Author Topic: Electrical igniter for gas engines A keystone to understanding by Magluvin (Read 250334 times)

Sprocket · « **Reply #525 on:** December 08, 2010, 12:15:24 AM »

Well, I finally managed to wade my way through this thread - it started out great, but went astray somewhere. Very interesting nonetheless, imo largely the result of Magluvin and delboy...

@delboy - you seem to have a masterful grasp of this stuff. Question, what do you think would be the effect of using the same setup (pancake coil with a few-turn primary on the perimeter) except when building the pancake, you use a different gauge wire every few turns - start outside with heavy-gauge, decreasing gauge as it winds its way towards the center?

Magluvin · « **Reply #526 on:** March 25, 2011, 01:10:17 AM »

Ok, were going to be doing some more on this subject for a bit.
Im finding more and more that if I think of the inductor as a flywheel and the capacitor as a spring, I get a better feel for what is going on in these circuits. We were in Rosemarry's thread so we will bring it here as its appropriate.

I think that with the igniter circuit, I need to tune it a bit. Like a lot of circuits, the desired output can be compromised by 1 off value part.

So Woopy has a couple vids that show that an inductor can be thought of as an electronic flywheel, and he does some tests on shorting the cap to energize the inductor.
http://www.youtube.com/watch?v=XrwgEb5ac_w
http://www.youtube.com/watch?v=RQFD1cDlEUU

Now I am suggesting the possibility of being able to use this flywheel to get more out than in. Ill be working on the circuits after the weekend as Im finishing some test setup of the bitoroid transformers right now.

Here is an example

If you connect a 1uf cap across a 5v battery then disconnect it, the cap should have 5v across the terminals. Im sure we agree. =]
The cap hold a determined amount of energy, correct? 5v on 1uf cap has a finite calculable amount of energy.

And that amount of energy is how much we took from that battery, correct? An amount we could count on every time.
I have another theory on this, but will get to that next week. ;]

So Im going to use the flywheel in my circuit to use only the amount of energy that is in a 1uf cap at 5v, but charge a 1uf cap to above 5v.
Its not that much, but it is something. ;]

mags

Goat · « **Reply #527 on:** March 25, 2011, 01:44:40 AM »

@ Magluvin

Fantastic work you're doing, I can see from your posts lately that you're on a high of creativity...keep the juices flowing and keep it up!

Regards,
Paul

Magluvin · « **Reply #528 on:** March 25, 2011, 01:59:07 AM »

Hey Goat

Thanks. Ya know, when I say flywheel in reference to and inductor, Im serious. ;]
And it helps some that never thought of it that way, see it in a new light. ;] It helps imagination take place where it may not have been before.

mags

Goat · « **Reply #529 on:** March 25, 2011, 05:30:10 AM »

Hey Magluvin

I think I know what you're talking about, I once tried charging a cap bank on a bedini wheel using milliamps and ended up with over 1,000 V on the cap bank in no time and man could it deliver an instantaneous charge...so if you add the inductor as you say then yes big bang for the milliamp buck

My problem was always returning that big bang to the source but you're way seems like it could do it....I look forward to your experiments and results.

Regards,
Paul

Magluvin · « **Reply #530 on:** March 25, 2011, 05:59:37 AM »

Hey Goat

Well that is an issue, getting it back to the source. Especially all of it and or more. ;] At least with what we have so far. Thats the way we have to think about it. Use the kick in the pants to at least know its real and we are not wasting out time. ;]

Ive read that a cop of at least more than 1.8 - 2 would be needed.
So once we get just above cop>1 we still have a ways to go.

But we are at the least getting more to a load than what the battery could put out by itself over time. Like an electric bike could go 50% further just on a regular charge, etc. Its not useless. These are just beginning things, but at least we are here. ;]

I gotta get some sleep.

Mags

forest · « **Reply #531 on:** March 25, 2011, 08:21:38 AM »

Mags

Just ground both capacitor and battery negative and touch positive of capacitor to positive of battery, you will see a back current flow to battery if capacitor if of correct value and enough charged (of cource voltage above battery).
The ultimate source for better HV devices is Earth ground ! Don Smith

woopy · « **Reply #532 on:** March 25, 2011, 08:02:37 PM »

OK magluvin

Good idea to transfer here.

I made a circuit with what i understood from your recent post

The result is impressive, but i wonder if i missed something

Thank's for your apreciation

good luck at all

Laurent

http://www.youtube.com/watch?v=Auv_66Ke-iw

nul-points · « **Reply #533 on:** March 25, 2011, 10:20:00 PM »

Quote from: woopy on March 25, 2011, 08:02:37 PM

OK magluvin

...

I made a circuit with what i understood from your recent post

The result is impressive, but i wonder if i missed something

...

Laurent

hi Laurent

your calculation of energy supplied by the input cap has a mistake

you must find the energy on the input cap for its start voltage and then subtract the energy remaining on the same cap for its final voltage - this will give you the correct input energy

if you need to be more precise about energy transfers between capacitors then you must measure the actual value of the capacitors and not rely on the case marking (which can be around +/- 20% of the actual value!)

here is a good online capacitor energy calculator to save time with cap energy calculations:
http://www.electronics2000.co.uk/calc/capacitor-charge-calculator.php

the page at that link can also be saved to your computer to use offline

hope this helps!

cordialement
np

http://docsfreelunch.blogspot.com

Magluvin · « **Reply #534 on:** March 27, 2011, 09:15:54 AM »

Ok this is the general idea.
There are 4 pics below that show where Im going with this. It is not an idea that I just think works. It just does.

The first pic is the same circuit Woopy made the first video of. With the addition of the diode across the batt and switch.

1 The first pic shows the circuit idle no current flowing and no charge in the cap. The scope shots are Source, Inductor and Cap in order from left to right. Source is 5v dc. The resistor is added to the inductor to give its real world values in the circuit. Woopys inductor.

2 The next pic shows the switch closed and a charge building in the cap, through the inductor and diode at upper left with current flowing clockwise. The diode on the right is not conducting anything yet.

3 This pic shows me open the switch when the cap reaches 5v, equal to source. Now the diode on the right is conducting, current is still going clockwise, just not through the battery, but now the diode.

4 Now the currents have finally stopped. The cap is charged to 6.32v.
The flywheel crammed that extra 1.32v into that cap after the battery was disconnected.

Is that extra 1.32 not free? When we charge a cap from a source, the cap when fully charged by the source, will contain and equal amount of energy that was taken from the source to charge it.

So when we cut off the battery here, when the cap reaches 5v, and we know the source is 5v, then we have to know that the 1.32v worth of energy in the cap came from where? ;]

Magluvin · « **Reply #535 on:** March 27, 2011, 10:05:04 AM »

Was thinking a one point if the 2 flywheels, a physical one and the inductor, what would be the differences. Could we get this same advantage in the physical world?

First, lets look at what is involved with getting them spinning.

The physical fly, tend to take a lot of energy to get them started, like an inductor has impedance. But I see them different. The real flywheel to get it spinning up consumes a lot of energy right from the start, and the excess input it took to get it going, will no longer be needed once up to speed.
But the inductor consumes less in the time it is switched on, and the energy consumed or let through the circuit as the inductors wheel gets spinning, allows more energy to be pulled from the source till the wheel is up to speed( field peak= current peak). So they have a bit of different attributes.

I broke off from these experiments early on when I figured that, at the time, how can I get 6.32 v in a cap to deliver all of its energy back into the battery. I then knew that just getting over cop was not enough for looping.

But lately, I have bunches of ideas that stem from old projects. And I think I found the answer for the 6.32v conversion.

Charge 10 caps to that voltage from the battery with the circuit.
Put them in series and discharge into the batt. 63.2v Now when this series cap discharges into the batt, once the cap is discharged down to 5v, we have put back more than we took to fill all ten caps individually. The amount we figured we got for free was
1.32v x 10 =13.2v 13.2v-5v = 8.7 of the free got back to the battery + 100% of what was used for the process.. The more caps we charge and put in series for discharge, the more % of free we get to keep per discharge. That last 5 can be joul thieved in. ;] lol

Mags

nul-points · « **Reply #536 on:** March 27, 2011, 02:10:19 PM »

Quote from: Magluvin on March 27, 2011, 10:05:04 AM

...
But lately, I have bunches of ideas that stem from old projects. And I think I found the answer for the 6.32v conversion.
...
Charge 10 caps to that voltage from the battery with the circuit.
Put them in series and discharge into the batt. 63.2v Now when this series cap discharges into the batt, once the cap is discharged down to 5v, we have put back more than we took to fill all ten caps individually.
...
Mags

hi Mags

yes, looks feasible - when i tried the old variation on this (charge 2 caps in parallel - reconnect 2 in series to feed back to battery) the battery just discharged!

http://www.overunity.com/index.php?topic=10231.0

perhaps 10 caps in series will behave differently? looking forward to hearing how you get on

all the best
np

http://docsfreelunch.blogspot.com

forest · « **Reply #537 on:** March 27, 2011, 06:35:20 PM »

The problem is to KNOW exactly where is energy.If at 1/4 of resonant frequency period it is still in magnetic field of coil then we need to disconnect coil right then and got energy in capacitors in parallel.Many.
The dump excess into battery but not allowing to kill the dipole, meaning voltage in capacitors HAS TO be left at higher level then power source.
Our habits are against us,not circuits
http://www.freestacja.pl/wp-content/uploads/2011/02/radio_0.jpg

Magluvin · « **Reply #538 on:** March 28, 2011, 01:44:25 AM »

Hey Forest

Maybe you are missing my point..

lets say that 5v in the cap can be considered an amount of energy, the same amount at 5v, every time.

When we charge the cap to 5v, then disconnect the battery, the cap continues to load up to 6.32v. The extra 1.32 is, well extra, free.

But having a way to force all that 6.32v worth of energy back into the batt may be done with a joule thief. If the JT can use the charge in the cap till the cap reaches .5v, then we are charging the battery with more energy than was taken from the battery.

Or like in my previous way, charge 10 caps with the circuit. Then all caps in series is 63.2v of which 50v is what the battery produced, and the 13.2 was free. In this way we can just dump the series caps into the batt till the cap is down to 5v, equal to the batt. That would leave .5v in each cap. 6.32-5= 1.32 1.32-.5= .82v
So each cap will have delivered 5.82v from each cap back to the batt, while the battery only provided 5v to each to begin with.
If it were 20 caps, then we will have gotten more of our free 1.32v back to the batt. 50 caps, even closer to 100% of the 6.32v back into the batt.

Took me a while to get the grip on it, but I find it solid.

Mags

nul-points · « **Reply #539 on:** March 28, 2011, 02:39:22 PM »

Quote from: Magluvin on March 28, 2011, 01:44:25 AM

Hey Forest

Maybe you are missing my point..
...
When we charge the cap to 5v, then disconnect the battery, the cap continues to load up to 6.32v. The extra 1.32 is, well extra, free.
...
If the JT can use the charge in the cap till the cap reaches .5v, then we are charging the battery with more energy than was taken from the battery.
...
Mags

hi Mags

i think that both you & Forest are essentially making the same point about collecting the stored energy: ie., when you transfer energy back to the battery, leave some in the cap(s) so the source dipole isn't killed

but the extra 1.32 is not showing free energy, i think

it's some energy from the battery that got stored in the field of the coil (as Forest mentioned) - caused by the series current flowing round the circuit. whilst the cap was getting charged

when the current flow was stopped, the coil 'flywheel' effect (which you've mentioned earlier) tried to maintain the current by using the energy stored in its field

this 'discharge' of the coil field energy got added to the energy in the cap - and it shows up as an extra 1.32V on the cap

so all the energy now in the cap has come from the battery

this must be why that old 'charge 2 caps in parallel, discharge as 2 caps in series' version (mentioned a couple posts above) just depleted the battery...

...no extra energy was getting returned to the battery - because it all came from the battery in the first place!

it looked such a neat idea too

np

http://docsfreelunch.blogspot.com

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