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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. As vehicles reach exit velocity on a fixed linear track space launcher, even small variations in track straightness can produce large perpendicular accelerations. If that acceleration is coupled to payload and passengers, that means a noisy, bumpy ride.
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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 . Unlike familiar moderate speed track systems like a Shinkansen train, deflections cannot propagate forwards faster than the speed of sound in the material. If track material is too close, it will not be gradually pushed aside, it will be collided against. Lateral accelerations are proportional to the curvature times the velocity squared, so for the same curvature, the accelerations are 21,000 times higher for a 11000 m/s launch vehicle than they are for a 75 m/s Shinkansen train. Minimum curve radius on the Tōkaidō Shinkansen line is 2500 meters, and at the normal 75 m/s (270 km/h) the lateral forces on the partly banked turns make standing in the aisles difficult. Turn radiuses much be much higher for high orbit launch, and variations in curvature must be much smaller than terrestrial high speed trains.
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|| $ \lambda $ || $ A_{max} $ ||
|| 1 meter || 0.2 nm ||
|| 10 m || 20 nm ||
|| 100 m || 2 $\mu$m ||
|| 1 kilometer || 0.2 mm ||
|| 10 km || 2 cm ||
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.

||<-3> Wavelengths, straightness requirements, and effects ||
||<-3> 1m/s^2^ and 11 km/s ||
|| $ \lambda $ || $ A_{max} $ || effects ||
|| 1 meter || 0.2 nm || 11 KHz sound, loud ||
|| 10 m || 20 nm || 1.1 KHz sound, very loud! ||
|| 100 m || 2 $\mu$m || 110 Hz hum, loud ||
|| 1 kilometer || 0.2 mm || 11 Hz throbbing ||
|| 10 km || 2 cm || 1.1 Hz shaking ||

How Straight Must the Track Be?

As vehicles reach exit velocity on a fixed linear track space launcher, even small variations in track straightness can produce large perpendicular accelerations. If that acceleration is coupled to payload and passengers, that means a noisy, bumpy ride.

Unlike familiar moderate speed track systems like a Shinkansen train, deflections cannot propagate forwards faster than the speed of sound in the material. If track material is too close, it will not be gradually pushed aside, it will be collided against. Lateral accelerations are proportional to the curvature times the velocity squared, so for the same curvature, the accelerations are 21,000 times higher for a 11000 m/s launch vehicle than they are for a 75 m/s Shinkansen train. Minimum curve radius on the Tōkaidō Shinkansen line is 2500 meters, and at the normal 75 m/s (270 km/h) the lateral forces on the partly banked turns make standing in the aisles difficult. Turn radiuses much be much higher for high orbit launch, and variations in curvature must be much smaller than terrestrial high speed trains.

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.

Wavelengths, straightness requirements, and effects

1m/s2 and 11 km/s

\lambda

A_{max}

effects

1 meter

0.2 nm

11 KHz sound, loud

10 m

20 nm

1.1 KHz sound, very loud!

100 m

2 \mum

110 Hz hum, loud

1 kilometer

0.2 mm

11 Hz throbbing

10 km

2 cm

1.1 Hz shaking

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!

Straightness (last edited 2012-03-23 17:34:38 by KeithLofstrom)