Re: [EVGRAY] Russians just do it

I'm  glad the work goes on and many are getting into the problems but light
bulbs are not a good measure of power unless you you use a light  meter to
ckeck luminosity .then do the math .with hi rf lighting up light bulbs is
easy .and easily miss read.

On Nov 28, 2017 11:31, "Lutz Liebers [email protected] [EVGRAY]" <
[email protected]> wrote:



Interesting video even if you do not speak the language:
Tesla transformer at 100 to 500kHz up to 300W. Circuit diagram and
properties
https://youtu.be/ItgjWqoSppE

Min 4:47:
Estimate the output-power at the incandescent *) light bulb rated 12V, 5W
on the
receiver side maybe not shining at full brightness ->




*) the pupil tests the temperature of the lampatshku prior to touching it,
since it
    might be hot

Min 4:53:
-> and compare that with the power consumption of the transmitter shown on
the
display of the connected power supply. It reads 30.5V, 0.04 Amp, equal to
1.22W

In order to clear things up, connect such a light bulb via a blue china
power meter
to a regulated power supply, adjusted to 1.22W output and just film the
bulb with
your mobile phone.

I am quite sure that it would appear much less bright than the bulb in the
video.
So the COP must be considerably beyond unity.

I did a new run with a logging voltmeter on the battery. I calculate that the energy dumped from the battery to the resistor was 1643Wt  (where a tick (t) is 255s) and the energy going into the battery during charging was 1068Wt (in the video I wrongly calculated it to be 1188Wt). For unity the charging should have taken 4 days but was finished 14h or so ahead.

https://youtu.be/WBOPQOEBEIU https://youtu.be/WBOPQOEBEIU

/Hob

Hi Hob,

How do you measure the power going into or out of the battery at each tick? Is it read from a power meter or calculated from volt and ampere readings?

Regards
Ole
 

---In [email protected], <hob.nilre@...> wrote :

 I did a new run with a logging voltmeter on the battery. I calculate that the energy dumped from the battery to the resistor was 1643Wt  (where a tick (t) is 255s) and the energy going into the battery during charging was 1068Wt (in the video I wrongly calculated it to be 1188Wt). For unity the charging should have taken 4 days but was finished 14h or so ahead.

https://youtu.be/WBOPQOEBEIU https://youtu.be/WBOPQOEBEIU

/Hob

Hi Ole,

Power out is simply Vbat²/R where Vbat is the voltage across the battery and R is the dump-resistor.

Power in is Vbat*f*C*(Vcap-Vbat) where f is the AC frequency, 50Hz where I am, and Vcap is the voltage across the cap just before the dump to the battery and C is the capacitance of the cap. C*(Vcap-Vbat) is the charge Q being dumped from the cap as in Q = C*V and so f*Q is the total charge during a second which in turn is the average current. Work done is Vbat*Q which for one second is the same as the avg power.

Does it make sense or am I making a mistake somewhere?

/Hob

Hi Hob
 I think using the voltage across the cap just before it discharges to battery in a power equation is not quite right, since the cap has that voltage only for a split millisecond, and as it discharges the voltage quickly drops, down to the level of the battery and then the voltage stops its decline in voltage at that time....perhaps also the cap goes below the battery voltage as it discharges with very stron force too....I think this why Bedini and BroMikey recommend for a 12V battery to charge from cap discharge, the cap would be for optimum charging atg 36V during discharge, and make the pulse width to the cap discharge just the right amount of time, so that the cap discharges to 23V, then stops right there....so all those joules coming from cap are doing good charging-work and this is supposed to be most efficient way to do it.
 

 Also to calculate watts from a cap discharge, this is the formula:
 

 FARAD size of capacitor / 2
 X
 (Cap voltage before discharge MINUS cap voltage after discharge)
 This voltage-value difference is then SQUARED
 X 
 cap discharge events per second
 = watts
 So anyways, I don't think the capacitance of capacitor and voltage in capacitor is how best to calculate watts from a cap discharge - you need to know more, such as the actual voltage drop, and time in seconds, and also note the voltage-drop measurement is Squared too...
 

 For charging battery, then whatever watts you have does not necessarily mean the battery charges up according to those "watts calculated" of course, so really long term battery charging tests are only real way to know if you are charging battery fast and as well as you think it is.....use battery internal resistance meter, and wet-cell batteries with hydrometer to check specific gravity to know for sure the state of charge, or drain battery into known load after each charge to calculate how many watts the battery has accumulated during charge is another way to do it.
 ciaoKone

Hi Kone,

You are right about the energy coming out from the cap is the formula you write (although I believe its voltage-value squared then difference rather than voltage-value difference then squared), but I'm calculating the energy going into the battery. There are losses between the cap and the battery that I dismiss at this time. I calculate the energy going into the battery as Vbat*Q which is the same as Vbat*I*t since I is Q/t.

The discharging takes 24h and the charging takes up to 4 times that depending on Vcap and C, so there is no fast charging going on. A discharge/charge-cycle takes almost a week to perform.

Right now I'm doing it all over again without any changes to any parameters to compare results.

/Hob

HobYou are doing good workLater, with this same battery, try doing cycles with a Bedini School Girl charger, the primitive one, with the wheel, just to compare your cycling time.A very large diameter non magnetic wheel is best, running on fine skate board bearings.The coil should be at least 400 turns giving you about 300 volt spikes, if I remember correctly.Anyway, the freaky thing about this type of charging on a lead acid battery after several cycles, is the time to achieve full charge gets less, and less. This defies the normal accepted charging practice,
as there are almost no amps going in, just a 300 volt spike.In my own experience, it worked on flooded type deep cycle lead acid batteries.There is no heating during charge.As Kone does, use a hydrometer to check state of charge.At first, it is painfully slow to cycle, then starts to improve after each cycle.It is a real test of patience, but a very interesting experiment.The batteries, so charged, seem to out perform the conventional charged batteries.A proper serious study should be performed by people interested in using batteries.You don't need to use the Bedini wheel, as his solid state unit is better.The Rene seems to do the same type of signal though I have not scoped them , myself.Cheers Warren
Sent from Yahoo Mail on Android 
 
  On Tue, Dec 5, 2017 at 5:57 AM, [email protected] [EVGRAY]<[email protected]> wrote:       
Hi Kone,

You are right about the energy coming out from the cap is the formula you write (although I believe its voltage-value squared then difference rather than voltage-value difference then squared), but I'm calculating the energy going into the battery. There are losses between the cap and the battery that I dismiss at this time. I calculate the energy going into the battery as Vbat*Q which is the same as Vbat*I*t since I is Q/t.

The discharging takes 24h and the charging takes up to 4 times that depending on Vcap and C, so there is no fast charging going on. A discharge/charge-cycle takes almost a week to perform.

Right now I'm doing it all over again without any changes to any parameters to compare results.

/Hob
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Hi Hob,

If you have access to a digital storage oscilloscope with mathematical functions it can be rigged up as a power meter to actually calculate the power. Only problem is that the memory depth isn't high enough to run for hours. But it can be used to confirm your measurements at different moments during the experiment or to see if excess energy comes out during the experiment. If the mean power going out of a component is greater than the mean power going into that component it extracts energy from somewhere (or uses up itself like a battery).

If you make curves for the current and power a curve representing the energy can be made too by integrating the power curve. I.e. it's done by summing up all the points representing power along the time axis and drawing a curve with the summed up numbers. This is the energy.

One thing to be aware of is to be sure that there is no phase shift between the voltage potential and the current as this represents energy changing direction (reactive power). This can be checked with an oscilloscope. Dumping a capacitor directly into a low impedance battery will make ringing (or spikes) in the circuit because of the parasitic inductance of the wires. I.e. they form loops which are actually parasitic inductors.

Charging a lead acid battery takes more power in the beginning (high voltage X high current) than later during charging (high voltage X low current). This is revealed if plotting the power curve or the energy curve.


Source: http://batteryuniversity.com/learn/article/charging_the_lead_acid_battery http://batteryuniversity.com/learn/article/charging_the_lead_acid_battery

I don't know if your formula is good enough but it ought to be confirmed with a proper power measurement.

Regards
Ole

 

---In [email protected], <hob.nilre@...> wrote :

 Hi Ole,

Power out is simply Vbat²/R where Vbat is the voltage across the battery and R is the dump-resistor.

Power in is Vbat*f*C*(Vcap-Vbat) where f is the AC frequency, 50Hz where I am, and Vcap is the voltage across the cap just before the dump to the battery and C is the capacitance of the cap. C*(Vcap-Vbat) is the charge Q being dumped from the cap as in Q = C*V and so f*Q is the total charge during a second which in turn is the average current. Work done is Vbat*Q which for one second is the same as the avg power.

Does it make sense or am I making a mistake somewhere?

/Hob

Hi Warren,

The purpose with the circuit and the experiments with it are to eliminate the Bedini wheels to make it easier to calculate what exactly goes into the battery and what goes away as losses. I will do a few charge/discharge-cycles to verify that actual charging are happening. With the cap I use now, a motor-cap, nothing gets hot, just slightly above room-temperature.

/Hob

HobThat sounds excellentGo man, go.Cheers Warren

Sent from Yahoo Mail on Android 
 
  On Tue, Dec 5, 2017 at 12:10 PM, [email protected] [EVGRAY]<[email protected]> wrote:       
Hi Warren,

The purpose with the circuit and the experiments with it are to eliminate the Bedini wheels to make it easier to calculate what exactly goes into the battery and what goes away as losses. I will do a few charge/discharge-cycles to verify that actual charging are happening. With the cap I use now, a motor-cap, nothing gets hot, just slightly above room-temperature.

/Hob
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Hi Ole,

Nice plot, thanks. I'm quite confident that my math is ok. If I compare it to mechanics Vbat*Q would correspond to F*x which is work done. Similarly, comparing the capacitor to a spring, although not as obvious, Q=C*V would correspond to x=F/k, or to make the mechanical side familiar, F=k*x corresponds to V=Q/C since C corresponds to 1/k, a hard spring corresponds to a small cap. So, all put together, I think my formulas adds up. I'm fully aware that lots of energy in the cap gets lost, but right now I'm only focusing on whats going into the battery, not how effective the whole circuit is and what the losses are. I just want to know if pulsing a battery is something to pursuit, and it seems it is. Next step is to make it efficient.

/Hob

Hi Gerry,

Yes, I think so too, thanks.

BTW, I've just completed another discharge and its almost exactly as the previous one, so the charging seems OK.

/Hob

Hi Hob,

Looking at your formula again I think there is an error in the calculation of the power going into the battery. Power can't be correctly calculated using the average current. A better way to do it is by using the RMS current when doing the calculation. This requires knowing the shape of the curve describing the current as any change of the curve shape changes the RMS current.

Take a look at the attached document with some examples taken from this document 'Waveform RMS Equations' (direct link): http://www.convertertechnology.co.uk/index.php? s=file_download&id=57 http://www.convertertechnology.co.uk/index.php? s=file_download&id=57.
 The source document was found here: http://www.convertertechnology.co.uk/index.php?c=technical_useful&s=50Technical http://www.convertertechnology.co.uk/index.php?c=technical_useful&s=50Technical

Regards
Ole 

---In [email protected], <hob.nilre@...> wrote :

 Hi Ole,

Nice plot, thanks. I'm quite confident that my math is ok. If I compare it to mechanics Vbat*Q would correspond to F*x which is work done. Similarly, comparing the capacitor to a spring, although not as obvious, Q=C*V would correspond to x=F/k, or to make the mechanical side familiar, F=k*x corresponds to V=Q/C since C corresponds to 1/k, a hard spring corresponds to a small cap. So, all put together, I think my formulas adds up. I'm fully aware that lots of energy in the cap gets lost, but right now I'm only focusing on whats going into the battery, not how effective the whole circuit is and what the losses are. I just want to know if pulsing a battery is something to pursuit, and it seems it is. Next step is to make it efficient.

/Hob

The attachment didn't show up at Yahoo. Here it is again. Maybe.

Regards
Ole
 

---In [email protected], <onielsen@...> wrote :

 Hi Hob,

Looking at your formula again I think there is an error in the calculation of the power going into the battery. Power can't be correctly calculated using the average current. A better way to do it is by using the RMS current when doing the calculation. This requires knowing the shape of the curve describing the current as any change of the curve shape changes the RMS current.

Take a look at the attached document with some examples taken from this document 'Waveform RMS Equations' (direct link): http://www.convertertechnology.co.uk/index.php? s=file_download&id=57 http://www.convertertechnology.co.uk/index.php? s=file_download&amp;id=57.
 The source document was found here: http://www.convertertechnology.co.uk/index.php?c=technical_useful&s=50Technical http://www.convertertechnology.co.uk/index.php?c=technical_useful&s=50Technical

Regards
Ole 

---In [email protected], <hob.nilre@...> wrote :

 Hi Ole,

Nice plot, thanks. I'm quite confident that my math is ok. If I compare it to mechanics Vbat*Q would correspond to F*x which is work done. Similarly, comparing the capacitor to a spring, although not as obvious, Q=C*V would correspond to x=F/k, or to make the mechanical side familiar, F=k*x corresponds to V=Q/C since C corresponds to 1/k, a hard spring corresponds to a small cap. So, all put together, I think my formulas adds up. I'm fully aware that lots of energy in the cap gets lost, but right now I'm only focusing on whats going into the battery, not how effective the whole circuit is and what the losses are. I just want to know if pulsing a battery is something to pursuit, and it seems it is. Next step is to make it efficient.

/Hob

Hi Hob,
I have never done this test manually myself but your math looks correct to
me.
Sometimes just multiplying by F doesn't take into account the duty cycle or
pulse width on/off ratio.
In your case using straight C*V it's a specific quantity of charge pulsed
in at each single F point.

So in my humble opinion your system seems correct.

Gerry

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On Wed, Dec 6, 2017 at 4:26 AM, [email protected] [EVGRAY] <
[email protected]> wrote:

>
>
> Hi Ole,
>
> Nice plot, thanks. I'm quite confident that my math is ok. If I compare it
> to mechanics Vbat*Q would correspond to F*x which is work done. Similarly,
> comparing the capacitor to a spring, although not as obvious, Q=C*V would
> correspond to x=F/k, or to make the mechanical side familiar, F=k*x
> corresponds to V=Q/C since C corresponds to 1/k, a hard spring corresponds
> to a small cap. So, all put together, I think my formulas adds up. I'm
> fully aware that lots of energy in the cap gets lost, but right now I'm
> only focusing on whats going into the battery, not how effective the whole
> circuit is and what the losses are. I just want to know if pulsing a
> battery is something to pursuit, and it seems it is. Next step is to make
> it efficient.
>
> /Hob
>
> 
>

Hi Ole,

Good reading!

Using average or RMS comes down to context. 
For a resistor work done is ∫R*I²*dt which can be tricky but becomes easy with RMS.
For a battery work done is simply V*Q, no need for RMS here.

/Hob