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Will 1280' (feet) coil the, 26 awg wire

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Wat dia in feet ?

Then one coil of #26 awg wire diameter o.015940" is a o.00133' added to the tube  results in a true o.33466' diameter which multiplied by Pi uses 1.05137' of wire per turn.

A 1280' length of wire divided by 1.0514' will allow 1217 turns of wire and
multiplied by wire diameter will need tube at least as high as 1.62' or 19.5"..
This assumes a perfect winding.  In practice add 12% slop to 1.82' or  21.8" tube height.

A.

Did you say I'll only get half of it done with this spool

1/3' dia =o.33333333333333' + o.00133 = o.33466' dia × 3.14159 = 1.0514

1280 / 1.0514 = 1217 turns

1217 × o.00133 = 1.62'  tube high perfect wind

1.62' × 1.12 = 1.82' actual tube winding height

2' tall, 3.5" diameter, 26 awg wire that is 1280' long will I have enough?

Where is the question here?

Whereas I can accept that English isn't all peoples first language google translate can do better than that.