I was recently given a maths question as part of test.=0A=
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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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My answer along with crap diagram was - find how much wire one turn =0A=
required. =0A=
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So circumference =3D pi * diameter =3D 3.142 * 10mm =3D 31.42mm =3D 1 turn =
length.=0A=
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5 turns =3D 31.42 * 5 =3D 157mm of wire.=0A=
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Their answer was completely different, so obviously my thinking is awry.=0A=
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They worked it out using 3-D triangulation.  =0A=
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Imagine a rectangle with the vertical side =3D L =3D 50mm, horizontal side =
=3D W =0A=
=3D 10mm=0A=
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They then drew a right angled triangle top left of rectangle and bottom =0A=
right to represent an unwrapped cylinder.=0A=
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They then stated the wire runs between two opposing corners, therefore the =
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wire length (one turn) =3D the standard sqr (L^2 + C^2), they then went on =
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show a series of equidistant triangles along the length of the core, 5 in =
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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? =
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Slightly rhetorical question there.=0A=
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Colin=0A=
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cdb,   25/03/2015=0A=
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colin@btech-online.co.uk=0A=
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