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Author Topic: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?  (Read 144790 times)

MileHigh

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #330 on: July 10, 2016, 04:41:49 AM »
MH

You really are lost,and it is becoming more apparent each day as to how little you understand the simple JT circuit.
The battery is also a resistor,and as the voltage drops,the internal resistance grows,and so do the I/R losses associated with that battery.

In the second circuit,these losses are only had when the transistor is on.
In the first circuit(your circuit) these battery I/R losses are not only included when the transistor is on,but they are also included when the transistor is off.

During the off time of the transistor in circuit 2,the current loop excludes the battery,and thus the losses associated with the batteries internal resistance.
In the first circuit(your circuit) the battery becomes part of the current loop during the off time of the transistor,and so also includes the I/R losses associated with the battery.

Why you find this so hard to understand-i guess we will never know.\

Brad

Okay Brad, here we go.

For starters, you are just playing a game when you say, "You really are lost,and it is becoming more apparent each day as to how little you understand the simple JT circuit."  It's all complete crap and you know it and everybody knows it.  It's a defensive measure by mounting a fake offense, nothing more than that.  It's you making a shameless spectacle of yourself.

Secondly, it's the old cliche at this point.  What I said to you passed right though you like you weren't even there.  That puts you and your understanding in question, again.  It's like you have an aerogel brain, and an idea can travel for 14 light-years though your head before it strikes a neuron.

Quote
In the second circuit,these losses are only had when the transistor is on.
In the first circuit(your circuit) these battery I/R losses are not only included when the transistor is on,but they are also included when the transistor is off.

Yes, but if the two circuits have the same average current draw, that means by definition that there must be a higher current draw when the transistor is ON in the second circuit.  Higher current draw equals higher internal losses when the transistor is ON.

So you have a case of lower internal losses with continuous current flow and higher internal losses when the current flow toggles ON and OFF.  They balance out.  It's like six of one and half a dozen of the other.  How many light-years will this concept have to travel in Brad's brain before it hits a neuron?

Quote
Why you find this so hard to understand-i guess we will never know.

I think we know why you find this so hard to understand.  You have to work on improving your ability to analyze and think beyond just your first impressions.  You have to work on applying the knowledge that you already possess.   You have to work on reading and understanding what people say to you.  And you most certainly have to work on your language skills so you can express yourself properly.

MileHigh

tinman

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #331 on: July 10, 2016, 06:04:42 AM »
Okay Brad, here we go.

For starters, you are just playing a game when you say, "You really are lost,and it is becoming more apparent each day as to how little you understand the simple JT circuit."  It's all complete crap and you know it and everybody knows it.  It's a defensive measure by mounting a fake offense, nothing more than that.  It's you making a shameless spectacle of yourself.

Secondly, it's the old cliche at this point.  What I said to you passed right though you like you weren't even there.  That puts you and your understanding in question, again.  It's like you have an aerogel brain, and an idea can travel for 14 light-years though your head before it strikes a neuron.







MileHigh

Quote
Yes, but if the two circuits have the same average current draw, that means by definition that there must be a higher current draw when the transistor is ON in the second circuit.  Higher current draw equals higher internal losses when the transistor is ON.

Only not ::)

Quote TK:==PSU
So, Circuit 1 ran at an average input power of 90 mW and produced 63.9 lux at the sensor
Circuit 2 ran at an average input power of 40 mW and produced 30.0 lux at the sensor

Quote TK:==battery
Circuit 1 gave 49.3 Lux at an average input power of 54.6 mW for an efficiency of 903 Lux/Watt.
Circuit 2 gave 26.1 Lux at an average input power of 28 mW for an efficiency of 932 Lux/Watt.

Now why do you suppose that is MH?  ::)

Quote
So you have a case of lower internal losses with continuous current flow and higher internal losses when the current flow toggles ON and OFF.  They balance out.  It's like six of one and half a dozen of the other.  How many light-years will this concept have to travel in Brad's brain before it hits a neuron?

Yep--you are lost.
In fact,i cant even believe i am reading this from you,it is like you went from some sort of reality,to a land of MH dreams. Do you have any idea as to how high the internal resistance go's in a AA battery when the voltage is down around .9 volts MH ?--lets see your mighty pen work that one out without the bench. Well i do know the answer to that MH,and i know it because i can measure it on the bench. So give it a shot MH,show us how mighty your pen is without the bench,and give us a round about internal resistance value of a AA battery at around 900mV--lets not get to carried away with different types of batteries ATM,just use an eveready heavy duty battery as an example.

Quote
I think we know why you find this so hard to understand.  You have to work on improving your ability to analyze and think beyond just your first impressions.  You have to work on applying the knowledge that you already possess.   You have to work on reading and understanding what people say to you.  And you most certainly have to work on your language skills so you can express yourself properly.

My language skills are just fine that you MH,and the fact that you keep bringing that up,is just a sign of weakness.
It is also true that it is you that dose not listen to others--you know,the guys on the bench.
It is also becoming very apparent that i could indeed run rings around you on the bench.

MH,i have made this very easy for you to follow and understand.
Below are the two circuits,where in both cases the same amount of energy will be delivered to the inductor--lets say 100mJ of energy. I have included the batteries internal resistance that i measured !on the bench! of a normal alkaline battery with 900mV across it,and subtracted 10% of that calculated value for error margin.
Now we have the inductor in both cases storing the same amount of energy.
Can you please tell everyone here,which circuit will deliver the greatest amount of that stored energy to the LED?.

It is extremely clear MH that you throw all common sense out the window,when you need to be right outways your intelligence.
I can also see now,why you will not take me on in a simple little JT challenge--even when some one else was going to do the building for you. All you had to do,was put your !claimed! knowledge to paper-->you know,that mighty pen ::)

So MH,i am afraid that i have clearly shown that the bench just kicked your ass ;),and that my learning's on the bench just showed how your pen was not even in the race.
I am still amazed at how you made such a blunder,and as to how you came up with circuit 1 being more efficient(in any way) than circuit 2.

For me, there are only two outstanding issues and I will mention them again and I will put them in a better sequence this time:

1.  MH gets up the learning curve and understands as to how the JT circuit work's,and can be made more efficient,and clearly demonstrates that he understands what he is doing.
2.  MH admits that he is wrong when he stated that circuit 1 is more efficient than circuit 2.


Brad

tinman

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #332 on: July 10, 2016, 06:17:44 AM »
author=MileHigh link=topic=16225.msg488167#msg488167 date=1468116506]








Quote
I seriously doubt that TK made a mistake in his power measurements.  It doesn't make any sense.  All that he had to do was scope the voltage and current from the power supply for both circuits.

Indeed,and when he carried out the power measurements while using the power supply,circuit 2 was more efficient at various voltages.
BUT-when he carried out those same measurements while using a battery,the outcome was the reverse :o
Now,why do you suppose that was MH?
I wonder if TK took into account as to how,or what the current was traveling through during the off time in each test in circuit 1  ;)
If the two test were carried out where the supply voltages where the same,then what changed between the PSU and battery?
You say i should use my brain a bit better?,well perhaps you should use yours a little better,and think before plastering such rubbish all over this thread.

Quote
your logic is flawed and makes no sense.

My logic is just fine,but seems your has gone missing.

Quote
I was not talking crap at all.  I tried desperately to explain to you what the true reason for having a base resistor was but you would have nothing to do with that, it was pure willful ignorance on your part.

And it is rubbish MH.
The less resistive losses you have in a circuit,the more efficient that circuit becomes.
I so wish you would have taken up the challenge presented to you MH,but know i see why you did not want to go there.

Quote
You are just talking stupid gratuitous foolish idiocy, you need to blow off some steam and thar she blows!

It would seem ATM MH,that your boiler has exploded.

Quote
My pen is absolutely fine.  You need to work on thinking straight and being able to string five sentences together that make sense.  Those are some of your biggest problems.

Only you have a problem understanding MH--no one else dose.


Brad

MileHigh

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #333 on: July 10, 2016, 07:27:43 AM »
Brad:

Quote
Only you have a problem understanding MH--no one else dose.

You don't know that.  If nobody says anything you don't know one way or the the other.  Critical thinking skills, Brad, say something that makes logical sense.

Look at this mangled pathetic excuse for the English language:

Quote
As  i pointed out to MH,the second circuit eliminates the losses associated with charging the battery also,when a battery is used in place of your PSU. The battery will produce more waste heat,when being discharged and charged continuously,and also there are the internal resistive losses that grow as the battery voltage drops.

You've got problems.  You can't even get "charging" vs. "discharging" right in your head.  The second sentence is an illogical disaster.

I asked you how you would measure a supercapacitor on your bench and you had nothing to say except for one obtuse and near-ridiculous statement.  So that is telling me you can't conceptualize by yourself how to measure the size of a supercapacitor and any other possible parameters.  So much for your "bench smarts," you are coming up goose eggs for the capacitor question.

So, knowing that, please tell me how you measured the output resistance of your battery.  I take nothing for granted with you and you will have to state how you did it so I can verify that you got it right.  And there is another challenge for you.  Don't you dare rattle off one vague and ambiguous sentence.  You have to properly articulate how you go about measuring the output resistance of a battery.  You have to put a sequence of sentences together that actually make proper sense.

MileHigh

MileHigh

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #334 on: July 10, 2016, 07:57:38 AM »
Okay Brad.

Let's just sample some approximate numbers and work out a very simple problem for illustrative purposes.

Let's say we have a source voltage of 1.5 volts and an output impedance of seven ohms.
Let's say that we have circuit #1 that draws 50 milliwatts from the power supply and draws continuous DC current.
Let's say that we have circuit #2 that draws 50 milliwatts from the power supply and draws current with an 80% ON time and a 20% OFF time.

Let's examine these two circuits.

Circuit #1:

The current is 0.050/1.5 = 33.3 milliamps
The power lost in the internal resistance of seven ohms is 0.0333^2 x 7 = 7.78 milliwatts

Circuit #2:

We know from above that the average current is 33.3 milliamps.
Therefore the ON current for 80% of the time is 0.0333 x 5/4 = 41.7 milliamps.
The power lost to the internal resistance of seven ohms is 0.0417^2 x 7 x 4/5 = 9.72 miliwatts

Well look at that Brad.  When you put the two circuits on an even playing field where they draw the same amount of power from the fixed 1.5 volt power supply, circuit #2 that has the 80% ON, 20% OFF duty cycle has more losses due to the internal resistance of seven ohms.

Brad, I have made this very easy for you to follow and understand.

MileHigh

tinman

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #335 on: July 10, 2016, 07:58:30 AM »
 author=MileHigh link=topic=16225.msg488175#msg488175 date=1468128463]



Quote
You've got problems.  You can't even get "charging" vs. "discharging" right in your head.  The second sentence is an illogical disaster.

As i said MH,only you think there is a problem with how i word thing's,as you have no idea as to how the JT work's when the battery is included in the inductive kickback circuit.

Quote
You don't know that.  If nobody says anything you don't know one way or the the other.  Critical thinking skills, Brad, say something that makes logical sense.

A classic example where my thinking is spot on,and yours is Miles off.

Quote
Look at this mangled pathetic excuse for the English language:

As  i pointed out to MH,the second circuit eliminates the losses associated with charging the battery also,when a battery is used in place of your PSU. The battery will produce more waste heat,when being discharged and charged continuously,and also there are the internal resistive losses that grow as the battery voltage drops.

Everything i said in that statement is correct MH,and even a grad 1 school kid would understand it.
The reason that you do not understand it,is because you have no understanding as to how !your! JT circuit works.

Quote
So, knowing that, please tell me how you measured the output resistance of your battery.  I take nothing for granted with you and you will have to state how you did it so I can verify that you got it right.

Lol
I most certainly do not need your verification MH on anything,and more so now that i have seen just how limited your understanding on a simple circuit really is.

Quote
  And there is another challenge for you.  Don't you dare rattle off one vague and ambiguous sentence.  You have to properly articulate how you go about measuring the output resistance of a battery.  You have to put a sequence of sentences together that actually make proper sense.

And again--making demands.
Like i said MH,i in no way have to bow down to your demands,or have you judge me on if i have it right or not.
You clearly have lost the race,and are now struggling to keep your head above water.
You know how it looks ,when an amature hobbyist like myself has to explain a simple circuits operation to some one like your self--a self acclaimed guru,who is now struggling for air,as he slowly sinks to the bottom.

I posted two schematics that represent the two circuit's,and gave a valid reason as to why circuit 2 was more efficient,but i see you avoided that like the plague,and we all know why.

Like i said,there is two reasons you did not take me up on my challenge.
1-you know the bench is mightier than the pen,as you said your self,you would need a bench before putting pen to paper.
2- You know that i would wipe the floor with you,and this much has become apparent from the foolishness of your claim that circuit 1 would be more efficient than circuit 2.

TK's power measurements are off when the battery is used,and when he has a good think about it,i bet you he will be back,and tell us what he found,or overlooked when he made those power measurements-on the assumption that he reads these last few posts.
This is something you would not do,as you would have no idea as to where to look--why the efficiency swap between the two circuits,when the battery was used in place of the PSU.
These are things i know,due to time on the bench MH,and things you are blind toward--as is evident by your insistence that circuit 1 will be more efficient.
And so that is the difference between bench workers and pen pushers MH--the bench man will go searching as to why the two efficiency differences between PSU and battery exist,and will find that reason,while you(the pen pusher) will just keep on wallowing in your own self pity,and continue to peddle rubbish-like you are here and now.


Brad

tinman

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #336 on: July 10, 2016, 08:18:51 AM »
I am going to shift this conversation back to the JT 101 thread,where it belongs.

Lets go have a recap on that thread MH,at some of the things you have argued against your self about.


Brad

MileHigh

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #337 on: July 10, 2016, 08:20:42 AM »
Quote
The reason that you do not understand it,is because you have no understanding as to how !your! JT circuit works.

Brad, back in the days when we were discussing the Joule Thief I found the following YouTube clip that describes how a Joule Thief works quite well:

https://www.youtube.com/watch?v=0GVLnyTdqkg

You were flustered and confused by that clip.  You couldn't understand it, and you disagreed with it and you obstinately refused to accept what it said.  This went on for a considerable amount of time.  So that means the whole time we were getting into the discussion about the Joule Thief, you didn't even have a clue how one really worked.  For sure you can build one on the bench and get it to run.  However, at the very same time you can still not have a clue about how one really works.  You were in that boat, and for all I know you still might be there.

So you can be a poser all you want, but that's the way the cookie crumbles.

MileHigh

MileHigh

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #338 on: July 10, 2016, 08:30:46 AM »
Brad:

Quote
I most certainly do not need your verification MH on anything,and more so now that i have seen just how limited your understanding on a simple circuit really is.

I think every single time I have asked you to explain one of your procedures you play the bullshit "I won't bow to your demands" card.  You clam up and freeze up.

Right now I am operating under the assumption that either you don't know how to measure the output impedance of a battery or you think that you do and whatever you do has some hapless tragic mistake in it.  I have seen things like this before.

Or, you can wipe the ridiculous attitude away, and simply explain how you measure the output impedance so we can check if if makes sense.

Why are you freezing up?

MileHigh

TinselKoala

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #339 on: July 10, 2016, 08:52:47 AM »
@Brad:
The measurement points are indicated on one of the photos of the apparatus. I am measuring input current by the voltage drop across a 0.1 ohm non-inductive resistor on the negative side of the circuit, and input voltage simply across the input terminals. I have done both measurements with oscilloscope and DMMs. We have seen from Poynt99's work that the DMMs do a very good job of averaging pulsed inputs, and I also used the oscilloscope's "average" measurements of each channel's raw readings to confirm the DMM readings. I used the scope's Math function to multiply the raw (not averaged) instantaneous voltage and current inputs, then had the scope compute the "average" of this power measurement,which I then used in the Lux per Watt calculation. The current traces are very different between the two circuits, as you probably know yourself.

Later on I'll do a comparison between the input current, and the output (Collector wrt emitter) voltage, for each circuit configuration. This will show the duty cycle of the output light pulse and its relationship to the input current waveform.

MileHigh

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #340 on: July 10, 2016, 09:53:07 AM »
Brad:

Okay, here we go with the deal on your two diagrams.

Quote
Below are the two circuits,where in both cases the same amount of energy will be delivered to the inductor--lets say 100mJ of energy. I have included the batteries internal resistance that i measured !on the bench! of a normal alkaline battery with 900mV across it,and subtracted 10% of that calculated value for error margin.
Now we have the inductor in both cases storing the same amount of energy.
Can you please tell everyone here,which circuit will deliver the greatest amount of that stored energy to the LED?.

Taking your diagrams at face value, the second circuit will deliver the greatest amount of energy to the LED.

The problem is that if you are going to model the setup, you at least have to provide the right level of detail.  Taking the first diagram at face value the inductor discharges into the LED and the 7-ohm resistor.  However, the 7-ohm resistor is the output impedance of the battery.  So the question of where the energy burning off in the 7 ohm resistor comes from is a bit more complicated than that.

I marked up the drawing and swapped out the discharging inductors for voltage sources in series with resistors.

We know the the LED clamps the voltage.  Let's say that VLED is 3 volts.

Looking at the marked up diagram you can see that most likely the voltage generated by the inductor L1 is going to be relatively low.

This can all be measured on the bench to get some numbers. The energy is supplied to the circuit by the battery voltage source and the inductor as a voltage source, it is split between the two.  You can see that the load has three parts, the battery internal impedance, the L1 resistance, and the LED.  And of course we know that the inductor voltage varies with time.  All that being considered, I am not sensing any special advantage with respect to going with the lower circuit where you eliminate the battery impedance.

I have stated many many times that when you get into low-level details like this you need to do a full power/energy audit for a circuit to know exactly where the power/energy goes.  However, I have never seen you do this once, and I am sure you have read me state this before.

MileHigh

d3x0r

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #341 on: July 14, 2016, 09:50:13 AM »

@all

World's first ‪#‎graphene‬ battery product unveiled in Beijing. The portable battery can be fully recharged within 15 minutes, about 20 times faster than a Li-ion battery.


https://www.facebook.com/XinhuaNewsAgency/videos/1322484261112348/ 


Ya...so this makes my opinion that....  Mile High is Soooo HIGH!   Even *I* don't smoke THAT much.
ya, a minor thing just outside your personal preference and way in the area of gross error... it IS still a good thing.

MileHigh

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #342 on: July 15, 2016, 12:30:41 AM »
@all

World's first ‪#‎graphene‬ battery product unveiled in Beijing. The portable battery can be fully recharged within 15 minutes, about 20 times faster than a Li-ion battery.


https://www.facebook.com/XinhuaNewsAgency/videos/1322484261112348/ 


Ya...so this makes my opinion that....  Mile High is Soooo HIGH!   Even *I* don't smoke THAT much.
ya, a minor thing just outside your personal preference and way in the area of gross error... it IS still a good thing.

I like to have xes in airplanes you fool.

It's not a graphene battery product unveiling.  It's just a fancy press release technology announcement but no product.  If you have ever looked at the Revolution Green web site over the past three years there have probably been 50 very similar announcements about new battery technologies.  Unfortunately none have come to market yet.

The simple facts are that Robert Murray Smith made a clip where he claimed that he had a capacitor the size of a credit card that was 2000 Farads but offered no measurements at all to prove his claim.  He has made some technical gaffes in the comments on some of his clips that make you wonder if he even knows how to properly measure the capacitance.  Hence, he runs a small electric motor as a diversion to deflect away from the measurement issue.

Here is what I said to RMS:

<<<
A Maxwell K2Series BCAP 2000 Farad ultracapacitor is in a can that is 10 cm long x 6 cm in diameter.  In one of your clips you claim that you made a home-brew 2000 Farad capacitor that is roughly the size of a credit card.  Let us be conservative and say that you are claiming 10X the energy density by volume with your credit card sized capacitor that you claim is 2000 Farads. The onus is on you to prove that is true - that your credit-card-sized capacitor is 2000 Farads because right now I do not believe it.
>>>

You agreed with me that RMS made a ridiculous and unacceptable mistake where he measured the current flow over time but ignored the voltage measurement.

You said this:

Quote
so ya pretty high estimate...

How about you have the courage to state that RMS made an unacceptable measurement error when he measured the current but did not measure the voltage?

I also challenge you to retract your statement alleging that I am a troll.

tinman

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #343 on: July 15, 2016, 06:28:57 AM »
I like to have xes in airplanes you fool.

It's not a graphene battery product unveiling.  It's just a fancy press release technology announcement but no product.  If you have ever looked at the Revolution Green web site over the past three years there have probably been 50 very similar announcements about new battery technologies.  Unfortunately none have come to market yet.

The simple facts are that Robert Murray Smith made a clip where he claimed that he had a capacitor the size of a credit card that was 2000 Farads but offered no measurements at all to prove his claim.  He has made some technical gaffes in the comments on some of his clips that make you wonder if he even knows how to properly measure the capacitance.  Hence, he runs a small electric motor as a diversion to deflect away from the measurement issue.

Here is what I said to RMS:

<<<
A Maxwell K2Series BCAP 2000 Farad ultracapacitor is in a can that is 10 cm long x 6 cm in diameter.  In one of your clips you claim that you made a home-brew 2000 Farad capacitor that is roughly the size of a credit card.  Let us be conservative and say that you are claiming 10X the energy density by volume with your credit card sized capacitor that you claim is 2000 Farads. The onus is on you to prove that is true - that your credit-card-sized capacitor is 2000 Farads because right now I do not believe it.
>>>

You agreed with me that RMS made a ridiculous and unacceptable mistake where he measured the current flow over time but ignored the voltage measurement.

You said this:

How about you have the courage to state that RMS made an unacceptable measurement error when he measured the current but did not measure the voltage?

I also challenge you to retract your statement alleging that I am a troll.

MH

Forget about RMS-there are bigger fish to fry.

I would like you to have a look at Mark Dansies latest thread,and let us know what you think.

Time to leave all the !will never work!stuff behind,and look at something that dose have potential.

What do you know about this ion transfer stuff?


Brad

seychelles

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Re: A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?
« Reply #344 on: July 15, 2016, 11:56:13 AM »
TINMAN you just do not need JUST salt water but valuable SWEAT WATER..
TO EVERY ACTION THERE IS AN EQUAL REACTION REMEMBER..
EXAMPLE THE NORTHERN AND SOUTHERN HEMISPHERES   WINDS ARE
NOW CROSSING OVER THE EQUATOR SOMETHING THAT NEVER HAPPEN UNTIL THIS YEAR
WHY MY REASON THEY ARE TAKING TOO MUCH ENERGY FROM THE NORTHERLY WIND FOR
WIND POWER GENERATION.. BEES AND INSECTS OF ALL TYPES ARE VANISHING ..
 WHY BECAUSE OF MICROWAVE TRANSMISSION FROM MOBILE PHONES,RADAR AND MILITARY,
 SAME FREQUENCY AS MICROWAVE
OVENS 4.7 GIGA HERTZ..