Neodymium Iron Boron Magnets for Launch Sleds
http://e-magnetsuk.com/neodymium_magnets/neodymium_grades.aspx
An N52 magnet has a maximum energy product of 50 mega-gauss Oersteds, which is 50*7958kJ/m3 or 398 MJ/m³ . the density is 7.5, so that works out to 53 kJ/kg.
A high strength NIB magnet should be kept below 80C; launch loop magnet sleds should be refrigerated, perhaps to liquid nitrogen temperatures, before use, as they will be heated by hypersonic passage through the thin upper atmosphere. Some grades (N38 EH) can tolerate 200C, but are 36 mega-gauss Oersteds, 38 kJ/kg.
A reusable launch sled containing one tonne of NIB might be 10 square meters and 1.3 cm thick, 38 MJ. Derate by a factor of 4, and assume 20% flux efficiency in a gap 4 centimeters across. That is 2 cubic meters of effective flux area filled with 4.75 MJ of energy per cubic meter. B² = 2 μ₀ × 4.75e6 = π × 8e-7 × 4.75e6, B = 3.4 Tesla??? No, I'm doing something wrong. Assume 0.5 Tesla in the gap.
If the one tonne sled is pushing itself and 5 tonnes of payload with an acceleration of 30 m/s², the total thrust on the sled is 150 KN, or 15 KN/m².
Linear Motor Guesstimates
(probably BS for now)
The sled is the "rotor", and the field windings of the launch track are the "stator". Assume 1 kg/m^2 of primary aluminum in the track, which is an average thickness of 3.7e-4 m, 80 μΩ/□. This will be in multiple windings with gaps, so lets increase that by a factor of 12.5 to 1 mΩ/□ .
Misusing the analysis on page 57 of Linear Electric Motors" by Nasar and Boldea (1987), let's fake some numbers: < F > ~ = ~ { \Large { { \sigma B^2 s V \lambda l } \over { 4 | \Delta |^2 Re( \alpha ) } } ~ N } Newtons N = number of windings \lambda = "twice the pole pitch" = winding repeat distance in meters l = pole width The area of the poles is Area = N \times \lambda \times l , so this simplifies to: < F > ~ = ~ { \Large { { \sigma B^2 s U Area } \over { 4 | \Delta |^2 Re( \alpha ) } } } Newtons < F > = thrust = 1.5e5 Newtons \sigma = 1000 mhos s = slip (which we will compute) B = 0.5 Tesla U = synchronous speed, assume small slip and vehicle speed What is \Delta and \alpha ? g = gap in meters \omega = frequency Oops, rethinking those equations, they assume a solid conductor, not high frequency Litz style windings. I'll need to rework this. My guess is that the slip is very small and the efficiency very high.