I think you basically answered your question, except that you just popped up yet another one.
Basically you need to rotate your 1 feet diameter rod while at the same time a mechanism moves laterally 5 feet. I know how to make this electronically. For example, I would use 1 stepper motor to move 1 revolution (200 steps if direct driven) and then another stepper motor to actuate a lead screw to move the 5 feet in unison. With gear reduction this can get easier or harder.
For example, lets assume you use a 5 feet long 1/4-20 screw as your linear actuator (the one that moves the wire sideways). To move 5 feet linearly you would need to move 60 inches, or 1200 turns (you need 20 turns of the screw to move an inch).
Since the rotational motor has 200 steps (or 1.8 degrees per step), you would move the linear motor 6 full turns (0.3 inches) every rotational step.
Headache about to pop a vein? Yes, I realize this is ubberly complicated, but I don't think helices are kindergarden technology. I am myself ready for a couple Tylenols... ;-)
Now, a stepper motor is hardly going to have the strength to bend wire of this diameter, but what you can do is create your own 200 steps index head. By the way, you can create it with how many steps you want, and the math will follow along. But if you follow the 200 steps per revolution example, then you can click the rotational head, move the screw through 6 revolutions, another click on the head, six revolutions on the screw... and so on!
Hope this math is correct... I kind of winged it, so there could be mistakes on the numbers. I think the logic should remain true, though.
Good luck!
