u as √{長さ/質量.時間}
出典: フリー百科事典『ウィキペディア（Wikipedia）』 (2021/04/26 22:47 UTC 版)
「数理電子」の記事における「u as √{長さ/質量.時間}」の解説
u , u n i t = L M T = u − 13 − 15 + 30 = 2 = u 1 {\displaystyle u,\;unit={\sqrt {\frac {L}{MT}}}={\sqrt {u^{-13-15+30=2}}}=u^{1}} x , u n i t = M 9 T 11 L 15 = u 0 = 1 {\displaystyle x,\;unit={\sqrt {\frac {M^{9}T^{11}}{L^{15}}}}=u^{0}=1} y , u n i t = M 2 T = u 0 = 1 {\displaystyle y,\;unit=M^{2}T=u^{0}=1} 結果; u 3 = L 3 / 2 M 3 / 2 T 3 / 2 = A , {\displaystyle u^{3}={\frac {L^{3/2}}{M^{3/2}T^{3/2}}}=A,\;} (アンペア) u 6 ( y ) = L 3 / T 2 M , ( G ) {\displaystyle u^{6}(y)=L^{3}/T^{2}M,\;(G)} u 13 ( x y ) = 1 / L , ( 1 / l p ) {\displaystyle u^{13}(xy)=1/L,\;(1/l_{p})} u 15 ( x y 2 ) = M , ( m P ) {\displaystyle u^{15}(xy^{2})=M,\;(m_{P})} u 17 ( x y 2 ) = V , ( c ) {\displaystyle u^{17}(xy^{2})=V,\;(c)} u 19 ( x y 3 ) = M L 2 / T , ( h ) {\displaystyle u^{19}(xy^{3})=ML^{2}/T,\;(h)} u 20 ( x y 2 ) = L 5 / 2 M 3 / 2 T 5 / 2 = A V , ( T P ) {\displaystyle u^{20}(xy^{2})={\frac {L^{5/2}}{M^{3/2}T^{5/2}}}=AV,\;(T_{P})} u 27 ( x 2 y 3 ) = M 3 / 2 T L 3 / 2 = 1 / A T , ( 1 / e ) {\displaystyle u^{27}(x^{2}y^{3})={\frac {M^{3/2}{\sqrt {T}}}{L^{3/2}}}=1/AT,\;(1/e)} u 29 ( x 2 y 4 ) = M 5 / 2 T L = M L / A T , ( k B ) {\displaystyle u^{29}(x^{2}y^{4})={\frac {M^{5/2}{\sqrt {T}}}{\sqrt {L}}}=ML/AT,\;(k_{B})} u 30 ( x 2 y 3 ) = 1 / T , ( 1 / t p ) {\displaystyle u^{30}(x^{2}y^{3})=1/T,\;(1/t_{p})} u 56 ( x 4 y 7 ) = M 4 T L 2 = M L T 2 A 2 , ( μ 0 ) {\displaystyle u^{56}(x^{4}y^{7})={\frac {M^{4}T}{L^{2}}}={\frac {ML}{T^{2}A^{2}}},\;(\mu _{0})} β (単位 = u), i (xから), j (yから); R = P = Ω r , u n i t s = u 8 {\displaystyle R={\sqrt {P}}={\sqrt {\Omega }}r,\;units=u^{8}} β = V R 2 = 2 π R 2 M = A 1 / 3 α 1 / 3 2 . . . , {\displaystyle \beta ={\frac {V}{R^{2}}}={\frac {2\pi R^{2}}{M}}={\frac {A^{1/3}\alpha ^{1/3}}{2}}\;...,\;} 単位 = u i = 1 2 π ( 2 π Ω ) 15 , {\displaystyle i={\frac {1}{2\pi {(2\pi \Omega )}^{15}}},\;} 単位 = 1 j = r 17 v 8 = k 2 t = k 8 r 15 . . . , u n i t = u 17 ∗ 8 u 8 ∗ 17 = u 15 ∗ 2 u − 30 . . . = 1 {\displaystyle j={\frac {r^{17}}{v^{8}}}=k^{2}t={\frac {k^{8}}{r^{15}}}...,\;unit={\frac {u^{17*8}}{u^{8*17}}}=u^{15*2}u^{-30}...=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}} M = 2 π β 15 ( i j 2 ) = r 4 v , u 15 {\displaystyle M=2\pi \beta ^{15}(ij^{2})={\frac {r^{4}}{v}},\;u^{15}} P = β 16 ( i j 2 ) = Ω r 2 , u 16 {\displaystyle P=\beta ^{16}(ij^{2})=\Omega r^{2},\;u^{16}} V = β 17 ( i j 2 ) = 2 π Ω 2 v , u 17 {\displaystyle V=\beta ^{17}(ij^{2})=2\pi \Omega ^{2}v,\;u^{17}} h = π β 19 ( i j 3 ) = 8 π 4 Ω 4 r 13 v 5 , u 19 {\displaystyle h=\pi \beta ^{19}(ij^{3})=8\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}},\;u^{19}} T P ∗ = 2 3 β 20 π α ( i j 2 ) = 2 7 π 3 Ω 5 α v 4 r 6 , u 20 {\displaystyle T_{P}^{*}={\frac {2^{3}\beta ^{20}}{\pi \alpha }}(ij^{2})={\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }}{\frac {v^{4}}{r^{6}}},\;u^{20}} e − 1 = α π β 27 ( i 2 j 3 ) 4 = α 128 π 4 Ω 3 v 3 r 3 , u 27 {\displaystyle e^{-1}={\frac {\alpha \pi \beta ^{27}(i^{2}j^{3})}{4}}={\frac {\alpha }{128\pi ^{4}\Omega ^{3}}}{\frac {v^{3}}{r^{3}}},\;u^{27}} k B = α π 2 β 29 ( i 2 j 4 ) 4 = α 32 π Ω r 10 v 3 , u 29 {\displaystyle k_{B}={\frac {\alpha \pi ^{2}\beta ^{29}(i^{2}j^{4})}{4}}={\frac {\alpha }{32\pi \Omega }}{\frac {r^{10}}{v^{3}}},\;u^{29}} T − 1 = 2 π β 30 ( i 2 j 3 ) = 1 2 π v 6 r 9 , u 30 {\displaystyle T^{-1}=2\pi \beta ^{30}(i^{2}j^{3})={\frac {1}{2\pi }}{\frac {v^{6}}{r^{9}}},\;u^{30}} μ 0 ∗ = π 3 α β 56 2 3 ( i 4 j 7 ) = α 2 11 π 5 Ω 4 r 7 , u 56 {\displaystyle \mu _{0}^{*}={\frac {\pi ^{3}\alpha \beta ^{56}}{2^{3}}}(i^{4}j^{7})={\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}}r^{7},\;u^{56}} ϵ 0 ∗ − 1 = π 3 α β 90 2 3 ( i 6 j 11 ) = α 2 9 π 3 v 2 r 7 , u 90 {\displaystyle \epsilon _{0}^{*-1}={\frac {\pi ^{3}\alpha \beta ^{90}}{2^{3}}}(i^{6}j^{11})={\frac {\alpha }{2^{9}\pi ^{3}}}v^{2}r^{7},\;u^{90}} j のSIの数値は、SI定数が持つことのできる値の限界（境界）を示しています。 r 17 v 8 = k 2 t = k 17 / 4 v 15 / 4 = . . . = .812997 . . . x 10 − 59 , {\displaystyle {\frac {r^{17}}{v^{8}}}=k^{2}t={\frac {k^{17/4}}{v^{15/4}}}=...=.812997...x10^{-59},\;} 単位 =1 SI用語では、単位「β」はこの値を持ちます。 a 1 / 3 = v r 2 = 1 t 2 / 15 k 1 / 5 = v k . . . = 23326079.1... ; {\displaystyle a^{1/3}={\frac {v}{r^{2}}}={\frac {1}{t^{2/15}k^{1/5}}}={\frac {\sqrt {v}}{\sqrt {k}}}...=23326079.1...;\;} 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}}
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