Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
It concerns me that an observer on Earth will notice that the mass and thus energy of the stationary car held up by the drive is becoming lower with time. He will not find where that energy is being deposited as the mass drops. The heat due to cavity loss can be calculated directly, but any other energy due to mass conversion will not be accounted for. This is a major issue with regard to accepting the reality of EM Drives. Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Sun, May 10, 2015 10:48 pm Subject: Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit In reply to Craig Haynie's message of Sun, 10 May 2015 18:07:28 -0400: Hi, [snip] It doesn't cost any energy at all to support a car. The ground does this just fine with no energy expenditure. E = F . d. If d = 0, then E = 0. I'm not sure how this applies to an EM drive (if at all), but perhaps it needs to be taken into consideration? Hello! I was hoping the Vorts could help me with this. Roger Shawyer, at minute 2:56 in this video, claims that the next generation EM Drive could generation 1 tonne of thrust per kilowatt of power. This means that a 1 tonne car should be able to hover above the ground for the price of one kilowatt. However, my calculation shows that to be about 48 times a theoretical maximum. Here is the video where he makes the claim at 2:56. http://tinyurl.com/ko5v6h7 But here is my calculation for a theoretical maximum, calculated two different ways: - A joule is a watt-second - A watt is a joule / second - The power required to hover an object is the same power required to increase the speed of the object from rest, in a weightless environment, to 9.8 m/s in one second. We know this because the pull of gravity is 9.8 meters/second2. - The kinetic energy in an object travelling at 9.8 m/s = 1/2 * m * v2. So for a car of 1000 kg, the energy = 1000 / 2 * 9.82 = 48,020 joules = 48 kilowatts to do this in one second. - This power should be 1/2 the power to raise an object of the same mass, to a height of 9.8 meters in one second, since it would require twice as much energy to do this. - The formula to determining how much energy it takes to raise something to height = E = m * g (gravitational constant) * h = 1000 * 9.8 * 9.8 = 96,040 watts-seconds = 96 kilowatts to do this in one second. So it agrees with the previous result. So, I don't understand how any device could hover an object with the mass of a tonne for less than a theoretical 48 kilowatts. Any thoughts on this would be appreciated. Craig Haynie ( Manchester, NH) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
In reply to Craig Haynie's message of Sun, 10 May 2015 23:19:42 -0400: Hi, I'm suggesting that in theory no energy is required as long as there is no movement. IOW he creates a force, but as long as that force doesn't act over a distance, then it need do no work. E = F x d; F = m x a. E = m x a x d. You have calculated the mass times acceleration part of it. OTOH a rocket would most definitely expend energy just to hover, as do helicopters etc. but they also accelerate mass downward to produce the thrust (air in the case of helicopters). So I think it just depends on exactly how the thrust is generated, i.e. how the drive interacts with the space-time continuum. His claim is 1 tonne of thrust per kilowatt. One tonne of thrust will accelerate an object. An object under the acceleration of gravity will be countered by the thrust, costing 48 kilowatts of power in the process. This is not the same as suspending an object by a rope or something. Are you suggesting that there is no theoretical limit as to how much power, applied as thrust, is needed to suspend an object weighing a tonne? Or are you suggesting that my math is wrong and that there is a lower number? If the number is lower, then how do you arrive at it? Craig On Sun, May 10, 2015 at 10:48 PM, mix...@bigpond.com wrote: In reply to Craig Haynie's message of Sun, 10 May 2015 18:07:28 -0400: Hi, [snip] It doesn't cost any energy at all to support a car. The ground does this just fine with no energy expenditure. E = F . d. If d = 0, then E = 0. I'm not sure how this applies to an EM drive (if at all), but perhaps it needs to be taken into consideration? Hello! I was hoping the Vorts could help me with this. Roger Shawyer, at minute 2:56 in this video, claims that the next generation EM Drive could generation 1 tonne of thrust per kilowatt of power. This means that a 1 tonne car should be able to hover above the ground for the price of one kilowatt. However, my calculation shows that to be about 48 times a theoretical maximum. Here is the video where he makes the claim at 2:56. http://tinyurl.com/ko5v6h7 But here is my calculation for a theoretical maximum, calculated two different ways: - A joule is a watt-second - A watt is a joule / second - The power required to hover an object is the same power required to increase the speed of the object from rest, in a weightless environment, to 9.8 m/s in one second. We know this because the pull of gravity is 9.8 meters/second2. - The kinetic energy in an object travelling at 9.8 m/s = 1/2 * m * v2. So for a car of 1000 kg, the energy = 1000 / 2 * 9.82 = 48,020 joules = 48 kilowatts to do this in one second. - This power should be 1/2 the power to raise an object of the same mass, to a height of 9.8 meters in one second, since it would require twice as much energy to do this. - The formula to determining how much energy it takes to raise something to height = E = m * g (gravitational constant) * h = 1000 * 9.8 * 9.8 = 96,040 watts-seconds = 96 kilowatts to do this in one second. So it agrees with the previous result. So, I don't understand how any device could hover an object with the mass of a tonne for less than a theoretical 48 kilowatts. Any thoughts on this would be appreciated. Craig Haynie ( Manchester, NH) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
IOW he creates a force, but as long as that force doesn't act over a distance, then it need do no work. I'm the one who suggests that the thrust created by the EM Drive could be used to levitate an object. Shawyer is saying that the EM Drive could create 1 tonne of thrust for 1 kilowatt of power, implying that this thrust would be used to accelerate a spacecraft. He's not siting these numbers as an example of levitation. So he's implying that the thrust will be used to do work, and therefore should not be able to violate a theoretical amount of power needed to do that work. On Sun, May 10, 2015 at 11:35 PM, mix...@bigpond.com wrote: In reply to Craig Haynie's message of Sun, 10 May 2015 23:19:42 -0400: Hi, I'm suggesting that in theory no energy is required as long as there is no movement. IOW he creates a force, but as long as that force doesn't act over a distance, then it need do no work. E = F x d; F = m x a. E = m x a x d. You have calculated the mass times acceleration part of it. OTOH a rocket would most definitely expend energy just to hover, as do helicopters etc. but they also accelerate mass downward to produce the thrust (air in the case of helicopters). So I think it just depends on exactly how the thrust is generated, i.e. how the drive interacts with the space-time continuum. His claim is 1 tonne of thrust per kilowatt. One tonne of thrust will accelerate an object. An object under the acceleration of gravity will be countered by the thrust, costing 48 kilowatts of power in the process. This is not the same as suspending an object by a rope or something. Are you suggesting that there is no theoretical limit as to how much power, applied as thrust, is needed to suspend an object weighing a tonne? Or are you suggesting that my math is wrong and that there is a lower number? If the number is lower, then how do you arrive at it? Craig On Sun, May 10, 2015 at 10:48 PM, mix...@bigpond.com wrote: In reply to Craig Haynie's message of Sun, 10 May 2015 18:07:28 -0400: Hi, [snip] It doesn't cost any energy at all to support a car. The ground does this just fine with no energy expenditure. E = F . d. If d = 0, then E = 0. I'm not sure how this applies to an EM drive (if at all), but perhaps it needs to be taken into consideration? Hello! I was hoping the Vorts could help me with this. Roger Shawyer, at minute 2:56 in this video, claims that the next generation EM Drive could generation 1 tonne of thrust per kilowatt of power. This means that a 1 tonne car should be able to hover above the ground for the price of one kilowatt. However, my calculation shows that to be about 48 times a theoretical maximum. Here is the video where he makes the claim at 2:56. http://tinyurl.com/ko5v6h7 But here is my calculation for a theoretical maximum, calculated two different ways: - A joule is a watt-second - A watt is a joule / second - The power required to hover an object is the same power required to increase the speed of the object from rest, in a weightless environment, to 9.8 m/s in one second. We know this because the pull of gravity is 9.8 meters/second2. - The kinetic energy in an object travelling at 9.8 m/s = 1/2 * m * v2. So for a car of 1000 kg, the energy = 1000 / 2 * 9.82 = 48,020 joules = 48 kilowatts to do this in one second. - This power should be 1/2 the power to raise an object of the same mass, to a height of 9.8 meters in one second, since it would require twice as much energy to do this. - The formula to determining how much energy it takes to raise something to height = E = m * g (gravitational constant) * h = 1000 * 9.8 * 9.8 = 96,040 watts-seconds = 96 kilowatts to do this in one second. So it agrees with the previous result. So, I don't understand how any device could hover an object with the mass of a tonne for less than a theoretical 48 kilowatts. Any thoughts on this would be appreciated. Craig Haynie ( Manchester, NH) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
In reply to Craig Haynie's message of Sun, 10 May 2015 23:43:04 -0400: Hi, IOW he creates a force, but as long as that force doesn't act over a distance, then it need do no work. I'm the one who suggests that the thrust created by the EM Drive could be used to levitate an object. Shawyer is saying that the EM Drive could create 1 tonne of thrust for 1 kilowatt of power, implying that this thrust would be used to accelerate a spacecraft. He's not siting these numbers as an example of levitation. So he's implying that the thrust will be used to do work, and therefore should not be able to violate a theoretical amount of power needed to do that work. ...but he isn't stating how much work is done, and hence how much power would be required. He is just saying that his device even at it's most efficient still requires that some power be expended to create a force, even though in theory no power expenditure is required to create a force, see e.g. gravity , or even a simple spring, which will happily create a constant force, without expending any energy. IOW the (in)efficiency of the device is what causes the power requirement. What I am trying to say is that the power requirement that he gives, is for a device doing no work. If it has to do work as well, then the power requirement will increase accordingly. Consider for a moment the ultimate form of the drive, which is constructed from a perfect superconductor with a consequent infinite Q. As the Q increases so does the force. Or looked at from a different perspective, the power requirement to obtain a given force decreases as the Q increases. IOW in a perfect device, the power requirement would approach zero (as long as no additional work need be done). Which is exactly what a spring does. (And also a current in a superconducting loop BTW.) BTW, IIRC (it was some time ago that I read this) he does say somewhere that the power consumption changes as work is done, and that consequently the limits on the input power also limit the amount of work that can be done. Note also that the tests to date, have been done on stationary devices, i.e. anchored to the work bench, so that they could not move (as I understand it), and hence did no work. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
Ok, well if it is used for static thrust only, it is then a coin toss if it would work opposing gravity as static on the surface of the earth experiences 1G of acceleration. According to the equivalence principle... On Mon, May 11, 2015 at 4:27 PM, Craig Haynie cchayniepub...@gmail.com wrote: Thanks Robin. You're right. He does say that this force of 1 tonne per kilowatt is for 'static thrust'. I found an answer from the website. He is referring specifically to a 'static thrust', not used to do work. The static thrust/power ratio is calculated assuming a superconducting EmDrive with a Q of 5 x 109. This Q value is routinely achieved in superconducting cavities. Note however, because the EmDrive obeys the law of conservation of energy, this thrust/power ratio rapidly decreases if the EmDrive is used to accelerate the vehicle along the thrust vector. (See Equation 16 of the theory paper). Whilst the EmDrive can provide lift to counter gravity, (and is therefore not losing kinetic energy), auxiliary propulsion is required to provide the kinetic energy to accelerate the vehicle. Craig On Mon, May 11, 2015 at 12:19 AM, mix...@bigpond.com wrote: In reply to Craig Haynie's message of Sun, 10 May 2015 23:43:04 -0400: Hi, IOW he creates a force, but as long as that force doesn't act over a distance, then it need do no work. I'm the one who suggests that the thrust created by the EM Drive could be used to levitate an object. Shawyer is saying that the EM Drive could create 1 tonne of thrust for 1 kilowatt of power, implying that this thrust would be used to accelerate a spacecraft. He's not siting these numbers as an example of levitation. So he's implying that the thrust will be used to do work, and therefore should not be able to violate a theoretical amount of power needed to do that work. ...but he isn't stating how much work is done, and hence how much power would be required. He is just saying that his device even at it's most efficient still requires that some power be expended to create a force, even though in theory no power expenditure is required to create a force, see e.g. gravity , or even a simple spring, which will happily create a constant force, without expending any energy. IOW the (in)efficiency of the device is what causes the power requirement. What I am trying to say is that the power requirement that he gives, is for a device doing no work. If it has to do work as well, then the power requirement will increase accordingly. Consider for a moment the ultimate form of the drive, which is constructed from a perfect superconductor with a consequent infinite Q. As the Q increases so does the force. Or looked at from a different perspective, the power requirement to obtain a given force decreases as the Q increases. IOW in a perfect device, the power requirement would approach zero (as long as no additional work need be done). Which is exactly what a spring does. (And also a current in a superconducting loop BTW.) BTW, IIRC (it was some time ago that I read this) he does say somewhere that the power consumption changes as work is done, and that consequently the limits on the input power also limit the amount of work that can be done. Note also that the tests to date, have been done on stationary devices, i.e. anchored to the work bench, so that they could not move (as I understand it), and hence did no work. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
Thanks Robin. You're right. He does say that this force of 1 tonne per kilowatt is for 'static thrust'. I found an answer from the website. He is referring specifically to a 'static thrust', not used to do work. The static thrust/power ratio is calculated assuming a superconducting EmDrive with a Q of 5 x 109. This Q value is routinely achieved in superconducting cavities. Note however, because the EmDrive obeys the law of conservation of energy, this thrust/power ratio rapidly decreases if the EmDrive is used to accelerate the vehicle along the thrust vector. (See Equation 16 of the theory paper). Whilst the EmDrive can provide lift to counter gravity, (and is therefore not losing kinetic energy), auxiliary propulsion is required to provide the kinetic energy to accelerate the vehicle. Craig On Mon, May 11, 2015 at 12:19 AM, mix...@bigpond.com wrote: In reply to Craig Haynie's message of Sun, 10 May 2015 23:43:04 -0400: Hi, IOW he creates a force, but as long as that force doesn't act over a distance, then it need do no work. I'm the one who suggests that the thrust created by the EM Drive could be used to levitate an object. Shawyer is saying that the EM Drive could create 1 tonne of thrust for 1 kilowatt of power, implying that this thrust would be used to accelerate a spacecraft. He's not siting these numbers as an example of levitation. So he's implying that the thrust will be used to do work, and therefore should not be able to violate a theoretical amount of power needed to do that work. ...but he isn't stating how much work is done, and hence how much power would be required. He is just saying that his device even at it's most efficient still requires that some power be expended to create a force, even though in theory no power expenditure is required to create a force, see e.g. gravity , or even a simple spring, which will happily create a constant force, without expending any energy. IOW the (in)efficiency of the device is what causes the power requirement. What I am trying to say is that the power requirement that he gives, is for a device doing no work. If it has to do work as well, then the power requirement will increase accordingly. Consider for a moment the ultimate form of the drive, which is constructed from a perfect superconductor with a consequent infinite Q. As the Q increases so does the force. Or looked at from a different perspective, the power requirement to obtain a given force decreases as the Q increases. IOW in a perfect device, the power requirement would approach zero (as long as no additional work need be done). Which is exactly what a spring does. (And also a current in a superconducting loop BTW.) BTW, IIRC (it was some time ago that I read this) he does say somewhere that the power consumption changes as work is done, and that consequently the limits on the input power also limit the amount of work that can be done. Note also that the tests to date, have been done on stationary devices, i.e. anchored to the work bench, so that they could not move (as I understand it), and hence did no work. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
A reactionless drive tends to break the conservation of energy by just existing. Since there is no equal and opposite energy does not balance, double the velocity would be achieved with double the energy but yield 4 times the stored energy, eventually that leads to excess energy out. Now in the case of a non-moving hover, a reactionless thrust against gravity would not build up any energy, it gains no velocity or height and would not be entirely dissimilar to a superconducting hover, or orbital velocity. Except if it operates the same in free space or in any other direction then yes it would breach the conservation of energy but by default this tends to occur anyway. John On Mon, May 11, 2015 at 3:35 PM, mix...@bigpond.com wrote: In reply to Craig Haynie's message of Sun, 10 May 2015 23:19:42 -0400: Hi, I'm suggesting that in theory no energy is required as long as there is no movement. IOW he creates a force, but as long as that force doesn't act over a distance, then it need do no work. E = F x d; F = m x a. E = m x a x d. You have calculated the mass times acceleration part of it. OTOH a rocket would most definitely expend energy just to hover, as do helicopters etc. but they also accelerate mass downward to produce the thrust (air in the case of helicopters). So I think it just depends on exactly how the thrust is generated, i.e. how the drive interacts with the space-time continuum. His claim is 1 tonne of thrust per kilowatt. One tonne of thrust will accelerate an object. An object under the acceleration of gravity will be countered by the thrust, costing 48 kilowatts of power in the process. This is not the same as suspending an object by a rope or something. Are you suggesting that there is no theoretical limit as to how much power, applied as thrust, is needed to suspend an object weighing a tonne? Or are you suggesting that my math is wrong and that there is a lower number? If the number is lower, then how do you arrive at it? Craig On Sun, May 10, 2015 at 10:48 PM, mix...@bigpond.com wrote: In reply to Craig Haynie's message of Sun, 10 May 2015 18:07:28 -0400: Hi, [snip] It doesn't cost any energy at all to support a car. The ground does this just fine with no energy expenditure. E = F . d. If d = 0, then E = 0. I'm not sure how this applies to an EM drive (if at all), but perhaps it needs to be taken into consideration? Hello! I was hoping the Vorts could help me with this. Roger Shawyer, at minute 2:56 in this video, claims that the next generation EM Drive could generation 1 tonne of thrust per kilowatt of power. This means that a 1 tonne car should be able to hover above the ground for the price of one kilowatt. However, my calculation shows that to be about 48 times a theoretical maximum. Here is the video where he makes the claim at 2:56. http://tinyurl.com/ko5v6h7 But here is my calculation for a theoretical maximum, calculated two different ways: - A joule is a watt-second - A watt is a joule / second - The power required to hover an object is the same power required to increase the speed of the object from rest, in a weightless environment, to 9.8 m/s in one second. We know this because the pull of gravity is 9.8 meters/second2. - The kinetic energy in an object travelling at 9.8 m/s = 1/2 * m * v2. So for a car of 1000 kg, the energy = 1000 / 2 * 9.82 = 48,020 joules = 48 kilowatts to do this in one second. - This power should be 1/2 the power to raise an object of the same mass, to a height of 9.8 meters in one second, since it would require twice as much energy to do this. - The formula to determining how much energy it takes to raise something to height = E = m * g (gravitational constant) * h = 1000 * 9.8 * 9.8 = 96,040 watts-seconds = 96 kilowatts to do this in one second. So it agrees with the previous result. So, I don't understand how any device could hover an object with the mass of a tonne for less than a theoretical 48 kilowatts. Any thoughts on this would be appreciated. Craig Haynie ( Manchester, NH) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
[Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
Hello! I was hoping the Vorts could help me with this. Roger Shawyer, at minute 2:56 in this video, claims that the next generation EM Drive could generation 1 tonne of thrust per kilowatt of power. This means that a 1 tonne car should be able to hover above the ground for the price of one kilowatt. However, my calculation shows that to be about 48 times a theoretical maximum. Here is the video where he makes the claim at 2:56. http://tinyurl.com/ko5v6h7 But here is my calculation for a theoretical maximum, calculated two different ways: - A joule is a watt-second - A watt is a joule / second - The power required to hover an object is the same power required to increase the speed of the object from rest, in a weightless environment, to 9.8 m/s in one second. We know this because the pull of gravity is 9.8 meters/second2. - The kinetic energy in an object travelling at 9.8 m/s = 1/2 * m * v2. So for a car of 1000 kg, the energy = 1000 / 2 * 9.82 = 48,020 joules = 48 kilowatts to do this in one second. - This power should be 1/2 the power to raise an object of the same mass, to a height of 9.8 meters in one second, since it would require twice as much energy to do this. - The formula to determining how much energy it takes to raise something to height = E = m * g (gravitational constant) * h = 1000 * 9.8 * 9.8 = 96,040 watts-seconds = 96 kilowatts to do this in one second. So it agrees with the previous result. So, I don't understand how any device could hover an object with the mass of a tonne for less than a theoretical 48 kilowatts. Any thoughts on this would be appreciated. Craig Haynie ( Manchester, NH)
Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
His claim is 1 tonne of thrust per kilowatt. One tonne of thrust will accelerate an object. An object under the acceleration of gravity will be countered by the thrust, costing 48 kilowatts of power in the process. This is not the same as suspending an object by a rope or something. Are you suggesting that there is no theoretical limit as to how much power, applied as thrust, is needed to suspend an object weighing a tonne? Or are you suggesting that my math is wrong and that there is a lower number? If the number is lower, then how do you arrive at it? Craig On Sun, May 10, 2015 at 10:48 PM, mix...@bigpond.com wrote: In reply to Craig Haynie's message of Sun, 10 May 2015 18:07:28 -0400: Hi, [snip] It doesn't cost any energy at all to support a car. The ground does this just fine with no energy expenditure. E = F . d. If d = 0, then E = 0. I'm not sure how this applies to an EM drive (if at all), but perhaps it needs to be taken into consideration? Hello! I was hoping the Vorts could help me with this. Roger Shawyer, at minute 2:56 in this video, claims that the next generation EM Drive could generation 1 tonne of thrust per kilowatt of power. This means that a 1 tonne car should be able to hover above the ground for the price of one kilowatt. However, my calculation shows that to be about 48 times a theoretical maximum. Here is the video where he makes the claim at 2:56. http://tinyurl.com/ko5v6h7 But here is my calculation for a theoretical maximum, calculated two different ways: - A joule is a watt-second - A watt is a joule / second - The power required to hover an object is the same power required to increase the speed of the object from rest, in a weightless environment, to 9.8 m/s in one second. We know this because the pull of gravity is 9.8 meters/second2. - The kinetic energy in an object travelling at 9.8 m/s = 1/2 * m * v2. So for a car of 1000 kg, the energy = 1000 / 2 * 9.82 = 48,020 joules = 48 kilowatts to do this in one second. - This power should be 1/2 the power to raise an object of the same mass, to a height of 9.8 meters in one second, since it would require twice as much energy to do this. - The formula to determining how much energy it takes to raise something to height = E = m * g (gravitational constant) * h = 1000 * 9.8 * 9.8 = 96,040 watts-seconds = 96 kilowatts to do this in one second. So it agrees with the previous result. So, I don't understand how any device could hover an object with the mass of a tonne for less than a theoretical 48 kilowatts. Any thoughts on this would be appreciated. Craig Haynie ( Manchester, NH) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Nextgen EM Drive's Potential seems way above the Theoretical Limit
In reply to Craig Haynie's message of Sun, 10 May 2015 18:07:28 -0400: Hi, [snip] It doesn't cost any energy at all to support a car. The ground does this just fine with no energy expenditure. E = F . d. If d = 0, then E = 0. I'm not sure how this applies to an EM drive (if at all), but perhaps it needs to be taken into consideration? Hello! I was hoping the Vorts could help me with this. Roger Shawyer, at minute 2:56 in this video, claims that the next generation EM Drive could generation 1 tonne of thrust per kilowatt of power. This means that a 1 tonne car should be able to hover above the ground for the price of one kilowatt. However, my calculation shows that to be about 48 times a theoretical maximum. Here is the video where he makes the claim at 2:56. http://tinyurl.com/ko5v6h7 But here is my calculation for a theoretical maximum, calculated two different ways: - A joule is a watt-second - A watt is a joule / second - The power required to hover an object is the same power required to increase the speed of the object from rest, in a weightless environment, to 9.8 m/s in one second. We know this because the pull of gravity is 9.8 meters/second2. - The kinetic energy in an object travelling at 9.8 m/s = 1/2 * m * v2. So for a car of 1000 kg, the energy = 1000 / 2 * 9.82 = 48,020 joules = 48 kilowatts to do this in one second. - This power should be 1/2 the power to raise an object of the same mass, to a height of 9.8 meters in one second, since it would require twice as much energy to do this. - The formula to determining how much energy it takes to raise something to height = E = m * g (gravitational constant) * h = 1000 * 9.8 * 9.8 = 96,040 watts-seconds = 96 kilowatts to do this in one second. So it agrees with the previous result. So, I don't understand how any device could hover an object with the mass of a tonne for less than a theoretical 48 kilowatts. Any thoughts on this would be appreciated. Craig Haynie ( Manchester, NH) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]: Rossi annnounces a new Hot Cat long term test --
It seems Rossi has learned from the Lugano test and now is running a 1-year Hot Cat test alongside the commercial 1Mw unit in NC or whereever it is (per comments today on Rossi Blog Reader. Rossi indcicates that the SSM is easier to achieve with the Hot Cat as has been suggested by Parkhomov and Lugano. He suggests that the domistic units will be easier to design and operate since they can be made smaller and still achieve a SSM. Commercial development is expanding it seems. Bob Cook
[Vo]:the dispute re LENR lasted too long!
See please: http://egooutpeters.blogspot.ro/2015/05/the-longest-lenr-dispute-unexpected.html Please tell me your own conclusions. Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
