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It is also critical to bear in mind that this shielding is calculated to help because it is so thin. Because that means that most particles will simply pass right through the whole ship practically as though it wasn't even there. Once one of those particles hits something, that's when the real trouble begins. They smash new particles off the atomic nucleus they hit, multiplying the problem. From Appendix E of NASA's "Space Settlements: A Design Study" (which is an excellent introduction):

There are three mechanisms that are important in mass shielding. First, a charged particle excites electrons for many hundreds of angstroms about its trajectory. This excitation extracts kinetic energy at a roughly constant rate for relativistic particles and acts as a braking mechanism. For relativistic protons in low-Z matter this "linear energy transfer" is 2 MeV/g-cm^-2 of matter. If the thickness of the mass shield is great enough a particle of finite kinetic energy is stopped. This is the least effective shielding mechanism in matter for relativistic particles.

The second mechanism is nuclear attenuation. For silicon dioxide the average nuclear cross section is 0.4 barn ($10^{-24}\ cm^2$). Thus if a charged particle traverses far enough in the shield (composed of silicon dioxide) it collides with a nucleus and loses energy by inelastic collisions with the nuclear matter. The measure of how far a particle must travel to have a substantial chance of nuclear collision is the mean free path, which for silicon dioxide is 106 g/cm^2 . This mechanism is an exponential damper of primary beam particles.

Opposing the beam clearing tendency of nuclear attenuation is the creation of energetic secondary particles. For each nuclear collision there is beam loss from nuclear excitation, and beam enhancement (though with overall energy degradation through the increase of entropy) from the secondaries emitted by the excited nuclei. These secondary particles are, of course, attenuated themselves by further nuclear collisions with roughly the same mean free path as the primary particles

To see what they mean, consider this table of high energy protons penetrating solid iron (on Earth, meaning the number of particles was a tiny fraction of numbers in interplanetary space), from this study :

See how the number of protons goes up at different thicknesses through the iron? Those are secondary particles produced by collisions with iron atoms. So, in all of the parts of the hypothetical ship where there is more stuff than the 3 g/cm² of aluminum between you and space, there is a much higher chance that you will be struck by damaging particles from those directions, unless the solid stuff in between is at least a couple of meters thick.

It is also critical to bear in mind that this shielding is calculated to help because it is so thin. Because that means that most particles will simply pass right through the whole ship practically as though it wasn't even there. Once one of those particles hits the nucleus of an atom in its path, that's when the real trouble begins. They smash new particles off the that nucleus, multiplying the problem. From Appendix E of NASA's "Space Settlements: A Design Study" (which is an excellent introduction):

There are three mechanisms that are important in mass shielding. First, a charged particle excites electrons for many hundreds of angstroms about its trajectory. This excitation extracts kinetic energy at a roughly constant rate for relativistic particles and acts as a braking mechanism. For relativistic protons in low-Z matter this "linear energy transfer" is $2 MeV/g*cm^-2$ of matter. If the thickness of the mass shield is great enough a particle of finite kinetic energy is stopped. This is the least effective shielding mechanism in matter for relativistic particles.

The second mechanism is nuclear attenuation. For silicon dioxide the average nuclear cross section is 0.4 barn ($10^{-24}\ cm^2$). Thus if a charged particle traverses far enough in the shield (composed of silicon dioxide) it collides with a nucleus and loses energy by inelastic collisions with the nuclear matter. The measure of how far a particle must travel to have a substantial chance of nuclear collision is the mean free path, which for silicon dioxide is $106 g/cm^2$ . This mechanism is an exponential damper of primary beam particles.

Opposing the beam clearing tendency of nuclear attenuation is the creation of energetic secondary particles. For each nuclear collision there is beam loss from nuclear excitation, and beam enhancement (though with overall energy degradation through the increase of entropy) from the secondaries emitted by the excited nuclei. These secondary particles are, of course, attenuated themselves by further nuclear collisions with roughly the same mean free path as the primary particles

So, in all of the parts of the hypothetical ship where there is more stuff than the 3 g/cm² of aluminum between you and space, there is a much higher chance that you will be struck by damaging particles from those directions, unless the solid stuff in between is at least a couple of meters thick. It might take 5 meters of stuff before the cascade of particles generated by such collisions peters out due to its kinetic energy having been dispersed.

There are three mechanisms that are important in mass shielding. First, a charged particle excites electrons for many hundreds of angstroms about its trajectory. This excitation extracts kinetic energy at a roughly constant rate for relativistic particles and acts as a braking mechanism. For relativistic protons in low-Z matter this "linear energy transfer" is 2 MeV/g-cm^-2 of matter. If the thickness of the mass shield is great enough a particle of finite kinetic energy is stopped. This is the least effective shielding mechanism in matter for relativistic particles.

The second mechanism is nuclear attenuation. For silicon dioxide the average nuclear cross section is 0.4 barn (10^-24 cm^2)**. Thus if a charged particle traverses far enough in the shield (composed of silicon dioxide) it collides with a nucleus and loses energy by inelastic collisions with the nuclear matter. The measure of how far a particle must travel to have a substantial chance of nuclear collision is the mean free path, which for silicon dioxide is 106 g/cm^2 . This mechanism is an exponential damper of primary beam particles.

Opposing the beam clearing tendency of nuclear attenuation is the creation of energetic secondary particles. For each nuclear collision there is beam loss from nuclear excitation, and beam enhancement (though with overall energy degradation through the increase of entropy) from the secondaries emitted by the excited nuclei. These secondary particles are, of course, attenuated themselves by further nuclear collisions with roughly the same mean free path as the primary particles

**Note: either the width of atomic nuclei is actually measured in 'barns', or there was an error in the OCR of the scan here. Please advise.

There are three mechanisms that are important in mass shielding. First, a charged particle excites electrons for many hundreds of angstroms about its trajectory. This excitation extracts kinetic energy at a roughly constant rate for relativistic particles and acts as a braking mechanism. For relativistic protons in low-Z matter this "linear energy transfer" is 2 MeV/g-cm^-2 of matter. If the thickness of the mass shield is great enough a particle of finite kinetic energy is stopped. This is the least effective shielding mechanism in matter for relativistic particles.

The second mechanism is nuclear attenuation. For silicon dioxide the average nuclear cross section is 0.4 barn ($10^{-24}\ cm^2$). Thus if a charged particle traverses far enough in the shield (composed of silicon dioxide) it collides with a nucleus and loses energy by inelastic collisions with the nuclear matter. The measure of how far a particle must travel to have a substantial chance of nuclear collision is the mean free path, which for silicon dioxide is 106 g/cm^2 . This mechanism is an exponential damper of primary beam particles.

Opposing the beam clearing tendency of nuclear attenuation is the creation of energetic secondary particles. For each nuclear collision there is beam loss from nuclear excitation, and beam enhancement (though with overall energy degradation through the increase of entropy) from the secondaries emitted by the excited nuclei. These secondary particles are, of course, attenuated themselves by further nuclear collisions with roughly the same mean free path as the primary particles

Current magnetic shield designs are adequate to protect against ionizing radiation from the sun. They aren't sufficient to protect against galactic cosmic radiation, which has a lot more energy in each particle. To effectively block that would take a shield with energy 100x greater. If Bamford's shield parameters from TildalWave's answer are used, then 500 kW of power would be needed. It can't be modulated according to current needs, it has to always be at that power level, because GCR is constant, it doesn't fluctuate like solar ionizing radiation. The mass and energy needs of such a system are prohibitive. The effect over time on human health of exposure to a magnetic field of such strength is unknown.

Current magnetic shield designs are adequate to protect against ionizing radiation from the sun. They aren't sufficient to protect against galactic cosmic radiation, which has a lot more energy in each particle. To effectively block that would take a shield with energy 100x greater. If Bamford's shield parameters from TildalWave's answer are used, then 500 kW of power would be needed. It can't be modulated according to current needs, it has to always be at that power level, because GCR fluctuates only very slowly, over the solar cycle, it isn't a matter of occasional storms like solar ionizing radiation. The mass and energy needs of such a system are prohibitive. The effect over time on human health of exposure to a magnetic field of such strength is unknown.