I agree that a capacitor is very useful in with high speed bulbs or when flashbulbs are shot at high shutter speeds. My post was in reference to the cameras and flashes I use most often. I own over a 100 cameras. Most (but not all) of the ones I use flashbulbs with are old simple cameras from the 1940s, 50s and 60s. Theses cameras have slow shutter speeds of 1/60th of a second or less.
Also to be clear to future readers of APUG. I have flash guns that can be used with OR WITHOUT a capacitor. Do not think you can cut a capacitor out of your flash gun and it will work unless you know what you are doing.
As far as Marriottworld is concerned, I have know of Stephanie untimely passing for a year or so. She had a passion that her husband can not be expected to continue.
I have over a 1,000 M3 and M3B flashbulbs in stock right now. If anyone on APUG would like to trade M3 (or other bulbs) for Phillips style bulbs PM me. That way we can do a trade and cut the ePRAY and PayPal (where is the pal ?) out of the equation.
Sirius Glass, you are smart and I have learned a lot from your posts in the past. We are both smart people. We just need to speak and we will respect each other. Let us not do this on a public forum trying to prove who is better. Neither of us is better. We are trying to post useful information. If you feel I am wrong let us not argue but come up with a proper post. Let us inform, NOT fight. PM me if you would like to talk and we could come up with a proper post that would be useful and informative to APUG.
For those who might not be sure how a capacitor flash gun circuit works, I post below the circuit diagram of my Leica CEYOO. When you plug in a flash bulb, it completes the outer circuit and a small current, limited by the 10K resistor to just 0.00225 amp (V=IR and assuming the resistance of the bulb is effectively nil compared to the 10K resistor), so the bulb does not fire. This current charges up the capacitor. So how long do you need to leave the bulb in for the capacitor to charge? If my maths is correct and I have remembered my electricity from physics 50 years ago, at 0.00225 amps, a 100 μf capacitor would require just 0.04 seconds to charge up. When the flash contact switch is closed, the battery is effectively out of circuit and it is the capacitor only which discharges rapidly through the flash bulb. The current rise rate of the capacitor would be much faster than a battery, particularly an old fashioned zinc carbon multi cell small battery, which was all that was available in the 1950's. If you have a leaky capacitor and leave a bulb in the flash, the battery will run down quite quickly, which is why it is always a good idea, if you are leaving a bulb in a flash for display purposes, take out the battery.
I have now found a source for 100 μf axial wire capacitors and should have 2 waiting for me, when I get back to the UK next week.
A little bit more physics if you can bear with me. I have a Chinese made 22.5V zinc carbon battery (which I am putting nowhere near my precious Leica and Rollei flashes). I have just been playing with my multimeter and 490 Ω resistor and as far as I can work out, the internal resistance of the battery is in round numbers 20 Ω. A #26FP flashbulb has a resistance of 5 Ω. Therefore if you were to fire with battery rather than capacitor, the flash current would in theory be 0.9 amps (22.5v / 25 Ω) but due to the limiting current of these very small cells, probably around half of this number but let's be generous and say it's 0.6 amp. The capacitor has effectively no internal resistance and the current would be 4.5 amps (22.5v / 5 Ω) i.e. nearly 7 times the amount via the battery. This is the attraction of using a capacitor fired flash.
I have a PDF file of the GE Flashbulb data booklet up at:
It is under the title
GE Flashbulb Data 1961
I don't get any money for this. But do donate to Mike if you use his site. He as 1,000s of photo related instruction manuals and you can see how much time he has dedicated to his site.