User:Brian P. Josey/Notebook/2010/03/29

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Detailed Neodymium in FEMM
With the new magnets for the yoke coming in sometime this week, I now have exact measurements to represent the yoke in FEMM. Ideally, I would be able to insert an arbitrary shape into FEMM that will give a nearly perfect representation of the magnetic field density, but the program is limited in the shapes that it can represent. To simplify the calculations, FEMM requires that the model is symmetrical either along one (axisymmetric) or two axis (planar symmetric). I am using axisymetric, and as a result, all the pieces have to be converted from arbitrary shapes to hollow cylinders, where the cross-sectional area of the hollow cylinder is equal too the cross-sectional area of the original piece. This ideally should gives a close approximation of the volume, and in turn magnetic strength, of the magnets. As a result, in the model there are some very bizarre looking geometries. In the interest of saving time in the event that I have to remake the model, here is where all the nodes belong in (r-coordinate, z-coordinate) fashion, and the reasoning behind them:

Conical Tip
Nodes:
 * Magnet Grade: N50
 * Hc Value: 1.125*106 A/m
 * Approximation Used: 52MGOe
 * Physical Dimensions: Radius 1/4", Height 1/2"
 * Tip: (0,0)
 * Center of Base: (0,0.5)
 * Edge of Base: (0.25,0.5)

This one is the simplest one to recreate. It is centered with the tip on the origin, and the r-axis is drawn from the tip to the center of the base, giving a perfect representation.

Upper Cylinder
Nodes:
 * Magnetic Grade: N42
 * Hc Value: 1.031*106 A/m
 * Approximation Used: 40MGOe
 * Physical Dimensions: Radius 1/4", Length 2"
 * Lower Center: (0,0.5)
 * Lower Edge: (0.25,.5)
 * Upper Center: (0,2.5)
 * Upper Edge: (0.25,2.5)

This one is also very simple to recreate. It is attached to the cone at the cone's base and is in the upper half of the model, z-coordinates are positive. There is another identical cylinder across the gap given by:

Lower Cylinder
Nodes:
 * Lower Center: (0,0.5)
 * Lower Edge: (0.25,.5)
 * Upper Center: (0,2.5)
 * Upper Edge: (0.25,2.5)

Upper Left Bend
Nodes:
 * Magnetic Grade: N45
 * Hc Value: 1.068*106 A/m
 * Approximation Used: 50MGOe
 * Physical Dimensions: Inner Radius 1" Outer Radius 2"
 * 0o Outer: (0,2.5)
 * 0o Inner: (0.5,2.5)
 * 30o Outer: (0.355,3.161)
 * 30o Inner: (0.634,3.0)
 * 60o Outer: (0.851,3.626)
 * 60o Inner: (1,3.366)
 * 90o Outer: (1.5,3.606)
 * 90o Inner: (1.5,3.5)

This one was very tricky, only the bends on the right side of the model will be any more difficult. This comes from the unique geometry of the horseshoe. While it creates a 90o arch, it is actually composed of three different pieces. Each piece carves out 30o, and where I used "cuts" above represents where two pieces meet. For the inner points, I used the fact that the inner radius is 1" at each of the points, and that they cut out a 300 angle to calculate the values of the coordinates.

The outer points were much more difficult. To find these I had to rely on my guiding principle for the model: keep the cross-sectional areas where two magnets meet consistent. To do this I imagined that the line connecting the inner and outer points was extended to the axis of symmetry so that when it is rotated around the axis, it traces out a cone without a base. Then I imagined that there would be some point along this line, that if I cut the cone in two, an upper cone, and a lower ring that is slanted inward, the lower ring would have the same surface area as the interface between the parts of the magnetic bend.

I couldn't find a formula for what I was looking for, so I re-derived what I imagined the formula would look like. If someone really needs to see it, let me know, just remind me that it's in my Japanese 102 notebook under March 8th. Here is the final formula for the radius:
 * $$r_2=\sqrt{{r_1}^2-\frac {A\cos(\theta)}{\pi}}$$

Where:
 * r2 is the radius of the smaller cone's base
 * r1 is the radius of the larger cone's base
 * A is the desired area
 * $$ \theta $$ is the angle between line I drew and the r-axis

Using the above formula, and the coordinates of the inner points that I had already determined, I was able to find the r-coordinates of the outer points. Using this, I figured the change in height, $$\Delta h$$, by plugging in the difference between the inner and outer radius, $$ \Delta r$$ into the following formula


 * $$\Delta h = \Delta r \tan{\theta} \,$$

This resulted in a unique geometry at the bend. Here the overhang is due to the fact that I am trying to keep the surface areas and lengths consistent along the interfaces between two magnets. While it is a little different, this geometry keeps my criteria consistent. I also simplified the model even further by assuming that on the left side, the outer 30o and 0o nodes would connect. This however is not the case in reality, as the magnets will be placed in as central of a location as possible, and I will probably change this in the near future.



Lower Left Bend
Nodes:
 * Magnetic Grade: N45
 * Hc Value: 1.068*106 A/m
 * Approximation Used: 50MGOe
 * Physical Dimensions: Inner Radius 1" Outer Radius 2"
 * 0o Outer: (0,-2.5)
 * 0o Inner: (0.5,-2.5)
 * 30o Outer: (0.355,-3.161)
 * 30o Inner: (0.634,-3.0)
 * 60o Outer: (0.851,-3.626)
 * 60o Inner: (1,-3.366)
 * 90o Outer: (1.5,-3.606)
 * 90o Inner: (1.5,-3.5)

These are the exact same points as for the "Upper Left Bend" except the z-coordinate has its sign flipped to indicate the change in location.

Note: All of the Hc values are calculated from the formula given in the manual. This is given by plugging in the numerical value of the grade (ie. 50 MGOe for a N50 magnet) into E in the following formula:
 * $$H_c=\frac{5(10^5)\sqrt E}{\pi}$$

Here Hc is the coercivity of the magnetic material. To simplify the model, if there is already a material that is a close approximation to the actual magnet in the software's library, I used it instead. This is denoted as "approximation used."


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