Length calculation

[lugligino]

Consider a toroidal solenoid with circular sections http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html. This item has an internal diameter ID an external diameter OD and an height H. If a wire with diameter d is wound up the solenoid in a turn that covers an angle t of the solenoid, how long is the wire?
Thanks a lot.

 

[HallsofIvy]

One way of estimating it is to take the volume filled by the wire, \(\displaystyle \pi H(OD^2- ID^2)/4\), and divide by the cross section area of the wire, \(\displaystyle \pi d^2/4\).

 

[lugligino]

One way of estimating it is to take the volume filled by the wire, \(\displaystyle \pi H(OD^2- ID^2)/4\), and divide by the cross section area of the wire, \(\displaystyle \pi d^2/4\).

Thanks for your reply.
In your approximation don't you think could be better to consider the volume occupied by the wire as the surface of the toroid(*) multiplied by the wire diameter d: \(\displaystyle \pi^2 (OD-ID) H d\)?

Any ideas for a less approximated result?

(*) see http://en.wikipedia.org/wiki/Pappus...9CzBjVGQ7wQHGAAAAAwUAAO6WCwAIBwAAOwAAANWjAQA=

 

[DrPhil]

Consider a toroidal solenoid with circular sections http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html. This item has an internal diameter ID an external diameter OD and an height H. If a wire with diameter d is wound up the solenoid in a turn that covers an angle t of the solenoid, how long is the wire?
Thanks a lot.

What is h? I don't identify it from the picture of the toroid. My interpretation is a=ID/2, b=OD/2.

The number of turns is limited by spacing on the ID:
N = t (ID/2)/d

Each turn has length pi*(OD - ID)

EDIT: a second interpretation is that the turns are not tightly packed, but rather are spaced such that each turn covers an angle t. In that case each turn is an ellipse rather than a circle. The minor axis of the ellipse is (OD-ID), and the major axis, tilted by angle t, I am guessing would be (OD-ID)/cos(t). Find the circumference of the ellipse. The number of turns would be 2pi/t.

2nd EDIT: Now I think the shape of the wire for one turn would be portions of two ellipses, each with major axis (OD-ID)/cos(t/2).

 

[lugligino]

About approximations: if you consider a toroid with condition \(\displaystyle (OD-ID)/2 >> p\) where p is the pitch of the coil, you can consider that the wire is a coil around a cylinder with lenght that can be calculated easily
\(\displaystyle 2 \pi sqrt((H+d)^2/4 + (p/2 \pi)^2)\) (*)

but what if the condition is not so true?

(*) see http://mathforum.org/library/drmath/view/55156.html

 

[DrPhil]

About approximations: if you consider a toroid with condition \(\displaystyle (OD-ID)/2 >> p\) where p is the pitch of the coil, you can consider that the wire is a coil around a cylinder with lenght that can be calculated easily
\(\displaystyle 2 \pi sqrt((H+d)^2/4 + (p/2 \pi)^2)\) (*)

but what if the condition is not so true?

(*) see http://mathforum.org/library/drmath/view/55156.html

The use of the Pythagorean theorem should be the same as my use of cosine. I like the statement "The hypotenuse follows the path of the coil."

 

[lugligino]

What is h? I don't identify it from the picture of the toroid. ...

Ehm, H is the toroid height that is [FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]O[/FONT][FONT=MathJax_Math]D[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]I[/FONT][FONT=MathJax_Math]D[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]2[/FONT] because the toroid section is circular ...

Are coils ellipse sectors? I don't think so because, if it was so, the tangent at the conjunctions of two near ellipses was perpendicular to toroidal solenoid central axis and it's not the case because the coil goes constantly forward.
If wire is around a cylinder, it becomes a helix. Around a toroidal solenoid it is different but not an ellipse.

 

[DrPhil]

Ehm, H is the toroid height that is [FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]O[/FONT][FONT=MathJax_Math]D[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]I[/FONT][FONT=MathJax_Math]D[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]2[/FONT] because the toroid section is circular ...

Are you sure that coils are ellipse sectors?

"Pretty" sure - at least to a pretty good approximation. If the turn is in a plane, then the intersection of a plane with a cylinder is an ellipse. The approximation is to consider the torus to be a cylinder - good if t is small. Consider three points that define a turn:
1) on the ID at angle=0.
2) on the OD at angle = t/2
3) back at the ID again, at angle t.

The distance between points 1 and 2 (and also between points 2 and 3) is H/cos(t/2). I am calling that the major axis of the ellipse, and the minor axis is H.

Another possible way to approximate. There are N = 2pi/t turns.
The wire runs back and forth in the radial direction N times, so the sum of the radial components of length is N(pi H).
At the same time, there is an orthogonal component around the torus, of total length pi(OD - ID)/2.
That ism there are N turns around the surface of the torus, and one turn around the centerline of the torus.
Combine those two lengths by Pythagorean theorem.

 

[lugligino]

The number of turns is limited by spacing on the ID:
N = t (ID/2)/d ...

See this clear discussion http://www.had2know.com/technology/calculate-toroidal-coil-inductor-wire.html . It seems clear that \(\displaystyle N=\pi/arcsin(r/(A-r))\) where \(\displaystyle r=d/2\) and \(\displaystyle A=ID/2\). Also the coil length is discussed but in the cylindric wound approssimation only.