Jump to content

Could a Gyroscopic inertial thruster ever work?


FREEFALL1984

Recommended Posts

"' There might be a small displacement to the right ' And ' I may have an explanation for the displacment to the right '"

All I'm saying is that the dot only measures the center of gravity (which is what we're trying to measure here) when there are no cycles being performed. If you account for the dot when cycles are being performed one will get faulty results. I don't think this is too controversial a statement.

"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."

I'll admit math isn't my strong subject, and I have no idea where "±1.4" and "±1.8" come from, or even what they mean. I simply calculated how many more frames the dot spent on the left side versus the right one, and the number that came up was 9.3% more. I'd like to know how to calculate... uhm, sigmas, statistics and whatnot, but I don't know how just yet.

"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?"

It is a pretty weak machine, but I also intentionally had it do very small swings so the dot would be easily tracked. I can easily set up an experiment where the swings are large and I fit 2-3 cycles just on the left and right side, trying not to do a cycle just as the dot passes the middle line (meaning it'll be easy to track the center of gravity at that point).

"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."

Well, I don't really agree with your comparison of a crippled kid. If the center of gravity moves identically, then it is my belief that the machine will behave identically, at least when it comes to starting and stopping a swing. In the end, it's just my opinion. Notice that I'm using the word "suggests", meaning I don't really claim it (I would've just said "This proves it's a functioning reactionless drive!"), just that I think it's interesting.

*Ahem* Getting back to the subject. You drew some lines on that graph, which wouldn't really be analyzable, but is still interesting for the sake of discussion. No, you don't clearly see a deviation, but you do see that the wave gets bigger after each cycle (push) on the bottom of the graph. This means the backwards stroke will be almost (99%?) as big, meaning it'll be harder to analyze by simply looking at it. The left swings (bottom waves) on the graph should be generally wider than the right swings (top waves), at least when there are cycles being performed. Something you can check?

Edited by M Drive
Link to comment
Share on other sites

I'm still amazed by video 18 and 21 in the video I posted above. The difference is significant.
Alchemist: Except I don't only do cycles when it's at the rightmost position. And the center of mass does shift when the gyros are off if you check out video 21. However, the fact that it can swing isn't the proof I'm after. Also, the fact that it can't get swing momentum nearly as well as when the gyros are on (see video 21) is pretty interesting.

There's the difference between what happens to center of mass in 18 and 21:

In 18 when you turn it on the gyroscopes swing forward, when you turn it off the gyroscopes swing backward. When you switch on during forward swing and off during backward swing it amplifies the swinging.

In 21 when you turn it on the gyroscopes swing backward, when you turn it off the gyroscopes swing forward. When you switch on during forward swing and off during backward swing it dampens the swinging as when you did breaking in 18.

When in 21 you do cycles at the most forward position you actually keep it on (with backward-shifted CoM) for part of backward swing that's why it accelerates, but much less than if you'd kept it on for the entire backward swing

Link to comment
Share on other sites

All I'm saying is that the dot only measures the center of gravity (which is what we're trying to measure here) when there are no cycles being performed. If you account for the dot when cycles are being performed one will get faulty results. I don't think this is too controversial a statement.

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 admit math isn't my strong subject, and I have no idea where "±1.4" and "±1.8" come from, or even what they mean. I simply calculated how many more frames the dot spent on the left side versus the right one, and the number that came up was 9.3% more. I'd like to know how to calculate... uhm, sigmas, statistics and whatnot, but I don't know how just yet.

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 :D ) 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.

It is a pretty weak machine, but I also intentionally had it do very small swings so the dot would be easily tracked. I can easily set up an experiment where the swings are large and I fit 2-3 cycles just on the left and right side, trying not to do a cycle just as the dot passes the middle line (meaning it'll be easy to track the center of gravity at that point).

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.

Well, I don't really agree with your comparison of a crippled kid. If the center of gravity moves identically, then it is my belief that the machine will behave identically, at least when it comes to starting and stopping a swing. In the end, it's just my opinion. Notice that I'm using the word "suggests", meaning I don't really claim it (I would've just said "This proves it's a functioning reactionless drive!"), just that I think it's interesting.

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.

*Ahem* Getting back to the subject. You drew some lines on that graph, which wouldn't really be analyzable, but is still interesting for the sake of discussion. No, you don't clearly see a deviation, but you do see that the wave gets bigger after each cycle (push) on the bottom of the graph. This means the backwards stroke will be almost (99%?) as big, meaning it'll be harder to analyze by simply looking at it. The left swings (bottom waves) on the graph should be generally wider than the right swings (top waves), at least when there are cycles being performed. Something you can check?

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.

Edited by N_las
Link to comment
Share on other sites

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.

First off, thanks for typing all that. I now have a much better understanding of what you were talking about.

What I meant was that the bottom "hills" should be generally wider than the top hills. (Drew a picture). The blue spaces should be generally wider than the green spaces. http://i.imgur.com/OOKQXHr.jpg

Oh and I got to thinking about how the graph doesn't show any displacement that you can see with the naked eye. What it does show is that the swing motion can gain or lose about 20-30mm for a single cycle. So, if I did cycles like in the following example (laugh all you want :P), wouldn't it show a significant displacement to the left (bottom?). http://i.imgur.com/58utqrO.jpg

The red squiggly lines are cycles being performed. You'd start off with a fairly big swing, say, 400-500mm, so you could fit a cycle like in the picture. After the dot has passed the middle line, but before the machine changes direction. The first cycle would (hopefully) cause the top hill to shrink by 20-30mm. The second one would cause it to grow by 20-30mm. You keep this up and you have a graph where you can clearly see the top "hills" on the bottom sides are larger than the ones on the top side.

Link to comment
Share on other sites

What I meant was that the bottom "hills" should be generally wider than the top hills. (Drew a picture). The blue spaces should be generally wider than the green spaces.

I can check that, but propably not before wednesday.

Oh and I got to thinking about how the graph doesn't show any displacement that you can see with the naked eye. What it does show is that the swing motion can gain or lose about 20-30mm for a single cycle. The red squiggly lines are cycles being performed. You'd start off with a fairly big swing, say, 400-500mm, so you could fit a cycle like in the picture. After the dot has passed the middle line, but before the machine changes direction. The first cycle would (hopefully) cause the top hill to shrink by 20-30mm. The second one would cause it to grow by 20-30mm. You keep this up and you have a graph where you can clearly see the top "hills" on the bottom sides are larger than the ones on the top side.

I am convinced such a movement pattern would be impossible to generate. But if you can generate it, it will have a very good change of passing the pendulum test.

Link to comment
Share on other sites

"I am convinced such a movement pattern would be impossible to generate"

Yet, I kind of already have. :)

tlgRpjK.jpg

QeuhtU0.jpg

There's a difference of about 5cm (50mm) in the most successful "pushes" and "brakes".

Link to comment
Share on other sites

Have you made an experiment, just doing cycles continiously? Because: In all your experiments, you try to coincide the cycle frequency with the swing frequency in the right why, to generate higher or lower swings. But nobody is really interested in that. Pushing a swing higher is nice and all, but ultimately has nothing to do with gyroscopic propulsion. If you machine could produce thrust, simply doing cycles continioulsy will pass the pendulum test.

Link to comment
Share on other sites

Yeah, on April 7th, in the video you first tried to analyze (00009.mts from April 7). For some reason I didn't do continuous cycles 3 days ago when I went back. Maybe because the results from the first try were so shaky.

The idea was to gain access to one of the companys bigger halls, that have up to 15 meters to the ceiling, but instead I got a room with 5 meters. I'm thinking I want to do continuous cycles in a room with at least 10 meters to the ceiling.

Link to comment
Share on other sites

A few things. Measure the weight of your machine. On your rail system, if you turn the machine off, and then push it by hand and film how it moves, from it slowing down a bit, we can calculate the drag from the rail. If you now make a video of the machine turned on, moving on the rails, we can measure its median velocity. IF WE PRETEND, the forward movement was caused by real thrust, we can calculate the force of this thrust. With this force, we can calculate how far the mean deflection on the pendulum should be. If it turns out this deflection is in the range of mm, then we know you will never be able to confirm the thrust using the pendulum. If it turns out this deflection should be in the range of several cm, then we know we should have seen the effect by now on the pendulum, and therefore it isn't real.

Link to comment
Share on other sites

I came around to measure the length of each 'hill' during the 'machine on' phase. The left (bottom) hills are 46±56 ms longer than the right (top) hills. Thats a diference of 0.8 ÃÆ’ from zero. So no evidence for any difference, if measured over the whole 'machine on' phase.

Edited by N_las
Link to comment
Share on other sites

  • 2 weeks later...

Looks like I'm going back on Wednesday, and I got access to a hall with 15 meters to the roof. I have some questions though.

I'm doing some things differently this time around. I'm going to orient the machine right side up, not up-side-down like last time, so I can attach the laser pointer directly beneath the center of gravity this time (when no cycles are being performed). Doing so with the machine hanging up-side-down would be too much work.

The gyros are also in the "forward" position this time, meaning, if I turn off the gyros and rotate them (do a cycle), they'll move exactly like when they're on. They'll be pulled out by centrifugal forces, then as the rotation stops, they'll be pulled back by the springs.

Now, what type of experiments can/should I do? I'm of course going to try and do continuous cycles, in an effort to make the dot (even when cycles are being performed) remain on one side at all times. That should be "impossible", right?

I'm also going to do more control experiments with the gyros off to try and see if there's any difference. Since the masses that are the gyros moves exactly like when they're on, there shouldn't be any differences. I wanted to do more control experiments (gyros off) last time, but the battery to my camera ran out unexpectedly.

Suggestions? Larger paper for the dot? Maybe one with a grid drawn out? Stronger laser pointer (won't be able to use the strong one I got to borrow last time, so I have some time to buy a new one)?

Link to comment
Share on other sites

Now, what type of experiments can/should I do? I'm of course going to try and do continuous cycles, in an effort to make the dot (even when cycles are being performed) remain on one side at all times. That should be "impossible", right?

That would be very good evidene, that your machine works. And it would help you to convince sceptic physicist to look at you machine. But I hope you understand, that people would not accept that as definitiv proof. For a sceptic, there is always the possibility that the video is faked, or that other effects are at work (like moving air in the room acting on the wires). Because of that, it is important to document everything of the experiment.

I mean, if the analysis shows that your machine passes the pendulum test, document everything, from the used camera and laser-pointer, a detailed plan of your machine, usw...

Suggestions? Larger paper for the dot? Maybe one with a grid drawn out? Stronger laser pointer (won't be able to use the strong one I got to borrow last time, so I have some time to buy a new one)?

To make the analysis way easier:

- be sure to get a laser pointer as bright as the one you borrowed last time. Automatic tracking wasn't possible with your first one.

- if possible, try to use a very big paper, so that the dot does't leave it. If the dot leaves the paper, the tracking software has to be paused, and adjusted to the new background. This takes ages. If the dot doesn't leave the paper, it's possible to do all tracking automaticaly.

- http://imgur.com/kgeehdD make those lines on your paper, and align it with the camera. If the paper is aligned like that, taking the perspective into account is much easier.

As before, have always a bit of free swinging at the end of your video.

Edit: To test if the machine behaves the same with gyros on or gyros off may be interesting for you personaly. But i want to emphasize that this has nothing to do with the pendulum test, and it convince no sceptic that your machine works. It would ony show, that the gyros do play a role in the swing movement, not that the gyros provide propellantless propulsion.

Edited by N_las
Link to comment
Share on other sites

I'm shocked this thread got to 37 pages in the "science" section of this forum. Because this isn't science, right?

Well, it kind of is. He's got a hypothesis, and he's testing it pretty rigorously. I'm actually pretty impressed, he's not just gone down the route of confirmation bias and rudeness most free energy people seem to, and for that, he has my respect.

Link to comment
Share on other sites

Well I'm only being rigorous so I can either 1: get something out of this damn invention, or 2: finally be able to prove to myself that it's all bunk, so I'll be able to clear my apartment of gyro stuff and continue with my life. :P Ironically I'm the one who wants to prove it's a fake most of all. Been kind of conflicting, trying to keep up motivation when you have that mindset.

Link to comment
Share on other sites

Keep at it until you figure it out!

I've got an invention of my own that has been in the works for over 5 years now. The furthest it has gone is it makes noise and blows smoke. I'm currently hammering out the blueprints for the second prototype, taking lessons from what the first one has done in its 4 years of testing.

Have you been able to test your device in a vacuum? The diagram shown on the first post actually would produce a force when tested while exposed to the air. Compare the layout of the mechanism to that of a gear-type pump. What you have happening is air being dragged by the components is being ejected in one direction only, making this a very crude sort of fan. I would have to sit down with your diagram and do a force-table to tell you if there would be any forces coming from the mechanism itself if operated in a frictionless vacuum.

Also to make a really really sensitive force measuring system, attach the mechanism to a floating block of wood on a tray of water that does not have a lot of room to turn side to side. Put a laser pointer somewhere on this block of wood, and aim it at a nearby target.

Your device can then operate, and in theory should push the float along the water making the laser dot move along the target. If it doesn't, there is no force being produced.

Edited by OdinYggd
Link to comment
Share on other sites

I can't believe people are actually debating this on KSP forums.

Gyroscopes aren't magical, and can't magically produce thrust with no reaction mass

Nah, leave them be.

Link to comment
Share on other sites

Well, it's Wednesday, but it's also snowing. We were supposed to open up a roof hatchet to lower the wires, but that's not happening today. Maybe next Wednesday.

Damn. Really wanted to do this experiment today...

Link to comment
Share on other sites

Oh for the love of....

So the machine And the camera broke. It was a lot of time and effort just to hang it from the 4 ~16m (54 feet) wires, only to have one of the points where I attached one break and make the machine crash into the floor. The middle axle broke in half (at a point where it had a hole drilled through it). Tried to fix it temporarily and actually got some pretty interesting results. Only, the camera only recorded 5 seconds before breaking down, so those results are lost in time.

Only thing to do is fix the machine and try again. At least now I know what to expect. Still pretty pissed though. :mad:

Link to comment
Share on other sites

Guest
This topic is now closed to further replies.
×
×
  • Create New...