Lets do a little brain blitzing.
Turn off the Sun; turn off the gravity.
Wrap the Sun with a string with a length of 35.3 AU (astronomical units).
Attach the end of the wrapped string to Halley's comet that is at perigee.
At perigee Halley's comet is moving 54.6 km/sec and it is at a distance of .587 AU from the Sun.
Let the string unwrap until the comet reaches the end of the string at 35.3 AU, at apogee.
What is the velocity of the comet when it reaches the end of the string?
energy producing experiments
Moderator: scott
re: energy producing experiments
The answer is 54.6 km / sec. There is nothing to slow it down, The mere new length of the string does not calculate a new velocity and make the comet slow down.
What makes it slow down is gravity; when it is on.
Gravity reduces the linear velocity of the comet to .908 km /sec at apogee.
Angular momentum is true if gravity is on. .908 km / s * 35.3 AU = .587 AU * 54.6 km /sec
With gravity off angular momentum conservation is a false statement.
There is no gravitational acceleration caused by the Dawn satellite or any other man made satellite. Gravity is turned off; so to speak.
They mention tether lengths of Kilometers; but let’s do the math on a short one, 12.5 m. Each particular tether length works with a certain mass, but these numbers should be reasonably close. Only a few yo-yo de-spin experiments are actually accompanied by numbers; as mentioned before. I don’t know of any that give actual release velocity, it is never measured. Like the heat they can’t find.
Lets let the radius of Dawn be .5 m and for its tether length they gave us; we round to 12.5 m. That is 25 to one. Dawn is 1422 kilograms stopped by 2.88 kg. For mv to be conserved the small 2.88 kg masses would have to be moving 493 (2.88 * 493) times faster than the average mass of the spinning satellite (1422 kg * 1 m/sec). For mv * r to be conserved the 2.88 kg would have to be only moving 19.7 times faster (25 * 2.88 * 19.7), and (2.88 * 19.7 /1422) only 4% of the original momentum remains.
At some point in the disclosure NASA states that when the small masses re-wrap around the satellite the original motion would be restored. But how can you get 1422 units of momentum from 56.7 units.
NASA is acting as if gravity will slow the 2.88 kg and store gravitation potential energy in the position of the 2.88 kg, but there is no gravity working. All motion has to be stored as motion.
Some may mistakenly believe that angular momentum conservation also conserves energy. Angular momentum does not conserve energy unless there is gravitational potential energy at apogee, and the thrown masses do not have gravitational potential energy. If you turn off the power to an electric motor none of the formulas work. Turn off the gravitational attraction and angular momentum conservation no longer works.
Bessler did have one major advantage over current physicists. He did not have the same rules.
The Law of Conservation of Energy was not yet a law. Leibniz clearly believed in the Law of Conservation of Energy, and he lived it, but it was not yet a law.
The Law of Conservation of Momentum (linear Newtonian) was fresh and not yet explained away and not yet corrupted. Newtonian linear momentum conservation was separate from Kepler's angular momentum conservation.
Kepler's momentum could only be correct if gravity changed the linear momentum of the comet by massive amounts from apogee to perigee. No good scientist would ever think that Kepler's momentum could be conserved in the lab.
In Bessler's time these three separate and distinct concepts of motion were keep separate and distinct. Today these three concepts are all rolled up together to form a nebulous ball of mush. Today they fit hand and glove together and they are all conserved; which is hogwash. Bessler was not living in a time of such misguided hogwash.
If you demand that the rules be keep; keep the rules of Bessler's time.
What makes it slow down is gravity; when it is on.
Gravity reduces the linear velocity of the comet to .908 km /sec at apogee.
Angular momentum is true if gravity is on. .908 km / s * 35.3 AU = .587 AU * 54.6 km /sec
With gravity off angular momentum conservation is a false statement.
There is no gravitational acceleration caused by the Dawn satellite or any other man made satellite. Gravity is turned off; so to speak.
They mention tether lengths of Kilometers; but let’s do the math on a short one, 12.5 m. Each particular tether length works with a certain mass, but these numbers should be reasonably close. Only a few yo-yo de-spin experiments are actually accompanied by numbers; as mentioned before. I don’t know of any that give actual release velocity, it is never measured. Like the heat they can’t find.
Lets let the radius of Dawn be .5 m and for its tether length they gave us; we round to 12.5 m. That is 25 to one. Dawn is 1422 kilograms stopped by 2.88 kg. For mv to be conserved the small 2.88 kg masses would have to be moving 493 (2.88 * 493) times faster than the average mass of the spinning satellite (1422 kg * 1 m/sec). For mv * r to be conserved the 2.88 kg would have to be only moving 19.7 times faster (25 * 2.88 * 19.7), and (2.88 * 19.7 /1422) only 4% of the original momentum remains.
At some point in the disclosure NASA states that when the small masses re-wrap around the satellite the original motion would be restored. But how can you get 1422 units of momentum from 56.7 units.
NASA is acting as if gravity will slow the 2.88 kg and store gravitation potential energy in the position of the 2.88 kg, but there is no gravity working. All motion has to be stored as motion.
Some may mistakenly believe that angular momentum conservation also conserves energy. Angular momentum does not conserve energy unless there is gravitational potential energy at apogee, and the thrown masses do not have gravitational potential energy. If you turn off the power to an electric motor none of the formulas work. Turn off the gravitational attraction and angular momentum conservation no longer works.
Bessler did have one major advantage over current physicists. He did not have the same rules.
The Law of Conservation of Energy was not yet a law. Leibniz clearly believed in the Law of Conservation of Energy, and he lived it, but it was not yet a law.
The Law of Conservation of Momentum (linear Newtonian) was fresh and not yet explained away and not yet corrupted. Newtonian linear momentum conservation was separate from Kepler's angular momentum conservation.
Kepler's momentum could only be correct if gravity changed the linear momentum of the comet by massive amounts from apogee to perigee. No good scientist would ever think that Kepler's momentum could be conserved in the lab.
In Bessler's time these three separate and distinct concepts of motion were keep separate and distinct. Today these three concepts are all rolled up together to form a nebulous ball of mush. Today they fit hand and glove together and they are all conserved; which is hogwash. Bessler was not living in a time of such misguided hogwash.
If you demand that the rules be keep; keep the rules of Bessler's time.
re: energy producing experiments
Oh; with a 19.7 times faster linear velocity and a 25 times larger radius, the angular velocity slows. That is not what you see when you do an experiment. The motion is violent.
re: energy producing experiments
Here is a question someone asked.
The wrong answer takes you to lollypop-land; and it seems that people never come back.Â
But here is the correct answer; and if Bessler was not a fraud this is probably what he used.
A two meter mass-less bar has a one kilogram mass on each end. The bar is in a vertical plane.
Around the center of mass and also around the point of rotation is fixed a mass-less pulley with a radius of .1 meter. The bearing is considered to be friction free.
On the pulley is hung a .5 kilogram mass.
The question was; How long will it take to achieve a speed of 50 rpm?
The correct answer:
First we change rpm into a velocity in linear (arc) meters per second.Â
50 rpm is 50/60 .833 rps.
The diameter of the pulley is .2 meters and the circumference is .2 m * 3.14 = .628 m. which means that the target velocity is .833 rps * .628 m/rotation = .5236 m/sec
The applied force is at .1 m and that is where the acceleration occurs.
The free fall acceleration is shared and slowed by two kilograms; but the 2 kilograms have a Laws of Levers advantage of 10 to one so they act like 20 kilograms. So the acceleration is (.5 kg * 9.81 N / kg) / 20.5 kg = .2393 m/sec². F = ma; a = F/m
We know that v = at. Solving for t with have v / a = t; .5236 m/sec / .2393 m/sec² = 2.19 seconds.
Those with sims are headed for lollypop-land; and they think it is 21.9 seconds, or there about. They use the same acceleration but they apply it to the outside of the bar at the masses. That changes the target velocity to 5.236 m/sec and it takes 21.9 second and 5.728 m of drop for the .5 kg to get there.
I think it takes .5730 m to get to the target velocity because the drive force is at .1 m not 1 meter.
If you want to know whom is correct; here is an experiment to test it. I won't make you by a sim.
Buy a 4' * 8' * 2'' sheet or ridged poly-foam.
Place two one kilogram masses on the end of a 4 meter fishing line. This fishing line is tied in the center with a second fishing line.
Place a portion of the Styrofoam sheet between the two masses so as to keep the two masses apart at a distance of 2 meters.
Suspend the system from the second string that had been tied to the middle of the string that is connecting the two masses. This will give you an upside down Y with one kilogram masses on the tips of the Y, and the base of the Y is ties to the ceiling. The bar will spin in a horizontal plane.
Cut a .2 m circle of Styrofoam and glue it to the bottom in the middle of the Styrofoam bar that separates the masses. This .2 m Styrofoam circle is the pulley.
Arrange two lab type pulleys in a counter balancing position that will deliver two tangent strings onto the outside of the Styrofoam pulley. Suspend a .250 kg mass from each of the lab pulleys. The lab pulleys are pointing up; the masses are pulling with 4.905 N down, and they are applying their force to the Styrofoam pulley at 180° to each other.
We are looking for the quantity of acceleration at the edge of the Styrofoam pulley.
Photo gates will be of use but not necessary. Air resistance will soon set in.
But there is a whole lot of room between 21.9 sec and 2.19 seconds. If it is 2.19 sec (as all my experiments prove) the system makes energy.
The wrong answer takes you to lollypop-land; and it seems that people never come back.Â
But here is the correct answer; and if Bessler was not a fraud this is probably what he used.
A two meter mass-less bar has a one kilogram mass on each end. The bar is in a vertical plane.
Around the center of mass and also around the point of rotation is fixed a mass-less pulley with a radius of .1 meter. The bearing is considered to be friction free.
On the pulley is hung a .5 kilogram mass.
The question was; How long will it take to achieve a speed of 50 rpm?
The correct answer:
First we change rpm into a velocity in linear (arc) meters per second.Â
50 rpm is 50/60 .833 rps.
The diameter of the pulley is .2 meters and the circumference is .2 m * 3.14 = .628 m. which means that the target velocity is .833 rps * .628 m/rotation = .5236 m/sec
The applied force is at .1 m and that is where the acceleration occurs.
The free fall acceleration is shared and slowed by two kilograms; but the 2 kilograms have a Laws of Levers advantage of 10 to one so they act like 20 kilograms. So the acceleration is (.5 kg * 9.81 N / kg) / 20.5 kg = .2393 m/sec². F = ma; a = F/m
We know that v = at. Solving for t with have v / a = t; .5236 m/sec / .2393 m/sec² = 2.19 seconds.
Those with sims are headed for lollypop-land; and they think it is 21.9 seconds, or there about. They use the same acceleration but they apply it to the outside of the bar at the masses. That changes the target velocity to 5.236 m/sec and it takes 21.9 second and 5.728 m of drop for the .5 kg to get there.
I think it takes .5730 m to get to the target velocity because the drive force is at .1 m not 1 meter.
If you want to know whom is correct; here is an experiment to test it. I won't make you by a sim.
Buy a 4' * 8' * 2'' sheet or ridged poly-foam.
Place two one kilogram masses on the end of a 4 meter fishing line. This fishing line is tied in the center with a second fishing line.
Place a portion of the Styrofoam sheet between the two masses so as to keep the two masses apart at a distance of 2 meters.
Suspend the system from the second string that had been tied to the middle of the string that is connecting the two masses. This will give you an upside down Y with one kilogram masses on the tips of the Y, and the base of the Y is ties to the ceiling. The bar will spin in a horizontal plane.
Cut a .2 m circle of Styrofoam and glue it to the bottom in the middle of the Styrofoam bar that separates the masses. This .2 m Styrofoam circle is the pulley.
Arrange two lab type pulleys in a counter balancing position that will deliver two tangent strings onto the outside of the Styrofoam pulley. Suspend a .250 kg mass from each of the lab pulleys. The lab pulleys are pointing up; the masses are pulling with 4.905 N down, and they are applying their force to the Styrofoam pulley at 180° to each other.
We are looking for the quantity of acceleration at the edge of the Styrofoam pulley.
Photo gates will be of use but not necessary. Air resistance will soon set in.
But there is a whole lot of room between 21.9 sec and 2.19 seconds. If it is 2.19 sec (as all my experiments prove) the system makes energy.
re: energy producing experiments
http://orion.neiu.edu/~pjdolan/rotational.html
An aluminum bar, with pairs of holes, spaced at 2 cm intervals out from the center, is attached to the top of the spool; the spacing of the holes has been chosen so that the actual rotational inertia of the bar is 95 % of the value of a solid bar of the same size and masses. Pairs of masses (about 400 g) fit into these holes. Moving the positions of these masses allows one to change I, without changing the total mass.
Torque may be applied via string is wrapped around the spool (radius 3 cm), which is attached to a standard 50 g weight hangar.
See table; These are mr numbers.
An aluminum bar, with pairs of holes, spaced at 2 cm intervals out from the center, is attached to the top of the spool; the spacing of the holes has been chosen so that the actual rotational inertia of the bar is 95 % of the value of a solid bar of the same size and masses. Pairs of masses (about 400 g) fit into these holes. Moving the positions of these masses allows one to change I, without changing the total mass.
Torque may be applied via string is wrapped around the spool (radius 3 cm), which is attached to a standard 50 g weight hangar.
See table; These are mr numbers.
re: energy producing experiments
A 10 kilogram mass on dry ice that is accelerated by .5 kilograms draped over a pulley has an acceleration of .4671 m/sec/sec. The total mass accelerated is 10.5 kilograms.
A 20.5 kilogram mass on dry ice that is accelerated by .5 kilograms draped over a pulley has an acceleration of .2336 m/sec/sec. (1/2 * .4671). The total mass accelerated is (20.5 + .5) = 21 kilograms (which is equal to 10.5 * 2).
The inertia 10.5 kg / 21 kg is inversely proportional to the accelerations .2336 m/sec² / .4671. m/sec².
In Table II the mass at 11 cm is roughly 11/3 harder to accelerate than the mass at 3 cm.
The inertia is mr: m*3 and m*11. The mass is the same so the proportion of the radii 3/11 is inversely proportional to the acceleration 5.858 / 1.231. The numbers are not real good; which I think is mainly because there is other mass being accelerated. But compare the same numbers to the other theory of mr².
The other theory of mr² would have the proportion be 3² / 11² = 1.231 / 5.858 which is about 300% off.
The theory of mr being the inertia is 30 % off and shows that the experiment needs further improvements. The 300% off means that the theory needs to be thrown out.
This means of course that it will take 2.19 sec not 21.9 seconds in the previous experiment that had been discussed.
A 20.5 kilogram mass on dry ice that is accelerated by .5 kilograms draped over a pulley has an acceleration of .2336 m/sec/sec. (1/2 * .4671). The total mass accelerated is (20.5 + .5) = 21 kilograms (which is equal to 10.5 * 2).
The inertia 10.5 kg / 21 kg is inversely proportional to the accelerations .2336 m/sec² / .4671. m/sec².
In Table II the mass at 11 cm is roughly 11/3 harder to accelerate than the mass at 3 cm.
The inertia is mr: m*3 and m*11. The mass is the same so the proportion of the radii 3/11 is inversely proportional to the acceleration 5.858 / 1.231. The numbers are not real good; which I think is mainly because there is other mass being accelerated. But compare the same numbers to the other theory of mr².
The other theory of mr² would have the proportion be 3² / 11² = 1.231 / 5.858 which is about 300% off.
The theory of mr being the inertia is 30 % off and shows that the experiment needs further improvements. The 300% off means that the theory needs to be thrown out.
This means of course that it will take 2.19 sec not 21.9 seconds in the previous experiment that had been discussed.
re: energy producing experiments
Okay so it is 11/3 times harder to accelerate the same mass at 11 cm. But if you reduce the mass to 3/11ths, you are back to the same acceleration that you had with one unit of mass at one unit of distance. But the 3/11ths mass at 11/3rds the distance is moving 11/3rds faster.
So which is more energy the full mass at 3 cm; ½ 1kg * 1 m/sec *1 m/sec: or the partial mass at 11 cm, ½ 3/11 kg * 11/3 m/sec * 11/3 m/sec?
So which is more energy the full mass at 3 cm; ½ 1kg * 1 m/sec *1 m/sec: or the partial mass at 11 cm, ½ 3/11 kg * 11/3 m/sec * 11/3 m/sec?
re: energy producing experiments
http://www.animations.physics.unsw.edu. ... tation.htm
See Newton's laws of rotation; Check the relationship of the radius to the acceleration. It is mr
See Newton's laws of rotation; Check the relationship of the radius to the acceleration. It is mr
re: energy producing experiments
Peq look at the vid of the spinning weighted aluminium rod. It is mrr.
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Furcurequs
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Re: re: energy producing experiments
This link is great!pequaide wrote:http://www.animations.physics.unsw.edu. ... tation.htm
See Newton's laws of rotation; Check the relationship of the radius to the acceleration. It is mr
Thanx pequaide!
At a point they say: ''torque is vector''...
A real novel stuff to some here at BW forum!
Any intelligent comparison with 'avalanchedrive' will show that all PM turning wheels are only baby's toys!
re: energy producing experiments
I was looking at the aluminum rod with the added masses; it is almost exactly half the acceleration for the masses at twice the distance. This is mr
The mrr would require that the larger radius would be moving only one fourth as fast. It is clearly not moving one fourth as fast. When the short radius reaches 90° the large is at 45°etc.
We all know what the static forces at r and 2r will be. How can you imagine that the forces change when motion occurs? The forces will remain the same when motion occurs and force is what causes motion. F = ma works just the same in a circle as it does in a line.
I measure the pulley to be about an r and 4r relationship. And I see the rate of fall for the drive mass (acceleration) to be about the same ¼. This makes sense; the force applied by the drive mass is one fourth and the acceleration is one fourth. Both these forces are delivered to the same pulley.
For mrr to have been true for the pulley it would have been moving only 1/16th as fast. Clearly this is not the case.
Now this does not mean that the lesser force or the lesser radius can not achieve high speeds. The videos are turned off before the two drive forces, that cause less acceleration, have fallen as far as the faster.
The mrr would require that the larger radius would be moving only one fourth as fast. It is clearly not moving one fourth as fast. When the short radius reaches 90° the large is at 45°etc.
We all know what the static forces at r and 2r will be. How can you imagine that the forces change when motion occurs? The forces will remain the same when motion occurs and force is what causes motion. F = ma works just the same in a circle as it does in a line.
I measure the pulley to be about an r and 4r relationship. And I see the rate of fall for the drive mass (acceleration) to be about the same ¼. This makes sense; the force applied by the drive mass is one fourth and the acceleration is one fourth. Both these forces are delivered to the same pulley.
For mrr to have been true for the pulley it would have been moving only 1/16th as fast. Clearly this is not the case.
Now this does not mean that the lesser force or the lesser radius can not achieve high speeds. The videos are turned off before the two drive forces, that cause less acceleration, have fallen as far as the faster.
re: energy producing experiments
Peq. You are so blinded by your beliefs you cannot see the truth directly before you. You will never see the answer if it bit you in the arse. Your chosen ignorance and continuing proclamations in the face of solid contradictory evidence IMO is fraudulent.
Perhaps the University of New South Wales is part of a conspiracy, doctoring their results to stop you from showing everyone the truth.
Perhaps the University of New South Wales is part of a conspiracy, doctoring their results to stop you from showing everyone the truth.
re: energy producing experiments
You 79 have been asked many many times not to post on my thread, please take your insults and falsified experiments somewhere else.
Look; if you don't like what I say don't read.
I should have continued not reading your posts, but I wanted to show you same respect. But sadly: you do not deserve any scientific respect.
To others: The acceleration matches the radius changes and is therefore mr. All the opposition has is junior high insults, and falsified experiments. These are third party experiments and the outcome is obvious.
The mr is dangerous to the opposition because they know that energy can be made with it alone, and you would not need to throw. I make lots of machines and have known for some time it is mr. I use photo gates on mine.
Look; if you don't like what I say don't read.
I should have continued not reading your posts, but I wanted to show you same respect. But sadly: you do not deserve any scientific respect.
To others: The acceleration matches the radius changes and is therefore mr. All the opposition has is junior high insults, and falsified experiments. These are third party experiments and the outcome is obvious.
The mr is dangerous to the opposition because they know that energy can be made with it alone, and you would not need to throw. I make lots of machines and have known for some time it is mr. I use photo gates on mine.
re: energy producing experiments
Instead of moving the full mass to the end of the rod: what if you moved the mass to the end of the rod and reduced the mass by one half. The full mass on the end rotates half as fast; but half the mass on the end would rotate just as fast as the full mass at half the radius.
Half the mass moving twice as fast has twice the energy; 1/2mv²
Half the mass moving twice as fast has twice the energy; 1/2mv²