Forum

[Alan R]

Hello SCALY,....Ah-ha !!------ there's that other automotive "Urban Myth," usually expounded by alleged "mechanics"...ie}---- "Your condensor is to stop the points from arcing, mate"... Agreed, that IS a useful by-product but I would ask you to try this simple test}---- With the Distributor cover removed, the engine set on the compression stroke, the points are closed and the ignition turned on..Remove the Spark Plug and earth it such that you can see the electrode...Now flick the points open and you will SEE a bright, blue spark and HEAR a sharp CRACK if all is well.... Now disconnect the capacitor and repeat the test...This time you will SEE a dull, red spark and probably HEAR nothing.....Bear in mind that this test is being done in Atmospheric Pressure conditions ( 15psi approx at sea level at 60 degrees F..) whereas the plug normally runs at pressures much higher than that, thus amplifying the "gap resistance" at the electrodes and the need for a high Voltage ( or Potential Difference ) to form the spark.........The purpose of the Capacitor within the spark-ignition circuit is to BOOST the strength of the spark being formed..........Just a thought but the Urban name of Condensor is probably quite apt as it would have the appearance to some of "condensing" the electricity to make it stronger, but it is a Capacitor and it's values are measured in Farads'. (.. http://en.wikipedia.org/wiki/Farad )............Incidentally, from an electrical circuit point of view it is wired in PARALLEL with the points and then the collective pair are wired in SERIES with the HT coil, ignition switch, fuse, battery and earth...For instance you could connect it to the non-battery side of the HT coil and earth it...it will still work...It doesn't HAVE to be inside the distributor...Again, on my 1975 Honda GL 1000 Gold Wing there are TWO sets of points inside a single distributor connected to 2 Capacitors which are mounted about one foot away on the battery mounting box !!...and that's as standard build, not some previous owners efforts......

[Count Johnny]

Hi Frank (and others). I asked the question because, in the course of an email conversation with Boyer, they advised me that, "once on the move, current draw from the [standard] coil would be around 1amp" which was contrary to my calculation that, at 12 volts, a 3.8 ohm coil would be drawing 3.158 amps. I then read an article telling me that "with the engine stopped and ignition on, a stock [non Enfield, on a different bike] coil drew 2.5 amperes of current. At idle, average current flow was 1.4 amps, and at 4500 rpm, flow was 1.2 amps. Revving the engine higher reduced current flow only slightly more, hence 4500 was used as the upper rpm limit". As I said, this was at odds with my calculations, and I wondered what (if anything) I was missing.

[Scalyback]

Thanks to ralph from Smokstak - Capacitors in ignition systems





When the points open the capacitor is connected in series with the coil. The voltage/current generated by the collapsing magnetic field charges the capacitor. This is the voltage/current source that is induced into the stepped up secondary winding. A coil opposes current flow while a capacitor enhances current flow. The enhancement of the capacitor balances out or cancels the opposition of the coil. The result is a faster collapse of the magnetic field and the highest possible voltage generated in the primary circuit. It's true that the capacitor reduces or prevents arcing at the points but its main function is to provide a circuit path for the coil after the points open and to speed up the collapse of the magnetic field. – Ralph




So it does the field collapse for the spark AND is a spark quench! WOW, all from that little thing... I'm going to take mine to dinner, it obviously deserves it!


Looks like we were both right!

[Count Johnny]

I have concocted a (probably half-ar*ed) theory. At zero rpm (with the points closed) the current draw per second would be (say) 3.1 amps. At 1000 rpm the points would be open for 17% of one second so the current draw, per second, would be 87% of 3.1 amps = 2.58 amps. At 2000 rpm, the points would be open for 33% of one second so the current draw, per second, would be 67% of 3.1 amps = 2.07 amps. Or am I just being silly?

[Count Johnny]

Sorry, I meant 83% at 1000rpm.

[Dennis C]

"When the points open the capacitor is connected in series with the coil. The voltage/current generated by the collapsing magnetic field charges the capacitor. This is the voltage/current source that is induced into the stepped up secondary winding. A coil opposes current flow while a capacitor enhances current flow. The enhancement of the capacitor balances out or cancels the opposition of the coil. The result is a faster collapse of the magnetic field and the highest possible voltage generated in the primary circuit. It's true that the capacitor reduces or prevents arcing at the points but its main function is to provide a circuit path for the coil after the points open and to speed up the collapse of the magnetic field. – Ralph"

Scalyback I would have thought that with your knowledge shown so far of how the ignition sytem works you would have known this, unless the current has somwhere to go back to it is impossible to create a momentary AC current to generate a voltage in the secondary winding, you will get a weak spark due to the primary voltage jumping back across the points as they open.

"My goal for this inquery is to get a coil with a hot and very strong spark for my Bullet, but without damaging the electrics (my rev counter blew up with the aftermarket coil).." Jacob a hot spark should not be your goal but the easy way is with a hotter spark plug, when tuning bikes for racing back in the 60s we would normally fit cooler plugs as the state of tune increased, as the engine temperature increases it is important to control the hot spot around the spark plug to prevent burning a hole in the piston. Bellet Whisperer, do you still do this in 2014?.

[Dennis C]

Oops Bellet Whisperer, damn where is the edit facility.

[Scalyback]

Hi Count Johnny



when moving, the points are closed for a percentage of the time of a 4 stroke cycle.

If the current when stationary and the points closed is 3A

and the points are closed for 20 percent of the time,

then your current draw whilst on the move will be 20 percent of 3A !



Electronic systems that have 'waisted spark', fore twice as often, so they use more current. The percentage that the 'points' are closed here, is the same as for a two stroke engine.



To calculate percent of points closed.



(it helps if you have a side cover removed so that you can see the crankshaft and timing gear.) Also handy to have an ohm meter connected across the points or a packet of Rizla blue cigarette papers!



Turn the crankshaft until the points close. this is when the ammeter goes to the left (ignition on) or... (ignition off) when your ohm meter swing to zero on the scale, or, whilst sliding the Rizla blue through the points gap, the point where it is suddenly gripped.

Make a visual note of the crankshaft or put a very small bit of white paint or tippex on it.

Now rotate the crankshaft until either :- the ammeter swings to the centre or your ohm meter swings back to infinity, or the fag paper becomes loose again.



The difference between where your crankshaft was, and where it is now is the angle of rotation. Try to get the percentage visually. If the crank has turned a half circle, then 50 percent, turned a quarter, then 25 percent. If it turned about 4 tenths of the way round, then 40 percent.



now pick the correct next step for your type of ignition



4 stroke points ignition - Let's say the points are closed for 4 tenths of a revolution, this is 40 percent of a revolution, but 4 stroke uses two revolutions, so half the percentage, points closed for 20 percent of the time, coil with engine running will draw 20 percent of the current when engine stationary and points closed.



2 stroke engines and 4 stroke electronic wasted spark systems - same as above except, you do not ned to half the percentage. The percentage you got from the crankshaft turn is the answer. so for points being closed for 40 percent of a revolution, your engine will use 40 percent of the power used when stationary with the point closed.



It's snowing like hell, its 1 degree C. No way I'm taking thunderbolt out of Switzerland to France! At least the nuclear bunker is somewhat secure!

[John L]

Scalyback - "hypo-pathetical" - is that a new word ?

[Gwilly]

Since the electrical resistance of a conductor such as a copper wire is dependent upon collisional proccesses within the wire, the resistance could be expected to increase with temperature since there will be more collisions. An intuitive approach to temperature dependence leads one to expect a fractional change in resistance which is proportional to the temperature change:

I trust that the increase in temperature of the coil and therefore the copper wired primary circuit has been factored in to your calculations using resistance/temp coefficient tables for copper wire… if not don't worry, I'm fast loosing the will to live…