[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

# Re: [femm] ring magnet

- To: femm@egroups.com
- Subject: Re: [femm] ring magnet
- From: dcm3c@aol.com
- Date: Tue, 28 Nov 2000 20:44:31 EST

    In a message dated 11/28/00 10:04:47 AM Eastern Standard Time,
    ac319@eng.cam.ac.uk writes:

    > Dear femm experts,
    > I am trying use femm to model a problem which I am not successful so
    > far: I have two disk shaped ring magnets, both of them are magnetized
    > along the symmetry axis. The outer diameter of small magnet is less than
    > the inner diameter of big magnet (about 1-2 mm) so that the small magnet
    > can fit inside the big magnet.
    > Question: Is it possible to model this configuration with femm. For
    > example I want to fix the position of small magnet and position the big
    > magnet in a way that both magnet's rotation axis coincides. If I move
    > outer magnet along to rotation axis and parallel to rotation axis is it
    > possible to measure relative vertical and lateral forces? Can femm do 3d
    > configuration, or can the existing one be modified to do 3d ?
    > best regards
    >
    > Ahmet Cansiz
    > Cambridge University
    > IRC in Superconductivity
    >

    It is possible to do part of what you are interested in with femm. As long
    as your magnet configuration is axisymmetric, you can do it with femm. I've
    attached a quick axisymmetric example with two concentric rings. When you
    are interested in computing forces, it is good to have a fine mesh in the
    region of the object upon which the force is to be measured--in this case,
    the magnets. In the example, I have defined a small region around the
    magnets with a pretty high mesh density. You'd then evaluate force in the
    postprocessor by drawing a contour around the magnet that you are interested
    in. Best results when the contour is through the air around the object
    rather than directly on the surface for numerical reasons.

    For lateral displacements of the magnets, however, you really need a 3-D
    code. This isn't axisymmetric any more. However, if you are only interested
    in small perturbations about the axisymmetric case, you can infer axial
    forces from the axisymmetric force results. A fun result for magnets that
    have a unit permeability acting upon one another (and NdFeB, SmCo, and most
    types of ceramic magnets are closely approximated as having a unit
    permeability) is that the stiffnesses all have to add up to zero (i.e. Kx +
    Ky + Kz = 0). For an axisymmetric system, if Kz is the axial stiffness and
    Kr is the radial stiffness about a certain point, they must satisfy: Kz + 2
    Kr = 0. Therefore, if you can evaluate the axial stiffness, the force for a
    perturbation of some distance dr from the centered position is:
    F = -(1/2) Kz dr
    However, this result is accurate only for relatively small displacements.

    Dave Meeker
    --
    http://members.aol.com/dcm3c

    --part1_41.407353f.2755b97f_alt_boundary
    Content-Type: text/html; charset="US-ASCII"
    Content-Transfer-Encoding: 7bit

    <HTML><FONT FACE=arial,helvetica><FONT SIZE=2>In a message dated 11/28/00 10:04:47 AM Eastern Standard Time, <BR>ac319@eng.cam.ac.uk writes:
    <BR>
    <BR>
    <BR><BLOCKQUOTE TYPE=CITE style="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px">Dear femm experts,
    <BR>I am trying use femm to model a problem which I am not successful so
    <BR>far: I have two disk shaped ring magnets, both of them are magnetized
    <BR>along the symmetry axis. The outer diameter of small magnet is less than
    <BR>the inner diameter of big magnet (about 1-2 mm) so that the small magnet
    <BR>can fit inside the big magnet.
    <BR>Question: Is it possible to model this configuration with femm. For
    <BR>example I want to fix the position of small magnet and position the big
    <BR>magnet in a way that both magnet's rotation axis coincides. If I move
    <BR>outer magnet along to rotation axis and parallel to rotation axis is it
    <BR>possible to measure relative vertical and lateral forces? Can femm do 3d
    <BR>configuration, or can the existing one be modified to do 3d ?
    <BR>best regards
    <BR>
    <BR>Ahmet Cansiz
    <BR>Cambridge University
    <BR>IRC in Superconductivity
    <BR></BLOCKQUOTE></FONT><FONT COLOR="#000000" SIZE=3 FAMILY="SANSSERIF" FACE="Arial" LANG="0">
    <BR>
    <BR>It is possible to do part of what you are interested in with femm. &nbsp;As long <BR>as your magnet configuration is axisymmetric, you can do it with femm. &nbsp;I've <BR>attached a quick axisymmetric example with two concentric rings. &nbsp;When you <BR>are interested in computing forces, it is good to have a fine mesh in the <BR>region of the object upon which the force is to be measured--in this case, <BR>the magnets. &nbsp;In the example, I have defined a small region around the <BR>magnets with a pretty high mesh density. &nbsp;You'd then evaluate force in the <BR>postprocessor by drawing a contour around the magnet that you are interested <BR>in. &nbsp;Best results when the contour is through the air around the object <BR>rather than directly on the surface for numerical reasons.
    <BR>
    <BR>For lateral displacements of the magnets, however, you really need a 3-D <BR>code. &nbsp;This isn't axisymmetric any more. &nbsp;However, if you are only interested <BR>in small perturbations about the axisymmetric case, you can infer axial <BR>forces from the axisymmetric force results. &nbsp;A fun result for magnets that <BR>have a unit permeability acting upon one another (and NdFeB, SmCo, and most <BR>types of ceramic magnets are closely approximated as having a unit <BR>permeability) is that the stiffnesses all have to add up to zero (i.e. Kx + <BR>Ky + Kz = 0). &nbsp;For an axisymmetric system, if Kz is the axial stiffness and <BR>Kr is the radial stiffness about a certain point, they must satisfy: Kz + 2 <BR>Kr = 0. &nbsp;Therefore, if you can evaluate the axial stiffness, the force for a <BR>perturbation of some distance dr from the centered position is:
    <BR>F = -(1/2) Kz dr
    <BR>However, this result is accurate only for relatively small displacements.
    <BR>
    <BR>Dave Meeker
    <BR>--
    <BR>http://members.aol.com/dcm3c</FONT></HTML>

    --part1_41.407353f.2755b97f_alt_boundary--

    [Format] = 3.0
    [Frequency] = 0.000000
    [Precision] = 1.000000e-008
    [LengthUnits] = inches
    [ProblemType] = axisymmetric
    [Coordinates] = cartesian
    [Comment] = "Add comments here."
    [PointProps] = 0
    [BdryProps] = 1
    <BeginBdry>
    <BdryName> = "ABC"
    <BdryType> = 2
    <A_0> = 0.000000
    <A_1> = 0.000000
    <A_2> = 0.000000
    <Phi> = 0.000000
    <c0> = 7832000.000000
    <c1> = 0.000000
    <Mu_ssd> = 0.000000
    <Sigma_ssd> = 0.000000
    <EndBdry>
    [BlockProps] = 2
    <BeginBlock>
    <BlockName> = "Strong Magnet"
    <Mu_x> = 1.000000
    <Mu_y> = 1.000000
    <H_c> = 1000000.000000
    <H_cAngle> = 0.000000
    <J_re> = 0.000000
    <J_im> = 0.000000
    <Sigma> = 0.000000
    <d_lam> = 0.000000
    <Phi_h> = 0.000000
    <LamType> = 0
    <LamFill> = 1.000000
    <BHPoints> = 0
    <EndBlock>
    <BeginBlock>
    <BlockName> = "Air"
    <Mu_x> = 1.000000
    <Mu_y> = 1.000000
    <H_c> = 0.000000
    <H_cAngle> = 0.000000
    <J_re> = 0.000000
    <J_im> = 0.000000
    <Sigma> = 0.000000
    <d_lam> = 0.000000
    <Phi_h> = 0.000000
    <LamType> = 0
    <LamFill> = 1.000000
    <BHPoints> = 0
    <EndBlock>
    [CircuitProps] = 0
    [NumPoints] = 12
    0.750000	0.250000	0	0
    0.750000	-0.250000	0	0
    1.250000	-0.250000	0	0
    1.250000	0.250000	0	0
    1.375000	0.250000	0	0
    1.375000	-0.250000	0	0
    1.875000	0.250000	0	0
    1.875000	-0.250000	0	0
    0.000000	4.000000	0	0
    0.000000	-4.000000	0	0
    0.250000	0.000000	0	0
    2.500000	0.000000	0	0
    [NumSegments] = 9
    0	1	-1	0	0	0
    0	3	-1	0	0	0
    3	2	-1	0	0	0
    2	1	-1	0	0	0
    4	5	-1	0	0	0
    4	6	-1	0	0	0
    6	7	-1	0	0	0
    7	5	-1	0	0	0
    8	9	-1	0	0	0
    [NumArcSegments] = 3
    9	8	180.000000	2.500000	1	0	0
    11	10	180.000000	2.500000	0	1	0
    10	11	180.000000	2.500000	0	1	0
    [NumHoles] = 0
    [NumBlockLabels] = 4
    1.000000	0.000000	1	2.500000e-002	0	90.000000	0
    1.625000	0.000000	1	2.500000e-002	0	-90.000000	0
    0.375000	1.375000	2	1.250000e-001	0	0.000000	0
    1.250000	0.750000	2	2.500000e-002	0	0.000000	0

- Prev by Date: ring magnet
- Next by Date: Fe2O3 magnetic data
- Previous by thread: Re: [femm] ring magnet
- Next by thread: Fe2O3 magnetic data
- Index(es):
  - Date
  - Thread