数理電子
1...;\;} 単位 = u
無単位比;
( A L ) 3 / T = A 3 T − 1 / ( L − 1 ) 3 ; u n i t = u 3 ( u 30 x 2 y 3 ) ( u 13 x y ) 3 = 1 / x {\displaystyle (AL)^{3}/T=A^{3}T^{-1}/(L^{-1})^{3};\;unit={\frac {u^{3}(u^{30}x^{2}y^{3})}{(u^{13}xy)^{3}}}=1/x}
T 2 T P 3 = T P 3 ( T − 1 ) 2 ; u n i t = ( u 20 x y 2 ) 3 ( u 30 x 2 y 3 ) 2 = 1 / x {\displaystyle T^{2}T_{P}^{3}={\frac {T_{P}^{3}}{(T^{-1})^{2}}};\;unit={\frac {(u^{20}xy^{2})^{3}}{(u^{30}x^{2}y^{3})^{2}}}=1/x}
M 9 ( L − 1 ) 15 / ( T − 1 ) 11 ; u n i t = ( u 15 x y 2 ) 9 ( u 13 x y ) 15 ( u 30 x 2 y 3 ) 11 = x 2 {\displaystyle {M^{9}(L^{-1})^{15}}/{(T^{-1})^{11}};\;unit={\frac {(u^{15}xy^{2})^{9}(u^{13}xy)^{15}}{(u^{30}x^{2}y^{3})^{11}}}=x^{2}}
..=1}
例えば; (r, v) に関して解かれた定数
β = V R 2 = 2 π Ω 2 v Ω r 2 , u {\displaystyle \beta ={\frac {V}{R^{2}}}={\frac {2\pi \Omega ^{2}v}{\Omega r^{2}}},\;u}
A = β 3 ( 2 3 α ) = 2 6 π 3 Ω 3 α v 3 r 6 , u 3 {\displaystyle A=\beta ^{3}({\frac {2^{3}}{\alpha }})={\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }}{\frac {v^{3}}{r^{6}}},\;u^{3}}
G = β 6 2 3 π 2 ( j ) = 2 3 π 4 Ω 6 r 5 v 2 , u 6 {\displaystyle G={\frac {\beta ^{6}}{2^{3}\pi ^{2}}}(j)=2^{3}\pi ^{4}\Omega ^{6}{\frac {r^{5}}{v^{2}}},\;u^{6}}
L − 1 = 4 π β 13 ( i j ) = 1 2 π 2 Ω 2 v 5 r 9 , u 13 {\displaystyle L^{-1}=4\pi \beta ^{13}(ij)={\frac {1}{2\pi ^{2}\Omega ^{2}}}{\frac {v^{5}}{r^{9}}},\;u^{13}}