heat resistant resistor?

It sound like you are exceeding the power rating of the resistor (by quite a bit). Without knowing how much current your resistor is having to pass (or the rest of your circuit and load) it is hard to say how high a rating the resistor needs to be (or if an inductor would be better).

I would advise having a look at the resistor manufacturer data sheets available on the manufacturer's websites (they contain lots of data on power-handling and de-rating factors for pulse applications like yours). You will VERY likely need multiple large wattage resistors (connected in series and parallel to give you the 8 ohms) to dissipate the heat.

Don't forget heatsinks for the resistors.

Can you supply more information on your application, your load and your pulse current?

I have no way of measuring the draw on the circuit during use. I have used a small motor winding to get the same result and it worked for a while, but it also overheated and burned through the wire eventually.

I have a fuse in line after the resistor and a 5 amp will burn, but a 10 amp fuse works fine.

Well as a very rough guide to give you an idea of just how much heat your likely dealing with

P=I^2 * R, so say 7A of current

P=7A^2 * 8 ohms = roughly 392W of heat.

This is fairly rough thou as the fuse wire element of standard glass cartridge fuses behaves differently when exposed to a pulse current, but in your case it is a close enough approximation (unless you want to delve into the math for a more accurate answer).

Say 14 x 20W resistors mounted on a heatsink should do the trick for short periods of operation. 50W and 100W resistors are a better option. There are other ways (like water-cooled resistive loads) but I would not advise them for 120V ac due to the danger of electrocution (especially if you don't know what you are doing).

So far the water filled wirewound resistor is the only thing that works for any period of time. 8 ohms of resistance will apparently take about 4 lbs of #18 ga wire though.

I agree; go with a choke rated @ 10 amps or series up 4 2ohm 20 watters.

The resistor works if the dissipated power at its level is under its capacity to convect/radiate heat to the ambiance.

The dissipated power is P= R*Im^2 where R is the resistor value (Ω) and Im the average current in (A). For you the problem is to determine Im which depends on :

- the amplitude of the applied sinus voltage

- the frequency of the voltage oscillation

- the value of the capacitor

Im= V/(Π*R)*K^2/(1+K^2)*( 1+ exp(-(1/f)/T)) where:

V- voltage amplitude

Π- the "pi" constant

K= T*ω = (R*C)* (2*Π*f)

f- frequency of your sinus function

If the dissipated power < dissipation possibility of the resistor it will last if not it will burn.

Jack is right, now you can compute and see which combination of resistors you should build up to obtain what you want. As you see the capacitor is one of the factors.

G.A. I like to use those and alos light bulbs.

While a light bulb will probably be a good solution to this problem it may make predictive calculations difficult. The problem with using a light bulb, the resistance of the filament varies greatly with temperature. The initial resistance of a room temperature 60 watt 120V light bulb is not 240 ohms. It achieves this resistance only once the filament reaches operating temperature.

Another possible solution though could be the collection of resistors designed for a heatsink. The heatsink could even have water flowing through it. Naturally there are limits to this but only a proper thermodynamic analysis with more data would reveal them.

Can you describe how to use the element as a resistor?

Just connect the wires at the two ends the same as you would a resistor. But if you are using it for spark control you might find the self-inductance becomes an issue.

I think self inductance is what is wrong with the 62 ohm resistor. If the element did the same thing then it would not work.

A standard electric heater is just resistance wire coiled inside a quartz tube surrounded by a metal heat radiator and sometimes equipped with some electronic switches (such as a power switch and safety cutoff tilt switch). The point is it's just a very big, very cheap resistor. We commonly use these at testing laboratories for loads due to their very low cost and high power dissipation (I even have 3 sets of 4x2400W heaters for three-phase load simulation). Hot water heater and kettle (or electric jug) elements also work well but are far messier due to requiring water or forced-air cooling.

Radiant heaters work wonderfully as loads if you do not need very specific or tight tolerance values of resistance and inductance is not an issue (ie- switching or ac frequency is low), because after all they are just big air-cored coils of resistance wire).

I like your thinking. The center wire would be used in series and the outer metal surface would dissipate heat. How much resistance do you think an electric range burner would have?

A close approximation for finding the resistance is simply using the element's rated voltage ,the power rating of the element (in Watts) and ohms law (the simple one not the complex math one).

P=V^2/R (thats V squared over R)

therefore R=V^2/P

or simply R=(V*V)/P

eg- V=240V, P=2400W, so R=24 ohms.

The light bulb filament will always act as a resistor. The value of the resistance will change with temperature. One of the reasons filaments are so light (no pun intended) in mass is so that they quickly reach operating temperature. The initial room temperature value of resistance will be typically an order of magnitude less than the operating resistance.

I got a 100 Amp, .1 Ohm resister on eBay for next to nothing. Sucker's big - over a foot long. But I knew how much current would be going through it, more than 60 Amps.

All indications suggest the resistor is well undersized for the job.

What about a number of 110V filament light bulbs instead of the resistor? One of the above posts suggests 392W - that would be, say, 4 x 100W lamps in parallel, then feed the capacitor in series with these.

What on earth is this circuit doing, for goodness' sake? There may be a more elegant way of doing whatever is needs to do that doesn't involve dissipating 392W in the process!

This may work fine but bear in mind that if this is a pulse application the resistance of the light bulbs will actually be about 1/10th the expected resistance when cold and will take several line cycles at constant voltage to heat up to the point where the resistance would be equivalent to a 8 ohm resistor.

Shawn

The capacitor should take care of that, though it would be good to know what such a strange circuit was doing, just so the forum could advise the OP better.

Regards.

It is a clear indication of under rated resistor or insufficient cooling.

Try a higher rated Alumimum clad resistor.

You need to rethink through your design, it appears to be wasting power, it may even possibly be dangerous!!

Something like even a drill speed controller for mains might be a better move.

If we were given an idea what you are trying to do, I am sure that we could come up with a simple method to achieve it probably WITHOUT the use of high power resistors, PWM for example.....and far less danger and less power use.....

geenandclean - it sounds as if you are trying to do this work using quick-and-dirty methods. This almost always takes far longer than planning things out - and it can be dangerous as well.

Can you answer some questions?

Is this a full-wave rectifier system, and the resistor used to reduce the peak pulses into the storage capacitor?
If so, what size is the capacitor, and what is the load at the output?
If not, what is your circuit configuration?

If the problem is due to requiring short pulses, ceramic-bodied power resistors may not be the most appropriate, because the film is quite thin and will will not be well-cooled during transients (but your post suggests you already have some idea of that). Paradoxically, series-parallel combinations of multiple small resistors with the same total rating could be better for pulses, particularly if you choose a thick-film, through-hole type (what matters here is at least in part the total surface area).
On the other hand, the constant-current rating of a 20-Watt, 8-Ohm resistor is only 1.6-Amps...

Quick and dirty: 20Watt wire-wound resistors should cost no more than US$0.8. Can you afford to use four of these in series-parallel? If single ceramic devices last a full minute, that should give you long enough to check the surface temperature is not excessive. (I think that the price roughly doubles if you need non-inductive windings).

Way over my head!

I like things simple!

I have done the water cooled resistor and it seems to work fine. The problem is that I had to use a lot of #18 ga winding wire (1/2 lb) to get about 2 ohms of resistance. The result was helpful, but still not enough.

I did find some interesting things happening with the new addition like the wires glowing in the dark. They were not hot. The winding wire under them does heat up, but these wires are wrapped around each other and one connects to ground. When the system is in operation the wires glow purple, but can only be seen in the dark and the camera will not pick it up.

Anyway, I may try multiple resistors in parallel when I can find some, I cleaned out the local radioshack.

As to questions;

I am not rectifying the ac voltage. I have tried using a variable resistor on the power supply. The dimmer switch will last about 1 second before failing. I used both types non-efi and efi reducing dimmers. The pulses are too fast or the on/off cycles are too much for this type of control. I do not have a working variable voltage power supply for testing and the application will have to utilize an inverter for power when installed. The resistor has to be used to control the voltage coming from the capacitor. I have diodes in series after the capacitor to stop high voltage dc backfeeding into the capacitor. Without the resistor I will melt a sparkplug in a couple minutes. With the right resistor my spark is maintained as a small ball between electrodes without (much) deterioration of the metal.

The capacitor is a motor starting cap rated for 110 vac approx 300uF. I cannot test the load at the output due to high voltage being present there.

I hope this information clears some things up. More likely it just raises more questions and lends ammunition to people to start bashing me for fringe science and witchcraft, lol. Thanks for the help.

Well I'm glad you found a configuration that works for you. I'm even happier to hear you have an understanding and respect for the power you're playing with. I suspect that the custom resistors that have multiple layers of winding are also acting as air core inductors in your configuration. I wouldn't be surprised if inductive heating maybe part of your power loss making the inner windings glow in the dark.

Good Luck and Be Safe.

You misunderstood my remark about the resistor glowing. It glows with plasma energy from the high voltage dc coil. The wavelength of light is too short for cameras to pick up which means it is not causing resistance heating which would show up on film as red heat and would also burn your fingers. The wires when removed from the coil still glow purple without ANY heat. The strong smell of ozone also indicates the high voltage present in the wire.

If you are wondering about the name change, it is because every time I identify myself with HHO or Plasma ignition my posting abilities are taken away. This is merely an inconvenience and will not stop me from posting my results or methodology.

It does go a long way in proving the suppression of technology rampant in the scientific community. If you are responsible for the supression of technology or ideas that disagree with what you understand or were taught, then this message applies to you. If you are curious about new findings and are openminded enough to try these experiments for yourself feel free to contact me at my onecraftydude youtube page as I am more than willing to help others replicate my experiments without accusation or persecution.

If you are wondering about the name change, it is because every time I identify myself with HHO or Plasma ignition my posting abilities are taken away. This is merely an inconvenience and will not stop me from posting my results or methodology.

Perhaps you should consider something - you are blatantly attempting to disrupt this forum. Why not go away and do whatever it is you do. Stop acting like a spoilt child, demanding that we listen to you. If a person forced their way into your home and demanded that you listen to them ramble about whatever, you'd be annoyed. If they persisted, you'd take more forceful measures. Having an audience appears to be your only goal. You would do better to go away and talk with like minded individuals.

Anything you have to say is now unworth reading - you have demonstrated quite clearly that you are an internet thug.

And you have identified quite clearly that you are unwilling to use you real name to refute me GUEST.

I have to agree with "Someguy." It's pretty hard to disrupt CR4 when you've started the thread yourself!!! All you as a reader has to do is simply un-subscribe. If everyone simply un-subscribed, Someguy would be free to ramble on about anything. So get over it!

There's nothing to get over - it's just an observation of his membership.

"my posting abilities are taken away"

The thread is interesting, but does that quote not concern you? Free speech is great, but just how far do you take that? If the people who run this place decide to block a user, they're exercising the right to say "we don't want to listen"/"you have stepped too far beyond forum rules"/whatever. Someguy/Greenthingumy is saying 'screw you, I'm going to carry on'.

Your observation about readers un-subscribing, I totally agree with as an approach. However, you seem to be saying that nobody should be blocked from posting on CR4, they're only exercising the right to free speech, yes? If you go to a party, is it not correct to follow the host's house rules, no matter what one thinks of them?

It looks as though Greenthing was asked to leave the party, but has simply returned in a change of clothes. That would be fine if he'd been asked to leave and return when sober to talk about something else. It's nice to give people another chance. Greenthing has comeback as Someguy and loudly announced, 'screw you, I'm going to continue with whatever I did that caused offence in the first place'.

My intent was merely to highlight this transgression. It's a matter for the people who run this site. Presumably, if I were to initiate a discussion/rant about politics, you would support me in having the right to do so. You would also defend me in returning under another guise if I were blocked.

Tell me where you live... I've got cap for your punk ass!!! And I won't hesitate to but that barrel down your ugly throat!!!!

I appreciate your input Guest. As you can see my name has changed again. I have never been told WHAT i did wrong. I would think that if a person was going to be removed for breaking rules they would at least tell him what rules were broken?

I could just post anonymously like you do to protect my identity, but that seems like a cowardly avenue to take.

Why you pompous... I am sorry now that I defended you and apologize to any and all I may have offended in my defense of you.

For the record: I indicated in a later post that you probably have a classic corona discharge from the high voltage, and not the inductive heating I originally thought in this post.

Your methodology of trying to communicate is severely lacking. Your mixed use of buzz words leads people to misinterpret what you are doing. You think an air core inductor is just a resistor and low inductance power resistor creates inductance. Clearly you do not know what you are doing.

You are incorrigible.

I apologize for my many faults. My understanding of electrical theory may be lacking, but my experiments prove what I am trying to convey.

I suppose the inductor coil I am using is not a resistor or an inductor or anything else. I don't know what to call it and have no intention of getting into a fight over terminology. I call it a smoothing coil now since that is what it does in my circuit. It is simply 2 wraps of #18 ga winding wire in one continuous direction with a smaller #28 ga wire (pair) soldered as a continuation of the winding on the last wrap. The 2nd wire of the #28 ga pair is grounded. When the ground is used the spark increases dramatically. I am not sure what to call the coil because I have never seen one designed this way.

I am sorry if I offended you since you are very helpful and have tried to help in the past.

I thank all of you who lend support.

Sounds a little like you've created a transformer... A little something like a Tesla coil. However, it is non-the-less an inductor.

So this is in the output of the capacitor to control a spark current. That means the current should be very peaky, but you should be able to judge both the source Voltage and the average current. From what you write (though the details are still not entirely clear) it looks very likely that there will not be dissipation with four resistors in series-parallel (to maintain the eight-ohm value). By the way, not knowing the full detail of the system, I'd be concerned that the glow you observe might be Voltage breakdown due to transients above 110-Volts. Transient sparking might contribute to the failure of your original resistors - so I'd be in favour of using them as two series pairs in parallel with each-other (four resistors total) to reduce the Voltage across them.

Presumably the resistors from RadioShack are the low inductance kind they advertise widely?

Maybe we could give better advice if we could see a circuit diagram...

Good luck and go carefully

Fyz

The resistor is non-inductor, 20W8ohmJ, TYH 05.28, NON-IND. I will work on a diagram and try and post it.

Can you answer a question for me? Why does the wire glow purple, but stay cold? I am not talking about the resistor wire which heats up, this is after the resistor between the capacitor and the high voltage lead.

Purple glow sounds a bit like arcing along the surface of the wire due to the high Voltage breaker.

A cool purple glow sounds like a corona discharge. But that would happen only at considerably higher voltages than have been discussed.

If the schematic was clearer on my screen I might be able to identify an inductor producing higher than anticipated voltages.

Wait a minute, this discussion mentioned a spark gap. According to the Wiki page http://en.wikipedia.org/wiki/Spark_gap the nominal gap length to arcing voltage ratio is 30kV/cm. This circuit just got dramatically more complicated and dangerous. Unless you used silicone insulated wire to configure the high voltage components of this circuit, current paths may exist between any two parts, including the operator.

The text needs to be at least twice as large to be readable, maybe more.....

I could not read most of the text, so I am unable to comment in any accurate manner....

Isn't that typical?

I can't make this out either - you have 110V AC and one terminal marked negative, the other positive. A 1kV output from a traditional coil... In this context I can't even determine the purpose of the 110-Volt source - though maybe a timing diagram and proper labelling would sort that out.

I hope this diagram is easier to read. I can't make it any bigger without redrawing the whole thing. The ac combines with the high voltage dc at the spark gap. The ground side of the dc is where the positive ac is attached. The negative ac(common) goes through the capacitor, diodes and the resistor before being attached to the high voltage dc lead or top of spark plug. It is a dipole ac/dc circuit.

The reason for it is the ability to ignite water in various forms as I have shown in my videos.

My problem has been destroying electrodes.

The resistor allows for control of the spark's destructive nature while maintaining a flame instead of a spark at the spark gap. When the resistance is right I can maintain a round glow in the spark gap.

Are sure it's a flame? High voltage 60 cycle AC looks and acts a lot like a flame - as opposed to a spark.

You can see this quite clearly with the discharge from a neon sign transformer.

refer to the picture if you doubt me. You can also watch video, but the camera does not pick up the entire flame like your eye does. It will burn hair off of my hand without shocking me. Try that with ac voltage.

You should compare the flame at my spark plug to the discharge from a neon sign transformer. I have done both and the neon transformer puts out a high voltage spark, but not a flame. I would not put my hand near a neon transformer output, but I can easily feel the heat from my plasma spark and burn hair without being shocked.

I for one am not doubting that you have a genuine flame*. I could readily reproduce this - though I would probably prefer to collect the Oxygen and Hydrogen separately (possibly using a Hoffman voltameter) so that I could mix them to burn at my convenience. But even a properly-constructed Hoffman system is relatively inefficient, as the heat available from burning the hydrogen will always be less than the energy input from the electrical source. Such a system could however be worthwhile if you want to use the hydrogen as an energy store (effectively a substitute for gasoline), and this is similar to some proposals that have been made for "hydrogen-powered transport" and energy stores for fuel cells.
*And PWSlack's comments indicate that he also understands only too well - as by now will vermin.

<...PWSlack...he...>

How abstruse!

he he he You do make life rather difficult for folkes like I wot can't construct sentences around you. 6W - PMSL ! Sorry, I'll get back in me sandpit

Without fully understanding the diagram, I think that the AC is electrolysing the water (creating hydrogen and oxygen) and the spark is reigniting it.

Assuming this is correct: is this supposed to improve the combustion of other fuels, or are you hoping that energy comes from the water? If the latter, I recommend that you stop wasting your (and other people's) time - the energy is transferred from the AC supply to the water (as chemical energy), and there will be both electrical and thermodynamic losses, so it will be much more efficient (and greener) to use an electric motor in the first instance.

The mauve glow is undoubtedly due to HV tracking along the surface of the wire. This implies quite large Voltages across the resistor as well, so it is may well be failing due to arcing.

By the way, if the diode did actually block current in its reverse direction the charge in the capacitor would build up until current through the diode stopped in both directions, so the AC would have no effect whatsoever.

You speak as if this can not work. I use this every day during testing and I guarantee it works as depicted, build it yourself if you like. I can invert the ac voltage to dc and it still works the same. The capacitor does not burn up unless I try to run the plasma spark too fast (glowing). While it is impressive when it glows, the plug will only last about 30 seconds with that kind of use.

This spark does enhance engine performance. I used it on my MG and it ran much better, but the timing was changed by the capacitor so starting was a problem. I plugged in the plasma after the engine was running then adjusted the timing till it had the best throttle response and the highest rpm at open throttle. The MG has no tach so I had to do adjustments by ear and the rpms were also confirmed by my listening.

If you feel like this is a waste of time please move on. I would not want to use up valuable time for something that does not work, can not work, or is just a dumb idea. I am glad nobody told Tesla 3 phase was a stupid idea.

greenandclean,

Physicist? is certainly free to correct me in my reply to you, greenandclean. (Phys, you've done an noble job of that in the past, particularly when I've deliberately stayed into questionable territory. ) But I believe Physicist?'s comment is that you are not creating, or more accurately releasing any stored chemical energy in this process. You are certainly releasing spectacular amounts of energy by "burning" water. But the electric power you are consuming in this spectacular display is the power source for this spectacular brilliant shower. Possibly though by the addition of a water spray with the plasma circuitry in discussion, you maybe getting a more efficient transfer of total power into mechanical power (steam expansion, ionization and recombination of water, etc.) carefully crafted experiments would have to be done to verify if more or less efficient transfer was occurring. Honestly, I think you are not taking into account all energy sources being utilized.

Looking at your latest diagram though, might you be able to swap the position of the load resistor with the charging capacitor? Normally in an RC series circuit the order of resistance and capacitance does not make a difference, but putting the multiple turn series resistor close to ground potential will prevent the corona arc discharge experienced by the windings.

No correction needed. But perhaps a little clarification on futility...
If the input is electrical energy, the most efficient way to convert it to mechanical energy is undoubtedly with an electric motor - converting it (by whatever means) to heat means that you then need to use a thermal engine to convert it to mechanical energy - and the most efficient such engines are below 60% - and engines using practical economic materials don't achieve more than 45%.

<...ignite water...>

Ah. How quaint!

"Ignite water"

Well, I think that may be possible - vapour in a pressurised high-temperature fluorine atmosphere.

With a bit of semantic juggling this might also describe the effect of throwing caesium into water

Does that mean it must be in a pressure chamber to ignite? Does anyone have knowledge of other people burning water? Just curious if this has ever been pursued by a university study or something like that.

GA from me.

If water "burns", why is it used worldwide as the most popular and readily-available fire extinguishing medium ever?

Except for electrical fires.

Water can be used to extinguish electrical fires, provided the jet is chopped into discrete globules, so that there isn't an electrical path from the burning equipment via the jet to earth via the jet nozzle and the body of the person operating it. The practice is not recommended for that reason.

Use de-ionized water instead.

<...Does that mean it must be in a pressure chamber to ignite?...>

No.

<...Does anyone have knowledge of other people burning water?...>

There are plenty of claims on the internet, most of which have been discredited as a simple method of extracting money from the gullible, with no prospect of a return, either of the money or from the supposed enhanced performance of the adapted equipment.

Or potassium, perhaps?

Positive and negative are terms. I am not going to get into a pissing contest over terminology. If you understand circuits you will understand the drawing even though you might be confused as to how it could work. I was. My drawing capabilities suck and should never be viewed by anyone ever, lol.

With this setup it is easiest to keep track of positive and negative ac and dc for me so I dont touch the wrong wire and make that loud yelling noise which always follows a capacitor discharge to my hand or worse my elbow (last week).

The 110 combines with the 1kvdc coil ouput at the spark plug. It enhances the spark a little bit.

Oh, look! It's another one of those....

Another one of what?

hahahahahaha, I hate it when it's another one of those!

Can't believe this thread is still going, sheesh, haven't beat this one to death yet guys?

It is toppling!

Cor! I wonder how good television reception is in your area?

Quite!

In the UK, such a large unsuppressed spark (392W? Good grief!) would obliterate the television reception over a wide area, a scenario likely to have an inspector round pretty quickly and an arrest made for unlicenced radio-spectrum interference!

That looks incredibly dangerous, Monsieur!

Pas de problem - mais on doit être prudent

<...dimmer switch will last about 1 second before failing...>

Many of these are rated only 350W. The calculations above suggest 392W dissipation at the resistance, therefore the dimmer switch is operating

1. above its design maximum
2. into a capacitance load.

Little wonder it pops early.

<...diodes in series after the capacitor to stop high voltage dc backfeeding into the capacitor...>

What voltage rating are these diodes?

It is unclear what the 12VDC circuit is doing from this diagram.

Think of the system as a 110-Volt AC source driving into a water cell to produce an oxygen-hydrogen mix that is subsequently ignited by the spark from an old-fashioned 12Volt-powered coil-circuit-breaker type of mechanism. The 1kV diode is supposed to isolate the spark from the AC system- but I suspect that either series inductance means the Voltage at that point is too high, or the diode is a high-power type and leaks anyway.

The diode is multiple diodes in series. It is a block diagram for ease of drawing. If you want specifics e-mail me.

I tried a new resistor today at the high voltage tap on the ac side. The new resistor is from a printing developer power supply. The identification on it reads GS10X50, 62ohm+/-5%, 587. I put it in series and tested with a/c on. The spark plug glowed until I shut the a/c off. I tried sparking the plug while it glowed and nothing happened. I ran it longer the third time and the resistor got red hot. The spark is the right intensity to keep from destroying the plug, but once it is initially sparked it continues to draw a/c accross the plug gap until the a/c is interrupted. The d/c can be disconnected with no effect on the glow. It is not a short circuit because the 10 amp fuse is fine.

Never had this effect before, just wondering what it was.

I'm not an expert on this sort of automotive side-effect -and there are almost certainly serious enthusiasts using CR4 who will have replicated the problem for other reasons. Plus not having seen it I'm sort of guessing what it really looks like.
But it sounds like contamination of the gap - once a filament is formed across the gap it will glow continuously, and short pulses of current from the spark coil won't change things much. (Lower resistor values may pass enough current to burn the filament out - or it may be the contamination levels have finally reached the critical level)

I think it is related to critical level contamination. There is no direct contact being made or it would blow the fuse.

What I find odd is that there is no pulse except for the initial firing of the plug, one spark starts the glow and the only thing that stops it is interrupting a/c power. I have no way yet of reading the draw on the a/c side. I presume it would be drawing 600 watts, but that's a guess.

I tried a diode on both sides of the resistor and it did not change anything.

I am stumped!

I thought you said you had now fitted a single 62-Ohm resistor. That would limit the current to a maximum of 1.8-Amp, so even a 2-Amp fuse would not blow. Oh, and with that resistor in place the power that could be drawn from the 110V supply could not be more than 200Watt.

If I have misunderstood, and you have connected 3 of these in parallel to give the 600-Watt, the current would still only be 5.5-Amps, which in practice is still insufficient to blow a 5-Amp fuse (after all, there does have to be some margine between the operational rating and the blowing current)
I think the first spark is setting up a filament, and the operating current is enough to maintain it but not not enough to blow it away. Whether the filament would survive the original 8-Ohms is another mater. However, I can't see that it matters a lot, as the real solution is to clean the plug and then adjust the gap.

By the way, at least one out of of the diode(s) and the capacitor must be shorted. If all of them have survived the capacitor will simply charge up to the point that the diode becomes permanently reverse-biased.
(Are you aware that the diode rating needs to be at least 330 Volts, or the combination of rectification (Voltage 'stored' on he capacitor) and peak supply Voltage will exceed the diode rating. Also, that diodes in series may not always provide the level of protection that you imagine - though diodes of similar type will generally be OK, and you could almost guarantee the performance of diodes from the same batch (except that marginal failure of a single diode could bring the lot down, so I would recommend having one or two diodes in excess of the number nominally required whatever you do.)

I explained the value of the parts used. The diodes are 1000 volts. They are there to prevent the high voltage d/c from overcharging the capacitor.

The 62 ohm resistor was placed inline just before the plug to limit the spark so I would not burn up my sparkplugs. If a filament is formed, the gap will be shorted and the fuse will blow or my diodes will be fried and my capacitor will fry. I know this from experience. The thing that is going on now with the addition of the new resistor is something I have never seen before. I can remove the resistor and things go back to normal, but I would like to understand what is going on before I move on.

There is obviously a field being set up that allows current to flow across the gap unimpeded. I am not an electrical engineer so this is new to me. The gap is almost 1/4 inch and there is no filament being formed.

My guess is that the resistor is self inducting. For my application that is a bad thing.

A 62-Ohm resistor that is in proper working order and in the position that you have drawn it will not draw enough current to blow the fuse. An 8 Ohm resistor which is what you reported before could of course pass sufficient current to blow the fuse.
It is also possible that a 62-Ohm resistor could fail in a manner that blows the fuse - some resistors can become low-resistance when failing. But a working 62-Ohm resistor at 110-Volts will only carry 1.8-Amps, which will not blow even a 2-Amp fuse.

.......though it will dissipate 201W!

Actually 195W, and already noted in post #86

What's 6W between friends?

6W ≈ 0.28 litre

.28 litre?

~ half an imperial pint.

If you know the actual voltage change on the capacitor the energy being stored is

Energy in Joules = 1/2 * C * Vmax^2 - 1/2 * C * Vmin^2

If the minimum Voltage is zero that reduces to 1/2 * C * Vmax^2

The power dissipated by the resistor is that energy multiplied by the frequency. Assuming the same resistor is charging and discharging the capacitor you must multiply by 2 again to acount for the electrical energy passing through the resistor twice.

Now we have

P in Watts = f * (C * Vmax^2 - C * Vmin^2)

Double that number will give you a good safe factor for the resitor power rating. Also remember that some power resistors need heat sinks and cooling to meet that specified ratings - read the data sheets.

That assumes that the configuration is a simple power supply, and is usually a reasonable over-estimate for a resistor charging a simple power supply capacitor that is charged through the resistor and discharged by an external load. It assumes that the resistor is the main limit to the rate of rise of capacitor Voltage, which is very rarely the case (if you are rectifying a sine wave, the Voltage across the resistor can never be equal to the full ripple Voltage). However, even here you would be assuming conventional design - it could be a very significant under-estimate if the load resistance is comparable with or lower than the protection resistor.

Non-power-supply situations could be very much higher or very much lower, depending on the configuration.

We all have to make assumptions when the information is lacking but I certainly did not assume a puower supply configuration. I try to base my response on the information given - I see many responses to posts that go in ten different directions because many quickly assume an application then respond based on that. The original post just claimed that the capacitor was being pulsed with 110V through a resistor at about 100Hz. As I thought I indicated in my response - If the same resistor is used for charging and discharging the capacitor then you multiply by 2 to account for the energy passing through the resistor twice. If the resistor is the primary power dissipation element then the assumption must be energy time frequency = Power. If their are more elements dissipating power then the original poster should be bringing that up in the discussion or the resistor will be oversized. The oversized resistor will solve the initial problem - it will not burn up.

"We all have to make assumptions when the information is lacking". Yes, I do that all the time. But I try to explain where such assumptions would fit and where not. I admit that I don't always succeed, and sometimes I make assumptions without realising it. My gripe with your response is that it did not apparently attempt to do this, and could therefore be extremely misleading.

In addition, the posting did say "110 VAC"; your values require that the capacitor is charged from zero to its final Voltage (and discharged vice-versa) in a time that is short compared with the period of the source*. You might (or not) have been addressing what was meant; however, given that your answer would already be fully known by someone skilled enough to recognise whether it applied, as presented it was incapable of providing useful information to the questioner. (Sorry to appear negative here - but you appear knowledgeable enough to be useful if you take the trouble - and my response was also intended to alert the questioner that the calculations might not fit the actual situation)

*for those not familiar with the issue: if the source potential increases significantly during charge the energy dissipated in the resistor will be lower; if the potential is decreasing it will be higher.

Daisy chain them... do SERIES PARALLEL with 4 resistors.