Triangle

Square

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

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} } } }

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} } } }