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#840 by PotomacNeuron on 14 Aug, 2017 19:51
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Hi PN,
Roger clearly shows now, using EmDrive reference frame velocity changes, CofE is conserved.
So no OU.
TT, I do not have time and incentive to read his new equations at this time. I will read when enough skeptical people say they are correct. Thanks. -
#841 by TheTraveller on 14 Aug, 2017 20:15
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Hi PN,
Roger clearly shows now, using EmDrive reference frame velocity changes, CofE is conserved.
So no OU.
TT, I do not have time and incentive to read his new equations at this time. I will read when enough skeptical people say they are correct. Thanks.
Hi PN,
The equations are very simple.
Just click here:
https://forum.nasaspaceflight.com/index.php?topic=42978.msg1712726#msg1712726 -
#842 by meberbs on 14 Aug, 2017 20:34
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Hi PN,
Roger clearly shows now, using EmDrive reference frame velocity changes, CofE is conserved.
So no OU.
TT, I do not have time and incentive to read his new equations at this time. I will read when enough skeptical people say they are correct. Thanks.
Hi PN,
The equations are very simple.
Just click here:
https://forum.nasaspaceflight.com/index.php?topic=42978.msg1712726#msg1712726
If you mean "simply wrong" then you would be correct.
The equation for Energy in is simply P*t, not what Shawyer wrote. What he wrote for Ein is just another way of writing the same formula as Eout, making the rest of which a fancy way of saying 1=1. (The final equation has 2 variables, Fd, and Fs that both refer to the same quantity.)
The equation at the end of the first slide you posted is a correct expression for Eout = 0.5*P*Ts*t*Vt.
Divide this by the actual input energy, P*t, and you get 0.5*Ts*Vt. This clearly shows there is a velocity for which any given value of Ts results in over unity. For the values of Ts Shawyer has claimed recently, the Vt that results in over unity is small.
(Relativistic analysis agrees with this conclusion as well, and it is easy to see how relativity allows photon thrusters, because you can never accelerate past c, so a sufficiently small Ts is acceptable.) -
#843 by PotomacNeuron on 14 Aug, 2017 20:49
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Hi PN,
Roger clearly shows now, using EmDrive reference frame velocity changes, CofE is conserved.
So no OU.
TT, I do not have time and incentive to read his new equations at this time. I will read when enough skeptical people say they are correct. Thanks.
Hi PN,
The equations are very simple.
Just click here:
https://forum.nasaspaceflight.com/index.php?topic=42978.msg1712726#msg1712726
TT, I took a look and I basically agree with meberbs' post below yours. -
#844 by moreno7798 on 14 Aug, 2017 20:57
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Hi PN,
Roger clearly shows now, using EmDrive reference frame velocity changes, CofE is conserved.
So no OU.
TT, I do not have time and incentive to read his new equations at this time. I will read when enough skeptical people say they are correct. Thanks.
I question your motives. -
#845 by moreno7798 on 14 Aug, 2017 21:06
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Hi PN,
Roger clearly shows now, using EmDrive reference frame velocity changes, CofE is conserved.
So no OU.
TT, I do not have time and incentive to read his new equations at this time. I will read when enough skeptical people say they are correct. Thanks.
Hi PN,
The equations are very simple.
Just click here:
https://forum.nasaspaceflight.com/index.php?topic=42978.msg1712726#msg1712726
If you mean "simply wrong" then you would be correct.
The equation for Energy in is simply P*t, not what Shawyer wrote. What he wrote for Ein is just another way of writing the same formula as Eout, making the rest of which a fancy way of saying 1=1. (The final equation has 2 variables, Fd, and Fs that both refer to the same quantity.)
The equation at the end of the first slide you posted is a correct expression for Eout = 0.5*P*Ts*t*Vt.
Divide this by the actual input energy, P*t, and you get 0.5*Ts*Vt. This clearly shows there is a velocity for which any given value of Ts results in over unity. For the values of Ts Shawyer has claimed recently, the Vt that results in over unity is small.
(Relativistic analysis agrees with this conclusion as well, and it is easy to see how relativity allows photon thrusters, because you can never accelerate past c, so a sufficiently small Ts is acceptable.)
If EMDrive is verified to work exactly as Roger explains it, then you will have to explain your "Simply wrong" conclusion. Are you prepared for that scenario? -
#846 by LowerAtmosphere on 14 Aug, 2017 21:10
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Shawyer has been proven wrong DOZENS of times throughout these threads. Especially regarding his hilarious lack of sidewall pressure. Newcomers: read the discussion or search through the old threads for keywords please!
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#847 by moreno7798 on 14 Aug, 2017 21:35
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Shawyer has been proven wrong DOZENS of times throughout these threads. Especially regarding his hilarious lack of sidewall pressure. Newcomers: read the discussion or search through the old threads for keywords please!
Proven what wrong? His equations? The EMDrive? . . . I'd suggest that you get your own paper disproving his equations peer reviewed and published. Talk is cheap. Get something tangible out there or just watch and learn. You know, in case you missed it, it is his invention (even if the math is flawed) he had the insight to conceive of it and bring it to ALL OF US - particularly here on this forum where we are trying to replicate HIS idea.
Again, get something out there or have a little respect for the man who conceived the machine you're trying to emulate. -
#848 by meberbs on 14 Aug, 2017 21:42
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If EMDrive is verified to work exactly as Roger explains it, then you will have to explain your "Simply wrong" conclusion. Are you prepared for that scenario?
The definition of input energy is input power times time. No experiment can change this, because it is the definition. What Shawyer wrote on those slides is wrong, and will always be wrong.
As LowerAtmosphere points out, Shawyer making wrong statements about high school physics is not new. One telling example is how he keeps flipping signs in simple force diagrams, you should go look that up to, I won't repeat the discussion here. -
#849 by moreno7798 on 14 Aug, 2017 22:05
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If EMDrive is verified to work exactly as Roger explains it, then you will have to explain your "Simply wrong" conclusion. Are you prepared for that scenario?
The definition of input energy is input power times time. No experiment can change this, because it is the definition. What Shawyer wrote on those slides is wrong, and will always be wrong.
As LowerAtmosphere points out, Shawyer making wrong statements about high school physics is not new. One telling example is how he keeps flipping signs in simple force diagrams, you should go look that up to, I won't repeat the discussion here.
Look, I am not going to waste my time arguing. I want to learn. That is why I am here. Disparaging other people gets the science nowhere soon. This forum should be about building and testing the hell out of EMDrive. That should be the focus. It's fine to disagree but the proof is in the pudding. If you feel so strongly about the math, then it should follow that EMDrive should not work. But if it does turn out to work, then you have to be prepared to explain your mathematical assertions.
Do you really believe that if EMDrive actually does work, it was conceived out of pure, one in a billionth, chance idea?
Let's wait and see. -
#850 by Rodal on 14 Aug, 2017 22:48
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If EMDrive is verified to work exactly as Roger explains it, then you will have to explain your "Simply wrong" conclusion. Are you prepared for that scenario?
The definition of input energy is input power times time. No experiment can change this, because it is the definition. What Shawyer wrote on those slides is wrong, and will always be wrong.
As LowerAtmosphere points out, Shawyer making wrong statements about high school physics is not new. One telling example is how he keeps flipping signs in simple force diagrams, you should go look that up to, I won't repeat the discussion here.
Look, I am not going to waste my time arguing. I want to learn. That is why I am here. Disparaging other people gets the science nowhere soon. This forum should be about building and testing the hell out of EMDrive. That should be the focus. It's fine to disagree but the proof is in the pudding. If you feel so strongly about the math, then it should follow that EMDrive should not work. But if it does turn out to work, then you have to be prepared to explain your mathematical assertions.
Do you really believe that if EMDrive actually does work, it was conceived out of pure, one in a billionth, chance idea?
Let's wait and see.
1) So, if tomorrow Shawyer posts a report stating that 1+1=5, would you think that it would be "disparaging other people" to point out that 1+1=2 and to state that 1+1=5 is simply wrong? I don't think that pointing out that what Shawyer posted is wrong, and stating the reasons why it is wrong is "disparaging other people". One has to make a difference between attacking an argument and attacking a person. One reaches the truth by constantly examining all assumptions. Feynman said that the easiest person to fool is oneself.
2) " If you feel so strongly about the math, then it should follow that EMDrive should not work. "
I think that is a non-sequitur. Whether something works or not is up to Mother Nature and not to the discoverer. The Discoverer may have no idea as to why what he discovered works. The first humans had no idea as to what was responsible for rain, many thought that it was the Gods pouring water on them, and they danced in order to enhance the chances of rain. Some people even had human sacrifices in order to please the Gods.
Countless things have been discovered by accident https://en.wikipedia.org/wiki/Kevlar for example.
3) " This forum should be about building and testing the hell out of EMDrive". Well, what is the purpose of posting Shawyer's slides on efficiency then? Why doesn't someone post a video of Shawyer's EM Drive Gen 3 self-levitating instead? Why doesn't someone show Shawyer building and testing Gen 3 EM drives instead of posting Shawer's latest slides on efficiency? I think that it is fine to have the forum also discuss issues like efficiency and overunity arguments. Disagreeing about them does not necessarily mean that one is disparaging the person.
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#851 by dustinthewind on 15 Aug, 2017 04:05
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From the FAQ on emdrive.com:
"Q. Does the theory of the EmDrive contravene the accepted laws of physics or electromagnetic theory?
A. The EmDrive does not violate any known law of physics."
How can this be?
Suppose you have a 1000 kg vehicle propelled by an EmDrive.
In one frame of reference the vehicle accelerates from 0 m/s to 1 m/s. An observer in this frame of reference concludes that the EmDrive must have consumed a minimum of ((1000)*(1)^2)/2 - ((1000)*(1)^2)/2 = 500 J.
In another frame of reference the vehicle accelerates from 10 m/s to 11 m/s. An observer in this frame of reference concludes that the EmDrive must have consumed a minimum of ((1000)*(11)^2)/2 - ((1000)*(10)^2)/2 = 10,500 J.
Energy is not being conserved.
I know this is the 10th thread and this has probably been covered previously but it seems that there are only two positions a reasonable person can take:
1.) The EmDrive does not work; the observed effects are all within experimental error.
2.) The EmDrive works; all errors have been accounted for; energy and momentum are not conserved.
Does anyone hold the position:
3.) The EmDrive works; all errors have been accounted for; energy and momentum are conserved.
If so, can you explain how?
Yes when calculating a change in velocity of a boost of 1m/s then 1/2*m*v^2 for 1/2*m*(0+1)^2 is much smaller than 1/2*m*v^2 for 1/2*m*(10+1)^2.
To truly get down to the mystery of why a constant boost can give such a difference in energy you have to consider the potential energy.
http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html
To simplify matters you can consider a ship in an elliptical orbit around the sun. At the aphelion of its orbit it has some decreased velocity due to its increased potential energy. If a burn is initiated here the burn will be less efficient. The reason being is because the fuel has a lot of wasted potential energy. As it is released from the ship it has a long ways to fall.
Rather if you let the ship fall into the gravity well the ship speed will increase and the fuel when released will be at a much lower potential energy.
There is another subtle matter here hidden in the physics. For the maximum efficiency you want to release your fuel at exactly the speed the ship is flying at. Why? So that the fuel gains no potential energy.
Letting the ship fall into a gravity well is two fold. It decreases the potential energy of the fuel and increases the velocity of the ship. Increasing the velocity of the ship increases the efficiency of fuel exhausted at high velocity (you must fire the fuel off at a higher specific velocity for it to come to a stop relative to the gravitational potential well). The larger the gravitational potential one falls into the better.
What about where there are no gravitational wells. How then does the ship change velocity with out accelerating its own fuel. Every bit of energy used to accelerate the fuel/ship increases the speed. The larger the speed the faster the fuel needs to be kicked out for the fuel to come to a rest. A rest you ask? With respect to what frame? Probably the same reason a photon rockets efficiency increases as the ship reaches c the photons kicked out increase in efficiency. Or the photons caught and reflected come to a complete stop or completely red-shifting.Relativistic solutions to directed energy
Neeraj Kulkarnia, Philip M. Lubina, and Qicheng Zhanga
aDepartment of Physics, University of California, Santa Barbara, CA, USA
...
Thus, a photon striking the sail of a spacecraft moving arbitrarily close to c transfers all its energy to the kinetic energy to the
spacecaft. The situation seems to indicate the efficiency approaches 1 as v → c...
By the way there are 2 types of gravitational assists. One is falling into a gravitational well and then using your thrusters (increasing their efficiency). The other is depending on how you approach a gravitational body and how you leave it you can swap kinetic energy between the two bodies i think.
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#852 by TheTraveller on 15 Aug, 2017 11:37
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Turned Roger's presentation into a spreadsheet as attached.
Example also attached using
326mN/kW (measured Flight Thruster data)
10kW Rf
3,000kg mass
2.5km/s dV
26.63 day burn time
Spreadsheet updated to version 2 to fix an incorrect input error message.
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#853 by Jim Davis on 15 Aug, 2017 13:08
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Turned Roger's presentation into a spreadsheet as attached.
I noticed that when the initial velocity is around 1420 m/s the efficiency becomes "OU" (over unity?).
So the EmDrive is a free energy machine? -
#854 by tdperk on 15 Aug, 2017 13:31
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Turned Roger's presentation into a spreadsheet as attached.
I noticed that when the initial velocity is around 1420 m/s the efficiency becomes "OU" (over unity?).
So the EmDrive is a free energy machine?
Only if you think the EM Drive can be built and operated for free and you don't think the OU energy is coming from somewhere else to be used locally. -
#855 by PotomacNeuron on 15 Aug, 2017 13:43
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Turned Roger's presentation into a spreadsheet as attached.
I noticed that when the initial velocity is around 1420 m/s the efficiency becomes "OU" (over unity?).
So the EmDrive is a free energy machine?
Only if you think the EM Drive can be built and operated for free and you don't think the OU energy is coming from somewhere else to be used locally.
If you can gain the energy as described, there are always ways to use it locally. I remembered the suggestion of an ultimate "weapon" made of EmDrive ( if it works). Accelerate weapon with EmDrive for 10 years into the space. Another 10 years to decelerate. Move back towards the Earth. Accelerate for 10+ years to impact the earth. -
#856 by TheTraveller on 15 Aug, 2017 14:10
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Version 2.
Now shows "Input Error" which is what the condition is and not "OU" which is not a real result.
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#857 by TheTraveller on 15 Aug, 2017 14:12
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Turned Roger's presentation into a spreadsheet as attached.
I noticed that when the initial velocity is around 1420 m/s the efficiency becomes "OU" (over unity?).
So the EmDrive is a free energy machine?
The message should say "Input Error", which it now does.
Thanks for the heads up. -
#858 by meberbs on 15 Aug, 2017 15:05
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Try it by just changing the final velocity to 6200 m/s. So what magic causes the emDrive to suddenly stop working when it reaches 6200 m/s? In what way is that an input error?Turned Roger's presentation into a spreadsheet as attached.
I noticed that when the initial velocity is around 1420 m/s the efficiency becomes "OU" (over unity?).
So the EmDrive is a free energy machine?
The message should say "Input Error", which it now does.
Thanks for the heads up.
Didn't Shawyer claim an emDrive could get to orbit, and with a higher specific force that results in overunity even sooner? -
#859 by Notsosureofit on 15 Aug, 2017 16:32