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Updated Feb. 16, 2009    Coil Winding Tips

 

 

 

 

 

 

 

1/4)  Securing the ends of a winding onto a coil form

Here's a method I developed using 3/16" to 4/16" wide strips of polypropylene.  I use 20 to 40 mil polyprop as the thickness of the band material should be at least 60% of the wire diameter.  I've used both black and white material and haven't noticed any problems with either.  If you're going to immediately apply some coating over the finished windings the second band won't be as necessary but will still prevent the turns from spreading until the coating hardens.  The bottom band should be used on all coils.

Photo 1 is a mock-up which shows the slant cut ends that almost meet.  This gap allows the wire to exit the main wound section at an angle.  Also in photo 1 is the small piece of black polypropylene with the rounded ends that is glued over those ends to secure the band.  When the coil is mounted on the winding jig and ready to wind, I draw two fine lines at each end of the coil where I want the winding to begin and end.  Then I cut these strips and glue the joining piece to one of the slant cut ends.  This strip is then positioned in place and glued with one end offset by the width of the magwire, making sure the angle of the gap and offset is in the correct direction.  The joining piece is then glued to the other slant cut end (secure it in place with tape while it dries).  Attach this first band on the end where you are going to begin winding.  The other band is attached once the coil is wound to the desired length.  Photo 2 is one end of a completed coil while photo 3 is that entire coil.  Usually I don't coat my coils after winding I cover them with Saran Wrap when not in use.

                                   

1                                 2                                3                                  4                                 5

Photo 4 - This is a coil winding tool that I developed.  The wooden part is a handle which has a piece of teflon attached to it with epoxy, I drilled two holes thru the teflon that allowed the epoxy to 'grasp' it.  There's a curve where the teflon joins the wooden handle that matches the diam. of the coil to be wound, this one is for a 6" form.

Photo 5 shows it resting on a piece of 6" acrylic form.  When winding a coil (motor driven winding jig) I put on a pair of light cotton photo lab gloves and guide the wire with one hand while the other is holding this tool against the turns on the coil form right where the wire is feeding up onto the form, this allows me to keep them snug against each other as the wire feeds up from below.  With one gloved hand guiding the wire I lightly pinch it as it feeds up which cleans the wire and that glove does pick up a fair amount of 'stuff' often greenish.  I may go through 3 or 4 gloves on that hand before the coil is wound, frequently exposing clean areas of the glove to the wire.  If I stop the motor while winding a coil I take special care to maintain a downward tension on the spool of magwire keeping the wire tight between the coil form and spool.  I immediately wrap new spools of wire with plastic.

I've begun using Red Glyptal to coat the forms prior to winding.  In photo 10 that's a 5" diam, 24" long clear acrylic form with one coat of Glyptal, next time I'll use two.  Glyptal also provides a nice surface to wind on, I used a foam brush and painted the Glyptal with the form rotating and left it running for about 30 minutes.  Be sure to let it dry 24 hours before winding.  I was recently informed of these Glyptal and Shellac Problems, read this and does anyone have any ideas for longer lasting form coating substances?

Don't use Behr 50  --- it cracks with temp changes and sometimes does not dry properly especially in states with high humidity's.  Use an enamel especially developed for coating xmfr wires --- Dolph's AC-43.  Do a web search and Dolph's will pop up.  It seals well, doesn't crack, and holds magnet wire in place.  You can buy it in quart sizes if you're on a budget.  It air dries overnight.  We usually apply 4-5 coats per coil over a 5 day period.  It can also be applied to the coil form as a sealant prior to winding.  Use 2 coats for best over lapping coverage.  Dr. Resonance.

                                               

   6                                   7                                    8                                     9                                10

2/4)  Photos 6 thru 10.  Motorized Winding Jig at work (the motor is spinning in these photos).  Here's some photos of a 5" diameter, 24" long, clear, acrylic form spinning with a coat of Red Glyptal drying on it.  The motor is 1725 rpm with the illustrated step-down pulleys.  I priced gear and chain components and went with the step-down pulleys ($80.00 compared to $250.00).  The belt is the 'softest' carried by Grainger who supplied all of the components.  I built the bases that anchor with bench dogs into my wood-working benches, each rotation takes about 3.9 seconds.  This will be tight-wound 23" with 24awg magwire for around 1050 turns for my twin parallel magnifier.  That's my 25kVA power distribution transformer in the background of photo 9.  August, 8, 2004, J. Cooper.

3/4)  Tesla list responses to someone asking about sealing their coil

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4/4)  Conical primary length formula

Original poster:  Godfrey L.

Imagine that the copper tubing is wound around a cone with bottom radius r. Take
P = angle of the cone edge with the horizontal,
d = the diameter of the copper tubing,
G = the distance between the turns, and
n = the number of turns.

The coil must start at the bottom of the cone. Take
R = r + d(1 + Cos[P])/(2Sin[P]),
A = R + (G + d)nCos[P], and
B = (G + d)/2Pi.

The formula for the length is Pi/((G + d)Cos[P]) times the below
A(A^2+B^2)^(1/2) - R(R^2 + B^2)^(1/2) +
((G + d)/(2Pi))^2Log[(A+(A^2+B^2)^(1/2))/(R+(R^2+B^2)^(1/2))].

Note that x^2 is x squared, x^(1/2) is the square root of x, and Log[x] is the natural logarithm of x.

Note that the length formula is the distance along the center of the copper tube.  This should be close enough. I suppose one could define the length to be an average of all the lengths through the tube, but is this really worth the effort?  The helix formula is gotten from the above with P = 90 degrees.

Q.  Does anyone know a formula to find the length of tubing required for a conical primary sloped at a given angle?

For a straight helix:
N * (sqrt( ( (pi*D)^2) + ( (d)^2) ) )

where:
N = Number-of-turns
D = diameter of helix
d = pitch, i.e., the spacing between the turns

Basically, it's like this (just in case I didn't get my parenthesis right ;)
1) Find the Diameter of the cylinder around which the helix is wrapped (either real or imaginary), multiply times pi, square the total.
2) Measure the center-to-center spacing between the turns of the helix and square that amount.
3) Add the totals from 1) and 2) together, and take the square root of the new total.
4) Multiply the result from 3) by the total number of turns to get the length of the helix.  (Multiply by 1 to get the length of a single turn)

Visualization:
Picture a right-triangle where the length of one 90-degree leg is the circumference of your form, and the length of the other 90-degree leg is the spacing between the turns.  The hypotenuse is the actual length of a single turn.  Apply Pythagorean Theorem to calculate the length, and multiply by number of turns :)

*******

In the above posting of a conical primary length formula, I wrote that you could take P = 90 degrees to get the formula for your straight helix, but this would lead to the indeterminate form 0/0.
http://www.pupman.com/listarchives/2004/May/msg00023.html
Sorry I did not notice this before. Rather, you would have to take the limit as P -> 90 degrees, using l'Hospital rule. This is too much work. Its easier to go back and do the calculus for your straight helix. The copper tube starts against the imaginary helix and the plane of the base of the helix.

For the length of the filament along the center of the copper tube,

L = n[4Pi^2(r+d/2)^2 + (G+d)^2]^(1/2).

For the length of the filament where the copper tube is in contact with the imaginary helix,

L =  n[4Pi^2r^2 + (G+d)^2]^(1/2).

Either one of these is a good estimator for the length of the copper tube. But for me there is a problem. If one deforms a straight copper tube into the shape of a ring, the copper metal would experience compression and stretching.  It seems like the copper on the outside radius of the ring would be stretched, while the copper on the inside radius would be compressed. I really don't know anything about the physics of the deformation of soft metals. I don't think its a problem worth working on for coiling.  Godfrey L.

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