tall chamber, your water level is now one floater's height lower, because you have not waved your magic wand, like Archer Quinn did in that silly video!
Magic is always an extra option. Mr. Quinn is free to try and argue how his scheme can buoy a float upward without replacing the the float's volume with the buoying fluid. It could be fun to see him try and do that.
Of course the water in the ocean is lifted when a submerged sub. extends it's telescope.
It is equal to volume of the telescope extension, spread over the entire surface of the container
holding the fluid and the sub.. Does this raise the pressure on the hull of the sub ?
YES, but it's utterly insignificant when the container is as broad as an ocean ! The amount of rise is
so small that it is practically immeasurable at this scale of proportion. The wider the container the
less the effect of the the rise.
Attached is PDF file it contains a greatly improved version of a "magic door" on the last page
of the file
CHEERS
floor
Quote from: MarkE on December 08, 2014, 02:51:12 PM
I read the document and the bad news still applies.
In this case can you please elaborate and explain why the machine is not going to work.
Regards
Addendum
The floater does not need to be hollow.
Please for the sake of all that is holy let this man alone. I believe he has never failed to access yet untapped energy because he has never actually tried. Stop encouraging him by acknowledging his existence.
I will bite my tongue and do penance if I end up being wrong and the almighty quinn is actually the vicar of unseen energy itself and it just so happens to be an utter asshole about it.
Quote from: telecom on December 09, 2014, 01:11:43 PM
In this case can you please elaborate and explain why the machine is not going to work.
Regards
It has been explained several times please read back through the answers and don't just blot out of your mind the stuff you don't want to exist.
- David
Even with a solar assist for the volumetric displacement mechanism(the thing that would make your hollow rock go up and down) you would be better of just using the solar power for your needs. I tried to make one of these when i was 16. Then I learned big boy physics. Quinn your a piece shit jesus or just admit your a kid who doesn't understand what things cost yet.
Quote from: dvy1214 on December 09, 2014, 01:25:36 PM
It has been explained several times please read back through the answers and don't just blot out of your mind the stuff you don't want to exist.
- David
Not it hasn't. Please provide at least one explanation.
Regards.
Ok as you wish. This is a respectful application of logic for your benefit. Quinns general theory is that buoyancy gives you access to excess forces on an object which seems true until you factor in the definition of buoyancy. Buoyancy: an upward force exerted by a fluid that opposes the weight of an immersed object. So, how does this happen? Relative Density per unit of area! IS MERCURY A FREE ENRGY DEVICE WHEN IT FLOATS A ROCK WITHOUT A GAS CASING?
Quote from: telecom on December 09, 2014, 01:39:44 PM
Not it hasn't. Please provide at least one explanation.
Regards.
Ok as you wish. This is a respectful application of logic for your benefit. Quinns general theory is that buoyancy gives you access to excess forces on an object which seems true until you factor in the definition of buoyancy. Buoyancy: an upward force exerted by a fluid that opposes the weight of an immersed object. So, how does this happen? Relative Density per unit of area! IS MERCURY A FREE ENRGY DEVICE WHEN IT FLOATS A ROCK WITHOUT A GAS CASING? no.
No matter what you do you have to equivocate the density of the object you want to move with buoyancy, which means that in most cases you would have had to create the circumstances or in the case of moving some thing in the ocean you have to do the work to vary the volume of some aspect of the object after modifying the object to allow this aspect to be varied. This might seem like a small about of energy but I assure you it is not in all cases whether creating the circumstances to equivocate the mass of the object you want to move(moving multiples of that mass) or modifying the object and exerting the force to modify its relative volumetric densities.
Any thoughts?
- David
Quote from: dvy1214 on December 09, 2014, 02:05:04 PM
Ok as you wish. This is a respectful application of logic for your benefit. Quinns general theory is that buoyancy gives you access to excess forces on an object which seems true until you factor in the definition of buoyancy. Buoyancy: an upward force exerted by a fluid that opposes the weight of an immersed object. So, how does this happen? Relative Density per unit of area! IS MERCURY A FREE ENRGY DEVICE WHEN IT FLOATS A ROCK WITHOUT A GAS CASING? no.
No matter what you do you have to equivocate the density of the object you want to move with buoyancy, which means that in most cases you would have had to create the circumstances or in the case of moving some thing in the ocean you have to do the work to vary the volume of some aspect of the object after modifying the object to allow this aspect to be varied. This might seem like a small about of energy but I assure you it is not in all cases whether creating the circumstances to equivocate the mass of the object you want to move(moving multiples of that mass) or modifying the object and exerting the force to modify its relative volumetric densities.
Any thoughts?
- David
Hi David,
if you look at the patent, the energy for changing the density is supplied by the weight.
In other words, he is using gravity to create buyouncy, and both are free.
Regards.
Quote from: telecom on December 09, 2014, 04:09:20 PM
Hi David,
if you look at the patent, the energy for changing the density is supplied by the weight.
In other words, he is using gravity to create buyouncy, and both are free.
Regards.
No as explained before what is declared free is anything but free. Buoyancy is the fluid equivalent of a counterweight. In order to get one thing to go up a greater mass (can be very slightly greater as long as you have lots of time) must go down. Manipulating the volume of a submerged object involves real work exchanged with the surrounding fluid. Make the object larger and surrounding fluid is displaced, lifting that fluid.
Quote from: telecom on December 09, 2014, 04:09:20 PM
Hi David,
if you look at the patent, the energy for changing the density is supplied by the weight.
In other words, he is using gravity to create buyouncy, and both are free.
Regards.
Exerted force by the system is dependent directly on the compression of the air. There is an american patent for this US patent 3,934,964(1974). As well as a British patent British patent No. 1330(1857). The idea arose in 1685. In the 1700s Bernoulli spoke out against them, cannot find his conclusion. Mine is that without the further exertion of enough energy to substantially alter the volumetric contents of the interior of a container you will not receive the needed differential to produce movement. I want to believe that a weight on a short throw piston plus 9.8 m/s^2 is enough energy to change the internal contents volumetrically to the point at which it will dramatically exert a buoyant force but I can't do to how little of the internal space is actually being modified compared to the whole size of the container.
So far the only two systems that I have seen recently are from Rosch Technology and James Kwok's. Rosch claims that their system makes 12KW and eats 4.4Kw in compressor energy, not a damn patent around tho. Kwok says his makes 30% more energy than it consumes. He did not renew his initial international patent and his world patent application hasn't been granted in 8 years now.
Quote from: MarkE on December 09, 2014, 04:58:51 PM
No as explained before what is declared free is anything but free. Buoyancy is the fluid equivalent of a counterweight. In order to get one thing to go up a greater mass (can be very slightly greater as long as you have lots of time) must go down. Manipulating the volume of a submerged object involves real work exchanged with the surrounding fluid. Make the object larger and surrounding fluid is displaced, lifting that fluid.
I have nothing against this explanation except that the work is done by the gravity - by the weight, which is always pushing down. He utilizes this in his patent to increase the dispacement on one branch and decrease it on another branch.
Gravity of the weight is doing the real work. Or I may be too stupid to understand the reasoning...
Regards
Quote from: dvy1214 on December 09, 2014, 06:06:20 PM
Exerted force by the system is dependent directly on the compression of the air. There is an american patent for this US patent 3,934,964(1974). As well as a British patent British patent No. 1330(1857). The idea arose in 1685. In the 1700s Bernoulli spoke out against them, cannot find his conclusion. Mine is that without the further exertion of enough energy to substantially alter the volumetric contents of the interior of a container you will not receive the needed differential to produce movement. I want to believe that a weight on a short throw piston plus 9.8 m/s^2 is enough energy to change the internal contents volumetrically to the point at which it will dramatically exert a buoyant force but I can't do to how little of the internal space is actually being modified compared to the whole size of the container.
So far the only two systems that I have seen recently are from Rosch Technology and James Kwok's. Rosch claims that their system makes 12KW and eats 4.4Kw in compressor energy, not a damn patent around tho. Kwok says his makes 30% more energy than it consumes. He did not renew his initial international patent and his world patent application hasn't been granted in 8 years now.
Hi David,
there is absolutely no compression of air in this patent - it works strictly on the dispacement. The Rosch is not going to work -air compressor is very inefficient.
Regards.
Dude, the piston is sealed with an O ring. The weight connected to the arm that pushes the piston down when the tank is properly oriented is most definitely compressing the air. That is what changes the volume of the air. If i am scuba diving, and i have my weight belt on, and I deflate my lungs i can sink because the compressed air in the tank on my back has a low volumetric profile. When I inflate my lungs I rise, because i have taken some of that air and decreased its compression by taking it from the tank with a valve and allowing it to expand into my lungs.
- David
Quote from: dvy1214 on December 09, 2014, 06:31:58 PM
Dude, the piston is sealed with an O ring. The weight connected to the arm that pushes the piston down when the tank is properly oriented is most definitely compressing the air. That is what changes the volume of the air. If i am scuba diving, and i have my weight belt on, and I deflate my lungs i can sink because the compressed air in the tank on my back has a low volumetric profile. When I inflate my lungs I rise, because i have taken some of that air and decreased its compression by taking it from the tank with a valve and allowing it to expand into my lungs.
- David
Most likely the chamber of the piston moving inward is connected to the chamber of the piston moving outward
in another branch, so there is no air compression involved - it simply moves back and forth not creating resistance.
As I said, this device works strictly on displacement.
Regards.
And as I said, the device in the French patent is just a variant of one that is fully analyzed on Simanek's page. How the patent ever was granted is just an illustration of the flaws in the French patent system, which are not unique to France.
Look at the diagram below.
The piston weights are pushing air from one side to the other, working by weight, moving displacement from one side to the other just as you said. Put the weights on lever arms... no difference, just more sources of drag.
https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm (https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm)
And of course it will not work.
@David: You should check out Wayne Travis's devices and claims. They put Kwok's claims and device to shame, and Travis actually has received a US patent recently.
http://mrwaynesbrain.com/
Try not to laugh too hard.... If you look on YT you can find a couple of video "demonstrations" of a couple of Travis's devices.
Quote from: telecom on December 09, 2014, 06:36:18 PM
Most likely the chamber of the piston moving inward is connected to the chamber of the piston moving outward
in another branch, so there is no air compression involved - it simply moves back and forth not creating resistance.
As I said, this device works strictly on displacement.
Regards.
That is a fantasy if we are still talking about the french patent but fine, sounds like a lot of mechanical resistance.
Quote from: TinselKoala on December 09, 2014, 07:11:57 PM
@David: You should check out Wayne Travis's devices and claims. They put Kwok's claims and device to shame, and Travis actually has received a US patent recently.
http://mrwaynesbrain.com/
Try not to laugh too hard.... If you look on YT you can find a couple of video "demonstrations" of a couple of Travis's devices.
You have made my day Mr. Koala.
Quote from: TinselKoala on December 09, 2014, 07:07:24 PM
And as I said, the device in the French patent is just a variant of one that is fully analyzed on Simanek's page. How the patent ever was granted is just an illustration of the flaws in the French patent system, which are not unique to France.
Look at the diagram below.
The piston weights are pushing air from one side to the other, working by weight, moving displacement from one side to the other just as you said. Put the weights on lever arms... no difference, just more sources of drag.
https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm (https://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm)
And of course it will not work.
In the patent weight is moving the piston outward/inward, not the air.Air volume between two chambers doesn't change,
it just flows between them just to maintain a free movement of the pistons.Semanek's machine is not described
completely, it misses important function of the weights.
Regards
Quote from: dvy1214 on December 09, 2014, 07:49:21 PM
That is a fantasy if we are still talking about the french patent but fine, sounds like a lot of mechanical resistance.
What kind of resistance you are talking about - hydraulic, pistons within the cylinders, or something else?
In any case, the weight can be chosen large enough to overcome it, IMHO.
Regards
As the weight increases as does the necessary strength of the mechanism.
Quote from: telecom on December 09, 2014, 06:13:30 PM
I have nothing against this explanation except that the work is done by the gravity - by the weight, which is always pushing down. He utilizes this in his patent to increase the dispacement on one branch and decrease it on another branch.
Gravity of the weight is doing the real work. Or I may be too stupid to understand the reasoning...
Regards
Work can be exchanged by increasing or decreasing gravitational potential energy. In order for a float to rise, a greater (can be only slightly greater) mass has to fall. Buoyancy schemes are therefore no better than rocks. Each is passive and does not act as a source of energy: Lift net mass gain GPE, lower net mass lose GPE.
Quote from: dvy1214 on December 09, 2014, 08:19:21 PM
As the weight increases as does the necessary strength of the mechanism.
Do you mean stresses on the mechanism?
This can be taken care of by using stronger elements using dormulas from the strength of materials/
Regards
Quote from: MarkE on December 09, 2014, 08:21:07 PM
Work can be exchanged by increasing or decreasing gravitational potential energy. In order for a float to rise, a greater (can be only slightly greater) mass has to fall. Buoyancy schemes are therefore no better than rocks. Each is passive and does not act as a source of energy: Lift net mass gain GPE, lower net mass lose GPE.
But from the gravitational point of view the whole mechanism is in equilibrium.
However, the work is done within the sealed chambers, within the system which is in equilibrium. The mass is falling,
without changing the balance.
Regards
Im talking about the mechanism that you think the french patent works by which is none existent. Go study the patent and form a better question around the working premise.
Quote from: telecom on December 09, 2014, 06:36:18 PM
Most likely the chamber of the piston moving inward is connected to the chamber of the piston moving outward
in another branch, so there is no air compression involved - it simply moves back and forth not creating resistance.
As I said, this device works strictly on displacement.
Regards.
And as i will say again the only thing that buoyancy has to do with displacement is the water which an object displaces. that objects ability to become more buoyant is entirely dependent on the compression of its volumes. Yes your right, buoyancy is dependent on displacement.
And to be more specific because I think you are a little dense, a mechanism linking two sides of this device would incur 1. Mechanical resistance when its shifts 2. Hydrodynamic resistance as the unit revolves
Didn't mean to call you dense just been drinking a little lol.
Quote from: telecom on December 09, 2014, 08:14:23 PM
In the patent weight is moving the piston outward/inward, not the air.Air volume between two chambers doesn't change,
it just flows between them just to maintain a free movement of the pistons.Semanek's machine is not described
completely, it misses important function of the weights.
Regards
Please explain in detail just how the French patent is different from the device described in the Simanek link.
In the diagram from Simanek's site:
The weight of the piston is moving the piston outward/inward.
Air volume between the two connected chambers doesn't change, it just flows between them.
In the diagram from the French patent application:
It is clear that the levered weight does nothing other than pull the piston down or push it up, depending on which side of the apparatus is considered.
It is also clear that the distance from the center of the apparatus to the weights is the same on either side of the apparatus, so the weights themselves do not act in the manner that the usual "gravity wheel" weights are supposed to act. So the weights could be dispensed with entirely and just the pistons themselves could be heavier to achieve the same pull or push that comes from the levered weights, as shown in Simanek's drawing.
So please explain, in detail, how the devices differ and just where you believe that Simanek's analysis goes wrong.
Quote from: TinselKoala on December 09, 2014, 09:25:52 PM
Please explain in detail just how the French patent is different from the device described in the Simanek link.
In the diagram from Simanek's site:
The weight of the piston is moving the piston outward/inward.
Air volume between the two connected chambers doesn't change, it just flows between them.
In the diagram from the French patent application:
It is clear that the levered weight does nothing other than pull the piston down or push it up, depending on which side of the apparatus is considered.
It is also clear that the distance from the center of the apparatus to the weights is the same on either side of the apparatus, so the weights themselves do not act in the manner that the usual "gravity wheel" weights are supposed to act. So the weights could be dispensed with entirely and just the pistons themselves could be heavier to achieve the same pull or push that comes from the levered weights, as shown in Simanek's drawing.
So please explain, in detail, how the devices differ and just where you believe that Simanek's analysis goes wrong.
He says:
Now reconsider the full version with piston chambers on a belt over two pulleys. Each pair of pistons gains energy moving on the straight portions of the belt, but loses the same amount of energy going around the pulleys to the other side of the apparatus.
Now we have to count how many pairs are loosing energy, and how many are gaining energy.
If each branch of the belt is longer than the transition part, we should get a net gain...from the pairs which are gaining energy.
Or I may be wrong because this subject is quite complex.
Regards
The whole unit in Semaneks is supposedly submerged ya?
In that case the whole thing barely make sense due to the fact that the air pressure will desire to move to the highest point. No matter how many belts and pulleys there are it would be fighting itself the whole time unless the weights of the pistons changes corresponding to their location on the track. Which, would be tasky and have its own issues in efficiency. Otherwise each piston moving over the top would want to keep it's air until equilibrium with those moving up the left leg fighting the forward progress.
Have had more to drink at this point so please correct as needed.
- David
Quote from: telecom on December 09, 2014, 09:55:28 PM
He says:
Now reconsider the full version with piston chambers on a belt over two pulleys. Each pair of pistons gains energy moving on the straight portions of the belt, but loses the same amount of energy going around the pulleys to the other side of the apparatus.
Now we have to count how many pairs are loosing energy, and how many are gaining energy.
If each branch of the belt is longer than the transition part, we should get a net gain...from the pairs which are gaining energy.
Or I may be wrong because this subject is quite complex.
Regards
The whole thing is quite simple: Buoyancy is the result of gravity acting on a fluid mass into which another mass or masses have been inserted. Gravity is conservative.
Quote from: TinselKoala on December 09, 2014, 09:25:52 PM
Please explain in detail just how the French patent is different from the device described in the Simanek link.
In the diagram from Simanek's site:
The weight of the piston is moving the piston outward/inward.
Air volume between the two connected chambers doesn't change, it just flows between them.
In the diagram from the French patent application:
It is clear that the levered weight does nothing other than pull the piston down or push it up, depending on which side of the apparatus is considered.
It is also clear that the distance from the center of the apparatus to the weights is the same on either side of the apparatus, so the weights themselves do not act in the manner that the usual "gravity wheel" weights are supposed to act. So the weights could be dispensed with entirely and just the pistons themselves could be heavier to achieve the same pull or push that comes from the levered weights, as shown in Simanek's drawing.
So please explain, in detail, how the devices differ and just where you believe that Simanek's analysis goes wrong.
@TinselKoala you get my PM?
Yes, I just sent you a reply with a lot of information. I can't attach anything to a PM but the zipped powerpoint file below belongs with it:
And just for fun, my Overunity U-Tube:
That's amazing! We need an Indiegogo campaign for the "Fix Inherent Buoyancy" movement.
the full sequence of events is listed, rough drawings and a basic cad drawing shows you why the water never leaves the tower. The cad guy is a bit slower than i have hoped. but everything is pretty much there except fall and generator which is not contested as producing power. simply whether someone could get a buoyant object in through the bottom and still leave the water in the tower.
Naturally any water loses from adherence etc are replaced by using a hollow block filled 80 percent with water, float up as per normal, lifter grabs it, it is emptied and empty container sent down outside.
TMQ
"The actual ram" No, Quinn, that is a photo of someone else's ram for a different purpose. But we are used to hearing and seeing your lies.
What happened to your video "demonstration" that demonstrated only that you can't build anything that works?
Hey, maybe Captain Zero can help you get a Canadian patent. They'll patent anything in Canada.
How's that "Sword of God" magnet wheel coming along? You should be running your house on it by now... shouldn't you? But of course it did not work then and does not work now, in spite of all the claims you made to get people to send you cigarette money and magnets.
You think you've got something? Build a "working" model and show it. Of course you cannot. You're all wet. Magic tubes and valves, objects that float but don't displace any volume, things that can be lifted without doing any work, towers that magically fill themselves without anyone pumping the water. You live in a fantasy world and if you weren't so obnoxious I'd almost feel sorry for you.
Tinsh!tTrolla quote:
" Hey, maybe Captain Zero can help you get a Canadian patent. They'll patent anything in Canada. "
Yeah Quinn...and if you ever feel you're fighting a losing battle in a forum exchange/debate Tinsh!t will gladly give you tips on how to discredit your unlucky opponent/victim by posting sh!t which implies they are a convicted criminal.
Kinda like how he implied that I do not own a US Patent, in order to make it possible for him to make an attempt to ridicule me...simply because he's just too dimwitted to come up with something imaginative.
But, as I have said, neither he nor the other dimwitt has ever had a thought or idea to call their own.
Regards...
Quote from: The Eskimo Quinn on December 14, 2014, 08:06:40 PM
the full sequence of events is listed, rough drawings and a basic cad drawing shows you why the water never leaves the tower. The cad guy is a bit slower than i have hoped. but everything is pretty much there except fall and generator which is not contested as producing power. simply whether someone could get a buoyant object in through the bottom and still leave the water in the tower.
Naturally any water loses from adherence etc are replaced by using a hollow block filled 80 percent with water, float up as per normal, lifter grabs it, it is emptied and empty container sent down outside.
TMQ
That's nice: a ram for another purpose some hand scribbles and part of a CAD diagram. Not one of those things or any of them in combination changes the basic fact that in order for a float to rise within a containing fluid, a greater mass of fluid must fall: ergo in any buoyancy scheme, the rising float results in a loss of of gravitational potential energy in the system. Ergo your scheme is dead before it starts.
Hey, Zero Captain of Nothing, how does it feel to be the most ignored person on three forums? You're a troll and a stalker and your last few posts are textbook examples of both. Why don't you try turning around, you are pissing into the wind and splattering all over yourself.
Quote from: TinselKoala on December 15, 2014, 04:05:23 AM
Hey, Zero Captain of Nothing, how does it feel to be the most ignored person on three forums? You're a troll and a stalker and your last few posts are textbook examples of both. Why don't you try turning around, you are pissing into the wind and splattering all over yourself.
That my friends, I consider high praise, coming from the most reprehensible troll on the forum.
Apparently this trained monkey troll doesn't have any Patents to present demonstrating his superior intellect, after ridiculing "stupid" ol' Cap.
Or is it that he's just frustrated that I don't have a wife that died a horrible death that he can disrespect, like he did to "stupid" Ken Wheeler, after losing a battle of wits with him.
And I guess I'm a troll for giving his #1 bumboy a taste of his own medicine...unlike me they "build things" doncha know. 'Things' being the operative term.
Troll fighting is a dirty job, but somebody had to do it.
Regards...
my Cad guy is slowly getting there his scale is a little out for the legs and platform (he keeps drawing it the size of the tank base.) But getting there, this should help you understand it better at least. skyping with him tomorrow, hope you got other pics including of the real existing legs and ram for the centre.
sorry i dont reply to any comments i do not read the thread at all i click on my posts and click on the original post of my own. I does my head in with people who talk about just rubbish, it works, that is all that matters. and in that we know beyond question that there is no second law of thermodynamics. I having proved that with no physics person in the wrold being able to beat the machine. I invite Archurians to vist the crowd fund page.
How do you still have so many Newtonians on here,?Newton;s second law says no overunity, so why do you not kick them off? isn't that the point and actual name of the site? If you believe (well believed now that it is crushed) why are you here other than undercover oil coal and government filth, oh and baby baby physics wankers in love with a dead idiot who was wrong, spewing their life beliefs were just bullshit.
anyway at least here i can delete all bullshit comments.
http://igg.me/at/ArchurianRocket/x/9027183
Oh and update 5 has the tank gen as well, just so that these will never be covered up, if you go to the bottom of the main page there is a link to the Mayernick rebuilt and all build instructions on video and breaking the wall in two video shots. So those who claim it was never done can now go crawl up in the corner, I am now 3 for 3
I am The Mighty Quinn
Have a nice day and welcome to the new world Archurians
Quote from: The Eskimo Quinn on December 15, 2014, 05:45:39 PM
my Cad guy is slowly getting there his scale is a little out for the legs and platform (he keeps drawing it the size of the tank base.) But getting there, this should help you understand it better at least. skyping with him tomorrow, hope you got other pics including of the real existing legs and ram for the centre.
sorry i dont reply to any comments i do not read the thread at all i click on my posts and click on the original post of my own. I does my head in with people who talk about just rubbish, it works, that is all that matters. and in that we know beyond question that there is no second law of thermodynamics. I having proved that with no physics person in the wrold being able to beat the machine. I invite Archurians to vist the crowd fund page.
I see so it is just a fundraising fraud now. You have never made this machine work as you claim it does and you never will.
Quote
How do you still have so many Newtonians on here,?Newton;s second law says no overunity, so why do you not kick them off? isn't that the point and actual name of the site? If you believe (well believed now that it is crushed) why are you here other than undercover oil coal and government filth, oh and baby baby physics wankers in love with a dead idiot who was wrong, spewing their life beliefs were just bullshit.
anyway at least here i can delete all bullshit comments.
http://igg.me/at/ArchurianRocket/x/9027183
Oh and update 5 has the tank gen as well, just so that these will never be covered up, if you go to the bottom of the main page there is a link to the Mayernick rebuilt and all build instructions on video and breaking the wall in two video shots. So those who claim it was never done can now go crawl up in the corner, I am now 3 for 3
I am The Mighty Quinn
Have a nice day and welcome to the new world Archurians
It's not encouraging when on your Indiegogo page you demonstrate that you do not understand energy or power. Good luck with the "294 megawatts an hour".
Quote from: telecom on November 28, 2014, 12:06:40 PM
This is a French patent I was referring too:
http://vitanar.narod.ru/files/02830575A1.pdf
I would like to hear what you, guys, think about its viability.
Regards This patent did work, It used gravity to change the displacement. I tried a similar thing in 2000. It worked all be it only to a depth of ten feet. Now if You could alter the density of a solid and not be effected by pressure, that would be the answer.
ANIMATION - central component
this is the section that makes the machine provide free energy, the full sequence of putting the block in the base of the tower, Have attached the full sequence list again i know it is painful to troll to find stuff, he has almost finished the full circuit run, but everyone knows stuff floats and the falling weight part. will post that one when done, Look like it really will be the perfect Christmas
click on pre render avi
TMQ
When i log in ther is usually one of the last replies up, i nearly died laughing when i read the webby response, what lower float he is talking about i have no idea, how it would push the weight down and out through a 50 ton ram is beyond me, oh, an i was wonder how many divers got pushed back into submarines after the pressure was equalized, or got pushed down and out back through the wall of the sub, clearly does not read instructions well, it has bee tested with floating weight blocks to several meters in water tanks, it rises faster, simply because the pressure difference increases, not decreases, and the gravity still stay as it always has at 9.81
Hilarious stuff, best call the submarine guys and tell them none of the airlocks work and divers can no longer go outside. It is 120 metres not the Marianas trench.
oh and to spare the forum your great science with some diver suit crap or crushing theories I have posted the pressure in the real world on a human with no gear at that depth. "NOTHING" I spose there may be ten ton marshmallow blocks????? HMMM yes i concede defeat to the ten ton marshmallow man, Webby
http://www.worldrecordacademy.com/sports/deepest_free_immersion_dive_world_record_set_by_William_Trubridge_101665.htm
Quote from: The Not So Mighty Quinnoh and to spare the forum your great science with some diver suit crap or crushing theories I have posted the pressure in the real world on a human with no gear at that depth. "NOTHING"
At 120 meters depth in a column of water the pressure is not "NOTHING". It is around 13 atmospheres, or about 190 pounds per square inch.
QuoteLike air, water causes pressure by its weight. But of course, water is considerably denser (i.e., heavier for a given volume) than air. As it turns out, a column of sea water one inch in cross-section would need to be only about 33 feet (10 meters) tall to weigh 14.7 pounds. Therefore, at a depth of 33 feet (10 meters) beneath the sea surface, the total ambient pressure is about 29.4 psi, or 2 ATM -- 1 ATM caused by the weight of the air in Earth's atmosphere, plus 1 ATM for the weight of 33 feet (10 meters) of seawater. To avoid confusion, when people discuss pressures underwater, the unit "ATA" (referring to "atmospheres absolute") is often used to represent the total, "absolute" pressure caused by both the water and the air above the water.
As is illustrated in the diagram at right, the ambient pressure increases underwater at an almost linear rate with increasing depth* (http://www.bishopmuseum.org/research/treks/palautz97/phys.html#Linear%20Pressure). For every 33 feet (10 meters) of depth in sea water, the ambient pressure increases by an an additional 14.7 psi (1 atm). At a depth of 99 feet (30 meters), the ambient pressure is 4 ATA -- one ATM caused by the Earth's atmosphere, plus 3 ATM for every 33 feet (10 meters) of depth. Similarly, the ambient pressure 297 feet (90 meters) beneath the surface is 10 ATA.
http://www.bishopmuseum.org/research/treks/palautz97/phys.html (http://www.bishopmuseum.org/research/treks/palautz97/phys.html)
or hundreds of other references.
Quote from: TinselKoala on December 22, 2014, 08:50:50 AM
At 120 meters depth in a column of water the pressure is not "NOTHING". It is around 13 atmospheres, or about 190 pounds per square inch.
http://www.bishopmuseum.org/research/treks/palautz97/phys.html (http://www.bishopmuseum.org/research/treks/palautz97/phys.html)
or hundreds of other references.
Quinn is most likely using His transporter and the source of energy is the dilithium crystals.
wrote for ten mins wouldn't load file then cleared everything when you go back, cant be bothered writing it all again, everything like full sequence is posted here as is lower section as a video on its own. last post on this machine.
You tube link is here https://www.youtube.com/watch?v=0RNuOMuJ57k
Merry Christmas
TMQ
Quote from: The Eskimo Quinn on December 22, 2014, 07:03:01 PM
wrote for ten mins wouldn't load file then cleared everything when you go back, cant be bothered writing it all again, everything like full sequence is posted here as is lower section as a video on its own. last post on this machine.
You tube link is here https://www.youtube.com/watch?v=0RNuOMuJ57k (https://www.youtube.com/watch?v=0RNuOMuJ57k)
Merry Christmas
TMQ
Nice cartoon animation. It cleverly leaves out the rise in water level when you push the floater in through the magic "loch" at the bottom, neglecting entirely the FACT that you are raising up a quantity of water that is equal to the volume of the floater, all the way up to the top of the water tank, when you push the floater in. In other words, your animation is a blatant lie. But we are used to that from you, Archer.
Don't listen to the Koala guy, he's probably a paid shill for big firewood or the Roochilds banksters.
I tried to submit a moneys with the kickfunding page but it gives my errors. Are there othere venues to fund this, like WesternUnion or something?
Quote from: webby1 on December 22, 2014, 07:58:34 PM
You have that wrong I think TK,, that is air that is being pushed out of the cylinder as the floater is being pushed up to the underside of the gate by the water entering into the cylinder,, then the piston moves up the cylinder and pushes the water back out and the gate closes,, where that water goes I am not sure,, but it can not go into the tank because then that piston would have to push the hole column of water up,, that would take more input force than what the floater can return I would think,, due to volume, surface area and distance.
I could be wrong,, but that is how it worked with my actual testbed,, it took a little more input than what I could get out of it.
Funny add,, why are the air bubbles not being caught in an upside down cup and also used for output? ???
Where that water goes you aren't sure? Yet you have completed the sentence by stating the exact thing that I and everyone else has been telling Quinn the whole time. Inserting the floater into the tank _displaces water_, that is why it floats! And the displaced water should be seen in a rise in the level at the top of the tank, as soon as the object is being inserted into the tank. But it is not shown in the animation, and if you go back and try to read what Quinn has "explained" previously it is clear that he always neglects this point.
You guys really should stop ridiculing the TEQ. He singlehandedly gangraped Newton, his silly 'laws' and his followers. Why don't you instead take some monies and upload it to the indigoogo website to fund this impotent project?
@Archie:
The crowd funding campaign is a bit slowish on the funding. Maybe you should consider a rebranding. Archurian Rocket sounds a little gayish, you know. Sexual connotation and etc. Just ask Cap'n heehoo. Something along the lines of 'The rectally applied Archurian fist of truth' may be more appropriate. Think about it. Marketing is a delicate business.
Quote from: orbut 3000 on December 22, 2014, 08:46:39 PM
You guys really should stop ridiculing the TEQ. He singlehandedly gangraped Newton, his silly 'laws' and his followers. Why don't you instead take some monies and upload it to the indigoogo website to fund this impotent project?
orbut 3000 the only thing that Archer Quinn has gangraped is reason. Buoyancy is the force exerted
back on an object of
any density as it displaces some volume, and therefore mass of surrounding fluid. This is even true under conditions of Archimedes' Paradox which do not apply to anything that Archer Quinn has proposed. The buoyant force counters the force of gravity on the object in question. Where the mass of the object is less than the mass of the displaced fluid volume, the net force is up. Where the mass is the same, the net force is neutral, and where the mass of the object is greater than the displaced fluid, the net force is down.
So, no big surprise, the only way that there is a net upward force on a submerged object is when that object displaces a greater mass of surrounding fluid than its own mass. IE, the object has a lower density less than the fluid. Taking an instant where the submerged object is at its most submerged, and any other point where the object is less submerged, as the object rises, an equal volume of fluid falls as the object and surrounding fluid change places. Denser surrounding fluid falls as the less dense object, IE the float, rises. The center of gravity for the combined fluid and object goes down. Ergo the gravitational potential energy goes down. Ergo, stored gravitational potential energy is expended raising the float with buoyancy. Energy is not created in this process. It is lost to the environment as waste heat.
No offense, markE, but you sound like a troll paid by the whale oil industry.
orbut 3000 I have explained to you the physics that you can easily verify. Verify, or don't it matters not to me. Make crazy accusations if you like, such behavior reflects only on you.
If you think you can stop the whaling sheiks you are sadly mistaken. What with the vast whaling fleets churning out trillions of dollars of slick gold each day no one can evade the treacherous reach of the blubber dollar. Do you think it is just coincidence that whenever Archer goes to the bar he finds himself surrounded by peg leg comrades all gripping harpoons?
Interesting how you never mentioned dishwasher repair procedures, MarkE. Do you possibly have something to hide? Just asking. Questions.
There is, however, a way to use the power of the Nile to float the blocks into place on the pyramid. I was looking for non-electric ways to move water uphill and stumbled onto this one.
It uses the same sealed column. The blocks were floated down the Nile from the quarry. The column is empty and the raft holding the block is moved into the bottom of the column and the column is sealed. Then a hydraulic ram pump is used to pump water uphill to the top of the column and into it to float the block to the top of the column. Drain the column by siphoning, lowering the raft to the bottom again, and repeat as necessary.
The hydraulic ram pump surprised me. I' had never heard of it before.
http://en.wikipedia.org/wiki/Hydraulic_ram (http://en.wikipedia.org/wiki/Hydraulic_ram)
https://www.youtube.com/watch?v=A9W0zx2D7aU (https://www.youtube.com/watch?v=A9W0zx2D7aU)
Scroll down on the following link to see the usage at the Nile river. They make a pretty convincing case that the Great Pyramid was a machine rather than a tomb.
http://sentinelkennels.com/Research_Article_V41.html (http://sentinelkennels.com/Research_Article_V41.html)
It may also be related to Tesla's Magnifying Transmitter. Pure speculation put Tesla did sometimes use hydraulic analogies to his electrical circuits. I think it was Leland Anderson who had drawings of the transmitter with two spark gaps. The first spark gap is quenched and the short turn off time would be like the first water valve closing and creating the back pulse. The second gap would be set a bit wider so the back pulse would fire that second gap and charge the condenser.
They poured concrete. Water, sand and lime. Concrete stands for [With Crete], Phoenicians invented it, sailed around and taught everybody how to make it.