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| $ R ~=~ \sqrt{ { \Large { \sqrt{ 3 } \over { 8 \pi } } } N } ~ L ~\approx~ 0.26252 \sqrt{ N } ~ L $ | $ R ~=~ \sqrt{ { \Large { \sqrt{ 3 } \over { 8 \pi } } } N } ~ L ~\approx~ 0.26252 \sqrt{ N } ~ L $ |
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| All those decimal places about to go to waste ... | If the thinsat sphere is 7842 thinsats (icosahedral geodesic sphere with tiling ratio V = 28) and L = 5 meters, then R = 116 meters . The capacitance of a square (pointy corner!) thinsat with side S is C = 0.367 × 4πε₀ × S; for a 16 cm thinsat, C₁ = 0.367 × 111.265 pF/m × 0.16 m = 6.530 pF. If the voltage on the thinsat is 800V, the charge is $Q_1 = C_1 V $ = 5223 picoColomb per thinsat |
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| If the voltage on each thinsat is 800V, the charge is 5223 picoColomb per thinsat. If the sphere is 7842 thinsats (icosahedral geodesic sphere with tiling ratio V = 28) and L = 2 meters then R = 46.49 M . The capacitance of a sphere is 4πε₀ R , so we will approximate the capacitance of this geodesic sphere as 5172.7 pF and the total charge as 40.965 μColomb so the "average voltage" of the sphere is 7919 V ... oh, really? Let's see where this goes ... | The charge on the whole sphere of thinsats creates a Colomb force on a single thinsat. This resembles the math that shows that the gravitational force at (or above) a spherical shell is like the force from a concentrated mass at the center; we can move the charge to the center, and the effect on a charge at the edge will be the same. This simplifies the math; the force on a thinsat is approximately the simple Colomb force between the charge of N thinsats in the center and the charge of one thinsat distance R away: |
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| The energy stored assembling this sphere is 1 | $ F ~=~ Q_1 Q_N / 4 \pi \epsilon_0 R^2 ~=~ Q_1^2 N / 4 \pi \epsilon_0 R^2 ~=~ { \Large {2 \over { \sqrt{3} \epsilon_0}} \left( Q_1 \over L \right)^2 } ~ \approx ~ 24.56 \epsilon_0 { \Large \left( { S V } \over L \right )^2 } $ |
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| . E-mail: Frank.Read@man.ac.uk | . E-mail: !Frank.Read@man.ac.uk |
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| . 16 cm thinsat ⇒ C = 0.366787 × 111.265006 × 0.16 = 6.530 pF | ------- Scaling: |
Thinsat Charging and Array Spreading Forces
Assume N thinsats are charged to voltage V_th and are spaced L units apart in an approximately triangular geodesic grid mapped on a sphere. The area of the unit cell is √3/2 L², the area of the sphere is √3/2 L² N = 4 π R², so
R ~=~ \sqrt{ { \Large { \sqrt{ 3 } \over { 8 \pi } } } N } ~ L ~\approx~ 0.26252 \sqrt{ N } ~ L
If the thinsat sphere is 7842 thinsats (icosahedral geodesic sphere with tiling ratio V = 28) and L = 5 meters, then R = 116 meters .
The capacitance of a square (pointy corner!) thinsat with side S is C = 0.367 × 4πε₀ × S; for a 16 cm thinsat, C₁ = 0.367 × 111.265 pF/m × 0.16 m = 6.530 pF. If the voltage on the thinsat is 800V, the charge is Q_1 = C_1 V = 5223 picoColomb per thinsat
The charge on the whole sphere of thinsats creates a Colomb force on a single thinsat. This resembles the math that shows that the gravitational force at (or above) a spherical shell is like the force from a concentrated mass at the center; we can move the charge to the center, and the effect on a charge at the edge will be the same. This simplifies the math; the force on a thinsat is approximately the simple Colomb force between the charge of N thinsats in the center and the charge of one thinsat distance R away:
F ~=~ Q_1 Q_N / 4 \pi \epsilon_0 R^2 ~=~ Q_1^2 N / 4 \pi \epsilon_0 R^2 ~=~ { \Large {2 \over { \sqrt{3} \epsilon_0}} \left( Q_1 \over L \right)^2 } ~ \approx ~ 24.56 \epsilon_0 { \Large \left( { S V } \over L \right )^2 }
More later ... not looking good
- Improved Extrapolation Technique in the Boundary Element Method to Find the Capacitances of the Unit Square and Cube
- F. H. Read
- Schuster Laboratory, University of Manchester, Manchester M13 9PL, United Kingdom
E-mail: !Frank.Read@man.ac.uk
- JOURNAL OF COMPUTATIONAL PHYSICS ARTICLE NO .133, 1–5 (1997). doi.org/10.1006/jcph.1996.5519
- square = 0.366787 × 4πε₀ × S ... cube = 0.8806785 × 4πε₀ × S ... 4πε₀ = 111.265006 pF/m
Scaling: