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Author Topic: SMOT TEST- can someone do this?  (Read 29799 times)

Offline nwman

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Re: SMOT TEST- can someone do this?
« Reply #45 on: January 13, 2008, 07:42:16 PM »
Here's an idea! How about someone make the same test as in the videos but with a foot of ramp before the smot. Would this stop this debate?

Tim

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Re: SMOT TEST- can someone do this?
« Reply #45 on: January 13, 2008, 07:42:16 PM »

Offline g4macdad

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Re: SMOT TEST- can someone do this?
« Reply #46 on: January 13, 2008, 08:37:32 PM »
Vidar,

Even though there can be a place near the entrance where the ball may feel a repulsion, the magnetic potential energy at that point and all points nearby is still less than it is far away.  There is nowhere you can place the ball where it will be ejected from the magnet's field, and no cheating to be had by placing the ball in the SMOT entrance vs. placing it without magnets.

The best you could hope to do is come out even, by placing the ball exactly at the null in the field where it feels no force.

Cheers,

Mr. Entropy

@Mr. Entropy,

This is arguable and not entirely necessary.
If there is an area where the magnetic field is essentially null (and there is such an area in SMOT), then the magnetic potential of that point is the same as for an infinite distance. (Magnetic field energy density is zero). That particular point is B by a very good approximation (not exactly, because the ball would remain in unstable equilibrium in that point but B is very, very close to it, as close as the hand can place the ball). So, B is the point of highest magnetic potential energy, although it seems hard to accept at a first glance. That?s why Mb>>Ma.

This is the main reason why conventional thinking about SMOT is not appropriate and why omnibus is obviously wrong.

Please comment.
Respectfully,
Tinu


This logic is ridiculous. Stop insisting that Omnibus is "obviously" wrong. If you want to prove his theory wrong, then do it! Stop patronizing the rest of us, and saying you are right without adequate proof. I don't think it would take a lot of effort to challenge what Omnibus is saying, but if you are unwilling to go through with it, please be quite.
« Last Edit: January 13, 2008, 09:04:54 PM by g4macdad »

Offline Low-Q

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Re: SMOT TEST- can someone do this?
« Reply #47 on: January 13, 2008, 10:18:19 PM »
Vidar,

Even though there can be a place near the entrance where the ball may feel a repulsion, the magnetic potential energy at that point and all points nearby is still less than it is far away.  There is nowhere you can place the ball where it will be ejected from the magnet's field, and no cheating to be had by placing the ball in the SMOT entrance vs. placing it without magnets.

The best you could hope to do is come out even, by placing the ball exactly at the null in the field where it feels no force.

Cheers,

Mr. Entropy

@Mr. Entropy,

This is arguable and not entirely necessary.
If there is an area where the magnetic field is essentially null (and there is such an area in SMOT), then the magnetic potential of that point is the same as for an infinite distance. (Magnetic field energy density is zero). That particular point is B by a very good approximation (not exactly, because the ball would remain in unstable equilibrium in that point but B is very, very close to it, as close as the hand can place the ball). So, B is the point of highest magnetic potential energy, although it seems hard to accept at a first glance. That?s why Mb>>Ma.

This is the main reason why conventional thinking about SMOT is not appropriate and why omnibus is obviously wrong.

Please comment.
Respectfully,
Tinu


This logic is ridiculous. Stop insisting that Omnibus is "obviously" wrong. If you want to prove his theory wrong, then do it! Stop patronizing the rest of us, and saying you are right without adequate proof. I don't think it would take a lot of effort to challenge what Omnibus is saying, but if you are unwilling to go through with it, please be quite.
Those prooves are easy accessible, and has ben presented several times - which he again are convinced are wrong. The thing is that someone does not give up on his SMOT, no matter what. Probably because he is pretty genuine in his conviction about it.

Vidar

Free Energy | searching for free energy and discussing free energy

Re: SMOT TEST- can someone do this?
« Reply #47 on: January 13, 2008, 10:18:19 PM »
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Offline Mr.Entropy

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Re: SMOT TEST- can someone do this?
« Reply #48 on: January 13, 2008, 10:49:04 PM »
@Mr.Entropy,

How come? There are direct measurements which prove I'm right. h1 can be measured directly, h2 can be measured directly, m can be measured directly. Think before posting.

How much energy is dumped into the ceramic dish at the end?  A lot means OU, a little means not.  There is no measurement.  What you say is spent there is just you talking.  And since your ridiculous analysis infers a CoE violation from mathematical laws that are known to forbid it, you are obviously in error.  Maybe you should learn the basics of vector calculus before you start going on about what happens in a linear superposition of conservative fields.

Cheers,

Mr. Entropy

Offline Mr.Entropy

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Re: SMOT TEST- can someone do this?
« Reply #49 on: January 13, 2008, 10:51:18 PM »
This logic is ridiculous. Stop insisting that Omnibus is "obviously" wrong. If you want to prove his theory wrong, then do it! Stop patronizing the rest of us, and saying you are right without adequate proof. I don't think it would take a lot of effort to challenge what Omnibus is saying, but if you are unwilling to go through with it, please be quite.

Describe something that you would accept as "proof", and I'll oblige if possible.

Cheers,

Mr. Entropy

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Re: SMOT TEST- can someone do this?
« Reply #49 on: January 13, 2008, 10:51:18 PM »
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Offline Mr.Entropy

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Re: SMOT TEST- can someone do this?
« Reply #50 on: January 13, 2008, 11:18:26 PM »
This is arguable and not entirely necessary.
If there is an area where the magnetic field is essentially null (and there is such an area in SMOT), then the magnetic potential of that point is the same as for an infinite distance.

Yes, that is correct.

Quote
That particular point is B by a very good approximation (not exactly, because the ball would remain in unstable equilibrium in that point but B is very, very close to it, as close as the hand can place the ball). So, B is the point of highest magnetic potential energy, although it seems hard to accept at a first glance. That?s why Mb>>Ma.

Still no, I think.  The point B you choose has to pull the ball against friction and gravity, which means it's reasonably far from the field null.  W.r.t. the Bus's video, point A in the dish looks to be far enough away to have a lower PE.  Note that I wouldn't feel the need to argue this if you'd said "theoreticaly, we can have Mb > Ma", instead of just "Mb >> Ma".  For those who don't know, >> means significantly greater.

W.r.t the SMOT test in this thread, we're comparing a placement at a point B with magents to a placement at point B without magnets, so even though the gravity+friction argument doesn't apply here (due to the downward tiliting of the ramp), there is no cheating going on, because the potential at that point with the magnets moved far away will be a very good zero.

Of course, if the magents really aren't far enough away when they are "removed", then there could indeed be "cheating" of the sort you mention.

Cheers,

Mr. Entropy

Offline tinu

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Re: SMOT TEST- can someone do this?
« Reply #51 on: January 14, 2008, 01:58:05 PM »
Vidar,

Even though there can be a place near the entrance where the ball may feel a repulsion, the magnetic potential energy at that point and all points nearby is still less than it is far away.  There is nowhere you can place the ball where it will be ejected from the magnet's field, and no cheating to be had by placing the ball in the SMOT entrance vs. placing it without magnets.

The best you could hope to do is come out even, by placing the ball exactly at the null in the field where it feels no force.

Cheers,

Mr. Entropy

@Mr. Entropy,

This is arguable and not entirely necessary.
If there is an area where the magnetic field is essentially null (and there is such an area in SMOT), then the magnetic potential of that point is the same as for an infinite distance. (Magnetic field energy density is zero). That particular point is B by a very good approximation (not exactly, because the ball would remain in unstable equilibrium in that point but B is very, very close to it, as close as the hand can place the ball). So, B is the point of highest magnetic potential energy, although it seems hard to accept at a first glance. That?s why Mb>>Ma.

This is the main reason why conventional thinking about SMOT is not appropriate and why omnibus is obviously wrong.

Please comment.
Respectfully,
Tinu


This logic is ridiculous. Stop insisting that Omnibus is "obviously" wrong. If you want to prove his theory wrong, then do it! Stop patronizing the rest of us, and saying you are right without adequate proof. I don't think it would take a lot of effort to challenge what Omnibus is saying, but if you are unwilling to go through with it, please be quite.

@g4macdad,

It seems you are quite revolted there, aren?t you?

Ok. Let?s shed some light by the means of a very short summary:

Firstly, it is necessary for you to understand that Omnibus? claim is in the first place a huge challenge to the actual electromagnetic theory (and not only).
His claim is very easy to dismiss by one word: conservative. But he went around and he said that a superposition of two or more conservative fields (magnetic and gravitational) may, under certain circumstances, be no longer conservative. Ok with me so far (although the challenge is as huge as the first one). Unfortunately, no proof was ever posted. Please keep this in mind: no proof, of any kind, was ever posted! The discussions went around the following equations: mgh1+Mb-Ma in relation to mgh1+Mb, in what it sounded like the following quote:

?Think about it, when you impart to the ball energy |(mgh1 - (Ma - Mb))| to raise it from A to B then, if CoE is to be obeyed the ball must lose that exact amount, that is |(mgh1 - (Ma - Mb))|, when it goes back to A and complete a closed loop. Not so, however, in our case of the ball completing a closed loop. The ball being at B (raised from A) and having energy (mgh1 + Mb) at B in our case doesn't lose energy |(mgh1 - (Ma - Mb)| when it goes back at A and closes the A-B-C-A loop. The ball in our case loses, as was said, energy Mb = (mgh2 + [KE + ...] ) as well as energy mgh1. This energy (that is the energy (mgh1 + Mb) which the ball loses in going from B back to A, closing the loop) is more than the energy |(mgh1 - (Ma - Mb)| that was imparted to it to raise it from A to B and that's a clear violation of CoE.?

Well, the above does not make sense to any scientists I know.
And consequently, the claim that SMOT is producing energy from nothing is not logically &scientifically supported by any proof. I wish it(energy out of nothing)  was true maybe as much as you do if not more, but in the absence of proof I have to point to the truth.

I hope you see that I don?t have to prove anything. I?m just asking for a pertinent and complete proof, following a ground-shaking claim for the entire scientific establishment. And the claim is not made by an amateur (as to be disregarded) but by a competent person.
But I?m also physicist. (It?s not about my non-significant person; most physicists would do the same). And when another person says that SMOT is overunity and he is also part of the scientific and civilized world, only one is possible:
A. he proves to the community he is right (which Omnibus hasn?t done yet, neither theoretically or experimentally).
OR
B. He withdraws his claims.

There is no in between.

My actual posts and arguments clearly support the alternative B above. But that?s just lately. I was as open-minded and forthcoming as I could be for over a year? I?ve seen some HUGE question marks, I?ve asked for details, posted my views (shared by many) but ?the dialogue? always ended in insults to me and to everyone else, as all of us except Omnibus being incompetent. Yet, no proof was ever posted and that?s the main aspect that bothers me most. Well, check the history for yourself, if interested?

The fact that you might not understand why the whole discussion is stuck at this point around Ma and Mb, is a different story that resides in you.

The mere issue that you are pissed of at me does not affect many except maybe yourself. No offense, but you don?t have to love me. Or I least I?m not here for that purpose. ;)

Last, but not least, don?t take it too personally. When I wrote the above I did it for everyone visiting here. The picture is not nice, indeed.

Cheers,
Tinu

Free Energy | searching for free energy and discussing free energy

Re: SMOT TEST- can someone do this?
« Reply #51 on: January 14, 2008, 01:58:05 PM »
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Offline tinu

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Re: SMOT TEST- can someone do this?
« Reply #52 on: January 14, 2008, 02:25:00 PM »
This is arguable and not entirely necessary.
If there is an area where the magnetic field is essentially null (and there is such an area in SMOT), then the magnetic potential of that point is the same as for an infinite distance.

Yes, that is correct.


Ok, thanks.
I saw your first post then started to slowly write in English and finally I saw your second post. What a relief! ;)

That particular point is B by a very good approximation (not exactly, because the ball would remain in unstable equilibrium in that point but B is very, very close to it, as close as the hand can place the ball). So, B is the point of highest magnetic potential energy, although it seems hard to accept at a first glance. That?s why Mb>>Ma.

Still no, I think.  The point B you choose has to pull the ball against friction and gravity, which means it's reasonably far from the field null.  W.r.t. the Bus's video, point A in the dish looks to be far enough away to have a lower PE.  Note that I wouldn't feel the need to argue this if you'd said "theoreticaly, we can have Mb > Ma", instead of just "Mb >> Ma".  For those who don't know, >> means significantly greater.

W.r.t the SMOT test in this thread, we're comparing a placement at a point B with magents to a placement at point B without magnets, so even though the gravity+friction argument doesn't apply here (due to the downward tiliting of the ramp), there is no cheating going on, because the potential at that point with the magnets moved far away will be a very good zero.

Of course, if the magents really aren't far enough away when they are "removed", then there could indeed be "cheating" of the sort you mention.

Cheers,

Mr. Entropy


One clarification, first: I was strictly talking about magnetic potential, in absence of any friction and in absence of gravity. It was my intention to analyze if Mb>Ma or Ma>Mb.

So, after seeing your agreement on the first statement (thanks for that again and please post a better written explanation that I can ever write, if needed), please eventually acknowledge that B is an inflexion point for magnetic potential (in a 2 dimensions graph) or a magnetic potential hill in 3dims. The chart of magnetic flux as provided by Vidar is an excellent aid to everyone interested.

Theoretically, in such a magnetic field the ball can be placed as to (magnetically) "fall" on whichever 360degree horizontally planar direction. But mechanical constraints in SMOT (the rail) make the ball possibly move either forward (toward C) or backward (toward the outside of SMOT). Please acknowledge that BOTH directions are possible. It?s only the user?s hand that places& finely adjusts the ball?s position as it moves forward (toward C), right?

If I?m not wrong in the above then, first conclusion: B is at least a local maximum of magnetic PE.
But: along A-B there are no other inflexion points on PE curve (except of course point C but that?s another story for later; let?s stick for now in the A-B region, where is the issue under discussion) because there is nothing that can cause such an additional inflexion.
From the two above it necessarily logically results that Mb>Ma for SMOT.

(I?ve wrote Mb>>Ma mainly to stress that B is similar to infinity while A is not, case in which is presumable that magnetic potential of B is much larger that those of A.)

There are many whats and ifs but I?d love to hear your thoughts on the above.
Whether you see any mistake, slap me hard to wake me up.
Thanks,
Tinu

Offline Low-Q

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Re: SMOT TEST- can someone do this?
« Reply #53 on: January 14, 2008, 06:41:32 PM »
Low-Q's claim can be tested easily, so there is no point arguing about it.  Just put a steel ball on a very very slight decline, so it rolls very slowly.  At one end of the decline is a SMOT.  If the ball rolls into the SMOT, never decelerating, then Low-Q is incorrect.  On the other hand, if the ball does not roll into the SMOT, or at least decelerates before entering, then he is correct.
Ok.

I have performed some small tests regarding this. I used two 30cm long neo-magnets (An array built up like bricks). The track is placed on the floor by using two wooden lists. The floor is a little bit uneven so the ball does not move smoothly, but the experiments went like this:

The ball is placed approx 30 cm in front of the magnets. In this horizontal level, it is no sign of attraction. I can bearly see some attraction when center of the ball is in line with the edge of the magnets enterance. There is bearly repelling forces right before this point, but not enough to stop the ball from entering, if some one accelerate the ball by hand in advance. However the repelling point clearly shows that the ball breaks down a bit in that area.

PS! What you'll see in the video, is that the ball trills backwards in the beginning of the experiment. This is caused by the floor, and not by magnetic repulsion.

The desired attreaction to the magnets before it enters the SMOT is next to nothing. The magnetic lines which is going from the magnets outer poles, are almost going perfectly angular to the direction of the ball, and the magnetic densidy is also very weak, so there is practically no magnetic attraction to the ball at all.

When the ball is pushed further into the SMOT, it accelerate, but stops approx. 3.5cm before the end of the magnet array.

Watch the 18Mb video I made 5 minutes ago, and let the experiment talk (Well, I don't say much though :))

http://www.lyd-interior.no/animations/P1140001.AVI

Vidar

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Re: SMOT TEST- can someone do this?
« Reply #53 on: January 14, 2008, 06:41:32 PM »
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Offline Omnibus

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Re: SMOT TEST- can someone do this?
« Reply #54 on: January 14, 2008, 07:12:56 PM »
As I said many times, you are confusing force and energy. Stop wasting everyone's time.

Offline Low-Q

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Re: SMOT TEST- can someone do this?
« Reply #55 on: January 14, 2008, 08:03:44 PM »
As I said many times, you are confusing force and energy. Stop wasting everyone's time.
Let me put it this way: Force is important to transfer energy. Without the force, you cannot move the mass (energy) in the ball, which has a product called transfered energy, or work. So in fact I'm talking about both force, mass and distance. The product is work - which I have understood is the sub-issue discussed in this thread.

Vidar

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Re: SMOT TEST- can someone do this?
« Reply #55 on: January 14, 2008, 08:03:44 PM »
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Offline Mr.Entropy

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Re: SMOT TEST- can someone do this?
« Reply #56 on: January 15, 2008, 03:39:05 AM »
please eventually acknowledge that B is an inflexion point for magnetic potential (in a 2 dimensions graph) or a magnetic potential hill in 3dims. The chart of magnetic flux as provided by Vidar is an excellent aid to everyone interested.

By B you mean the field null.  Yes, that is a potential energy maximum, as follows:  Assuming an ideal ball, i.e., very small and magnetically soft, the attractive force on the ball is proportional to the gradient of the magnetic field energy density.  This is a scalar field equal to the square of the magnetic field strength at every point, so:

F = k*grad(|B|^2), for some constant k.

We know that the force on object from a conservative force field is F = -grad(E), where E is the corresponding potential energy field, so the magnetic potential energy:

E = -k*|B|^2.

Isn't that convenient -- the magnetic potential energy in the ball at any position is proportional to the minus inverse square of the flux density.  Of course, then, the field zero at point B is a potential energy maximum because it its a field strength minimum.

Quote
Theoretically, in such a magnetic field the ball can be placed as to (magnetically) "fall" on whichever 360degree horizontally planar direction. But mechanical constraints in SMOT (the rail) make the ball possibly move either forward (toward C) or backward (toward the outside of SMOT). Please acknowledge that BOTH directions are possible. It?s only the user?s hand that places& finely adjusts the ball?s position as it moves forward (toward C), right?

Yes, at your point B, the ball feels no force, but that position is unstable (it's the top of a hill), so if you give it a little nudge either forward or backward, the magnets will push it further in that direction.

Quote
If I?m not wrong in the above then, first conclusion: B is at least a local maximum of magnetic PE.

yes

Quote
But: along A-B there are no other inflexion points on PE curve (except of course point C but that?s another story for later; let?s stick for now in the A-B region, where is the issue under discussion) because there is nothing that can cause such an additional inflexion.
From the two above it necessarily logically results that Mb>Ma for SMOT.

Ah, but there are other inflection points.  As you move away from the SMOT from point B, the magnetic field strength quickly increases to a maximum, which means a PE minimum.  This is the point where it switches from repulsion to attraction.  The field strength then goes down as you continue to move away, approaching a 1/r^3 drop off.  As the field drops off, the PE comes back up to zero as -1/r^6, and that gets you pretty close to zero pretty darned quickly.  By the time we get to point A, the PE is quite close to its maximum value of zero again.

So, we have Ma and Mb both close to the maximum value of zero.  Which is greater, do you think?  Theoretically, Mb might be bigger, but in real life:

- consider that you can't actually place point B, where you put the ball, in the field null to start a SMOT, because there's no force there.  You have to place it closer to the magnets -- a lot closer, because the magnets have to pull it uphill.

- consider the effect of the ball's radius on Mb -- you can't fit the whole ball into that single point where |B|=0.  There will still be a place where it feels no net magnetic force, and it will still be a magnetic PE maximum, but that maximum will be rounder and lower that the one a small ball would see.

In real life, because of these considerations, Ma > Mb.

Cheers,

Mr. Entropy

Offline Omnibus

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Re: SMOT TEST- can someone do this?
« Reply #57 on: January 15, 2008, 06:27:27 AM »
@tinu,

?His claim is very easy to dismiss by one word: conservative.?

No, that?s not only not a basis for an easy dismissal of my claim but isn?t a basis for its dismissal at all. On the contrary, as I?ve explained many times, violation of CoE is observed in this case not because the sum of all the energy terms in the closed loop gives zero mathematically but because some of these terms have appeared out of nothing, out of no source, which is a physical conclusion, not a mathematical one. It should be well understood that in such matters subject of discussion here physics makes mathematics and not vice versa.



It is also most curious to observe how someone like you, claiming to be a scientist, allows himself to say in the same breath:

??Unfortunately, no proof was ever posted. Please keep this in mind: no proof, of any kind, was ever posted!?

while at the same time posting my very proof which shortly states as follows:

?Think about it, when you impart to the ball energy |(mgh1 - (Ma - Mb))| to raise it from A to B then, if CoE is to be obeyed the ball must lose that exact amount, that is |(mgh1 - (Ma - Mb))|, when it goes back to A and complete a closed loop. Not so, however, in our case of the ball completing a closed loop. The ball being at B (raised from A) and having energy (mgh1 + Mb) at B in our case doesn't lose energy |(mgh1 - (Ma - Mb)| when it goes back at A and closes the A-B-C-A loop. The ball in our case loses, as was said, energy Mb = (mgh2 + [KE + ...] ) as well as energy mgh1. This energy (that is the energy (mgh1 + Mb) which the ball loses in going from B back to A, closing the loop) is more than the energy |(mgh1 - (Ma - Mb)| that was imparted to it to raise it from A to B and that's a clear violation of CoE.?

Therefore, your former statement that no proof was ever posted, let alone proof of any kind isn?t true.

Then, not satisfied by saying untruths you appoint yourself to be the speaker of an entity called by you ?any scientists I know?:

?Well, the above does not make sense to any scientists I know.?

continuing to push the falsity that:

?And consequently, the claim that SMOT is producing energy from nothing is not logically &scientifically supported by any proof.?

Speak for yourself, don?t hide behind that fictitious entity ?any scientist I know?. Also, if you?re really a scientist you should know better that it cannot stand as a scientific argument in a scientific discourse.

Therefore, I?m not only not withdrawing my claim despite your wish that claim to just somehow go away but I urge you to curb your enthusiasm to fill the thread with your misunderstanding and confusion, let alone cavalierly massaging the truth.

Offline hoptoad

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Re: SMOT TEST- can someone do this?
« Reply #58 on: January 15, 2008, 07:09:21 AM »
@Omnibus

Let's assume for a moment, that your claims for violation of C of E in a SMOT are true.
Next, let's assume that it's a "mere engineering problem" to close the loop in a SMOT, as you've claimed in so many posts already.

Such a "mere engineering problem" shouldn't be beyond your ability to solve with a practical working demonstration!

Instead of constantly telling us that it is a "mere" engineering problem to be overcome, someone with your "apparent" theoretical ability should have no problem in producing a working model that proves your claims.

If you are really not very good at practical hands on construction, then why not get a "mere" engineer to help you.

It is not a requirement of anyone to prove you are wrong. When you make an extraordinary claim, it is up to you to provide extraordinary proof of your claim !

A practical working model of a looped SMOT would end all debate. I'm sure if you spent half as much time producing a working model, as you have spent trying to convince everyone of your claims, you would have already succeeded in closing the loop, and all your detractors would have to admit they were wrong. But until someone, including yourself succeeds in getting the SMOT to work in a loop, thus proving in a practical way the correctness of your claims, then your claims will always be dismissed as theory not based in fact!




Offline Omnibus

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Re: SMOT TEST- can someone do this?
« Reply #59 on: January 15, 2008, 07:22:18 AM »
@hoptoad,

You've remained with the wrong impression that to prove CoE one needs to build a self-sustaining contraption. That isn't so. Therefore, your desire to see such contraption has nothing to do with what science puts forth as a requirement for a rigorous proof. Also, whatever seems to you extraordinary or non-extraordinary will always remain your own perception. Science doesn't divide claims into such categories and the criteria in science to prove the validity of a claim are uniform throughout.

 

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