Good to see this post inspiring thought and dialog. Here is some more to keep it going.
12.435V, is this Dc or Ac ? At 12V about every coil work well.
12.435V is the DC voltage powering the tester. The stimulus applied to the coil under test is derived from this DC power but is NOT a steady, fixed voltage DC. The test stimulus is a very fast pulse of electricity which emulates the magneto pulse at very high engine speed where ignition timing is most critical.
A common fallacy is: The Model T magneto puts out an AC voltage that operates the coils, therefore, coils should be adjusted using an AC voltage like that produced by a HCCT." Here are a few facts why this reasoning is flawed:
1. The magneto pulse that fires a Model T coil does NOT alternate polarity during firing. It will be either a positive pulse or a negative pulse depending upon spark lever setting. Accordingly, the coil current does not alternate between positive and negative when firing a particular spark. Hence, alternating current does not occur during coil firing.
2. The HCCT contains no timer function to permit the coil to rest in between firings and return to steady state. The coil is subjected to a continuous succession of positive and negative voltage pulses from the magneto during coil adjustment but will never occur during normal operation.
3. The HCCT is hand cranked at abnormally slow engine speeds (60-120 RPM) in which the car will never operate. The resulting magneto voltage rises abnormally slowly to an abnormally low in voltage value is used to adjust coils but the coils will never operate during normal use.
Very low amperage draw leads to easier successful starts when cranking by hand. The higher amperage draw setting, the fiercer the pull must be on the hand crank to achieve rotational speed to create a spark when starting on magneto.
This makes perfect sense but, is it true? Does anyone have hard data on this? Its something I have often pondered but have yet to actually test. I suspect this common reasoning is also a fallacy and here is why with the help of some math rather than just conjecture:
1. Spark energy depends upon the amount of coil current flowing in its primary winding at the moment the points open to generate spark (note this is NOT the average coil current read on the HCCT meter; typical coil firing current is ~6A). The spark energy can be approximated by the following equation:
- EQ1.JPG (8.56 KiB) Viewed 4258 times
Where E=Spark Energy, L=coil primary inductance and I = coil current flowing at the time the points open.
The lower the current flowing in the coil primary at the time spark fires, the lower the spark energy.
Note that the current term, I, is squared (I^2) which means spark energy decreases significantly as the points are adjusted to open at lower firing current. The lower the spark energy, the more difficult it may be to start the engine due to insufficient spark energy to reliably ignite the air/fuel mixture (ie more vulnerable to partial or incomplete combustion; misfire).
2. What coil current can we generate via brisk hand crank? The magneto voltage output is a triangular wave (not sine wave as is another common misconception). The resulting coil current rises exponentially with the linearly rising triangular magneto voltage as the crank is pulled swiftly. The coil current can be calculated by the formula:
- EQ2.JPG (9.03 KiB) Viewed 4258 times
Where L is the coil primary inductance, Vcoil is the stimulus voltage applied to the coil and dt is the change in time.
So to calculate the value of coil current, we need to know have fast the applied coil voltage rises with time (VL dt) term. Lets assume we can pull the crank a 1/4 turn in ~30ms (That's equivalent to completing 1 full rotation in 1 second or 60 in 1 minute; 60 RPM. Thus, 8 magneto voltage cycles produced per in 1 second revolution or 1/8 = 0.125 seconds to complete 1 complete magneto voltage cycle that of two voltage pulses (one positive and one negative) and lets say the voltage pulse is a relatively weak 2Volts cranked that slow. Here is what that would look like in time
- Magneto Voltage 1 cycle at 60 RPM.JPG (40.22 KiB) Viewed 4258 times
so the time it takes to produce a single magneto voltage pulse to rise 0 to 2V takes 1/4 of that time (1/4 crank pull) or 0.125/4 = 0.03125 seconds for the magneto to produce a weak 2V voltage pulse (see dashed red box) . So then the magneto voltage (VL) rises from 0V to 2V in 0.03125s so the linear slope (m) of that line is 2/0.03125 = 64. The voltage applied to the coil can be written as:
- EQ3.JPG (7.95 KiB) Viewed 4258 times
Where m is the slope (rate of rise) and t is time.
Plugging that back into our equation for coil current equation and solving:
- EQ4.JPG (12.34 KiB) Viewed 4258 times
9.47A is more than enough coil current to pull open the coil points to fire spark. A properly adjusted Model T coil firing current is about 6A on the ECCT so should have no problem starting an engine via hand crank.
Not sure what the correlation is between ease in engine hand crank starting and what the average HCCT coil current reading (0.8A vs 1A vs 1.3A vs 1.5A) measured while hundreds of sparks are firing continuously. There are so many variables involved with the average coil current reading on an HCCT, I don't know that you can correlate a solid relationship between average HCCT coil current reading and ease in engine starting. You would have to know the correlation between average coil current measured and the coil current flowing in the coil primary at the moment the coil points open to produce spark. That is the coil current that really matters.
