N_las

In this one - dimensional case: Gravitational Force: F = - G*m*M_earth/x^2 Newton: F = m*a acceleration and position can be connected by: a = (dx)²/(dt)² putting everything togheter: -G*M_earth/x² = (dx)²/(dt)² the start position: x(t=0s) = x_moon; and start velocity: dx/dt = 0; You have to learn how to solve differential equations to learn the general way. But I am just lazy and solve something like this numerically with my computer.

You are right, that is true. I only wanted to remind: one can't draw from that, that a non-faulty measurement would show a displacement to the left. Like if you measure the heigth of a kid compared to last week. You would expect that it would be taller, but if you see that the measurment shows a shrinking, and you find an uncertainty in the measurement, you can't know if the kid was actually shrinking, growing, or staying the same heigth. I'll give you a quick overview: Imagine an event like the flight of an airplain would be sceduled do be at the same time every day. You don't know the sceduled times of start, landing, and duration. But you want to find out: You measure the time of the plane lift-off every day. Your measurment of the first day shows a liftoff time of 55347 seconds (The time of the day in seconds from midnight). Naturally, one doesn't know from that, if this is the sceduled liftoff time. It could run late or early. Your measurment of the second day shows a liftoff time of 55529s. The third day shows 55498s. From these three measured values one could try to estimate the actual sceduled liftoff time. If you take the median of all three times, you get: 55458s. (I hope i didn't screw up my math ) Would you take someone serioulsy, if he would suggest, that This is the actual sceduled time? Not really. It may be nearer to the real result than just assuming that the first measured value is right, but it is still just an estimation. An there is a mathematical way to describe how exact the estimation could be. You have to caluclate the 'distance' of every measured start from the median: for the first fligth: 55347s - 55458s = -111 s second flight: 55529s - 55458s = 71 seconds third fligth:55498 - 55458s = 40 seconds Then square everyone of the values, and add the results: (-111s)^2 + (71s)^2 + (40s)^2 = 18962s² This sum has to be divided by the number of measurments minus one. In our case: 3 measurements -> we have do divide the value by 2. 18962s² / 2 = 9481s² now we have to take the square root of this, to get the 'corrected sample standard deviation' sqrt(9481s²) = 97.37 s This is a value, how far the measured times are spread around the median. To now get the 'standart error of the mean', you have to divide this by the squareroot of the number of measuremets: 97.37s / sqrt(3) = 56.22 s -> ca 56. seconds You can now say, that the median liftoff time was 55458±56 seconds. The error behind the plus-minus is a value for how sure you can be about the liftoff time. The real liftoff time can be anywere around 55458 seconds, but the probility for it decreases the farther away from it. It would be expected to be in a window of 55458 s plus or minus 3*56 seconds (±3ÃÆ’). If you measure the landing time in a similar fashion, and you get 67364±42 seconds, the fligth duration would be 11906 seconds. And the error of it would be sqrt(56² + 42²) = 70 seconds. So the measured fligth duration is 11906±70 seconds. The real sceduled flight duration can be anywere around 11906 seconds but probably in a window of plus or minus 3*70 seconds. In your case, if you have frame durations of 59.8±1.4 frames on the left side, and 55.7±1.8 on the right. If take the difference of both, the measured difference is 4.1±2.3 seconds. So the REAL differencs can be anywere around 4.1 seconds, but probalby in a window of plus or minus 3*2.3 seconds. The null hypothesis is, that there is no difference in frame numbers, so the difference should be zero. Zero is in the window of probablity of the real difference (±3ÃÆ’, so between -2.8s and 11s). So there is no evidence, that the null hypothesis is wrong. Fair point. But one can't really track the point if you make the swings that large. So you are right, if there would be a reactionless propulsion effect, it could be very small. But if it is so small, that it will always drown in the statistical error of the measurement, than you have to change the experiment somehow to minimize the error. Or make many many many measurments. For example, if i make one measurement shows a displacement of 3.0±2.0mm, and another measurment shows a displacement of 2.6±3.0mm, the resulting analysis shows a displacement of 2.8±1.8mm. Notice how the error got smaller? The more measurements, the smaller the error in the result. But you shouldn't change the experiment between those measurments. And it would be very very tedious and would take forever to anlyse that many measurments. I watched your videos with gyro off. You put the gyro in a weird position, that was different than during the 'gyro on' test. They were angled forward. I am not convinced, that the center of mass behaved the same in both cases. And I am not sure if only the movement of the center of mass can contribute a difference here. But for me it is a meaningless task to look at the difference in 'gyro off' and 'gyro on' behavior. I am only interessted if ANY machine can pass the pendulum test. But what you tried is repeating the same pattern of movement for 'gyro off' and 'gyro on'. Maybe, if you play a bit with 'gyro off' you can find out that it can be pushed to have the same big swing movement, it just needs a different control pattern. If this would be the case, than you would have evidence, that gyros aren't relevant for the machine. If this would not be the case, it wouldn't allow for any conclusion. I don't know what you mean, can you frame it differntly? But I don't know how i can check anything. I can easily draw lines were the bottom waves are wider or less wide. One can't say anything from the drawn red lines. I only put them there to have something nice to look at.

I don't now what to tell you, but thats not how evidence works. You can't start with: ' There might be a small displacement to the right ' And ' I may have an explanation for the displacment to the right ' And conclude: ' There is almost guaranteed a displacement to the left ' You have to make it more like this. Start with: ' There might be a small displacement to the right ' And ' I may have an explanation for the displacment to the right ' Now: ' I try to proof that the explanation for the displacement to the rigth is correct', ' I try to calculate the real average displacement if the explanation is taken into account.' If then a significant average displacement to the left is observed, you have good evidence that you passed the pendulum test. You said: 'Counting frames, that is, the amount of time the dot spends on the left side, is a viable method for calculating displacement.' But you lie to yourself. Counting frames CAN'T be a viable methode for calculating a displacement, since the time the dot spends on any side hasn't directely to do with the displacement. And even if counting the frames would be a viable methode. From the Data you provided a few posts earlier, the dot is 59.8±1.4 frames on the left side, and 55.7±1.8 frames on the right side. Yet in your post and in the describtion of the youtube video, you claim the dot spends 9.3% more time on the left side than the right. You do realize that the data doesn't show that, do you? The difference between left and right is 4.1±2.3 frames (ÃÆ’ = 1.8) That means there is NO evidence for a difference between the frame number. (For comparison, effects measured at CERN do need a ÃÆ’>5 to be consider real. That would mean the frame difference between left and right should be something like 4.1±0.8 to support your claim, that there is a difference in left and right side. But ÃÆ’>3 can be used as very weak evdence) If you counted the whole video, you can post the data and I can make the calculations for you, to set your mind at ease. But as it looks now, frame counting dosen't support your machine. K^2 made an error here. He calculated the corrected sample standard deviation and got the right result. http://en.wikipedia.org/wiki/Standard_deviation#Corrected_sample_standard_deviation But one have to divide by the square of the sample size again, to get the standart error of the mean. http://en.wikipedia.org/wiki/Standard_error_of_the_mean#Standard_error_of_the_mean But as shown above, that doesn't change that your frame counting doesn't show a difference between left and right EDIT: And don't you think, that if your machine works, there would be a huge displacement to the left, that would easily overshadow the incorrect movement of the dot in relation to the center of mass? The average displacement to the right was 6.3±2.0 mm. Don't you think that a real working machine would show very very clear results? As it stands, we are arguing over displacements in the range of mm. Do you really think the effect your machine has (if it works) would be so insignificant? Or are you saying to yourself the displacement to the left is easily like a few cm, but it is hidden behind the incorrect movement of the dot? If I understand you correctelly, the correction I made in this graph by hand would show the actual movement of the center of mass: http://imgur.com/ljhx6Oj An average position to the left with this correction isn't clearly obvious, and I can't easily do the analysation with this hand drawn picture. And even if I could make the analysation, it would be limited evidence, because this isn't real data, but just hand drawings. But look at the graph, are you really convinced, that there is an average displacement? It looks to me just like an ordinary swing movement, without any anomalies. In the describtion of your Youtube Video, you write: ' These experiments show that the machine is able to accelerate in one direction only. If you start a swing and accelerate as the machine is swinging backwards, the swinging motion will decelerate, that is, brake. This is interesting because it suggests this is a functioning reactionless drive. ' And in an earlier post, you wrote: ' It can both accelerate and decelerate swing speed by doing cycles. However, when you turn off the gyroscopes (video 21) that phenomenon more or less disappears. This means that the pendulum test should be pretty easy to pass, and I probably do pass it in one or more of these videos. ' Can you explain why? How does the observation suggest a functioning reactionless drive? Stopping and starting a swing motion doesn't lead to the conclusion of reactionless drives. So being better at stopping and starting the motion if the gyros are on hasn't anything to do with reactionless drive or the pendulum test. An comparison: Imagine for an object traveling to the north and south would be considered impossible. You test it, and find out that you can go to the west. And you measure, that the object can go to the east. From that you think it suggests it could go the the north. Then you cripple your machine. Now it can't really go to the west and east anymore. From that you conclude the north-south test would e pretty easy to pass.

I don't now how you come to the conclusion, that it may have passed the pendulum test. Don't you think, that if your machine was capable of producing thrust, that there was a very significant displacement to the left? For example, if I would place a hair dryer on the pendulum, that would produce thrust from its air blowing capacity, it would easily have a displacement of over 10 ÃÆ’. From my analysis, it isn't even clear if there IS any displacement to the right. There could be, but it may be just as well a statistical error. EDIT: Your whole taking about gyro off and on: It really doesn't matter. The effect in which the machine builds up the swing momentum might be a very complicated combination of things, that depend on the gyro. So it may be lost if the gyros are off. But thats like placing a disabled kid on a swing, and measure the swing movement. And then place an abled kid on the swing, and measure a greater movement. It dosen't have anything to do with reactionless propulsion. The one kid just is a better swinger.

Even if the antimatter would react with matter inside the black hole, the resulting energy is still traped in there. And since the energy kinda has a weight (E=mc^2) There is now way to tell from outside the event horizont, because the black hole as a whole will keep its mass

Ok, I am done with my analysis. The brigth laser dot in video 17 was easy to track. It was very hard to put the perspective into account and calculate accurate length values. Keep in mind that the distance between two drawn lines is 20 mm. I took the phase after the machine was turned off to calculate the rest position of the dot: 3.0±0.9 mm to the right of the middle line. (so 3.3 ÃÆ’ from the line) It could very well be that the rest position IS at the middle line, but the video doesn't allow for a better estimation. The median position of the laser dot (during the time the machine was on) was 9.3±1.8 mm TO THE RIGHT of the middle line. Taking both values into account, the average deflection from the rest position (during the time the machine was on) was 6.3±2.0 mm TO THE RIGHT. This is just 3.2 ÃÆ’ (borderline) from the rest position, so one could argue if the measurment is significant or if its just an statistical artifact. In summary: Maybe there is a very small effect that favors a position to the right of the line (maybe has something to do with the cable of the controls). But there is no clear proof, that there even is a deflection. There is clearly no evidence whatsoever (in video 17) that the machine has an average deflection to the left (the machine is intended to deflect to the left). So there is no evidence for a reactionless propulsion. There is a small margin of error, because i didn't tracked the dot until the very end, because even with autotracker it would take to much time. ( the autotracker isn't reliable and needs babysitting). But I tracked EVERY frame of the machine-on-phase and every frame of a good portion of the machine-off-phase. Sometimes the dot leaves the picture. This also creates a small error. If anybody is interested, i can upload a table containing all tracked x-pixel and y-pixel values, the corresponding time values and my calculated real-x values (in mm, taking the perspective into account). EDIT: Here a link of my tracking Data, if you compare with the video in real time, you see it is accurate. http://imgur.com/a/yLBvD/embed

'Stay on one side more than the other' is a simplification. The average position is the important point. And as K^2 pointed out, you have to take statistics into account. The correct analysis of an experiment is usually much more work than the performing the experiment. I will post my results in a few hours.

Well M-drive it isn't sufficent to only count the frames. For the average position, the distance from the middle is important to. Imagine the dot is 1 frame 10 cm left of the middle, and 2 frames 5 cm right of the middle. If you only see the number of frames, you think the dot wasn't in the middle on average, but of course it was. The real analysis is loads more work than counting frames.

Ok. Today I can analyse video 17. I do it as good as I can, and I will send you all results. If video 17 shows a pass of the pendulum test, I can do the same for other videos next week. If not, than i won't put more work into it. Edit: I only have the free software 'tracker'. But thats only to get the Data from pixel position of the dot in the video. To make the analysis of the data I use Matlab.

A kid can stop the movement of the swing by shifting its center of mass, the same way it can increase the swing movement. But maybe thats not what you mean. I take a look at the videos. Can you do a quick overview which video contains what?

That would be great. The dot should start static, and should stop moving at the end of the video. If you use paper with crossed lines on it (like for math class) than it is much easier to take the perspective into account. The paper in the first video was folded and a little bend. It shouldn't have much of an effect, but try to avoid it if possible. It is not so bad, that the dot leaves the paper from time to time, but the dot should never leave the camera picture. Do another video, with the machine off, but hold your controls and stay in the same position. Now your friend should give the machine a bit of momentum, and film the movement of the dot as the machine moves just like a pendulum. From that we can do an analysis how strong the influence of the your control cable is. If you can make the videos, i will do the analysis again. After that, i can send you the files and diagrams.

It looked at the first video again. - I tracked every 5th frame - took the perspective into account - corrected for the small camera movement at the beginning. - I always tracked the 'middle' of the laser-streak I don't now the length of the paper, so I use '% paper-length' as unit of length. Now the median of the dot position in x-direction is (3.6±0.7) %paperlength from the marked line. Thats a difference of 5 ÃÆ’ from zero. So it looks like the median position is clearly not on the marked line. It would be nice if someone could do another analysis, if the values are in the same ballpark. 3.6% the length of the paper isn't much, but one can't ignore that. A few questions MDrive: - Is the drawn line on the paper perpendicular to the paper? Or did you just draw a roughly straight line? - Is the drawn line the position the dot has at rest? It would be nice if the video started with the point at rest position with no movement, and would last until the dot is at rest poistion again with no movement. Or couldn't the tremble be stoped? - You control the device through a cable. What was the deflection if you picked up your controls? Could you observe the dot moving from that, before you started the machine? The deflection from my analysis does go to the right (from the cameras point of few). I don't no how exactly your machine is orientated, but if there is a deflection, in which direction would you expect it to be? Does a deflction to the right conform with your observations from the machine movement on the track? EDIT: Ok, dosen't matter if the video is flawed.

Hi M Drive, I tock the video behind the first link (00009.mts) and tracked the laser point with the free program 'Tracker'. The video had approx. 2440 frames. The autotracker didn't work good, so i tracked it manually every 10th frame. In some frames the laser point wasn't visible. I then had a list with 233 x-positions of the point. Using statistics, the median is 12.0±8.6 pixels to the right of the line. So it is only 1.4 ÃÆ’ from zero. This means there is no evidence in the first video, that there is an average deflection. Weak evidence for a deflection would be, if the median was more than 3 ÃÆ’ from zero. My methode isn't perfect, several things could create small errors. - Framerate is to low to allow for a perfect tracking (instead of a point, there is a laser streak, so no accurate position, but maybe good enough) - The camera was't completely static, i adjusted the zero x-position to be at the drawn line on the paper. But at the end of the video the line was at a different position. - I could give a better result, if i would track the point at every frame, not every 10th frame. But that would be to much work. - Perspective! A x-position value for a point at the far end of the paper is really a different x-position value for a point at the near end of the paper. But i ony measured pixels. - I didn't watched the other videos If this first video would have shown a deflection of more than 3 ÃÆ’, than it would be worth to try to be more thorough. But as it is I have more real life work to do.

The problem is that you have to cool the animatter down enough. If its to hot, the velocity of the positrons is way to high to see any effect of gravity. To cool positrons down, one shots them into a metal. It is a misconception, that they will annihilate instantaniously. A large number of positrons will leave the metal at a lower temperature, because they lost energy by colissions with regular matter in the metal. Imagine they would have room temperature after that. That would mean a velocity of 1.1*10^5 m/s. If the free fall along a 100 m horizontal vacuum tube, they will have a vertical displacement of 4 µm. But after 100 m the positron beam would have a very big radius in the order of meter. So it is impossible to measure if they went up or down. Especially if you consider that the smallest electric charge in the area of the experiment would displace the beam much farther.

If it works like that, precongnition isn't usefull anymore. To use the lotto example. When do you think the certainty of 1337 crosses the 90% boundary? A day before the drawing? An hour? A minute? I think there are so many infinitly interwoven factors in how the numbers roll, that the threshold to make a reasonable prediction is a fraction of a second before the drawing of each number. And most things so static, that a precognition is possible days in advance, can be predicted without precognition (like: two days from now is my final exam).

Your Formula calculates the magnitude of the instantaneous acceleration, but it ignores the direction of it completely. You are right, that instantaneous and average acceleration of an object with constant velocity shoud be identical, but the watch hand doesn't have a constant velocity, since it changes direction all the time. For example: the average acceleration of the clock hand during 60 seconds is zero. Because after 60 seconds, the clock hand has exactly the same velocity (magnitude and direction) as at the beginning, so in 60 seconds there was no change in velocity. That means zero average acceleration. In your Problem: You have to calculate the velocity of the clock hand at 30 seconds as vector, because the direction is important. Then you have to calculate the velocity at 50 seconds as vector. Now you substract the 50s-vector from the 30s-vector, to get the vector of the change in velocity. If you divide this vector by 20 seconds you get the vector of the average acceleration. Since the magnitude is wanted, you then have to calculate the magnitude of this vector.

That could work, but how is decided what events are 100% occuring? Is there a predetermined destiny list written by god? This would limit the free will of at least some people. Imagine the Lotto numbers are 100% destined to be 1337. Some force now has to forbid me to call the lotto studio at the time of the number drawing, and fake a bomb thread, thus stopping the lotto numbers from beeing drawn. And if this 100% events are predestined on gods list, since when are they on the list? I mean: If the lotto numbers are destined to be 1337, was every aspect of human history since the stone age somehow shaped to lead some people to invent the game called lotto? So these people coulnd't have had free will. I am fairly certain one can not find a consistend set of rules that allow such 100% occurences to exist.

So, in case of the car crash, in the moment i get the information, the future will be changed? Or will it only be changed if i base a desicion to drive or not drive on that day? If the future is changed in the moment i get the information, than that will ruin your lottery, because a changed future is a changed future. It has nothing to do with choosing 1234 or 1337 having any effect. If the future is only changed because i decide to base a descion on the information, than that will create all sorts of paradoxes. And you said, that the precognition only shows one of many possible futures. So I will just see a random lottery number, because all numbers are possible. I can't win the lottery with that. Edit: Or are you saying, one can see every possible future? Than I will see a future where 1234 is winning, one wherer 1337 is winning, one were 9999 is winning... and million more possible futures.

So immediately upon taking information to the present time, this information becomes useless, because the future will be different. Now the precongnition is as useless as a guess.

What are you talking about? Imagine a gyro with a mass of 101g, and a tower with a mass of 100. This would be 'a heavy gyro and a light tower'. The CoM is now nearly at the middle between the gyro and the tower. If we have a 'light gyro and heavy tower', the tower has a mass of 101g and the gyro of 100g. The CoM is now shifted by a few mm. It doesn't reverses anything. And any weird gyro effect will be seen in both situations. So the relative masses doesn't matter. If there are unexplained gyro effects, why would they disapper, only because the gyro is lighter than the tower?

If he would be right, and there is anomalous behavour, the gyro would spin around the tower no matter what is heavy and light. A Light tower would just make for a better presetation, but the same effect should be seen with a heavy tower: It would stay still in the middle with the gyro rotation around it. The replication of the experiment isn't to the letter. They don't use a lighter tower because it doesn't matter. To show that Laithwaite is wrong, they just have to show that the tower doesn't stay still. Watch the video again: The tower is clearly rotating.

It doesn't matter if they used a light or heavy gyro. If Laithwaites would be right, the tower would stay still. And it doesn't.

To support your claim, you cite a site that disproves your claim with an experiment. We have absolutely no info about the mass of the gyro or the mass of the tower. You just assume, that the gyro is heavier. The people who made the experiment in video 5 do come to the conclusion, it is rotation around its center of mass, and one can assume they know what the involving masses were. And as explanied on the site, placing the tower on ice, like Laithwaite did, doesn't provide a frictionless surface, if the tower has legs.

Maybe the birthday of some more popular character than Jesus: Harry Potter - July 31st, 1980 or Skynet (Terminator) - August 12, 1997

For me, it looks like it is rotating around the center of mass. But that is hard to tell, because we don't now the mass of the gyro and the mass of the support structure. But the people who made the site and the experiment also comes to the conclusion, that it is rotation arount its center of mass. So i fail to see your point. How is it 'clearly not rotationg about the centre of mass'?