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Size: 877
Comment:
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Size: 1203
Comment:
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| || ||<:-2> {{attachment:Tri.png| |width=320}} ||<:-2> {{attachment:Quad.png| |width=320}} || || Area ||<:-2> $ A = \sqrt{3} T^2 / 4 $ ||<:-2> $ A = Q^2 $ || || Side ||<:-2> $ T = 2 \sqrt{A / \sqrt{3} } $ ||<:-2> $ Q = \sqrt{A} $ || || Ratio ||<:-2> $ T = ( 2 /\sqrt[4]{3} ) Q ~ \approx ~ 1.51967 ~ Q $ ||<:-2> $ Q = ( \sqrt[4]{3}/2 ) T ~ \approx ~ 0.65804 ~ T $ || || Rotational<<BR>>Center ||<:>{{attachment:TriA.png| |width=160}}||<:>{{attachment:TriB.png| |width=160}}||<:>{{attachment:QuadA.png| |width=160}}||<:>{{attachment:QuadB.png| |width=160}}|| || Moment<<BR>>of<<BR>>Inertia|| || || || || |
|| ||<:-2> {{attachment:Tri.png| |width=240}} ||<:-2> {{attachment:Quad.png| |width=240}} || || Area ||<:-2> $ A = \sqrt{3} C^2 / 4 $ ||<:-2> $ A = Q^2 $ || || Side ||<:-2> $ C = 2 \sqrt{A / \sqrt{3} } $ ||<:-2> $ Q = \sqrt{A} $ || || Height ||<:-2> $ B = \sqrt{3} C / 2 = \sqrt{3\sqrt{A}} $ || || || Ratio ||<:-2> $ C = ( 2 /\sqrt[4]{3} ) Q ~ \approx ~ 1.51967 ~ Q $ ||<:-2> $ Q = ( \sqrt[4]{3}/2 ) C ~ \approx ~ 0.65804 ~ C $ || || Rotational<<BR>>Center<<BR>><<BR>>Average<<BR>>Force $F/2$ ||<:>{{attachment:TriA.png| |width=160}}||<:>{{attachment:TriB.png| |width=160}}||<:>{{attachment:QuadA.png| |width=160}}||<:>{{attachment:QuadB.png| |width=160}}|| || Moment of<<BR>>Inertia $ I $ || $ $ || $ $ || $ $ || $ $ || || Torque $ T $ || $ $ || $ $ || $ $ || $ $ || || Angular<<BR>> acceleration<<BR>>$ \dot \omega $ || $ $ || $ $ || $ $ || $ $ || |
Triangle or Square?
What is the optimum shape for a thinsat?
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Triangle |
Square |
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Area |
A = \sqrt{3} C^2 / 4 |
A = Q^2 |
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Side |
C = 2 \sqrt{A / \sqrt{3} } |
Q = \sqrt{A} |
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Height |
B = \sqrt{3} C / 2 = \sqrt{3\sqrt{A}} |
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Ratio |
C = ( 2 /\sqrt[4]{3} ) Q ~ \approx ~ 1.51967 ~ Q |
Q = ( \sqrt[4]{3}/2 ) C ~ \approx ~ 0.65804 ~ C |
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Rotational |
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Moment of |
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Torque T |
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Angular |
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