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The deflection field "entering" at the bottom exits at the two diagonal pole faces, with less flux density. The deflection force at the bottom is $ F_A ~=~ A ~×~ B_{max}^2 / 2 \mu_0 = ~$ A × 1.29 MPa . The deflection field "entering" at the bottom exits at the two diagonal pole faces, with less flux density. The (atttracting) deflection force at the bottom is $ F_A ~=~ A ~×~ B_{max}^2 / 2 \mu_0 ~=~ $ A × 1.29 MPa .  The flux at each diagonal face is $ B_D = ( A / 2 D ) B_{max} $ so the force at each face is $ F_D ~=~ D ~×~ B_D^2 / 2 \mu_0 ~=~ D ~×~ (A/2D)^2 B_{max}^2 / 2 \mu_0 ~=~ A^2/D × B_{max}^2 / 8 \mu_0 ~=~ $ A²/D × 0.16 MPa .

Trapezoid Bolt

attachment:BoltCross.png

The launch loop rotor is composed of separable bolts, mostly made of thin laminations of transformer steel, along with carbon fiber stiffeners, embedded aluminum induction motor conductors, and a central spine. The bolts are normally used in two modes; assembled into a multilobed rotor for the main tracks and inclines, and separated into separate bolts for minimum radius deflection.

The number of lobes in the rotor (and the number of bolts in parallel) remains to be determined; probably between 4 and 6. The length of the bolts may also change. The size and cross section is constrained by the linear density of the rotor.

For a 4.32 kg/m rotor with six lobes, each lobe will mass 7.2 grams per centimeter of length. If 30% of the bolt mass is iron mass M₂, and 70% of the bolt is M₁ then M₁ is approximately 5 g/cm = 5e-3 kg/m.

The density of 4% silicon transformer steel is 7.65 g/cm³, so A₁, the cross section of the area M₁, is A₁ = 0.65 cm² = 6.5e-5 m².~~~ A₁ = (A-C)×B .

Deflection Mode

In deflection mode, an electromagnet pole (shown at the bottom of this diagram) pulls the rotor towards it, with near-saturation flux density B_{max} , perhaps 1.8 Tesla. The material will be strongly saturated and with considerable hysteresis at this field strength; that actually helps regulate the deflection force.

The deflection field "entering" at the bottom exits at the two diagonal pole faces, with less flux density. The (atttracting) deflection force at the bottom is F_A ~=~ A ~×~ B_{max}^2 / 2 \mu_0 ~=~ A × 1.29 MPa . The flux at each diagonal face is B_D = ( A / 2 D ) B_{max} so the force at each face is F_D ~=~ D ~×~ B_D^2 / 2 \mu_0 ~=~ D ~×~ (A/2D)^2 B_{max}^2 / 2 \mu_0 ~=~ A^2/D × B_{max}^2 / 8 \mu_0 ~=~ A²/D × 0.16 MPa .

MoreLater

TrapezoidBolt (last edited 2018-08-17 22:55:16 by KeithLofstrom)