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I read where he powered up the coil positive but not the ign box. If im wrong sorry. Just trying to help.



Jumping power to the coil positive "should" provide power the the ignition ECU box in the same manner as having the key in the start position. Which means the power has to flow (back) through the ballast resistor to power the box. Because the ballast resistor should be 1-ohm or less this should not be a problem unless there is excessive current draw on the same "run" circuit which also the the power ti the alternator field and voltage regulator (on some models also the electric choke?)
I guess you could isolate that circuit be disconnecting the alternator field terminal, the voltage regulator and the ECU, then with the ignition switch in the "off" position, jumper power to the coil positive again and verify no current is flowing through the ballast resistor. You could wire an ampmeter in series with the ballast resistor, or measure the voltage drop across the ballast resistor and see how re-attaching the various loads previously disconnected draws power. It would be a good way the check out the wiring in that circuit. Also, you can check that power is getting the the ECU connector. In fact, you should see power on two of the ECU terminals, the normal power terminal and the wire connected to the coil negative terminal. The other two terminals are the magnetic pickup connections to the distrubitor.
Re-connecting the ECU connector should result in the ECU grounding the coil negative terminal if the ECU box is grounded.
That is all there is the ignition and charging circuitry. As I mentioned before this should work even if the bulkhead connector or ignition switch are removed... They are just supplying the power that the jumper wire is now doing.
The only other required circuit is the starter circuit, to spin the engine. The major difference here is when cranking the engine it places a large current load through the battery and battery cables and any resistance in these paths or a weak battery will result in voltage drops seen in all the circuits.
A small resistance of even 0.5-ohm with 50 amps through it is a 2.5volt voltage drop, and a 5-volt voltage drop at 100 amps.

When the engine wants to start only after releasing the key from the start position to the run position, it is because the high current starter load is removed and with less current draw, the voltage drops are reduced resulting in more voltage, and to the ECU. Also because the starter and the ECU are really grounded in two totally different areas (starter to engine block, and ECU to body) there is also a chance of a ground loop voltage problem that can interfere with the ECU sensing the magnetic pickup signal from the distrubitor.