Alpha Centauri Explorer II

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vidargander

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> <span><font face="Times New Roman"><font size="2">In step 3 it says "Here's the magic".&nbsp; That is correct, it is magic and not true physics.&nbsp; The spring is 3x as efficient in transmitting the force ?&nbsp; Nope.&nbsp; &hellip;&hellip;.&nbsp; There will be a net force that will cancel the motion as the weights are brought back to center. </font></font></span></p><p><span><font face="Times New Roman"><font size="2">But let me tease you with one, similar in nature, that's not so easily disproven. &hellip;&hellip;. &nbsp;This goes on ad nasuem turning electricity into motion w/o expelling mass.&nbsp; Violates conservation of momentum but why won't it work .....</font></font></span><br />Posted by mee_n_mac</DIV><span><font face="Times New Roman" size="2">&nbsp;</font></span></p><p><span><font face="Times New Roman"><font size="2">Please explain &hellip;.. </font></font></span><span><font face="Times New Roman"><font size="2">An image would help to see the answer.</font></font></span><span><font face="Times New Roman" size="2">&nbsp;</font></span></p><p><span><font face="Times New Roman"><font size="2">Anyhow, I still think that it is possible to make some of the thrust convert to other kinds of energy, like vibrations and heat. Still I think that it&rsquo;s possible to make it larger in the backwards thrust than the forward thrust. Then the net result will be forward motion in the frictionless space.</font></font></span><span><font face="Times New Roman" size="2">&nbsp;</font></span> </p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><span><font face="Times New Roman"><font size="2">Can you also argue why any such differential thrust is impossible?</font></font></span></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal">&nbsp;</p><span><font face="Times New Roman"><font size="2">&nbsp;</font></font></span> <p><br /><br />&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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MeteorWayne

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;Please explain &hellip;.. An image would help to see the answer.&nbsp;Anyhow, I still think that it is possible to make some of the thrust convert to other kinds of energy, like vibrations and heat. Still I think that it&rsquo;s possible to make it larger in the backwards thrust than the forward thrust. Then the net result will be forward motion in the frictionless space.&nbsp; Can you also argue why any such differential thrust is impossible?&nbsp;&nbsp; &nbsp; <br />Posted by vidargander</DIV><br /><br />It's really rather simple, comservation of momentum</p><p>You get nothing for free in the universe, that just seems to be the rule. </p><p>In your example, remember compressing a spring is not lossless either. It heats the spring. So whatever energy you use to compress the spring (even assuming that energy is delivered at 100% efficiency, which it won't be) will be lost.</p> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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vidargander

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>It's really rather simple, comservation of momentumYou get nothing for free in the universe, that just seems to be the rule. In your example, remember compressing a spring is not lossless either. It heats the spring. So whatever energy you use to compress the spring (even assuming that energy is delivered at 100% efficiency, which it won't be) will be lost. <br /><font size="2">Posted by MeteorWayne</DIV><br /><br /></font><span><font face="Times New Roman"><font size="2">Yes, I have heard it&nbsp;many times before: Newton's Third Law states that for every action there is an equal and opposite reaction.</font></font></span><span><font face="Times New Roman" size="2">&nbsp;</font></span></p><p><span><font face="Times New Roman"><font size="2">But let&rsquo;s face it. Isaac Newton was no rocket scientist either. In fact, he was rather ignorant when it comes to electro magnetism too.</font></font></span></p> <div class="Discussion_UserSignature"> </div>
 
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MeteorWayne

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Yes, I have heard it&nbsp;many times before: Newton's Third Law states that for every action there is an equal and opposite reaction.&nbsp;But let&rsquo;s face it. Isaac Newton was no rocket scientist either. In fact, he was rather ignorant when it comes to electro magnetism too. <br />Posted by vidargander</DIV><br /><br />Which has nothing whatsoever to do with the universal fact that momentum is consereved, and all energy transfers are less than 100% efficient.</p><p>Give me a "P", errr "U"</p><p>You have gone off the deep end my friend. When this thread was about ACE, it belonged here. I was happy to contribute.Now that it has descended into perpetual motion, and lossless energy, it belongs in the dustbin.</p><p>Wayne</p> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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hal9891

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<p>All these reactionless drives remind me of baron Munchausen raising himself (and his horse!) out of water by pulling up his hair <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" /></p><p><br /> <img src="http://sitelife.space.com/ver1.0/Content/images/store/6/7/468301ab-2da4-4c71-873c-dda644a83085.Medium.jpg" alt="" /><br />&nbsp;</p> <div class="Discussion_UserSignature"> <div style="text-align:center"><font style="color:#808080" color="#999999"><font size="1">"I predict that within 100 years computers will be twice as powerful, 10000 times larger, and so expensive that only the five richest kings of Europe will own them"</font></font><br /></div> </div>
 
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MeteorWayne

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>All these reactionless drives remind me of baron Munchausen raising himself (and his horse!) out of water by pulling up his hair &nbsp; <br />Posted by hal9891</DIV><br /><br />Or pulling oneself up by&nbsp;one's own bootstraps ;) <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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geegel

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<p>I'm not an engineer, but the idea of fuelless drive seems the most compelling to me.</p><p>The best option I think would be to make this ship exit the solar system at a near of light speed. You could achieve this through a combination of gravitational tugs and laser propulsion. For the latter I envision a series of sun powered laser platforms orbiting the sun. They would make sense in the long run since they could very well help other space projects as well. Imagine how much money would be saved if you wouldn't have to pack the fuel to get around the solar system. I envision the acceleration stage to last somewhere between 3 to 5 years (this is still going to be a long mission) </p><p>The problem though will be in keeping the ship/probe within the solar system. But again I'm not an engineer, so maybe you guys could give some input.</p><p>Once the probe leaves the solar system, it would go driven by inertia. Once the Alpha Centauri system is reached, the probe would unfurl a huge set of solar sails. Given that the probe is moving towards the star. they would act as a brake. The speed with which the probe would orbit the Beta Centauri (I think this was the target in the Longshot project), would be very high, but still much lower than the peak speed achieved by the probe. The goal here is to slow down to a speed so measurements can be made. If solar sails are not enough to brake, ion drives could be used in a dual system.</p><p>I think that the absence of fuel would transform this into a much more viable project, less mass equals less energy needed. Also all the technologies described are well within reach of our current stage of technological evolution.</p><p>Thoughts anyone?</p><p>P.S. This has been my very first post. A big hello to the community here from a long time lurker. &nbsp; &nbsp; &nbsp; </p>
 
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keermalec

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I can.&nbsp; In step 3 it says "Here's the magic".&nbsp; That is correct, it is magic and not true physics.&nbsp; The spring is 3x as efficient in transmitting the force ?&nbsp; Nope.&nbsp; Assuming the mass of the tube setup is />> larger than any one of the moving weights. When one weight (#1)&nbsp;hits the spring near it's end, it'll start to force the tube to move. The spring will arrive at some point of compression and then send the weight moving back. So right here not all the energy in that moving weight is transmitted to the tube. At the the other end the weight smacks the tube and let's say imparts all it's energy. This will stop the tube and start it moving back the other way. Thus it comes to a stop, like a billiard ball hitting another. The tube is now in motion as the forces weren't balanced. However the weight is still stuck at the far end of the tube. The energy and forces needed to get both weights back to the center and recompress the spring aren't equal as one is moving and one isn't.&nbsp; There will be a net force that will cancel the motion as the weights are brought back to center. But let me tease you with one, similar in nature, that's not so easily disproven.&nbsp; I think I saw it on the old Uplink.&nbsp; Let's have 2 spaceships separated by some fair distance, a few light seconds at least. Each one has a magnetic feild generator that can turn the ship into the equivalent of a bar magnet with north and south poles. We line the ships up nose to tail and have the trailing ship turn on it's electromagnet for 2 seconds.&nbsp; The feild propagates out towards the lead ship who, just a microsecond before the feild arrives, turns on his electromagnet (for 2 secs) such that he gets a push for the 2 seconds the feild exists (he makes his tail south to oppose the north side of the trailing ship). The lead ship moves forward.&nbsp;Now that 2'nd feild propagates back towards the trailing ship but his electromagnet switches polarity so instead of getting pushed, he gets a pull. This new, reversed feild propagates towards the lead ship again &nbsp;and catches it. But just before it does, he switched polarty as well and gets a push, not a pull.&nbsp; This goes on ad nasuem turning electricity into motion w/o expelling mass.&nbsp; Violates conservation of momentum but why won't it work .....&nbsp; <br />Posted by Mee_n_Mac</DIV></p><p>I think I can argue with this one. Changing polarity will not make the two ships accelerate each other, it will make them move apart and together again, alternately. Their common center of mass, however, will not budget a milimeter...</p><p>Actually, its easy to work out why near-lightspeed is simply not attainable with present-day technology.&nbsp;&nbsp;</p><p>1. To accelerate 1 kg to lightspeed requires 4.5x10E16 Joules at least (E = 1/2.mv^2) assuming 100% energy conversion efficiency.</p><p>2. The energy contained in 1 kg of matter is mv^2. Therefore, you will need to convert at least half the ship's mass to pure energy to accelerate it to lightspeed.</p><p>3. Therefore you will need as much antimatter as matter on your ship to approach lightspeed, still assuming perfect conversion efficiency.</p><p>4. Therefore you cannot attain lightspeed, even with antimatter, because efficiency is never 100%.</p><p>I would say the only way of even daring to dream of attaining near light speeds would be with an external energy source, such as a gigantic laser on the Earth and a light sail on the ship. However, even giant lasers&nbsp;decrease in&nbsp;power with distance and therefore initial acceleration must be enormous to obtain an interesting final speed.</p><p><br /><br />&nbsp;</p> <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>
 
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vidargander

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Actually, its easy to work out why near-lightspeed is simply not attainable with present-day technology.&nbsp;&nbsp;1. To accelerate 1 kg to lightspeed requires 4.5x10E16 Joules at least (E = 1/2.mv^2) assuming 100% energy conversion efficiency.2. The energy contained in 1 kg of matter is mv^2. Therefore, you will need to convert at least half the ship's mass to pure energy to accelerate it to lightspeed.3. Therefore you will need as much antimatter as matter on your ship to approach lightspeed, still <font size="2"><font size="1">assuming perfect conversion efficiency.4. Therefore you cannot attain lightspeed, even with antimatter, because efficiency is never 100%.<br />Posted by keermalec</DIV></font><br /><br /></font><p style="margin:0cm0cm0pt" class="MsoNormal"><span><font face="Times New Roman"><font size="2">That true, if E=mc^2 represent the total energy of the mass.</font></font></span></p><p style="margin:0cm0cm0pt" class="MsoNormal"><span><font face="Times New Roman"><font size="2">However, there is no proof of that.</font></font></span></p><span><font face="Times New Roman"><font size="2">(here we go again ... )</font></font></span> <div class="Discussion_UserSignature"> </div>
 
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MeteorWayne

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>That true, if E=mc^2 represent the total energy of the mass.However, there is no proof of that.(here we go again ... ) <br />Posted by vidargander</DIV></p><p>&nbsp;</p><p>It has been repeatedly proven in accelarators all over the planet, and in those nuclear weapons, if you want a powerful example.<br /><br />If you are going to argue that, maybe you should open a thread in the Unexplained....</p> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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baulten

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;It has been repeatedly proven in accelarators all over the planet, and in those nuclear weapons, if you want a powerful example.If you are going to argue that, maybe you should open a thread in the Unexplained.... <br /> Posted by MeteorWayne</DIV></p><p>Yeah seriously, if you're not going to debate this under the assumptions that what we know about conservation and energy laws is true, there's not much point.&nbsp;</p>
 
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Mee_n_Mac

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I think I can argue with this one. Changing polarity will not make the two ships accelerate each other, it will make them move apart and together again, alternately. Their common center of mass, however, will not budget a milimeter... Posted by <strong>keermalec</strong></DIV><br /><br />Not saying your right nor wrong but (absent a diagram) I think you've grasped the basic idea. Now tell me how, if all forces are limited to the speed of light, the leading ship's reaction to the magnetic feild affects the trailing ship if/when it reverses polarity.</p><p>Let me simplify the scenario.&nbsp; As before the trailing ship turns on it's magnetic generator.&nbsp; Some 2 seconds later the feild reaches the lead ship which results in a push.&nbsp;Now 1.9+&nbsp;seconds later the trailing ship shuts off it's generator and becomes magnetically inert. In this scenario it appears that the leading ship got a push but before Newton could give the trailing ship it's equal and opposite response, it went neutral.&nbsp; Thus momentum seems to have come from nowhere .....</p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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MeteorWayne

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Not saying your right nor wrong but (absent a diagram) I think you've grasped the basic idea. Now tell me how, if all forces are limited to the speed of light, the leading ship's reaction to the magnetic feild affects the trailing ship if/when it reverses polarity.Let me simplify the scenario.&nbsp; As before the trailing ship turns on it's magnetic generator.&nbsp; Some 2 seconds later the feild reaches the lead ship which results in a push.&nbsp;Now 1.9+&nbsp;seconds later the trailing ship shuts off it's generator and becomes magnetically inert. In this scenario it appears that the leading ship got a push but before Newton could give the trailing ship it's equal and opposite response, it went neutral.&nbsp; Thus momentum seems to have come from nowhere ..... <br />Posted by Mee_n_Mac</DIV></p><p>In any case, lots of energy got used at less than 100% efficiency, so it ain't free.</p><p>Is it the moste efficient? I think not.</p> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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dryson

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> <p>You would have to consider the total mass required to do this. The thrust of an ion engine is infantisimal. It's the steady push that builds speed over time, but to build speeds to the realm you are interested in would take thousands of years and many thousand pounds or reactant. The Bussard collector is a prime example, it collects the ions needed as it goes, the problem then becomes how do you inonize the Hydrogen and various atomic particals scooped in? Solar would work out to the heliopause, though how well outside the major Planets is a question. A scoop the size of&nbsp; Earth and solar panels the size of Jupiter might work, but you still have to get them into Space to begin with.</p><p>I see no reason a Solar collector can't also act as a sail, but the scale required to get to even a relatively close system is beyond reason. What we need is a leap in technolgy akin to the Spanish introducing horses to the Americas. That they failed to introduce the wheel at the same time is another thing.</p><p></DIV></p><p>&nbsp;</p><p>Try this on for size then with the solar sail. What about using a magnet that would pull the elements in space that are being ejected by the sun? I am looking at the paramagentic propeties here. If an external magnetic force is applied to an element then it's paramagnetic properties must follow these rules:</p><p>1.An externaly applied magentic force must be present. </p><p>2.Paramagnetic elements are attracted to magnetic fields.</p><p>3.Elements must have a permeability greater then one to be effected.</p><p>Imagine this. On the left you have a magnet a certain measure from the solar sail in the center. On the right you have the elements being ejected from solar bodies at the speed of light and also a certain measure from the&nbsp;sail in the center.&nbsp;The magnet would attract the elemets in atom form to the magnetic field. The&nbsp;solar sail would catch the&nbsp;"wind". The sail&nbsp;would have to have properties that would not allow the paramagneticly pulled atoms to pass through the sail. This would cause the effect of the sail being filled with paramagnetic atoms that are traveling at the speed of ligh or (SOL). The craft would then start to pick up velocity based on the&nbsp;strength of the magnetic&nbsp;and the length of the EM field being generated.&nbsp; A longer EM wavelength would gather more paramagnetic atoms to the sail at a higher rate thus creating more velocity, a shorter EM wavelength would not gather as much paramagnetic atoms to the sail, thus reducing the foward velocity of the craft.</p><p>In essence it would like the Bussard Collector but would pull atoms to a sail instead of conditioning the&nbsp;atoms to be used for&nbsp;fuel.&nbsp;</p><p>Going in reverse? That's a whole other story.</p>
 
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vidargander

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> <span style="font-size:7.5pt;font-family:Verdana">It has been repeatedly proven in accelarators all over the planet, and in those nuclear weapons, if you want a powerful example.If you are going to argue that, maybe you should open a thread in the Unexplained.... <br />Posted by MeteorWayne</span> <p>Yeah seriously, if you're not going to debate this under the assumptions that what we know about conservation and energy laws is true, there's not much point.&nbsp; <br />Posted by baulten</DIV></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><span><font face="Times New Roman"><font size="2">The argument was a respond to keermalec on how much the total energy can be when released&nbsp;in a matter/anti-matter reaction.</font></font></span></p><span><font face="Times New Roman" size="2">&nbsp;</font></span><span><font face="Times New Roman" size="2">It doesn&rsquo;t make sense that nuclear weapons are proof that E=mc^2, and represent the total energy in atoms too. Take a look at the images showing the fission and fusion chain reactions; </font><font face="Times New Roman" size="2">http://en.wikipedia.org/wiki/Nuclear_fission#Chain_reactions</font>&nbsp;&nbsp; </span><span><font face="Times New Roman" size="2">http://en.wikipedia.org/wiki/Nuclear_fusion#Important_reactions</font></span> <p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><span><font face="Times New Roman"><font size="2">How much of the matter was converted to energy? None at all. Fission and fusion reaction is about releasing binding energy to some atom&rsquo;s elements.</font></font></span></p><span><font face="Times New Roman" size="2">&nbsp;</font></span> <p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><span><font face="Times New Roman" size="2">Still my point is: If nuclear weapons are the proof of E=mc^2, then the total energy released from matter to pure energy&nbsp;must be far greater than E=mc^2.</font></span></p> <div class="Discussion_UserSignature"> </div>
 
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keermalec

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>That true, if E=mc^2 represent the total energy of the mass.However, there is no proof of that.(here we go again ... ) <br />Posted by vidargander</DIV><br /><br />There never is absolute proof in science, but this equation has been demonstrated to be very very reliable. For example, in a nuclear explosion, the energy released is equal to the mass lost according to this equation. Coincidence?</p><p>With all due respect, <font color="#000000">Vidargander,</font> Einstein's equation is infinitely less questionable than some of the hairbrained theories you have posted on this thread. If you want to&nbsp;question the complete shared heritage of scientific knowledge since Newton all by yourself, good luck to you but IMHO you should do some reading first to understand what you're trying to disprove.</p><p>&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>
 
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keermalec

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>How much of the matter was converted to energy? None at all. <br />Posted by vidargander</DIV><br /><br />Quote from Wikipedia:</p><p><em>The total rest masses of the fission products (<strong>Mp</strong>) from a single reaction is less than the mass of the original fuel nucleus (<strong>M</strong>). The excess mass <strong>&Delta;m</strong> = <strong>M</strong> - <strong>Mp</strong> is the </em><em>invariant mass</em><em> of the energy that is released as </em><em>photons</em><em> (</em><em>gamma rays</em><em>) and kinetic energy of the fission fragments, according to the </em><em>mass-energy equivalence</em><em> formula E&nbsp;=&nbsp;mc&sup2;.</em></p><p>&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I found an image that might explain a similar mechanical impulse effect that cause forward motion. It&rsquo;s called the &ldquo;Henry Bull&rsquo;s Impulse Engine of 1935&rdquo; http://jnaudin.free.fr/html/hbimp35.htmIn the ElectroMagnetic propulsion I propose, the magnet in the front should be attached to several springs, or elastic material, that will compress by the trust. That should make a net forward thrust that will move the ship forwards. Can anyone argue with that? <br />Posted by vidargander</DIV></p><p>Easily.&nbsp; You cannot change the net momentum of a closed, isolated&nbsp;system.&nbsp; Period.&nbsp; The notion that a spring is somehow "more efficient' that impact of a flat plate is utterly meaningless.<br /></p> <div class="Discussion_UserSignature"> </div>
 
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vidargander

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Quote from Wikipedia:The total rest masses of the fission products (Mp) from a single reaction is less than the mass of the original fuel nucleus (M). The excess mass &Delta;m = M - Mp is the invariant mass of the energy that is released as photons (gamma rays) and kinetic energy of the fission fragments, according to the mass-energy equivalence formula E&nbsp;=&nbsp;mc&sup2;.&nbsp;&nbsp; <br />Posted by keermalec</DIV></p><p><span><font face="Times New Roman" size="2">&nbsp;</font></span><span><font face="Times New Roman"><font size="2">Both the terms &lsquo;Binding Energy&rsquo; and &lsquo;Rest Mass&rsquo;/&rsquo;Invariant Mass&rsquo; are used on the same energy release. It&rsquo;s calculated to be about 0.1184% of the total mass. However, that is the release of both the binding energy and conversion of mass to energy.</font></font></span><span><font face="Times New Roman" size="2">http://en.wikipedia.org/wiki/Binding_energy#Specific_quantitative_example:_a_deuteron</font></span><span><font face="Times New Roman" size="2">&nbsp;</font></span></p><p><span><font face="Times New Roman"><font size="2">None is not an exact number. In this case None is next to nothing, or to be more precise; 0.1184%. </font></font></span><span><font face="Times New Roman"><font size="2">Still, that tiny fraction is far from proof how much energy all the whole&nbsp;mass can release in a matter/anti-matter reaction.</font></font></span> </p><p><br /><br />&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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keermalec

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;Both the terms &lsquo;Binding Energy&rsquo; and &lsquo;Rest Mass&rsquo;/&rsquo;Invariant Mass&rsquo; are used on the same energy release. It&rsquo;s calculated to be about 0.1184% of the total mass. However, that is the release of both the binding energy and conversion of mass to energy.http://en.wikipedia.org/wiki/Binding_energy#Specific_quantitative_example:_a_deuteronNone is not an exact number. In this case None is next to nothing, or to be more precise; 0.1184%. Still, that tiny fraction is far from proof how much energy all the whole&nbsp;mass can release in a matter/anti-matter reaction. &nbsp; <br />Posted by vidargander</DIV><br /><br />0.1184% of 1 kg is 0.001184 kg, right? 0.001184 x 300000000^2 = 1.0656 x 10^14&nbsp;Joules (E = mc^2). That is the amount of energy liberated by fissioning 1 kg of fission mass, as predicted by Eisntein's equation. It also happens to be what is measured when fission really occurs. Which part of this proof do you not believe in?</p><p>&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>
 
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vidargander

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>0.1184% of 1 kg is 0.001184 kg, right? 0.001184 x 300000000^2 = 1.0656 x 10^14&nbsp;Joules (E = mc^2). That is the amount of energy liberated by fissioning 1 kg of fission mass, as predicted by Eisntein's equation. It also happens to be what is measured when fission really occurs. Which part of this proof do you not believe in?&nbsp;&nbsp; <br />Posted by keermalec</DIV><br /><br /><span><font face="Times New Roman"><font size="2">The documentation - that&rsquo;s missing.</font></font></span></p><p><span><font face="Times New Roman"><font size="2">It is like the other stories about the atom clocks, the GPS satellites and Mercury&rsquo;s orbit etc. Those so-called proofs fade away when confronted with scientific demand for documentation.</font></font></span><span><font face="Times New Roman" size="2">&nbsp;</font></span></p><p><span><font face="Times New Roman"><font size="2">Wikipedia isn&rsquo;t any 100% reliable sources. </font></font></span><span><font face="Times New Roman"><font size="2">Actually, I could write in Wikipedia something about stating my scepticism that E=mc^2 represent the total energy of mass. </font></font></span><span><font face="Times New Roman"><font size="2">You wouldn&rsquo;t blindly believe that, just because Wikipedia tells you so, - would you?</font></font></span></p><p><span><font face="Times New Roman"><font size="2">I could even add that&nbsp;E=mc^2 is&nbsp;merely Newton&rsquo;s formula for kinetic energy at light-speed; E=(1/2)mc^2, multiplied with 2.</font></font></span></p> <div class="Discussion_UserSignature"> </div>
 
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DrRocket

Guest
<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The documentation - that&rsquo;s missing.It is like the other stories about the atom clocks, the GPS satellites and Mercury&rsquo;s orbit etc. Those so-called proofs fade away when confronted with scientific demand for documentation.&nbsp;Wikipedia isn&rsquo;t any 100% reliable sources. Actually, I could write in Wikipedia something about stating my scepticism that E=mc^2 represent the total energy of mass. You wouldn&rsquo;t blindly believe that, just because Wikipedia tells you so, - would you?I could even add that&nbsp;E=mc^2 is&nbsp;merely Newton&rsquo;s formula for kinetic energy at light-speed; E=(1/2)mc^2, multiplied with 2. <br />Posted by vidargander</DIV></p><p>What documentation is missing ?</p><p>If you were, as you say to "add that E=mc^s is merely Newton's formula for kinetic energy at light-speed; E+(1/2)mc^2, multiplied by 2" you would simply demonstrating a complete lack of comprehension of the physics involved.&nbsp; Einstein's equation appllies even at rest, and has nothing to do with kinetic energy in Newton's sense.&nbsp; In fact, if you want to compute kinetic energy relativisticly what you do is compare energies using Einsteins equation, but apply it to the rest mass and the mass increased by the relativistic factor gamma and then take the difference.</p><p>These so-called proofs hardly fade in the face of demands for documentation.&nbsp; They are documented in the literature and in text books.&nbsp; If you can't find the documentation indicates that either you have not looked hard enough or you do not know enough to recognize the&nbsp;supporting evidence when you see it.<br /></p> <div class="Discussion_UserSignature"> </div>
 
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baulten

Guest
<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>What documentation is missing ?If you were, as you say to "add that E=mc^s is merely Newton's formula for kinetic energy at light-speed; E+(1/2)mc^2, multiplied by 2" you would simply demonstrating a complete lack of comprehension of the physics involved.&nbsp; Einstein's equation appllies even at rest, and has nothing to do with kinetic energy in Newton's sense.&nbsp; In fact, if you want to compute kinetic energy relativisticly what you do is compare energies using Einsteins equation, but apply it to the rest mass and the mass increased by the relativistic factor gamma and then take the difference.These so-called proofs hardly fade in the face of demands for documentation.&nbsp; They are documented in the literature and in text books.&nbsp; If you can't find the documentation indicates that either you have not looked hard enough or you do not know enough to recognize the&nbsp;supporting evidence when you see it. <br /> Posted by DrRocket</DIV></p><p>No matter what documentation we posted, I figure he'd still claim it isn't really proof. </p>
 
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vidargander

Guest
<span style="font-size:7.5pt;font-family:Verdana"><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>What documentation is missing ?If you were, as you say to "add that E=mc^s is merely Newton's formula for kinetic energy at light-speed; E+(1/2)mc^2, multiplied by 2" you would simply demonstrating a complete lack of comprehension of the physics involved.&nbsp; Einstein's equation appllies even at rest, and has nothing to do with kinetic energy in Newton's sense.&nbsp; In fact, if you want to compute kinetic energy relativisticly what you do is compare energies using Einsteins equation, but apply it to the rest mass and the mass increased by the relativistic factor gamma and then take the difference.These so-called proofs hardly fade in the face of demands for documentation.&nbsp; They are documented in the literature and in text books.&nbsp; If you can't find the documentation indicates that either you have not looked hard enough or you do not know enough to recognize the&nbsp;supporting evidence when you see it. <br />Posted by DrRocket</DIV></span> <font size="1">Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>No matter what documentation we posted, I figure he'd still claim it isn't really proof. Posted by baulten</DIV><br /></font><font size="2"><span><font face="Times New Roman">Really?</font></span><span><font face="Times New Roman">&nbsp;</font></span></font> <p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><span><font face="Times New Roman"><font size="2">I don&rsquo;t think it matter much if you can convince me or not. It should be more important for you&nbsp;if you could convince anybody at all without reference to any scientifically solid documentation.</font></font></span></p><span><font face="Times New Roman"><font size="2">We disagree, - and that&rsquo;s OK. <span>&nbsp;</span>I don&rsquo;t expect you to agree to my scientism that the total energy of mass can be accurately calculated just by speed.</font></font></span><span><font face="Times New Roman" size="2">&nbsp;</font></span><span><font face="Times New Roman"><font size="2">We know that there are many more forms of energy, - like mechanical, thermal, electric, electromagnetic, chemical and nuclear energy. I find it obvious that the total energy of mass is a far more complex formula including these energy forms too. I think we still have to search for the full formula for total energy, and not stay stuck in such a cold-war dogma. </font></font></span><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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MeteorWayne

Guest
Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> Really?&nbsp; I don&rsquo;t it matter much if you can convince me or not, - it should be more important if you could convince anybody without reference to any scientific documentation.We disagree, - and that&rsquo;s OK. &nbsp;I don&rsquo;t expect you to agree to my scientism that the total energy of mass can be accurately calculated just by speed.&nbsp;We know that there are many more forms of energy, - like mechanical, thermal, electric, electromagnetic, chemical and nuclear energy. I find it obvious that the total energy of mass is a far more complex formula including these energy forms too. I think we still have to search for the full formula for total energy, and not stay stuck in such a cold-war dogma. &nbsp; <br />Posted by vidargander</DIV><br /><br />That's it. Gimme a "P"....errr "U" <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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