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.

## Discussions

Best Answer 7 years ago

Wat dia in feet ?

Answer 7 years ago

1/3'

Answer 7 years ago

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.

Answer 7 years ago

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

Answer 7 years ago

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

Answer 7 years ago

Yeah you said half

7 years ago

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

7 years ago

Where is the question here?

Answer 7 years ago

You just asked it......

Answer 7 years ago

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