Re: Mallory tech answer to coil/ballast ?
Tom ,
Good post.
Lets do the math:
Using a 1.5 ohm coil AND a 1.5 ohm ballast would give a total of 3.0 ohms resistance.
3.0 ohms divided into 12 volts= 4 amps of primary current ( close to maximum using contact points)
OR volts x amps = watts of 4x12= 48 watts of energy to make spark.
Now : 1.5 ohms with no ballast divided into 12 volts= 8 amps or watts = volts x amps of 96 watts of energy.
Now the problem becomes another issue with the 8 amps being very close to what the electronics can take requiring a current limiting circuit ( for protection) in the control. This could be lower than the 8 amps we figured above.
Example , GM HEI limits the current to 5.5 amps etc......
NOW the discussion can begin regarding to how much spark do you actually need ???
And the fact that a coil will only put out what is required, and the fact that flatheads are fairly low compression requiring very little spark to ignite the air fuel ratio.
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