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On 25/03/2015 13:10, CDB wrote:=0A=
> The question.=0A=
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> You have a round coil core 10mm thick  and 50 mm long. You wrap a copper =
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> wire exactly 5 times around. How long is the wire? (don=92t consider the =
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> thickness of the wire)=0A=
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> They worked it out using 3-D triangulation.  =0A=
>=0A=
> Imagine a rectangle with the vertical side =3D L =3D 50mm, horizontal sid=
e =3D W =0A=
> =3D 10mm=0A=
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> They then drew a right angled triangle top left of rectangle and bottom =
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> right to represent an unwrapped cylinder.=0A=
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> They then stated the wire runs between two opposing corners, therefore th=
e =0A=
> wire length (one turn) =3D the standard sqr (L^2 + C^2), they then went o=
n to =0A=
> show a series of equidistant triangles along the length of the core, 5 in=
 =0A=
> total for the turns.=0A=
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> So finally they have N * sqr((L/N)^2 + (pi * W)^2) =3D 165mm.=0A=
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> Now I need to get myself a 10mm * 50mm (min) rod and test this out.=0A=
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> Is their version the only correct solution or just an over engineered one=
? =0A=
> Slightly rhetorical question there.=0A=
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If you are not required to "fill" the coil surface they are right.=0A=
If you must "pack" the windings (and ignore the thickness of wire) you are =
right.=0A=
       Nic=0A=
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