In electrical engineering, coil winding is the manufacture of electromagnetic coils. Coils are used as components of circuits,and to provide the magnetic field of electrical machines such as motors and generators, and in the manufacture of loudspeakers. The winding height can be estimated using the formula.
Let's say we have a magneticfield that's coming out of the right side of the screen. And it's not just along thescreen, it's actually three dimensional. So it's going above thescreen, below, but the direction of the magneticfield is from the right to the left. Let me just draw that.
And I'm not going to draw abunch of the field vector arrows because that'll justtake up a lot of valuable space on our- so that's thevector magnetic field. It's down here, too. If I could, I would draw itabove your screen and below your screen. But it's coming from theright to the left. And then in that magneticfield I have an electric circuit. I won't draw the wholecircuit right now. I'll do that in a second.
But let's say I have-part of the electric circuit is a loop. And the loop looks like this. I'm trying to drawit carefully. So that we can- because I thinka careful drawing will be more useful ina second than an uncarefully drawn drawing. So let's see. So it's a loop.
You can almost- you couldimagine taking a paper clip and putting it intothis shape. Oh, that's good enough,I think. And I have a current goingin this direction in this paper clip. So this is the positive,that's the negative. So the current isgoing like that. Current is going ina loop like that.
The current's coming outof this end, it's coming into this end. And let's say that loop of-it could be a paper clip or anything- let's saythat it can rotate. And that's important. What's going to happen? Well, my magnetic field iscoming in this direction. The current is goingdown here, up here.
What's going to be the netforce of the magnetic field on this loop? Well, let's try it out. And it turns out it's going tobe a different magnitude at different pointsof the current. And we don't worry about- allwe're worried about right now is direction. And then maybe a littleintuition of the magnitude. So we know that the force of themagnetic field is equal to the current times thelength vector cross the magnetic field.
Well, what would be the force ofthe magnetic field on this segment of wire? We could call this L. And that L goes in the samedirection as the current. Well, let's see. Current is just a scalar,but L is going down. Magnetic field goingto the left. Cross product.
Cross product, I take my righthand, put my index finger in the direction of the current,or in the direction of L, because that's the first termof the cross product. So that's the index finger. So my index finger goes down,because that's the direction of the current.
And then my middle finger- andremember, you have to do this with the right hand. If you do it with the left hand,you're going to get the opposite result. And now my middle finger isgoing to go in the direction of the field. So let me point mymiddle finger. My middle finger isgoing to go in the direction of the field.
I keep having to lookat my own hand. And then my other two fingersare just going to do what they need to do. So that's my third finger. That's my pinky.
And then what is mythumb going to do? What is my thumb going to do? Well, my hand- that'smy hand. This is what my hand is doing. I'm pointing downward.
And my palm is kind ofpointing at my body. So what is my thumb doing?
I know it's hard to see. This is my middle fingerright here. So my thumb is on the otherside of this drawing. And my thumb is pointingdownwards.
I hope you see that. And you try it withyour own hand. So my thumb is pointingdownwards.
So the direction of the forcecreated by the magnetic field on this current is goingto go downward. So let me draw that. So the force vector-I'll do it in this orangey brown color. The force vector on this segmentof the wire is going to be going down. Now what about this segmentof the wire?
Well, think about it. This segment of the wire, theL vector- this L vector- It's parallel to the magneticfield just in the opposite direction. And so when you take the crossproduct- remember, the cross product is you're multiplyingthe magnitude of the vectors that are perpendicularto each other. But if this is the L vectorright here, there's no component of it that it'sperpendicular to the magnetic field.
So the magnetic field and thecurrent are in the same plane. They're parallel. They're not orthogonal at all. There's no components of themthat are at 90 degrees. So when you take the crossproduct, you're going to see that the net force on thissegment of the wire is 0.
And likewise on this segment ofthe wire and this segment of the wire. Because they aren't in anyway perpendicular. No components of them areeven perpendicular. So fair enough.
So all we know right now is themagnetic field is exerting a downward force on thisside of our paper clip or of our circuit. Now what about this side?
Well, same thing. Take the cross product. If this is our L, L cross B. So take your index fingerin the direction-. So index finger goeslike that.
Your middle finger will go inthe direction of the field. So your middle finger is goingto look something like that. And then your othertwo fingers are going to be like that. And what is your thumbgoing to do? And this has to be yourright hand to work. Your thumb is going topoint straight up.
This is like the heelof your thumb. Your thumb is going to-. I don't know if that's a gooddrawing of a thumb. But your thumb is essentiallypointing out of the page. Middle finger in the directionof the current, or in the direction of our length.
Sorry, index finger in thedirection of the current. Middle finger in the directionof the field. Thumb points out of the page. Do that with your own right handand you'll see that the net force of the magnetic fieldon this segment of the wire is going to be upwards. Let me do it in a differentcolor just to get some contrast. So what's going to happen?
Assuming that this circuitcan rotate, what's going to happen? On this side there'sa downward force. On this side there'san upward force. So the magnetic field isactually exerting a torque on this wire. If you viewed this little dottedline as our axis of rotation, the whole coil isactually going to rotate around that line. And so there's some force overhere, along this whole line being applied downwards.
And it's actually perpendicular to our moment arm. If you remember what we hadlearned about torque. So it will actually exert allof that force- that force times this distancewill be the torque applied on this side. And then likewise there's atorque- it's really the same sign, in the same direction,because here on the other side of the arm it's pushingupwards. So they're not goingto cancel out. They're both goingto reinforce. And this whole coilis going to be turning in this direction.
Here it's going to be moving upout of your video screen. Here it's going to be movingdown into your video screen.
Now what happens? I'm going to try to not runout of time either. So it's going tostart rotating.
So the left hand side's goingto go below the page. The right hand side's goingto be above the page. I want to draw some perspective,that's why I'm just drawing it bigger. Maybe it looks like that. Maybe my circuit startsto look like that.
And I'll redraw myaxis of rotation. So this is my axisof rotation. And on the way I drew it-this part, the axis of rotation is still in theplane of our video. But this part of thecoil is, you could imagine it popping out. I wish you had 3D glasses on.
It's popping outof your screen. This part is goinginto your screen. And the current is still goingin the same direction. Current is going in thatdirection there. So using the same right handrule, on this side of the wire the magnetic field is goingto be exerting a net downward force. But the torque is actually goingto be less because our moment arm distance is going tobe like- I want to draw it with some perspective. It's going to look somethinglike that.
So it's going to be going to theleft and behind the page while the torque is stilljust into the page. So you would actually take thecomponent of the torque that's perpendicular.
So there's some component ofthe torque that's actually perpendicular. I don't want to confuseyou too much. But you could imaginethe torque lessens. Even though the net magneticforce is the same, the component of that force that'sperpendicular to your moment arm, that lessens. So there's still going to besome torque that's going to be causing it to rotate downwardsin that direction. You know what? I drew this wrong here.
We're pushing up on the righthand side, we're pushing down on the left hand side. So the direction is goingto be like that. Pushing up on the right hand,down on the left hand. So you're still going to bedoing the same thing here. You are going to bepushing up here.
But you're going to be pushingup directly out of the page. But that's not completelyperpendicular to the moment arm.
So the component that isperpendicular, that's actually creating rotational torque,that's going to be a little less. And then you could imagine thatall the way to the point, the coil's going to keeprotating with smaller torque.
At some point you'll belooking at it head on. So I can just draw it likea straight line, right? You can imagine. This arm is on top and thisarm is behind it.
And at this point, what'sgoing to happen? All the magnetic force onthis top arm is going to be popping up. It's going to be popping up outof your page but it's not going to be providingany torque. Because it's not perpendicularanymore to your moment arm. And likewise, on the bottombehind this, if you could visualize this, it would beexerting a net downward force.
And that's also not goingto be helpful. But maybe they have some angularmomentum so the wire will still rotate. But then when it still rotates, what's going to happen? And this is where I'llleave you with a little bit of a conundrum.
Actually, I don't want to goover the Youtube limit, so I'm going to continue this in thenext video and I'll show you the conundrum. See you soon.
Motor turns tables are limited to the lengths and cross sections listed. Those of us who cut and strip rubber to other than standard size need something more general.I use a formula.Caution; the breaking turns are affected by many things. Every batch of rubber has different properties, warmer temperature stresses the rubber, lubrication, stretching, rate of winding, cooling, previous stretching, braiding, contamination and other things affect how many turns a motor will take. Winding heats the rubber, it heats faster as torque increases and every turn puts in more work. Slow your winding as you get close to full to let it dissipate the heat.Formulas can be a useful guide to winding under standardized conditions.My formula is:T = 10.64/Sqrt(S)T is breaking turns per inch. Multiply that by the length of your motor to get breaking turns for that motor.S is motor cross section in square inches. Multiply strip thickness (0.042″) by motor width times number of strands.Sqrt( ) is square root.
Look on your pocket calculator. Most computers have a calculator window.10.64 is an empirical coefficient. It is the result of testing some pretty good Tan II.
Lou got exactly the same for some Tan SS. Purely coincidence.You can establish your own formula by winding a short test motor under standard conditions and using the formula to calculate a new coefficient. “Good” rubber will run from 9 to 13.Here is an example of a turns table I made up for a motor.Denny Dart II, 7″ NP Prop, 17″ loop of 0.083″, 127.4 tpi% Breaking Turns Number100 290 180 170 1516If you are using a 15:1 winder, divide those turn numbers by 15 to get the number of cranks on the winder.Gary Hinze. I am pleased that someone is reading this closely enough to try the numbers. It makes me feel that my effort is justified. Sometimes I make a mistake, so I carefully reviewed the calculation.Your discrepant number is the result of an easy oversight.
The width of 0.083″ is the width of a single strand of rubber. There will be two strands in the loop used in the motor. So the cross section is twice the width times the thickness.This number is a rough guide. Every batch of rubber is different and motors cut from the same batch may differ noticeably. The width and thickness may vary slightly.
When winding a motor, you can wind up to about 90% of this number, then you can stretch check. You will find that energy may be stored in two ways, by torsion and by tension. The sum has a limit. As you approach the limit in torsion, you will feel the motor stiffening in tension. As you approach maximum turns, you should wind slowly and stop occasionally.
Let the motor cool and then give light tugs to the motor to see how it responds to stretching. When it doesn’t want to stretch with a light tug it is a good time to stop winding.
The decision is a matter of judgement based on experience. You wind this hard only in competition. If you wind this hard you probably will prestretch the motor once and fly it once, then discard it. It will likely break the next time you wind that hard. For sport flying, 70% to 80% of maximum turns is preferred.
That will allow many flights from the same motor.
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