Differences between revisions 1 and 12 (spanning 11 versions)
Revision 1 as of 2015-07-14 18:29:14
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Revision 12 as of 2015-07-15 20:28:26
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#format jsmath
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|| '''Triangle''' || '''Square''' ||
|| {{attachment:Tri.png| |width=400}} || {{attachment:Quad.png| |width=400}} ||
|| || ||
Mass $M$, Density $M/A$

|| ||<:-2> '''Triangle''' ||<:-2> '''Square''' ||
|| ||<:-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}/2) C = \sqrt{\sqrt{3} ~ A} $ ||<:-2> ||
|| 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 $ || $ { \large { { 23 \sqrt{3} ~ M A } \over 486 } } ~\approx 0.081695 ~ M A $ || $ $ || $ $ || $ $ ||
|| Torque $ T $ || $ F \sqrt{\sqrt{3}~ A} / 6 ~\approx 0.21935 ~ F \sqrt{ A } $ || $ $ || $ $ || $ $ ||
|| Angular<<BR>> acceleration<<BR>>$ ~~~~~~~~~ \dot \omega $ || $ { \Large { { 81 F } \over { 23 \sqrt[4]{3} ~ M \sqrt{A} } } } ~\approx 2.0333 { \Large { F \over { M \sqrt{A} } } } $ || $ $ || $ $ || $ $ ||
----------
|| {{attachment:TriA.png| |width=160}} || $ x = 3 y / B $ <<BR>><<BR>> Balance: $ \int_{-B/3}^{2B/3}(2/3-y/B) y ~ dy ~= ( B^2 / 9 ) \int_{-1}^{2}(2-x) x ~ dx ~= ( B^2 / 9 ) ( (12-8)-(3+1) ) ~= 0 $ <<BR>> $ I ~= \int_{-B/3}^{2B/3}(M/A)C(2/3-y/B) y^2 ~ dy ~= { \large \left( { 2 M B^2 } \over 81 \right) } \int_{-1}^{2}(2-x)(x^2)~ dx ~= { \large { { 23 \sqrt{3} ~ M A } \over 486 } } ~\approx 0.081695 ~ M A $ <<BR>> $ T ~= ( F/3 ) ( 2B/3 ) - 2 ( F/12 )( 2B/3 ) ~= FB / 6 ~= F \sqrt{\sqrt{3} ~ A} / 6 ~\approx 0.21935 ~ F \sqrt{ A } $ <<BR>> $ \dot \omega ~= T / I ~= { \Large { { 81 F } \over { 23 \sqrt[4]{3} ~ M \sqrt{A} } } } ~\approx 2.0333 { \Large { F \over { M \sqrt{A} } } } $ ||
|| {{attachment:TriB.png| |width=160}} || ||
||{{attachment:QuadA.png| |width=160}} || ||
||{{attachment:QuadA.png| |width=160}} || ||

Triangle or Square?

What is the optimum shape for a thinsat?


Mass M, Density M/A

Triangle

Square

Tri.png

Quad.png

Area

A = \sqrt{3} C^2 / 4

A = Q^2

Side

C = 2 \sqrt{A / \sqrt{3} }

Q = \sqrt{A}

Height

B = (\sqrt{3}/2) C = \sqrt{\sqrt{3} ~ A}

Ratio

C = ( 2 /\sqrt[4]{3} ) Q ~ \approx ~ 1.51967 ~ Q

Q = ( \sqrt[4]{3}/2 ) C ~ \approx ~ 0.65804 ~ C

Rotational
Center

Average
Force F/2

TriA.png

TriB.png

QuadA.png

QuadB.png

Moment of
Inertia I

{ \large { { 23 \sqrt{3} ~ M A } \over 486 } } ~\approx 0.081695 ~ M A

Torque T

F \sqrt{\sqrt{3}~ A} / 6 ~\approx 0.21935 ~ F \sqrt{ A }

Angular
acceleration
~~~~~~~~~ \dot \omega

{ \Large { { 81 F } \over { 23 \sqrt[4]{3} ~ M \sqrt{A} } } } ~\approx 2.0333 { \Large { F \over { M \sqrt{A} } } }


TriA.png

x = 3 y / B

Balance: \int_{-B/3}^{2B/3}(2/3-y/B) y ~ dy ~= ( B^2 / 9 ) \int_{-1}^{2}(2-x) x ~ dx ~= ( B^2 / 9 ) ( (12-8)-(3+1) ) ~= 0
I ~= \int_{-B/3}^{2B/3}(M/A)C(2/3-y/B) y^2 ~ dy ~= { \large \left( { 2 M B^2 } \over 81 \right) } \int_{-1}^{2}(2-x)(x^2)~ dx ~= { \large { { 23 \sqrt{3} ~ M A } \over 486 } } ~\approx 0.081695 ~ M A
T ~= ( F/3 ) ( 2B/3 ) - 2 ( F/12 )( 2B/3 ) ~= FB / 6 ~= F \sqrt{\sqrt{3} ~ A} / 6 ~\approx 0.21935 ~ F \sqrt{ A }
\dot \omega ~= T / I ~= { \Large { { 81 F } \over { 23 \sqrt[4]{3} ~ M \sqrt{A} } } } ~\approx 2.0333 { \Large { F \over { M \sqrt{A} } } }

TriB.png

QuadA.png

QuadA.png

TriQuad (last edited 2015-07-18 07:25:18 by KeithLofstrom)