In this video, I show a fast, easy way to use Zen Magnets to build a strong double-walled tube. It is geometrically identical to, but magnetically different from, Damian OConnors perfect cylinder ( My tube looks exly the same as his. The difference is in the manner of construction, and how connections are made between magnets. \r
\r
In the video, I show how to build a 5-ring tube with 400 magnets and a 25-ring tube with 2000 magnets, using a sheet of parallel magnet chains to build the inner wall and pairs of antiparallel chains for the outer wall. It takes just 6 minutes to build the 25-ring tube from scratch. In general, an N-ring tube requires 80N magnets, 80 magnets for each ring.\r
\r
A lighthouse shape ( shown briefly in the video illustrates the strength of the double-walled tube. The lighthouse is a 4.4-pound (2.0 kg) double-walled tube with 4060 magnets supporting a 7.1-pound (3.2 kg) snub ball with 6590 magnets. Also shown briefly is a long tube with 10,640 magnets.\r
\r
Forty-column cylinders best match the natural curvature of the double wall. Heres the proof: The column spacing for the square-packed columns in the outer wall is just the magnet diameter, D. If there are n columns in this wall, its circumference will be given approximately by nD = 2πR, where R is its radius. The column spacing for hexagonally-packed columns in the inner wall is d = (√3/2)D. If there are n columns also in this wall, its circumference will be given approximately by nd = 2πr, where r = R - d is its radius. Solving these equations yields the ideal number of columns that best matches the natural curvature, n = 2(3+2√3)π = 40.615. But n must be an integer; you cant have 40.615 columns. In f, n must be an even number because of the alternating column offsets. Since 40 is the closest even number to 40.615, 40-column cylinders best match the natural curvature of the double wall.\r
\r
I now justify the approximations made in this derivation. The approximate circumference nD of the outer wall is the sum of the straight-line distances between the centers of its n columns. The ex but more complicated circumference, 2nR*arcsin(D/2R), is the distance around the arc of the circle of radius R. Replacing the approximate circumferences of the inner and outer walls with the ex circumferences gives a slightly smaller value for the ideal number of columns, n = π/arcsin(1/√3-1/2) = 40.575. Whether using the approximate circumferences or the ex ones, the conclusion is the same: 40-column cylinders best match the natural curvature of the double wall.
\r
In the video, I show how to build a 5-ring tube with 400 magnets and a 25-ring tube with 2000 magnets, using a sheet of parallel magnet chains to build the inner wall and pairs of antiparallel chains for the outer wall. It takes just 6 minutes to build the 25-ring tube from scratch. In general, an N-ring tube requires 80N magnets, 80 magnets for each ring.\r
\r
A lighthouse shape ( shown briefly in the video illustrates the strength of the double-walled tube. The lighthouse is a 4.4-pound (2.0 kg) double-walled tube with 4060 magnets supporting a 7.1-pound (3.2 kg) snub ball with 6590 magnets. Also shown briefly is a long tube with 10,640 magnets.\r
\r
Forty-column cylinders best match the natural curvature of the double wall. Heres the proof: The column spacing for the square-packed columns in the outer wall is just the magnet diameter, D. If there are n columns in this wall, its circumference will be given approximately by nD = 2πR, where R is its radius. The column spacing for hexagonally-packed columns in the inner wall is d = (√3/2)D. If there are n columns also in this wall, its circumference will be given approximately by nd = 2πr, where r = R - d is its radius. Solving these equations yields the ideal number of columns that best matches the natural curvature, n = 2(3+2√3)π = 40.615. But n must be an integer; you cant have 40.615 columns. In f, n must be an even number because of the alternating column offsets. Since 40 is the closest even number to 40.615, 40-column cylinders best match the natural curvature of the double wall.\r
\r
I now justify the approximations made in this derivation. The approximate circumference nD of the outer wall is the sum of the straight-line distances between the centers of its n columns. The ex but more complicated circumference, 2nR*arcsin(D/2R), is the distance around the arc of the circle of radius R. Replacing the approximate circumferences of the inner and outer walls with the ex circumferences gives a slightly smaller value for the ideal number of columns, n = π/arcsin(1/√3-1/2) = 40.575. Whether using the approximate circumferences or the ex ones, the conclusion is the same: 40-column cylinders best match the natural curvature of the double wall.
Category
📺
TV