Well fellahs
I spoke with Bill this evening,He's got it running down in the basement making power with no input ,hooked to nothing ??
Is this it??
http://www.overunity.com/index.php?topic=11303.0
I sure hope so!
He says he can build one in 5 minutes now [even sipping Coffee]
And will build one on a live link while he's running the Demmo.
Chet
PS
No he doesn't want your money ,your wife or the keys to the city!
He wants what we want.
To leave this world a better place then he found it.......................
PPS
I believe his Handle is Captainpecan
The output voltage he's indicated is in the hundreds of
milliVolts and he's not specified what the current produced
may be.
What comes to mind is something akin to a thermocouple
or a thermo-electric generator.
A solar heated home-made thermocouple is capable of
producing significant voltage at tens to hundreds of
milliAmperes for a small unit.
But then again, it may be something entirely different.
Guess we'll have to wait and see...
I tried...
http://www.overunity.com/index.php?topic=11303.msg298006#msg298006 (http://www.overunity.com/index.php?topic=11303.msg298006#msg298006)
Because:
#1: A proclamation.
#2: We must wait.
#3: Everything is safe.
#4: A visitor will show up.
Sadly, the love of money or the desire to become
wealthy is powerful bad medicine...
The way this thing is progressing is very strange.
I'll send my friend over there to witness it, he lives in Portland.
I'm prepared to be shocked, since voltage is very low, it must be a dirty solar panel! :o
I'm about half-way through the OU thread right now.
Something that costs about $20 in parts that can be purchased from your local hardware store.
Claims that it produces voltage and can charge capacitors. However, no indication of how much current it can output at a given voltage, hence no indication of a power output.
I don't know how technical "billmehess" is but that's already an issue. We will see......
MH surely this means the current is unlimited at any voltage and depends on the scale size of the build
QuoteI am currently generating a stable 850 mv this and amperage can be scaled to any level of output.
I think we need to wait this one out, with no details it's impossible to work out if it's feasible.
Personally i think he has done this the right way by not giving out advanced information before a public demonstration is held, although he may need to hire a football stadium to house us OU enthusiasts LOL
I asked him what the values of the caps he is loading but no answer. Problem is if the loading caps have high enough uF values, they could be loading the 800mv on their own or via a low voltage circuit in a loop with the loading cap and this could wind up being a total waste of time. As for the way he is disclosing this, it is juvenile and chock full of areas where Murphy's Law can just jump in and screw things up. So many things can go wrong until that date but men with plans think always that everything will go "as planned". Since when?
On a different angle, this may also be a good thing since if anyone else has a small device that is OU, they may be compelled to disclose it before Bill does and hence, beat him to the punch. Any takers? lol
wattsup
Bill certainly has raised our interest. He says he plans to make money by "selling plans". Perhaps OK -- but I for one wish to make sure this is not just a battery of some sort, that is, running off of chemical energy stored in dissimilar metals -- or pulling a small amount of electrostatic energy out of the air...
More on this concern: a friend noted to me that Bill gave his email address:
Quote"Not to worry all you good people all the necessry safeguards have been met.
I will have a location for the demo nailed down in the next few days.
I have received a number of request for direct contact- no problem for me
Bill Mehess
excel60@hotmail.com"
-- post at OU, 15 Aug 2011
Then you find at Youtube, searching for "excel60", the following "water battery" --
http://www.youtube.com/watch?v=TdWKfav5s_8
He claims "no degradation" of the metals, which is doubtful -- over time, this is a battery after all, and the energy comes from chemistry.
Is this a useful clue to his present claims?
Regrettably, I suspect an "air battery" for Bill's current claims ... which even if it works, will have a cost larger than what one can get today with solar cells (About $1.50 per watt).
Quote from: wattsup on 2011.08.16, 13:52:35
On a different angle, this may also be a good thing since if anyone else has a small device that is OU, they may be compelled to disclose it before Bill does and hence, beat him to the punch. Any takers? lol
Sure C.C
I have a set of 1F caps that will explode on their own without a shorting jumper. Give me a month after I remove the jumpers but I wasn't able to determine when they would explode before ;D
Another youtube video by excel60 (Bill Mehess, we surmise) --
http://www.youtube.com/watch?v=TvNCrDTQZNk&feature=mfu_in_order&list=UL
"It will just continue hour after hour, day after day, 24/7 - just by replacing water".
And -- http://www.youtube.com/watch?v=lvETzw0KkNc&feature=autoplay&list=ULaiA1FQ4iQGw&index=4&playnext=1
Copper cathode and zinc-coated iron anode.
"These water batteries do not wear out". Plus a "little bit of lemon juice".
"Doesn't wear out"
NO, Bill -- the metals will degrade.
Bill replied that -- Yes, he posted vids under excel60 on water batteries, etc.
My question at OU:
QuoteQuote from: JouleSeeker on Today at 06:03:40 PM
Ah -- "excel60" in you email (thanks NP for the observation), leads one to do a search on youtube for excel60 ...
And we find an intriguing discussion of "water batteries"
http://www.youtube.com/watch?v=TdWKfav5s_8
AND
http://www.youtube.com/watch?v=TvNCrDTQZNk&feature=mfu_in_order&list=UL
Copper cathode and zinc-coated iron anode (galvanized).
(These can be purchased at larger hardware stores.)
And a few other interesting vids by excel60.
Did you do these "excel60" videos, Bill? Thanks.
He said "Yes."
But then replied further --
QuoteI think I see where you may be going with this so let me clarify a bit
All batteries require some kind of electrolyte to work , even the so called dry cell battery use an electrolyte in paste form.
Batteries have materials that are consumable
Lazerlight did some great work with a dry cell which had mgnesium strips interacting with a carbon center no electrolytes but the magnesium was consumable
Batteries increase in voltage when connected in series and in current (amperage) when connected in parallel.
My device:
No electrolyte
No Consumables
New Information for all of you:
The individal modules increase in both amperage AND voltage when connected in parallel
There is a strong non moving mechanical component
I'm not sure how his reply/results fit with Ohm's law... curious reply, really.
Quote from: PhysicsProf on 2011.08.16, 15:22:22
...
NO, Bill -- the metals will degrade.
Absolutely correct! When very tiny current is drawn from
the cells the result of electrochemical deterioration is not
evident at all.
Few experimenters realize the amount of electrical power
available from even a small piece of zinc, or aluminum, or
magnesium.
Electroplaters know; but they're a vanishing breed nowadays.
PhysicsProf:
QuoteActually, batteries connected in parallel will have the same voltage (assuming batteries of the same voltage) and therefore the SAME CURRENT (across a resistor, for example) by Ohm's Law, I = V/R. Perhaps a minor point, but as a retired professor of physics, having taught EM for many years, I sort of jump in on points like that. Habit, I guess.
Sorry but you're incorrect here. The available current will double if you put two batteries in parallel.
At these low power levels putting two "batteries" (or whatever they are) in parallel is not really an issue but if you want to consider all of the angles when it comes to playing with batteries or unknown voltage sources, you would be best off to never put them in parallel.
Quoting Billmehess:
QuoteThe device absolutly works in the parallel mode. If I take say 4 modules that are putting out 100 mv ( they do not all put out exactly the same amount I am using this to simply) if I put them in series my voltage is around 170mv if I connect them in parallel the voltage is around 380mv.
One of the other devices setting next to it has 6 modules and in parallel they put out around 510 mv
and in series 185 mv ?????????????????????????
Also the current increases as it should. I know this because I can connect the device to a capacitor (I usually use a electrolytic 25 4900 mf for this purpose) and in one hour I can record the voltage in the cap. The voltage is always directly perportional to the number of modules in parallel, which of course is the way it should be.
It doesn't look good at all from reading Billmehess on OU today.
The Las Vegas odds-makers are upping the odds for the chances of this being real. They now sit at 1 in 1000, and those are odds within the frame of reference of the forums.
Don't shoot me, I am just sharing my honest opinion. This isn't real and it's pretty obvious that Bill is not at a technical level where he can demonstrate mastery over the subject matter at hand. If you make the claim then you are going to be subject to scrutiny whether you like it or not.
I will now wait for the post-demo fallout, not up to debating anything about this one personally.
MileHigh
I wonder if Bill is confused about the meaning of Series and Parallel, or we don't understand the context or geometry of his device. Placing plates in a battery next to each other can be considered "PARALLEL" but the plates can be connected either in series or parallel from a circuit stand point. Maybe he is talking about geometry here not electrical circuits.
@MH
QuoteSorry but you're incorrect here. The available current will double if you put two batteries in parallel
MH, the Profesor is correct and so are you. Try to understand what he is saying before jumping to conclusions. He is talking about the load current which depends on the voltage, that's why he quoted ohms law. And what you say is also correct, because all batteries have a source impedance and putting two in parallel lowers the source impedance to half, so we can draw twice as much current for the same battery voltage drop (under load), and of course the obvious, putting two batteries in parallel doubles the amphour rating.
EM
I doubt he is confused but I think almost everyone else is. Intentional?
Maybe this is a stretch for some... What has the result of twice the potential and twice the stored energy when side by side?
"The individal modules increase in both amperage AND voltage when connected in parallel
There is a strong non moving mechanical component"
When you use two magnets instead of one during induction you double the potential and the energy produced.
So this tells me he is using EM induction without obvious relative motion between the magnet and the coil. You can mount a magnet with springs inside a coil and produce several hundred millivolts with nothing more than ambient energy (vibration, sound, etc.).
Certainly, this is not new.
>>Edit...
Maybe I'm off base on this one. I just remembered placing two magnets in parallel with each other yields 4 times the magnetic force not double. Oh well, at least it still increases the amount of induction for relative motion ;D
Thanks, EMDevices -- I agree (above)..
Update from the OU thread: What strikes me is that I calculated the output power from Bill's data, and the result was a mere 53 MICROwatts for TEN modules -- And no one blinked. Not even here.
Even if the effect is real -- some kind of electrostatic antenna for example -- the power is TINY. Or am I wrong about this? ;)
I just posted this at OU; again, comments welcomed.
QuoteOK --
I've checked my math and found it correct, to recap:
53 MICROwatts for Bill's 10-module system (connecting the 6-modules with the 4-modules).
See below along with previous posts -- I have also checked from Bill's comments that the 280mV rise in the 4900 uF cap took one hour for "the two sets of modules together in series".
But please correct me if my calculations are wrong, because, Bill and other friends, this is a very VERY tiny effect:
1. 53 uW for 10 modules means about 5.3 microwatts/module (average)
2. To reach just ONE watt would thus require about 189,000 modules.
3. Earlier there was an indication that one could get the materials needed to build a module at a hardware store for less than $20. OK, let's say just $10.
4. Combining 2 and 3, we find that for ONE stinkin' watt it will cost roughly $1,890,000 -- nearly 2 million bucks !
5. Let's take roughly 1 kW to run a house (a low estimate I believe) -- the cost for the modules (with present data) would be $1,890,000,000 or nearly $2 Billion bucks. That's billion with a "B".
I can't afford it.
5. By comparison, solar panels are now available for about $1.50 per watt output, a factor of roughly a million CHEAPER than this device, at present.
Please correct me if I'm wrong. I hope I'm wrong. OR -- I hope the device can be SCALED up in output POWER by a factor of roughly a million without raising the price... Is this possible?
Quote from: billmehess on August 16, 2011, 10:34:53 PM
Yes this would certainly help me , and yes I meant to say uf not mf. When I connect the two sets of modules together in series I get again around a 700 mv out put , but the total of the summation of all the individuals is around 1400 mv, voltage loss?
Anyway at 700 mv connected to the cap. and using a multimeter and starting with the cap reading 0.00 the cap will charge at the rate of about 180 mv/hour.
Now I have a way to increase the current output substanially to radically improve the amperage available.
Based on these figures could you calculate the power output.
The physical size of the 10 modules is around 12" x 5" x 5" total for all 10. This can be scaled down greatly but that size is not really of concern to me at this point in the game.
Bill
I just re-read above the voltage on the cap after 1 hr is 280 mv not 180 mv.
PhysicsProf:
Assuming that we are talking about some sort of voltage source with a high-impedance output, then using the cap charging method would not be my preferred way to calculate the power output. The way you crunched the numbers was oversimplified because you did not attempt to derive the output impedance of the device using the provided data. Also, Billmehess' statement that the cap could be charged at "280 millivolts per hour" would seem to indicate that he is not familiar with the exponential waveform associated with capacitor charging and the RC time constant.
All that Billmehess has to do is connect something like a 1-megaohm potentiometer to his array and adjust the resistance until the voltage output drops by half, and then measure that resistance value with a mulimeter. That will give you the impedance match and therefore tell you the output impedance of his device and at the same time tell you the maximum possible output power from the device.
This is all based on the preliminary and presumably reasonable assumption that this device as a source of electrical power more or less resembles a high-impedance voltage source.
Unfortunately Billmehess is being led through this exercise as opposed to leading. If he is making the OU proposition and talking about registering patents one would expect that he or one of his associates with more technical knowledge would at least be able to characterize the device and be able to tell you it's power output capabilities before making the preliminary announcement.
The fact that it is 12" x 5" x 5" and only (allegedly) generates power on the order of microwatts is also a major issue. It could very well be an effect related to what WaveWatcher stated. You have a physically large device sitting in a bath of ambient thermal energy and it's producing microwatts of power. Not only would it be overwhelmingly expensive to get usable power from but it would also be overwhelmingly huge. Billmehess has made claims of delivering "considerably more output" after he incorporates some upgrades. We will have to see if he can deliver on that.
The bottom line is that even if the power output is low, it would still be an Earth-shattering event if it was indeed true. However, for me personally, this is all simply not adding up and there is a conventional explanation for all of this and my money is on WaveWatcher.
MileHigh
Guys:
I really don't get what this is all about. There is practically no output from what Bill is saying, so... Seams like another game to me, as always. Hopefully there is more to this. But, will it even light an led???
This morning before leaving for the office, I took a 4000 uF 25vdc capacitor and put my volt meter on it and it showed a steady 0.387 vdc, sometimes hitting the 0.388 then back. But it stayed there using nothing. I tried moving a magnet near the capacitors metallic casing and there is no effect. So, is that OU? Now if you put a very small circuit on that capacitor that can click on and off a small transistor to cause some pulsing in anyway through a parallel coil, would that increase the capacitors loading, don't know, but the capacitor itself seems to be doing a good job on its own.
wattsup
Quote from: MileHigh on 2011.08.17, 10:01:00
All that Billmehess has to do is connect something like a 1-megaohm potentiometer to his array and adjust the resistance until the voltage output drops by half, and then measure that resistance value with a mulimeter. That will give you the impedance match and therefore tell you the output impedance of his device and at the same time tell you the maximum possible output power from the device.
[snip]
MileHigh
MH -- would you spell that out, pls -- say the voltage drops by half with a resistance on the pot of 100Kohms -- what is "the maximum possible output power from the device"?
Here is a good link:
http://www.qsl.net/wa5bxo/power.html (http://www.qsl.net/wa5bxo/power.html)
The power out is the "half" voltage squared divided by the resistance value. Therefore the lower the output impedance of the source the higher the maximum power flow is.
MileHigh
Professor, I simulated this graph for you so you can see the maximum power delivery condition. See chart.
When the battery source impedance (Rs) = load impedance (Rl), we have a condition of maximum power delivery to the load. For reactive circuits or general impeances, the load has to be equal to the conjugate of the source impedance.
Note: This condition does not mean the battery's internal voltage source is delivering the maximum power it can. Obviously not. The maximum is when you short the terminals of the battery and then all that you have in the circuit is the internal voltage source in series with the internal battery source impedance. This produces the largest current possible. The battery will dissipate this power as heat generated in the internal source impedance or resistance.
EM
I must caution those few capable of determining the internal resistance of their batteries and other energy sources.
In the battery world it is safe to assume that applying a load impedance matching that of the source battery will probably end up as a disaster.
Impedance matching is not always the best thing to do between source and load. Keep in mind that when your load and battery have matched impedance 1/2 of the energy used is wasted by the battery as heat. It can be better to have a higher impedance in the load.
You want as much of the expended energy as possible doing useful work not boiling out your battery.
WaveWatcher:
Your points about the (big) battery testing are indeed good and we don't want anybody doing anything risky or dangerous with their batteries.
In the rarefied model for Billmehess we are sure that it's a very high impedance source and since the power is all allegedly "free" we don't care if we burn off half of the power inside the device, all that we care about is extracting the most possible power from the device.
We are making the assumption that we are not going to damage the device doing this because the power levels are so low from this "fountain of energy" so we go for broke and try to suck as much free power as possible.
Certainly we know that there should not be any thermal issues when we go for broke and try to get as much power as possible. On the other hand, my original point was to characterize the device and not necessarily do an impedance match when running the device. The important thing is to measure the output impedance itself so you can characterize the nature of the power source.
The implicit desire is to "elevate" the discussion on OU and get people up to speed with the concept of output impedance. I have seen a fair number of devices like this where the forum members only talk about how many millivolts it generates and how many hours it takes to charge a capacitor.
You never know....
MileHigh
Quote from: WaveWatcher on 2011.08.18, 01:48:54
...
Impedance matching is not always the best thing to do between source and load. Keep in mind that when your load and battery have matched impedance 1/2 of the energy used is wasted by the battery as heat.
...
Impedance matching is done for only one reason: to get the maximum power from the source to the charge.
Its utility is mainly for ac circuit, by nullifying the reactive component of the charge impedance.
Naturally it would be a disaster to apply it to the battery except if we want to obtain the maximum energy in the shortest time, at the price that as many power in the charge as in the battery will be dissipated, possibly leaving us with a melted battery... :(
MH,
We are in complete agreement except for one small detail.......
I would have suggested a 100 megohm pot :)
My concern was for parts of the discussion being appled elsewhere out of context. I've made that mistake before on some other subjects :-[
Very good points. Thanks for the refresher and the application to Bill Mehess's claims. (He seems to have gone rather silent lately.)
Bill
Quote:
I spent about 12 hours in my shop adding my upgrades here are this morning results, I will be running tests all day long and over the weekend to see if these hold up
First I am using just 3 modules and a 60,000 uf cap.
Time Start voltage end voltage Cap voltage
8:30 am 2.106 volts 2.256 volts .562 volts
9:00 am 2.115 2.256 .563
9.15 am 2.136 2.261 .573
9.45am 2.199 2.285 .555
10 am 2.202 2.290 .540
10:15 am 2.195 2.276 .539
10:30am 2.213 2.285 .538
The ending voltage is the voltage after 15 minutes when I disconnect the cap. I short out the cap during that time to restart from a 0.00 reading of the cap.
The test is now done for 60 seconds not 1 hr or 3600 sec.
The device has always showed a increase in voltage even after running the load.
using E= 1/2 cv (squared) and dividing by 60 sec. wattage is now .145 watts
Yikes, there are a lot of problems with this whole attempt to measure here the more I think about it.
Firstly, there is some uncertainty if the time period is 60 seconds or 15 minutes.
Let's assume it's 15 minutes for the sake of argument. Bill would have to set up a control test where he just has a cap and connects approximately the same amount of wire to the capacitor terminals. It's very possible that the big electrolytic cap is self-charging from the ions in the air. So if the control setup creeps up in voltage also that just about kills the proposition.
Going back to the testing as it is being done now, when you first start the test with the big cap at zero volts, then essentially 100% of the alleged free energy being generated by the device is being burned off inside the device itself. It's a crippling problem for the cap test and the larger the capacitor the worse off things are.
WaveWatcher also mentioned you probably need a 100 Mega-ohm potentiometer to do the impedance matching test and he might be right. If he is right then the typical 1 Mega-ohm input resistance of the multimeter would be disturbing your attempt to make the impedance matching measurement. If that's the case you need to do a differential voltage measurement between say a 1 kilo-ohm potentiometer and Billmehess' device and turn the potentiometer to zero-out the voltage differential. Then measure the voltage on the potentiometer output separately. All that just to measure the open-circuit voltage on what we assume is a super-high-impedance output device.
Then you have to repeat the process, but this time say use a 10 Mega-ohm resistor across the device, and use the potentiometer and voltage differential method to measure the voltage drop.
You do all that and you have a relatively undisturbed open-circuit voltage measurement, and a relatively undisturbed voltage measurement under a 10 Mega-ohm load. That gives you the ability to calculate the output impedance of the device with some confidence.
Once you do that you are done. You know the open-circuit voltage and you know the output impedance of the device. That's it, now you know it's power output capabilities under load. All of this is based on the presumption that the input resistance of your multimeter is too "low" and will disturb Billmehess' device.
Before I did any of this, I would do the control experiment with the capacitor to see if the whole thing is just due to hot (ionized) air and the nature of electrolytic capacitors.
But certainly this primitive big electrolytic cap test is junk, the device is seeing a short-circuit to ground or near-ground for a long time and most of the alleged free power is being burned off inside the device itself. Therefore the output power is not really being measured properly and the number crunching is showing power values that are much lower than they potentially are with an impedance-matched setup.
The critical thing is to figure out the output impedance of Billmehess' device. If the output impedance is very very high then you have to go to the extreme measures like I outline above.
Anyway, this posting will likely be ignored so what the hey....
MileHigh
MH, I'm going to let you handle this one. I'm too tired now. But we can calculate the source impedance from the data he supplied because:
He gave us the capacitance. 60 000 uF
He gave us the charge time = 60 sec.
He gave us the capacitor voltages, e.g. 0.573 V
He gave us the open source voltage, e.g. 2.26 V
So, work backwords and solve for the time constant, then solve for the resistance.
EM
EM:
You could very well be right, and you are bringing it back full circle, because that's what i said the other day. I suppose that WaveWatcher mentioning using a 100 Mega-ohm pot got me thinking. Not to mention the whole thing could primarily be due to the caps self-charging. He does mention 15 minutes and 60 seconds in the same posting though, which is confusing.
The bottom line is somebody should try to measure the output impedance. Assume for starters that your open-circuit voltage is not being disturbed by your multimeter. That would be a good start.
MileHigh
MH,
This is what I get by inverting the basic equation: V(t) = Vf exp(-t/tau), where 'tau' is the time constant.
tau = - t / ln (V(t)/Vf) = -60 / ln(0.573/2.26) = 43.7 s
tau = RC , therefore
R = 43.7 s / 0.06 F = 729 ohms
EM
PS. He said he has 3 modules, and probably they are in parallel like he mentioned, so the internal resistance of each module is 3 x 729 ohms = 2187 ohms What kind of process would have this kind of internal resistance? It has to be a chemical process and poorly designed battery geometry.
EM:
It sounds really low, I am suspecting that the time is actually 15 minutes. Or something else is not right.
Anyway, I should just wait for the big unveil.
MileHigh
PS: That sounding better but I'm still unsure of the data.
PPS: You are blowing my mind with your use of natural logarithms! lol
Bill
Quote:
Voltage in = 0
Voltage out = 800 mv
No electrolytes
No consumables
No heat produced
No emissions
-----------
You won't be able to wait MH!!
None of us is any good at waiting !!
Chet
Chet:
I have heard you on Karoke night doing an upbeat version of a twangy country song, "The Ballad of the Unhatched Chickens" many times now...
Careful!
MileHigh
Bill Mehess is this the one?, I'm thinking not just because of the low power output and a simple gut feeling which ain't very logical but it has usually served me well in the past, call it intuition. As well if he did have something would it matter? If the Kapanadze device can supposedly pull 100Kw out of thin air from something that could sit on my kitchen table we have an issue of practicality to contend with.
That's my 2 cents
Regards
AC
Bill
Quote:
Ok lets run the numbers
joule seeker says the wattage is 150 mw or .000015
voltage is 2.5 volts
we know that
wattage= voltage x amperage
then
.000015=2.5v(amps)
voltage divided into watts gives us a amperage of .000006 or 6 microamps. Google says that a LED can use very low amperage, with its norm being 20-25 ma. And on the low side 1 ma to 2 ma.
What am I missing here?
But the LED is lighting looks like real world application to me. And this is with only 3 modules that have been under a load for 11 hours and after the load (60000 mf cap.) was removed continued to raise in value.
My hat is off to all your really bright people out there so please explain how or why this is happening
Bill
----------------
Quote from: WaveWatcher on 2011.08.18, 09:54:59
...
My concern was for parts of the discussion being appled elsewhere out of context.
...
I understand. But Bill Mehess claim is not propped up by any evidence, so general concepts involved in this thread are much more constructive... :)
Quote from: allcanadian on 2011.08.19, 02:07:25
Bill Mehess is this the one?, I'm thinking not just because of the low power output and a simple gut feeling which ain't very logical but it has usually served me well in the past, call it intuition. As well if he did have something would it matter? If the Kapanadze device can supposedly pull 100Kw out of thin air from something that could sit on my kitchen table we have an issue of practicality to contend with.
That's my 2 cents
Regards
AC
I'm in agreement here, once again the tail is wagging the dog as the best minds on the forum are bogged down with something that will surely crash and burn before Aug 31. (my gut feeling based on Captain Pecans other bold claims of OU in the past). How many times have we seen this scenario play out before with Ainslie LTseung, etc.
Now Bill Mehess is asking forum members to answer questions that he poses without revealing anything of the construction of the device. Do we have to do his homework for him while he keeps all the marbles?
These minds would probably be better off discussing how OU might be achieved starting from scratch with a clean blackboard.
Meanwhile, the perpetual cheerleader wants to hitch his wagon to the latest long shot.
Hats off to you guys anyway.
ION
I bring my wagon here for a reality Check!,for the sole purpose of Vetting these claims ASAP so as not to waste the resources of the Community!
Its the reason I call people ,Its the reason my Name is no secret.
How can you have a credible conversation with a complete stranger If the Conversation starts out "My identity is a secret" ,or something of that nature............
As long as the members here and elsewhere remain anonymous it will always Stifle progress!,however I do see the Need for anonymity.
If you think I like putting my name and Email address up on a public forum when A man is screaming "I GOT IT I GOT IT".
You are Very Very Mistakin!!
I do this for this community!
I will gladly play the fool for this effort,nothing ventured nothing gained!
I respect your opinion.
Chet
BTW
I'm not as bad as I used to be.............
BTBTW
Anybody else having problems getting on OU.com?
A friend in UK also not getting thru for some time now??
I haven't been able to get onto OU for about the last 7 hours.
Chet
For the record, I think you are a good guy and bring a lot of things to be examined.
It sometimes puzzles me that you start with such zeal, you must have been disappointed many times.
Anyway better to have Enthusiasm (filled with God) than my rabid skepticism.
Still I keep hoping too.
Cheers, ION
Thank you ION
My Zeal does come from faith
Faith in a creator and faith in Our destiny ,which IMO includes a source of power for all!!
Your scepticism is founded in Knowledge,a very strong footing!,And I know if anyone knows his way around a
transformer its you.............
Thank you
@Prof
A good sign is Stefan put up the "I'll be back" ..........
Chet