Interstellar Medium Shielding - Revision history

Lwcamp at 19:02, 7 March 2026

2026-03-07T19:02:19Z

← Older revision		Revision as of 12:02, 7 March 2026
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	* To Kerr for the relativistic flux equation and derivation		* To Kerr for the relativistic flux equation and derivation

	[[Category:Physics & Engineering‏‎]][[Category:Engineering]][[Category:Astronomy & Cosmology‏‎]][[Category:Spacecraft Systems]][[Category:Transportation & Infrastructure]]		[[Category:Physics & Engineering‏‎]][[Category:Engineering]][[Category:Physics & Math & Engineering]][[Category:Astronomy & Cosmology‏‎]][[Category:Spacecraft Systems]][[Category:Transportation & Infrastructure]]

Lwcamp: /* Calculating the Temperature of the Forward Shield */

2026-02-25T05:32:08Z

Calculating the Temperature of the Forward Shield

← Older revision		Revision as of 22:32, 24 February 2026
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	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >
	<tr>		<tr>
	<td rowspan="2" nowrap="nowrap" align="right"> <i>T</i> = <span style="font-size: 250%;">(</span>		<td rowspan="2" nowrap="nowrap" align="right"> <i>T</i> =
			<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">(</span>
	<td nowrap="nowrap">   φ		<td nowrap="nowrap">   φ
	<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">)</span>		<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">)</span>

Lwcamp: /* Calculating Collision Effects */

2026-02-19T06:05:30Z

Calculating Collision Effects

← Older revision		Revision as of 23:05, 18 February 2026
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	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >
	<tr>		<tr>
	<td rowspan="2" nowrap="nowrap" align="right"> <i>r</i> = <span style="font-size: 250%;">(</span>		<td rowspan="2" nowrap="nowrap" align="right"> <i>r</i> =
			<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">(</span>
	<td nowrap="nowrap">   3 <i>V</i>		<td nowrap="nowrap">   3 <i>V</i>
	<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">)</span>		<td rowspan="2" nowrap="nowrap" align="right"> <span style="font-size: 250%;">)</span>

Lwcamp: /* Calculating Collision Effects */

2026-02-19T00:28:03Z

Calculating Collision Effects

← Older revision		Revision as of 17:28, 18 February 2026
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	The energy of a dust grain with mass <i>m</i> impacted at a relative (and non-relativistic) speed of <i>v</i> is <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>.		The energy of a dust grain with mass <i>m</i> impacted at a relative (and non-relativistic) speed of <i>v</i> is <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>.
	If we take a grain at the upper limit of our mass distribution from above, <i>m</i> = 10<sup>-14</sup> kg, and if we are moving at 0.1 c (3×10<sup>7</sup> m/s), it will have an energy of 4.5 J. At 0.9 c and using relativistic corrections the dust particle energy will be about 1 kJ. An individual dust grain will blast out a ~~spherical~~ crater with a radius		If we take a grain at the upper limit of our mass distribution from above, <i>m</i> = 10<sup>-14</sup> kg, and if we are moving at 0.1 c (3×10<sup>7</sup> m/s), it will have an energy of 4.5 J. At 0.9 c and using relativistic corrections the dust particle energy will be about 1 kJ. An individual dust grain will blast out a hemispherical crater with a radius
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >
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	<td nowrap="nowrap"> <sup>1/3</sup>		<td nowrap="nowrap"> <sup>1/3</sup>
	<tr>		<tr>
	<td style="border-top:solid 1px black;"> 4 π		<td style="border-top:solid 1px black;"> 2 π
	<td>		<td>
	</table>		</table>
	</div>		</div>
	In ice with a cavity strength of 50×10<sup>6</sup> Pa, the 0.1 c grain will make a crater 2.8 mm in radius while the 0.9 c grain will make a crater 1.8 cm in radius. As this is a quite large grain, most of the craters will be one to two orders of magnitude smaller and the craters in stronger shielding materials will be orders of magnitude smaller still. You would need to run into an unusually large dust grain to cause a serious problem to your shield.		In ice with a cavity strength of 50×10<sup>6</sup> Pa, the 0.1 c grain will make a crater 3.5 mm in radius while the 0.9 c grain will make a crater 2.2 cm in radius. As this is a quite large grain, most of the craters will be one to two orders of magnitude smaller and the craters in stronger shielding materials will be orders of magnitude smaller still. You would need to run into an unusually large dust grain to cause a serious problem to your shield.

	So instead of individual craters, let's look at the average erosive effect of all the combined craters from the trip. The energy density of the dust particles in the swept volume of the shield will be <i>U<sub>k</sub></i> = ½ ρ<sub>d</sub> <i>v</i><sup>2</sup>. As the dust particles are expected to land randomly from an unbiased uniform distribution across the front of the shield, we can expect that after moving a distance &ell; light years a shield of area <i>A</i> will intercept an energy of <i>E</i> = <i>U<sub>k</sub></i> <i>A</i> &ell; which is more-or-less uniformly spread across the shield. For an erosion depth of <i>d</i>, the eroded volume will be <i>V</i> = <i>d</i> <i>A</i>. We now know from the above relationship of volume to energy deposited that		So instead of individual craters, let's look at the average erosive effect of all the combined craters from the trip. The energy density of the dust particles in the swept volume of the shield will be <i>U<sub>k</sub></i> = ½ ρ<sub>d</sub> <i>v</i><sup>2</sup>. As the dust particles are expected to land randomly from an unbiased uniform distribution across the front of the shield, we can expect that after moving a distance &ell; light years a shield of area <i>A</i> will intercept an energy of <i>E</i> = <i>U<sub>k</sub></i> <i>A</i> &ell; which is more-or-less uniformly spread across the shield. For an erosion depth of <i>d</i>, the eroded volume will be <i>V</i> = <i>d</i> <i>A</i>. We now know from the above relationship of volume to energy deposited that

Lwcamp: /* Relativistic corrections */

2026-02-19T00:22:34Z

Relativistic corrections

← Older revision		Revision as of 17:22, 18 February 2026
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	With these two parameters, the above formulas can be corrected by substituting in the relativistic result for the kinetic energy <i>E</i> = <i>m</i> (γ - 1) c<sup>2</sup> for the non-relativistic <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>. This can be applied directly to the formula for the energy of individual dust grains. For erosion rates, the formula works out to		With these two parameters, the above formulas can be corrected by substituting in the relativistic result for the kinetic energy <i>E</i> = <i>m</i> (γ - 1) c<sup>2</sup> for the non-relativistic <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>. This can be applied directly to the formula for the energy of individual dust grains. For erosion rates, the formula works out to

	~~At relativistic speeds, this is modified to~~
	<div align=center>		<div align=center>
	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >

Lwcamp: /* Calculating Collision Effects */

2026-02-19T00:21:41Z

Calculating Collision Effects

← Older revision		Revision as of 17:21, 18 February 2026
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	where <i>K<sub>h</sub></i> is the cavity strength of the material. The cavity strength can be found from commonly reported material properties using <i>K<sub>h</sub></i> = (2/3) <i>K<sub>c</sub></i> × (1 + ln[2 <i>G</i>/<i>K<sub>h</sub></i>]), for compressive strength <i>K<sub>c</sub></i> and shear modulus <i>G</i>. However for approximate uses, <i>K<sub>h</sub></i> tends to be about 3 to 4 times <i>K<sub>c</sub></i>.		where <i>K<sub>h</sub></i> is the cavity strength of the material. The cavity strength can be found from commonly reported material properties using <i>K<sub>h</sub></i> = (2/3) <i>K<sub>c</sub></i> × (1 + ln[2 <i>G</i>/<i>K<sub>h</sub></i>]), for compressive strength <i>K<sub>c</sub></i> and shear modulus <i>G</i>. However for approximate uses, <i>K<sub>h</sub></i> tends to be about 3 to 4 times <i>K<sub>c</sub></i>.

	An individual dust grain will blast out a spherical crater with a radius		The energy of a dust grain with mass <i>m</i> impacted at a relative (and non-relativistic) speed of <i>v</i> is <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>.
			If we take a grain at the upper limit of our mass distribution from above, <i>m</i> = 10<sup>-14</sup> kg, and if we are moving at 0.1 c (3×10<sup>7</sup> m/s), it will have an energy of 4.5 J. At 0.9 c and using relativistic corrections the dust particle energy will be about 1 kJ. An individual dust grain will blast out a spherical crater with a radius
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >		<table style="margin-top:0.5em; margin-bottom:0.5em; text-align: center;" align="center" cellpadding="0" cellspacing="0" >
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	</table>		</table>
	</div>		</div>
	~~The energy of a dust grain with mass <i>m</i> impacted at a relative (and non-relativistic) speed of <i>v</i> is <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>.~~		In ice with a cavity strength of 50×10<sup>6</sup> Pa, the 0.1 c grain will make a crater 2.8 mm in radius while the 0.9 c grain will make a crater 1.8 cm in radius. As this is a quite large grain, most of the craters will be one to two orders of magnitude smaller and the craters in stronger shielding materials will be orders of magnitude smaller still. You would need to run into an unusually large dust grain to cause a serious problem to your shield.
	If we take a grain at the upper limit of our mass distribution from above, <i>m</i> = 10<sup>-14</sup> kg, and if we are moving at 0.1 c (3×10<sup>7</sup> m/s), it will have an energy of 4.5 J. At 0.9 c and using relativistic corrections the dust particle energy will be about 1 kJ. In ice with a cavity strength of 50×10<sup>6</sup> Pa, the 0.1 c grain will make a crater 2.8 mm in radius while the 0.9 c grain will make a crater 1.8 cm in radius. As this is a quite large grain, most of the craters will be one to two orders of magnitude smaller and the craters in stronger shielding materials will be orders of magnitude smaller still. You would need to run into an unusually large dust grain to cause a serious problem to your shield.

	So instead of individual craters, let's look at the average erosive effect of all the combined craters from the trip. The energy density of the dust particles in the swept volume of the shield will be <i>U<sub>k</sub></i> = ½ ρ<sub>d</sub> <i>v</i><sup>2</sup>. As the dust particles are expected to land randomly from an unbiased uniform distribution across the front of the shield, we can expect that after moving a distance &ell; light years a shield of area <i>A</i> will intercept an energy of <i>E</i> = <i>U<sub>k</sub></i> <i>A</i> &ell; which is more-or-less uniformly spread across the shield. For an erosion depth of <i>d</i>, the eroded volume will be <i>V</i> = <i>d</i> <i>A</i>. We now know from the above relationship of volume to energy deposited that		So instead of individual craters, let's look at the average erosive effect of all the combined craters from the trip. The energy density of the dust particles in the swept volume of the shield will be <i>U<sub>k</sub></i> = ½ ρ<sub>d</sub> <i>v</i><sup>2</sup>. As the dust particles are expected to land randomly from an unbiased uniform distribution across the front of the shield, we can expect that after moving a distance &ell; light years a shield of area <i>A</i> will intercept an energy of <i>E</i> = <i>U<sub>k</sub></i> <i>A</i> &ell; which is more-or-less uniformly spread across the shield. For an erosion depth of <i>d</i>, the eroded volume will be <i>V</i> = <i>d</i> <i>A</i>. We now know from the above relationship of volume to energy deposited that
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	</table>		</table>
	</div>		</div>

			=== Relativistic corrections ===

			At small fractions of the speed of light, relativity has a small effect and the results can be approximated by the methods described above. At 0.1 c, for example, non-relativistic formulas are still fairly close. But as speeds get to larger and larger fractions of light speed, relativity becomes more and more important.

			To include the effects of relativity, first find the parameters β and γ. β is the fraction of light speed of the motion – if you are moving at 0.1 c your β is 0.1. Once you have β you can find γ = 1/(1 - β<sup>2</sup>).

			With these two parameters, the above formulas can be corrected by substituting in the relativistic result for the kinetic energy <i>E</i> = <i>m</i> (γ - 1) c<sup>2</sup> for the non-relativistic <i>E</i> = ½ <i>m</i> <i>v</i><sup>2</sup>. This can be applied directly to the formula for the energy of individual dust grains. For erosion rates, the formula works out to

	At relativistic speeds, this is modified to		At relativistic speeds, this is modified to
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	</table>		</table>
	</div>		</div>
	where γ = 1/(1 - β<sup>2</sup>) and β = <i>v</i>/c. At relativistic speeds the kinetic energy of an individual dust grain is likewise modified to <i>E</i> = m (γ - 1) c<sup>2</sup>.

	==Prospects for Shield Materials==		==Prospects for Shield Materials==

Lwcamp: /* Derivation of the Erosion Depth Equation */

2026-02-19T00:11:05Z

Derivation of the Erosion Depth Equation

@@ Line 281: / Line 281: @@
 '''Reference for the Derivation of the Relativistic Flux Equation'''
 <references group="RFE"/>
 =Credit=

Lwcamp: /* Derivation of the Relativistic Flux Equation */

2026-02-18T17:19:48Z

Derivation of the Relativistic Flux Equation

← Older revision		Revision as of 10:19, 18 February 2026
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	<ref group="RFE">https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies		<ref group="RFE">https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies
	<br/> Reference for the relativistic kinetic energy equation.</ref>		<br/> Reference for the relativistic kinetic energy equation.</ref>
	# The kinetic energy of an amount of mass is given by ~~<math alt=>~~(\gamma -1)~~mc^~~2</~~math~~>. To get the power, the kinetic energy is differentiated against time and thus assuming constant velocity, obtain that with <~~math alt=~~>~~\dot{m}~~</~~math~~> (the mass flow rate); the power is given by ~~<math alt=>~~(\gamma -1)~~\dot{m}~~ c^2</~~math~~>.		# The kinetic energy of an amount of mass is given by (γ - 1) <i>m</i> c<sup>2</sup>. To get the power, the kinetic energy is differentiated against time and thus assuming constant velocity, obtain that with <i>ṁ</i> (the mass flow rate); the power is given by (γ - 1) <i>ṁ</i> c<sup>2</sup>.
	# The mass flow is given by the mass per volume encountered every second, doing this in the reference frame of the ship, the ISM density is length contracted to to <~~math alt=~~>~~\gamma \rho~~</~~math~~> ~~and multiply by~~ <~~math alt=~~>Av</~~math~~> where <~~math alt=~~>A</~~math~~> is area.		# The mass flow is given by the mass per volume encountered every second, doing this in the reference frame of the ship, the ISM density is length contracted to to γ ρ and multiply by <i>A</i> <i>v</i> where <i>A</i> is area.
	# This yields <~~math alt=~~>P = \gamma \rho Av(\gamma -1)c^2</~~math~~>, to obtain the flux per unit area divide by <~~math alt=~~>A</~~math~~> and thereby cancel the <~~math alt=~~> A</~~math~~> in the earlier expression.		# This yields <i>P</i> = γ ρ <i>A</i> <i>v</i> (γ - 1) c<sup>2</sup>, to obtain the flux per unit area divide by <i>A</i> and thereby cancel the <i>A</i> in the earlier expression.

	'''Reference for the Derivation of the Relativistic Flux Equation'''		'''Reference for the Derivation of the Relativistic Flux Equation'''

Lwcamp: /* Credit */

2026-02-18T16:12:46Z

Credit

← Older revision		Revision as of 09:12, 18 February 2026
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	=Credit=		=Credit=
	To Tshhmon for writing the article		To Tshhmon for writing the article
	* To ~~lwcamp~~ for helping with particle erosion calculation, editing, and physics and math checking and correction		* To Luke Campbell for helping with particle erosion calculation, editing, and physics and math checking and correction
	* To Rocketman1999 for helping with dust erosion distribution		* To Rocketman1999 for helping with dust erosion distribution
	* To Kerr for the relativistic flux equation and derivation		* To Kerr for the relativistic flux equation and derivation

	[[Category:Physics & Engineering‏‎]][[Category:Engineering]][[Category:Astronomy & Cosmology‏‎]][[Category:Spacecraft Systems]][[Category:Transportation & Infrastructure]]		[[Category:Physics & Engineering‏‎]][[Category:Engineering]][[Category:Astronomy & Cosmology‏‎]][[Category:Spacecraft Systems]][[Category:Transportation & Infrastructure]]

Lwcamp: /* Example Required Shield Thickness Table */

2026-02-18T16:10:06Z

Example Required Shield Thickness Table

← Older revision		Revision as of 09:10, 18 February 2026
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	{\| class="wikitable sortable"		{\| class="wikitable sortable"
	\|-		\|-
	! Ship Velocity !! Ice (m/ly) !! Steel (m/ly) !! Perfect ~~nanotube Fiber~~ (m/ly)		! Ship Velocity !! Ice (m/ly) !! Steel (m/ly) !! Perfect Nanotubes (m/ly)
	\|-		\|-
	\| 0.1c \|\| 1.9 \|\| 0.020 \|\| 0.0035		\| 0.1c \|\| 1.9 \|\| 0.020 \|\| 0.0035