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| Suppose we have a sinusoidal variation in vertical distance on the track, amplitude $ A $, wavelength $ \lambda $, wavenumber $ k = 2 \pi / \lambda $. The vertical position of the vehicle would be $ z = A sin( k x ) $. The vertical position given horizontal velocity $ v $ would be $ z = A sin( k v t ) $ and the vertical acceleration would be $ \ddot z = - A ( k v )^2 sin( k v t ) $. The peak vertical acceleration would be $ \ddot z_{max} = A ( k v ) ^2 = A ( 2 \pi v / \lambda )^2 $. If we limit the $ \ddot z_{max} $ acceleration to 1 m/s^2^ (pretty shaky!) at 11 km/s, then solve for $ A_{max}= (1 m/s^2) ( \lambda / 2 \pi v )^2 $ we get $ A_{max} $ = 2e-10 $ m^{-1} \lambda^2 $ . For a slower exit velocity of 7.5 km/s, we can tolerate twice as much variation . | Suppose we have a sinusoidal variation in vertical distance on the track, amplitude $ A $, wavelength $ \lambda $, wavenumber $ k = 2 \pi / \lambda $. The vertical position of the vehicle would be $ z = A sin( k x ) $. The vertical position given horizontal velocity $ v $ would be $ z = A sin( k v t ) $ and the vertical acceleration would be $ \ddot z = - A ( k v )^2 sin( k v t ) $. The peak vertical acceleration would be $ \ddot z_{max} = A ( k v ) ^2 = A ( 2 \pi v / \lambda )^2 $. If we limit the $ \ddot z_{max} $ acceleration to 1 m/s^2^ (pretty shaky!) at 11 km/s, then solve for $ A_{max}= (1 m/s^2) ( \lambda / 2 \pi v )^2 $ we get $ A_{max} $ = 2e-10 $ m^{-1} \lambda^2 $. For a slower exit velocity of 7.5 km/s, we can tolerate twice as much variation . |
How Straight Must the Track Be?
As vehicles reach exit velocity, even small variations in track straightness can produce large perpendicular accelerations. If that acceleration is coupled to payload and passengers, that means a bumpy ride.
Suppose we have a sinusoidal variation in vertical distance on the track, amplitude A , wavelength \lambda , wavenumber k = 2 \pi / \lambda . The vertical position of the vehicle would be z = A sin( k x ) . The vertical position given horizontal velocity v would be z = A sin( k v t ) and the vertical acceleration would be \ddot z = - A ( k v )^2 sin( k v t ) . The peak vertical acceleration would be \ddot z_{max} = A ( k v ) ^2 = A ( 2 \pi v / \lambda )^2 . If we limit the \ddot z_{max} acceleration to 1 m/s2 (pretty shaky!) at 11 km/s, then solve for A_{max}= (1 m/s^2) ( \lambda / 2 \pi v )^2 we get A_{max} = 2e-10 m^{-1} \lambda^2 . For a slower exit velocity of 7.5 km/s, we can tolerate twice as much variation .
\lambda |
A_{max} |
1 meter |
0.2 nm |
10 m |
20 nm |
100 m |
2 \mum |
1 kilometer |
0.2 mm |
10 km |
2 cm |
A launch loop rotor is difficult to deflect - it will enforce a straight launch path. For track loading variations with wavelengths of less than a kilometer or so, we can expect the rotor and proper spacing algorithms to keep the track straight enough to mimimize vehicle vibration.
Magnetic coupling will also act as a spring, and reduce high frequency acceleration of the vehicle. The coupling must be lossy, or there will be resonances, and periodic track variation may cause those resonances to build destructively at certain speeds. Below the first resonance, the track must be very straight - accurate measurement and active control is required.
Without a rotor, where the track straightness is controlled only by terrain, stiffness, and cable truss forces, expect a wild ride!