Size: px
Start display at page:

Download ""

Transcription

1 25 22

2

3 1 [1] 3

4 3 [1] 3 [9] [13]

5

6 θ φ ε ε + ( ) θ + ( ) AC OA O O

7 6 ε ϕ CD EF BC CD C θ ε, θ, ϕ 3 0 M f, M b [kg] M c [kg] OA, AB, AC, CD L d, L c, L a, L e [m] DE DF L h [m] g [m/s 2 ] F f, F b [N] K m [N/V] V f, V b [V] F f = K m V f 0, F b = K m V b 0 ε, θ, ϕ η ε, η θ, η ϕ [kg m 2 /s] OA BC CD EF J ε ε = (M f + M b ) g L a cos (ε δ a ) + M c g cos (ε + δ c ) η ε ε cos δ a cos δ c +K m L a (V f + V b ) cos θ (2.1) J θ θ = Mf g L h cos (θ δ h ) + M b g cos (θ + δ h ) η θ θ cos δ h cos δ h +K m L h (V f V b ) (2.2) J ϕ ϕ = ηϕ ϕ Km L a (V f + V b ) sin θ (2.3) L h L c δ a = tan 1 {(L d + L e )/L a } δ c = tan 1 (L d /L c ) δ h = tan 1 (L e /L h ) J ε J θ J ϕ ε θ ϕ OA BC CD EF

8 7 2.2 u = [u P 1 u P 2 ] T =[V f + V b V f V b ] T x = [x P 1 x P 2 x P 3 x P 4 x P 5 x P 6 ] T =[ε ε θ θ ϕ ϕ] T ẋ = + x P 2 p 1 cos x P 1 + p 2 sin x P 1 + p 3 x P 2 x P 4 p 5 cos x P 3 + p 6 sin x P 3 + p 7 x P 4 x P 6 p 9 x P p 4 cos x P p p 10 sin x P 3 0 y = [x P 1 x P 3 ] T u (2.4) ε, θ, ϕ 2 3 ε θ p i (i = 1,..., 10) p 1 = [ (M f + M b )gl a + M c gl c ]/J ε p 2 = [ (M f + M b )gl a tan δ a M c gl c tan δ c ]/J ε p 3 = η ε /J ε p 4 = K m L a /J ε p 5 = ( M f + M b )gl h /J θ p 6 = ( M f M b )gl h tan δ h /J θ p 7 = η θ /J θ p 8 = K m L h /J θ p 9 = η ϕ /J ϕ p 10 = K m L a /J ϕ

9 8 (2.4) D = L g1 L ν 1 1 f h 1 (x) L gm L ν 1 1 f h 1 (x)..... L g1 L ν m 1 f h m (x) L gm L ν m 1 f h m (x) (2.5) [2] [4] ν 3 ε θ ν 1 = ν 2 = 2 D [ ] p4 cos x P 3 0 D = (2.6) 0 p 8 D = 0 ε

10 ε θ [ ] [ ] [ ] [ ] d ξ i ξ i = v P i (3.1) dt 1 ξ i 2 a i0 a i1 v P i y 1 = ε y 2 = θ ξ 1 1 ξ2 1 a 10 a 11 a 20 a 21 A i = [ 0 1 a i0 a i1 ] ξ i 2 i = 1 2 (3.2) u P = [ ] [ ] up 1 p4 cos x P 3 0 D = u P 2 0 p 8 f = [ ] [ f1 p1 cos x P 1 + p 2 sin x P 1 + p 3 x P 2 = f 2 p 5 cos x P 3 + p 6 sin x P 3 + p 7 x P 4 h = [ ] [ ] h1 a10 x P 1 a 11 x P 2 + v P 1 = a 20 x P 3 a 21 x P 4 + v P 2 h 2 ] (2.4) (3.1) f + Du P = h (3.3)

11 10 D D = 0 u P = D 1 ( f + h) (3.4) [e 1 e 2 ] = [x P 1 ξ 1 1 x P 3 ξ 2 1] { ẋp 2 ξ 1 2 = a 10 (x P 1 ξ 1 1) a 11 (x P 2 ξ 1 2) ẋ P 4 ξ 2 2 = a 20 (x P 3 ξ 2 1) a 21 (x P 4 ξ 2 2) { ë1 + a 11ė 1 + a 10e 1 = 0 ë 2 + a 21 ė 2 + a 20 e 2 = 0 e i (3.4) u E P,k = D 1 k ( f k + h k ) (3.5) 3.3 (3.5) x(k + 1) = x(k) + T ẋ(k) + T 2 ẍ(k) (3.6) 2 T w k u R P,k = u E P,k + T w k (3.7)

12 11 ε θ [ ] [ ] [ ] xp 1 (k + 1) xp 1 (k) xp 2 (k) = + T + T 2 [ ] s11,k x P 2 (k + 1) x P 2 (k) s 11,k 2 s 12,k T d T d 12 T d 11 + T 2 u P (k) (3.8) [ xp 3 (k + 1) x P 4 (k + 1) ] = 2 δ 11 T d 12 + T 2 2 δ 12 [ ] [ ] xp 3 (k) xp 4 (k) + T + T 2 x P 4 (k) s 21,k 2 T d T d 22 T d 21 + T 2 2 δ 21 T d 22 + T 2 2 δ 22 [ s21,k s 22,k ] u P (k) (3.9) u P,k (k) u P (k) = 0 (3.8) (3.9) s 1,k = s 2,k = D = = [ ] [ ] s11,k p1 cos x P 1 + p 2 sin x P 1 + p 3 x P 2 = s 21,k p 5 cos x P 3 + p 6 sin x P 3 + p 7 x P 4 [ ] [ s12,k p1 x P 2 sin x P 1 + p 2 x P 2 cos x P 1 + p 3 (p 1 cos x P 1 + p 2 sin x P 1 + p 3 x P 2 ) = s 22,k p 5 x P 4 sin x P 3 + p 6 x P 4 cos x P 3 + p 7 (p 5 cos x P 3 + p 6 sin x P 3 + p 7 x P 4 ) [ ] [ ] d11 d 12 p4 cos x P 3 0 = d 21 d 22 0 p 8 [ ] [ ] δ11 δ 12 p3 p 4 cos x P 3 p 4 x P 4 sin x P 3 0 = δ 21 δ 22 0 p 7 p 8 ] [ ] [ ] [ ξ i 1 (k + 1) ξ i = 1 (k) ξ i + T 2 (k) ξ2(k i + 1) ξ2(k) i ξ 2(k) i ] + T 2 2 [ ξi 2 (k) ξ i 2(k) ] (3.10) (3.10) ξ i 2(k), ξ i 2(k) = a i0 ξ i 1(k) a i1 ξ i 2(k) + v P i t ξ i 2(k) = a i0 ξ i 2(k) a i1 ξi 2 (k) + v P i (3.11) = a i0 ξ i 2(k) a i1 ( a i0 ξ i 1(k) a i1 ξ i 2(k) + v P i ) + v P i = (a 2 i1 a i0 )ξ i 2(k) + a i1 a i0 ξ i 1(k) a i1 v P i + v P i ξ i 2 g 2 (k) = [ g 1 2 (k) g 2 2(k) ] = [ ] (a 2 11 a 10 )x P 2 + a 11 a 10 x P 1 + a 11 v P 1 + v P 1 (a 2 21 a 20 )x P 4 + a 21 a 20 x P 3 + a 21 v P 2 + v P 2

13 12 w k s 2,k + u E P,k + 2Dw k = g 2 (k) (3.12) (3.8) (3.9) (3.7) (3.10) O(T 2 ) w k = 1 2 D 1 {g 2 (k) (s 2,k + u E P,k)} (3.13) 3.4 p 1 = p 2 = p 3 = p 4 = p 5 = 0.0 p 6 = p 7 = p 8 = p 9 = p 10 = a 11 = a 21 = 3 a 10 = a 20 = 2

14 13 T = 0.2 [s] ξ (0) = 0 [ ε ε θ θ ϕ ϕ ] = [ ] [ v1 v 2 ] = [ 0.3 sin(0.1t) 0.2 sin(0.2t) ]

15 14 (1) Reference Input v v Reference Input v v 2

16 15 4 Input u u Input u u 2

17 Output x ε;, 0.2 Output x θ;,

18 Error x ε ξ Error x θ ξ 2 1

19 18 (2) Emu Input u u Emu Input u u 2

20 19 Emu Output x ε;, 0.1 Emu Output x θ;,

21 20 0 Error x ε ξ Error x θ ξ 2 1

22 21 (3) Re Input u u Re Input u u 2

23 Re Output x ε;, 0.1 Re Output x θ;,

24 23 0 Error x ε ξ Error x θ ξ 2 1

25 24 (4) Input u u 1 ;, 0.2 Input u u 2 ;,

26 25 0 Error x ε ξ1 1 ;, Error x ε ξ1 1 ( );,

27 Error x θ ξ1 2 ;, Error x θ ξ1 2 ( );,

28 27 T = 0.1[s] T = 0.2[s] 0 Error x ε ξ1 1 ;, Error x ε ξ1 1 ( );,

29 Error x θ ξ1 2 ;, Error x θ ξ1 2 ( );,

30 29 (5) T = 0.2[s] T = 0.1[s] (T = 0.2[s]) ε ξ θ ξ (T = 0.1[s]) ε ξ θ ξ O(T 2 ) T = 0.2[s] 10 1 T = 0.1[s] 20 1 T = 0.1[s] T = 0.2[s] 0

31 4 3 3

32 2.

33 [1] D. S. Laila, D. Nešić and A. Astolfi: Sampled-data control of nonlinear systems; Advanced topics in control systems theory, Springer Verlag, pp , [2] H K Khalil Nonlinear Systems 3rd ed. Prentice Hall [3],,,,,,, 1993 [4],,,,,,, 1997 [5] [6] M. Ishitobi, M. Nishi and K. Nakasaki, Nonlinear adaptive model following control for a 3-DOF tandem-rotor model helicopter, Control Engineering Practice, Vol.18(8), pp , [7] K. Takamuku, M. Ishitobi, T. Kumada, M. Nishi and S. Kunimatsu, Redesign implementation of a nonlinear sampled-data controller for a 3-DOF model helicopter, Proc. of 2011 SICE Annual Conference, pp , [8] [9] [10] [11] 3, 2012.

34 33 [12] [13]

35 cont main.m 35 state equation cont.m 38 emu main.m 40 state equation emu.m 43 stable input.m 44 redes main.m 45 state equation redes.m 48 stable input2.m 49 simu.mdl Simulink Emu Controller 51 v emu 52 Re Controller 52 add input w 53 51

36 35 cont main.m clc clear all global p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 a11 a10 a21 a20 dt v u x0 t; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Mf=0.69; Mb=0.69; Mc=1.67; La=0.62; Lc=0.44; Ld=0.05; Le=0.02; Lh=0.177; Km=0.5; g=9.81; % etae=0.01; etap=0.01; etat=0.05; % Je=(Mf+Mb)*(La^2+Ld^2)+Mc*(Lc^2+Ld^2); Jp=(Mf+Mb)*(Lh^2+Le^2); Jt=(Mf+Mb)*La^2+Mc*Lc^2; % tan a, c, h tana = (Ld+Le)/La;tanc = Ld/Lc;tanh = Le/Lh; % p1=(-(mf+mb)*la+mc*lc)*g/je; p2=(-(mf+mb)*la*tana-mc*lc*tanc)*g/je; p3=-etae/je; p4=km*la/je; p5=(-mf+mb)*g*lh/jp; p6=(-mf-mb)*g*lh*tanh/jp; p7=-etap/jp; p8=km*lh/jp; p9=-etat/jt; p10=-km*la/jt; % a11=3; a10=2; a21=3; a20=2; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

37 36 % x0=[-0.1; 0; 0.1; 0; 0.2; 0]; min=0; max=90; begin=min; dt=0.2; xout=x0 ; tout=0; rout=[0 0]; uout=[0 0]; % for i=min:max/dt [t, x]=ode45( state_equation_cont, [begin,begin+dt], x0); x0=x(length(x),:); xout=[xout;x0]; Tend=t(length(t),:); begin=tend ; tout=[tout; begin]; vout=[vout; v ]; uout=[uout; u ]; end % eout1=vout(:,1)-xout(:,1); eout2=vout(:,2)-xout(:,3); % figure plot(tout,vout(:,1)); xlabel( time[s] );ylabel( Reference Input v1 );xlim([min,max]); figure plot(tout,vout(:,2)); xlabel( time[s] );ylabel( Reference Input v2 );xlim([min,max]);

38 37 figure plot(tout,uout(:,1)) xlabel( time[s] );ylabel( Input u1 );xlim([min,max]); figure plot(tout,uout(:,2)) xlabel( time[s] );ylabel( Input u2 );xlim([min,max]); figure plot(tout,vout(:,1), :,tout,xout(:,1)); xlabel( time[s] );ylabel( Output x1 );xlim([min,max]); figure plot(tout,vout(:,2), :,tout,xout(:,3)); xlabel( time[s] );ylabel( Output x3 );xlim([min,max]); figure plot(tout,eout1, r ); xlabel( time[s] );ylabel( Error x1 );xlim([min,max]); figure plot(tout,eout2, g ); xlabel( time[s] );ylabel( Error x3 );xlim([min,max]);

39 38 state equetion cont.m function[dx]=state_equation_cont(t,x) global p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 a11 a10 a21 a20 v u ; % v1,v2 v1=0.3*sin(0.1*t); v1dot=0.3*0.1*cos(0.1*t); v1dot2=-0.2*0.1*0.1*sin(0.1*t); v2=0.2*sin(0.2*t); v2dot=0.2*0.2*cos(0.2*t); v2dot2=-0.2*0.2*0.2*sin(0.2*t); v=[v1; v2]; % u1,u2 u1=1/(p4*cos(x(3)))*(-p1*cos(x(1))-p2*sin(x(1)) -(p3+a11)*x(2)-a10*x(1)+a10*v1+a11*v1dot+v1dot2); u2=1/p8*(-p5*cos(x(3))-p6*sin(x(3)) -(p7+a21)*x(4)-a20*x(3)+a20*v2+a21*v2dot+v2dot2); u=[u1; u2]; % A=[x(2); p1*cos(x(1))+p2*sin(x(1))+p3*x(2); x(4); p5*cos(x(3))+p6*sin(x(3))+p7*x(4); x(6); p9*x(6)];

40 39 B=[0 0; p4*cos(x(3)) 0; 0 0; 0 p8; 0 0; p10*sin(x(3)) 0]; dx=a+b*u; end

41 40 emu main.m clc clear all global p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 a11 a10 a21 a20 dt r u t dt v; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Mf=0.69; Mb=0.69; Mc=1.67; La=0.62; Lc=0.44; Ld=0.05; Le=0.02; Lh=0.177; Km=0.5; g=9.81; % etae=0.01; etap=0.01; etat=0.05; % Je=(Mf+Mb)*(La^2+Ld^2)+Mc*(Lc^2+Ld^2); Jp=(Mf+Mb)*(Lh^2+Le^2); Jt=(Mf+Mb)*La^2+Mc*Lc^2; % tan a, c, h tana = (Ld+Le)/La;tanc = Ld/Lc;tanh = Le/Lh; % p1=(-(mf+mb)*la+mc*lc)*g/je; p2=(-(mf+mb)*la*tana-mc*lc*tanc)*g/je; p3=-etae/je; p4=km*la/je; p5=(-mf+mb)*g*lh/jp; p6=(-mf-mb)*g*lh*tanh/jp; p7=-etap/jp; p8=km*lh/jp; p9=-etat/jt; p10=-km*la/jt; % a11=3; a10=2; a21=3; a20=2; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % x0=[-0.1; 0; 0.1; 0; 0.2; 0]; x=[-0.1; 0; 0.1; 0; 0.2; 0]; u=[0;0];

42 41 v=[0;0]; min=0; max=90; begin=min; dt=0.2; xout=x0 ; tout=0; rout=[0 0];uout=[0 0];vout=[0 0]; u=stable_input(begin,x0); % for i=min:max/dt; [t, x]=ode45( state_equation_emu, [begin,begin+dt], x0); x0=x(length(x),:); u=stable_input(begin,x0); xout=[xout;x0]; Tend=t(length(t),:); begin=tend ; tout=[tout; begin]; rout=[rout; r ]; uout=[uout; u ]; vout=[vout; v ]; end % eout1=rout(:,1)-xout(:,1); eout2=rout(:,2)-xout(:,3); % figure plot(tout,uout(:,1)) xlabel( time[s] );ylabel( Input u1 );xlim([min,max]);

43 42 figure plot(tout,uout(:,2)) xlabel( time[s] );ylabel( Input u2 );xlim([min,max]); figure plot(tout,rout(:,1), :,tout,xout(:,1)); xlabel( time[s] );ylabel( Output x1 );xlim([min,max]); figure plot(tout,rout(:,2), :,tout,xout(:,3)); xlabel( time[s] );ylabel( Output x3 );xlim([min,max]); figure plot(tout,eout1, r ); xlabel( time[s] );ylabel( Error x1 );xlim([min,max]); figure plot(tout,eout2, g ); xlabel( time[s] );ylabel( Error x3 );xlim([min,max]);

44 43 state equation emu.m function[dx]=state_equation_emu(t,x) global p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 u; % A=[x(2); p1*cos(x(1))+p2*sin(x(1))+p3*x(2); x(4); p5*cos(x(3))+p6*sin(x(3))+p7*x(4); x(6); p9*x(6)]; B=[0 0; p4*cos(x(3)) 0; 0 0; 0 p8; 0 0; p10*sin(x(3)) 0]; dx=a+b*u; end

45 44 function[u]=stable_input(t,x) stable input.m global p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 a11 a10 a21 a20 r v; % v1,v2 r1=0.3*sin(0.1*t); r1dot=0.3*0.1*cos(0.1*t); r1dot2=-0.3*0.1*0.1*sin(0.1*t); r2=0.2*sin(0.2*t); r2dot=0.2*0.2*cos(0.2*t); r2dot2=-0.2*0.2*0.2*sin(0.2*t); r=[r1; r2]; v=[a10*r1+a11*r1dot+r1dot2; a20*r2+a21*r2dot+r2dot2]; % u1,u2 u1=1/(p4*cos(x(3)))*(-p1*cos(x(1))-p2*sin(x(1))-(p3+a11)*x(2)-a10*x(1)+v(1)); u2=1/p8*(-p5*cos(x(3))-p6*sin(x(3))-(p7+a21)*x(4)-a20*x(3)+v(2)); u=[u1; u2]; end

46 45 redes main.m clc clear all global p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 a11 a10 a21 a20 dt r u t v ru x0; %///////////////////////////////////////////////////////////////////// % Mf=0.69; Mb=0.69; Mc=1.67; La=0.62; Lc=0.44; Ld=0.05; Le=0.02; Lh=0.177; Km=0.5; g=9.81; % etae=0.01; etap=0.01; etat=0.05; % Je=(Mf+Mb)*(La^2+Ld^2)+Mc*(Lc^2+Ld^2); Jp=(Mf+Mb)*(Lh^2+Le^2); Jt=(Mf+Mb)*La^2+Mc*Lc^2; % tan a, c, h tana = (Ld+Le)/La;tanc = Ld/Lc;tanh = Le/Lh; % p1=(-(mf+mb)*la+mc*lc)*g/je; p2=(-(mf+mb)*la*tana-mc*lc*tanc)*g/je; p3=-etae/je; p4=km*la/je; p5=(-mf+mb)*g*lh/jp; p6=(-mf-mb)*g*lh*tanh/jp; p7=-etap/jp; p8=km*lh/jp; p9=-etat/jt; p10=-km*la/jt; % a11=3; a10=2; a21=3; a20=2; % x0=[-0.1; 0; 0.1; 0; 0.2; 0]; x=[0; 0; 0; 0; 0; 0]; v=[0;0]; ru=[0;0]; rv=[0;0];

47 46 min=0; max=90; begin=min; tout=0; vout=[0 0]; uout2=[0 0];vout2=[0 0]; %//////////////////////////////////////////////////////////////////// % dt=0.2; %for i=0 ru ru=stable_input2(begin,x0); vout=v ; xout2=x0 ; % for i=min:max/dt; %/////////////////////////////////////////////////////////// [t, x]=ode45( state_equation_redes, [begin,begin+dt], x0); x0=x(length(x),:); begin=t(length(t),:); xout2=[xout2; x0]; tout=[tout; begin]; %////////////////////////////////////////////////////////// ru=stable_input2(begin,x0); vout=[vout; v ]; uout2=[uout2; ru ]; end % ( ) eout3=vout(:,1)-xout2(:,1); eout4=vout(:,2)-xout2(:,3);

48 47 figure plot(tout,uout2(:,1)) xlabel( time[s] );ylabel( Input u1 );xlim([min,max]); figure plot(tout,uout2(:,2)) xlabel( time[s] );ylabel( Input u2 );xlim([min,max]); figure plot(tout,vout(:,1), :,tout,xout2(:,1)); xlabel( time[s] );ylabel( Output x1 );xlim([min,max]); figure plot(tout,vout(:,2), :,tout,xout2(:,3)); xlabel( time[s] );ylabel( Output x3 );xlim([min,max]); figure plot(tout,eout3, r ); xlabel( time[s] );ylabel( Error x1 );xlim([min,max]); figure plot(tout,eout4, g ); xlabel( time[s] );ylabel( Error x3 );xlim([min,max]);

49 48 state equation redes.m function[dx]=state_equation_redes(~,x) global p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 ru; % A=[x(2); p1*cos(x(1))+p2*sin(x(1))+p3*x(2); x(4); p5*cos(x(3))+p6*sin(x(3))+p7*x(4); x(6); p9*x(6)]; B=[0 0; p4*cos(x(3)) 0; 0 0; 0 p8; 0 0; p10*sin(x(3)) 0]; dx=a+b*ru; end

50 49 function[ru]=stable_input2(t,x) stable input2.m global p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 a11 a10 a21 a20 r v dr dt; % a11=3; a10=2; a21=3; a20=2; % v1,v2 v1=0.3*sin(0.1*t); v1dot=0.3*0.1*cos(0.1*t); v1dot2=-0.3*0.1*0.1*sin(0.1*t); v1dot3=-0.3*0.1*0.1*0.1*cos(0.1*t); v2=0.2*sin(0.2*t); v2dot=0.2*0.2*cos(0.2*t); v2dot2=-0.2*0.2*0.2*sin(0.2*t); v2dot3=-0.2*0.2*0.2*0.2*cos(0.2*t); v=[v1; v2]; r=[a10*v1+a11*v1dot+v1dot2; a20*v2+a21*v2dot+v2dot2]; dr=[a10*v1dot+a11*v1dot2+v1dot3; a20*v2dot+a21*v2dot2+v2dot3]; % u1,u2 u1=(-p1*cos(x(1))-p2*sin(x(1))-(p3+a11)*x(2)-a10*x(1)+r(1))/(p4*cos(x(3))); u2=(-p5*cos(x(3))-p6*sin(x(3))-(p7+a21)*x(4)-a20*x(3)+r(2))/p8; u=[u1; u2]; % invd=[1/(p4*cos(x(3))) 0; 0 1/p8]; xi=[(a11^2-a10)*x(2)+a11*a10*x(1)-a11*r(1)+dr(1); (a21^2-a20)*x(4)+a21*a20*x(3)-a21*r(2)+dr(2)]; S=[-p1*x(2)*sin(x(1))+p2*x(2)*cos(x(1))+p3*(p1*cos(x(1))+p2*sin(x(1))+p3*x(2)); -p5*x(4)*sin(x(3))+p6*x(4)*cos(x(3))+p7*(p5*cos(x(3))+p6*sin(x(3))+p7*x(4))];

51 50 d=[p3*p4*cos(x(3))-p4*x(4)*sin(x(3)) 0; 0 p7*p8]; % w=invd/2*(xi-s-d*u); % ru=u+dt*w; end

52 51 C.1 simu.mdl C.2 Emu Controller

53 52 C.3 v emu C.4 Re Controller

54 53 C.5 add input w

all.dvi

all.dvi 38 5 Cauchy.,,,,., σ.,, 3,,. 5.1 Cauchy (a) (b) (a) (b) 5.1: 5.1. Cauchy 39 F Q Newton F F F Q F Q 5.2: n n ds df n ( 5.1). df n n df(n) df n, t n. t n = df n (5.1) ds 40 5 Cauchy t l n mds df n 5.3: t

More information

Gmech08.dvi

Gmech08.dvi 51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r

More information

K E N Z OU

K E N Z OU K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................

More information

PSCHG000.PS

PSCHG000.PS a b c a ac bc ab bc a b c a c a b bc a b c a ac bc ab bc a b c a ac bc ab bc a b c a ac bc ab bc de df d d d d df d d d d d d d a a b c a b b a b c a b c b a a a a b a b a

More information

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0 1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45

More information

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e 7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z

More information

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

II ( ) (7/31) II (  [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier

More information

C:/KENAR/0p1.dvi

C:/KENAR/0p1.dvi 2{3. 53 2{3 [ ] 4 2 1 2 10,15 m 10,10 m 2 2 54 2 III 1{I U 2.4 U r (2.16 F U F =, du dt du dr > 0 du dr < 0 O r 0 r 2.4: 1 m =1:00 10 kg 1:20 10 kgf 8:0 kgf g =9:8 m=s 2 (a) x N mg 2.5: N 2{3. 55 (b) x

More information

all.dvi

all.dvi 5,, Euclid.,..,... Euclid,.,.,, e i (i =,, ). 6 x a x e e e x.:,,. a,,. a a = a e + a e + a e = {e, e, e } a (.) = a i e i = a i e i (.) i= {a,a,a } T ( T ),.,,,,. (.),.,...,,. a 0 0 a = a 0 + a + a 0

More information

支持力計算法.PDF

支持力計算法.PDF . (a) P P P P P P () P P P P (0) P P Hotω H P P δ ω H δ P P (a) ( ) () H P P n0(k P 4.7) (a)0 0 H n(k P 4.76) P P n0(k P 5.08) n0(k P.4) () 0 0 (0 ) n(k P 7.56) H P P n0(k P.7) n(k P.7) H P P n(k P 5.4)

More information

Part () () Γ Part ,

Part () () Γ Part , Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

More information

9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P

9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P 9 (Finite Element Method; FEM) 9. 9. P(0) P(x) u(x) (a) P(L) f P(0) P(x) (b) 9. P(L) 9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L)

More information

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2) 3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)

More information

Note.tex 2008/09/19( )

Note.tex 2008/09/19( ) 1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................

More information

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H 199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)

More information

211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,

More information

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi) 0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()

More information

( ) ( )

( ) ( ) 20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))

More information

pdf

pdf http://www.ns.kogakuin.ac.jp/~ft13389/lecture/physics1a2b/ pdf I 1 1 1.1 ( ) 1. 30 m µm 2. 20 cm km 3. 10 m 2 cm 2 4. 5 cm 3 km 3 5. 1 6. 1 7. 1 1.2 ( ) 1. 1 m + 10 cm 2. 1 hr + 6400 sec 3. 3.0 10 5 kg

More information

I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

I A A441 : April 15, 2013 Version : 1.1 I   Kawahira, Tomoki TA (Shigehiro, Yoshida ) I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17

More information

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2 2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6

More information

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ 4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ

More information

(1) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37

(1) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37 4. 98 () θ a = 5(cm) θ c = 4(cm) b = (cm) () D 0cm 0 60 D 99 () 0m O O 7 sin 7 = 0.60 cos 7 = 0.799 tan 7 = 0.754 () xkm km R km 00 () θ cos θ = sin θ = () θ sin θ = 4 tan θ = () 0 < x < 90 tan x = 4 sin

More information

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq 49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r

More information

f (x) x y f(x+dx) f(x) Df 関数 接線 x Dx x 1 x x y f f x (1) x x 0 f (x + x) f (x) f (2) f (x + x) f (x) + f = f (x) + f x (3) x f

f (x) x y f(x+dx) f(x) Df 関数 接線 x Dx x 1 x x y f f x (1) x x 0 f (x + x) f (x) f (2) f (x + x) f (x) + f = f (x) + f x (3) x f 208 3 28. f fd f Df 関数 接線 D f f 0 f f f 2 f f f f f 3 f lim f f df 0 d 4 f df d 3 f d f df d 5 d c 208 2 f f t t f df d 6 d t dt 7 f df df d d df dt lim f 0 t df d d dt d t 8 dt 9.2 f,, f 0 f 0 lim 0 lim

More information

dynamics-solution2.dvi

dynamics-solution2.dvi 1 1. (1) a + b = i +3i + k () a b =5i 5j +3k (3) a b =1 (4) a b = 7i j +1k. a = 14 l =/ 14, m=1/ 14, n=3/ 14 3. 4. 5. df (t) d [a(t)e(t)] =ti +9t j +4k, = d a(t) d[a(t)e(t)] e(t)+ da(t) d f (t) =i +18tj

More information

さくらの個別指導 ( さくら教育研究所 ) A AB A B A B A AB AB AB B

さくらの個別指導 ( さくら教育研究所 ) A AB A B A B A AB AB AB B 1 1.1 1.1.1 1 1 1 1 a a a a C a a = = CD CD a a a a a a = a = = D 1.1 CD D= C = DC C D 1.1 (1) 1 3 4 5 8 7 () 6 (3) 1.1. 3 1.1. a = C = C C C a a + a + + C = a C 1. a a + (1) () (3) b a a a b CD D = D

More information

K E N Z U 01 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.................................... 4 1..1..................................... 4 1...................................... 5................................

More information

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d A 2. x F (t) =f sin ωt x(0) = ẋ(0) = 0 ω θ sin θ θ 3! θ3 v = f mω cos ωt x = f mω (t sin ωt) ω t 0 = f ( cos ωt) mω x ma2-2 t ω x f (t mω ω (ωt ) 6 (ωt)3 = f 6m ωt3 2.2 u ( v w) = v ( w u) = w ( u v) ma22-9

More information

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P 6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P

More information

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)

More information

) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4

) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4 1. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev

More information

A 99% MS-Free Presentation

A 99% MS-Free Presentation A 99% MS-Free Presentation 2 Galactic Dynamics (Binney & Tremaine 1987, 2008) Dynamics of Galaxies (Bertin 2000) Dynamical Evolution of Globular Clusters (Spitzer 1987) The Gravitational Million-Body Problem

More information

x,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2 = ( 2, b 2, c 2 ) v

x,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2 = ( 2, b 2, c 2 ) v 12 -- 1 4 2009 9 4-1 4-2 4-3 4-4 4-5 4-6 4-7 4-8 4-9 4-10 c 2011 1/(13) 4--1 2009 9 3 x,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2

More information

.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,

.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =, [ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b

More information

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345

More information

2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP

2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP 1. 1 213 1 6 1 3 1: ( ) 2: 3: SF 1 2 3 1: 3 2 A m 2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP

More information

Untitled

Untitled II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j

More information

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11

More information

K E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................

More information

Gmech08.dvi

Gmech08.dvi 63 6 6.1 6.1.1 v = v 0 =v 0x,v 0y, 0) t =0 x 0,y 0, 0) t x x 0 + v 0x t v x v 0x = y = y 0 + v 0y t, v = v y = v 0y 6.1) z 0 0 v z yv z zv y zv x xv z xv y yv x = 0 0 x 0 v 0y y 0 v 0x 6.) 6.) 6.1) 6.)

More information

all.dvi

all.dvi 29 4 Green-Lagrange,,.,,,,,,.,,,,,,,,,, E, σ, ε σ = Eε,,.. 4.1? l, l 1 (l 1 l) ε ε = l 1 l l (4.1) F l l 1 F 30 4 Green-Lagrange Δz Δδ γ = Δδ (4.2) Δz π/2 φ γ = π 2 φ (4.3) γ tan γ γ,sin γ γ ( π ) γ tan

More information

xia2.dvi

xia2.dvi Journal of Differential Equations 96 (992), 70-84 Melnikov method and transversal homoclinic points in the restricted three-body problem Zhihong Xia Department of Mathematics, Harvard University Cambridge,

More information

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h) 1 16 10 5 1 2 2.1 a a a 1 1 1 2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h) 4 2 3 4 2 5 2.4 x y (x,y) l a x = l cot h cos a, (3) y = l cot h sin a (4) h a

More information

m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)

m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120) 2.6 2.6.1 mẍ + γẋ + ω 0 x) = ee 2.118) e iωt Pω) = χω)e = ex = e2 Eω) m ω0 2 ω2 iωγ 2.119) Z N ϵω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j 2.120) Z ω ω j γ j f j f j f j sum j f j = Z 2.120 ω ω j, γ ϵω) ϵ

More information

数学の基礎訓練I

数学の基礎訓練I I 9 6 13 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 3 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............

More information

第5章 偏微分方程式の境界値問題

第5章 偏微分方程式の境界値問題 October 5, 2018 1 / 113 4 ( ) 2 / 113 Poisson 5.1 Poisson ( A.7.1) Poisson Poisson 1 (A.6 ) Γ p p N u D Γ D b 5.1.1: = Γ D Γ N 3 / 113 Poisson 5.1.1 d {2, 3} Lipschitz (A.5 ) Γ D Γ N = \ Γ D Γ p Γ N Γ

More information

I

I I 6 4 10 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............

More information

さくらの個別指導 ( さくら教育研究所 ) A 2 P Q 3 R S T R S T P Q ( ) ( ) m n m n m n n n

さくらの個別指導 ( さくら教育研究所 ) A 2 P Q 3 R S T R S T P Q ( ) ( ) m n m n m n n n 1 1.1 1.1.1 A 2 P Q 3 R S T R S T P 80 50 60 Q 90 40 70 80 50 60 90 40 70 8 5 6 1 1 2 9 4 7 2 1 2 3 1 2 m n m n m n n n n 1.1 8 5 6 9 4 7 2 6 0 8 2 3 2 2 2 1 2 1 1.1 2 4 7 1 1 3 7 5 2 3 5 0 3 4 1 6 9 1

More information

http://www.ike-dyn.ritsumei.ac.jp/ hyoo/wave.html 1 1, 5 3 1.1 1..................................... 3 1.2 5.1................................... 4 1.3.......................... 5 1.4 5.2, 5.3....................

More information

( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e

( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e ( ) Note 3 19 12 13 8 8.1 (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R, µ R, τ R (1a) L ( ) ) * 3) W Z 1/2 ( - )

More information

2D-RCWA 1 two dimensional rigorous coupled wave analysis [1, 2] 1 ε(x, y) = 1 ε(x, y) = ϵ mn exp [+j(mk x x + nk y y)] (1) m,n= m,n= ξ mn exp [+j(mk x

2D-RCWA 1 two dimensional rigorous coupled wave analysis [1, 2] 1 ε(x, y) = 1 ε(x, y) = ϵ mn exp [+j(mk x x + nk y y)] (1) m,n= m,n= ξ mn exp [+j(mk x 2D-RCWA two dimensional rigoros copled wave analsis, 2] εx, εx, ϵ mn exp +jmk x x + nk ] m,n m,n ξ mn exp +jmk x x + nk ] 2 K x K x Λ x Λ ϵ mn ξ mn K x 2π Λ x K 2π Λ ϵ mn ξ mn Λ x Λ x Λ x Λ x Λx Λ Λx Λ

More information

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π 4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan

More information

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν k ω/c k x ωt ω k α c, k k x ωt η αβ k α x β diag + ++ x β ct, x O O x O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k

More information

7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa, f a g b g f a g f a g bf a

7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa, f a g b g f a g f a g bf a 9 203 6 7 WWW http://www.math.meiji.ac.jp/~mk/lectue/tahensuu-203/ 2 8 8 7. 7 7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa,

More information

29

29 9 .,,, 3 () C k k C k C + C + C + + C 8 + C 9 + C k C + C + C + C 3 + C 4 + C 5 + + 45 + + + 5 + + 9 + 4 + 4 + 5 4 C k k k ( + ) 4 C k k ( k) 3 n( ) n n n ( ) n ( ) n 3 ( ) 3 3 3 n 4 ( ) 4 4 4 ( ) n n

More information

( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) =

( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) = 1 9 8 1 1 1 ; 1 11 16 C. H. Scholz, The Mechanics of Earthquakes and Faulting 1. 1.1 1.1.1 : - σ = σ t sin πr a λ dσ dr a = E a = π λ σ πr a t cos λ 1 r a/λ 1 cos 1 E: σ t = Eλ πa a λ E/π γ : λ/ 3 γ =

More information

5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1

5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 4 1 1.1 ( ) 5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 da n i n da n i n + 3 A ni n n=1 3 n=1

More information

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 + ( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n

More information

通信容量制約を考慮したフィードバック制御 - 電子情報通信学会 情報理論研究会(IT) 若手研究者のための講演会

通信容量制約を考慮したフィードバック制御 -  電子情報通信学会 情報理論研究会(IT)  若手研究者のための講演会 IT 1 2 1 2 27 11 24 15:20 16:05 ( ) 27 11 24 1 / 49 1 1940 Witsenhausen 2 3 ( ) 27 11 24 2 / 49 1940 2 gun director Warren Weaver, NDRC (National Defence Research Committee) Final report D-2 project #2,

More information

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y [ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)

More information

IA 2013 : :10722 : 2 : :2 :761 :1 (23-27) : : ( / ) (1 /, ) / e.g. (Taylar ) e x = 1 + x + x xn n! +... sin x = x x3 6 + x5 x2n+1 + (

IA 2013 : :10722 : 2 : :2 :761 :1 (23-27) : : ( / ) (1 /, ) / e.g. (Taylar ) e x = 1 + x + x xn n! +... sin x = x x3 6 + x5 x2n+1 + ( IA 2013 : :10722 : 2 : :2 :761 :1 23-27) : : 1 1.1 / ) 1 /, ) / e.g. Taylar ) e x = 1 + x + x2 2 +... + xn n! +... sin x = x x3 6 + x5 x2n+1 + 1)n 5! 2n + 1)! 2 2.1 = 1 e.g. 0 = 0.00..., π = 3.14..., 1

More information

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,, 01 10 18 ( ) 1 6 6 1 8 8 1 6 1 0 0 0 0 1 Table 1: 10 0 8 180 1 1 1. ( : 60 60 ) : 1. 1 e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1,

More information

1 variation 1.1 imension unit L m M kg T s Q C QT 1 A = C s 1 MKSA F = ma N N = kg m s 1.1 J E = 1 mv W = F x J = kg m s 1 = N m 1.

1 variation 1.1 imension unit L m M kg T s Q C QT 1 A = C s 1 MKSA F = ma N N = kg m s 1.1 J E = 1 mv W = F x J = kg m s 1 = N m 1. 1.1 1. 1.3.1..3.4 3.1 3. 3.3 4.1 4. 4.3 5.1 5. 5.3 6.1 6. 6.3 7.1 7. 7.3 1 1 variation 1.1 imension unit L m M kg T s Q C QT 1 A = C s 1 MKSA F = ma N N = kg m s 1.1 J E = 1 mv W = F x J = kg m s 1 = N

More information

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 2 ), ϕ(t) = B 1 cos(ω 1 t + α 1 ) + B 2 cos(ω 2 t

More information

ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +

ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + 2.6 2.6.1 ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j

More information

ẍ = kx, (k > ) (.) x x(t) = A cos(ωt + α) (.). d/ = D. d dt x + k ( x = D + k ) ( ) ( ) k k x = D + i D i x =... ( ) k D + i x = or ( ) k D i x =.. k.

ẍ = kx, (k > ) (.) x x(t) = A cos(ωt + α) (.). d/ = D. d dt x + k ( x = D + k ) ( ) ( ) k k x = D + i D i x =... ( ) k D + i x = or ( ) k D i x =.. k. K E N Z OU 8 9 8. F = kx x 3 678 ẍ = kx, (k > ) (.) x x(t) = A cos(ωt + α) (.). d/ = D. d dt x + k ( x = D + k ) ( ) ( ) k k x = D + i D i x =... ( ) k D + i x = or ( ) k D i x =.. k. D = ±i dt = ±iωx,

More information

[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt

[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt 3.4.7 [.] =e j(t+/4), =5e j(t+/3), 3 =3e j(t+/6) ~ = ~ + ~ + ~ 3 = e j(t+φ) =(e 4 j +5e 3 j +3e 6 j )e jt = e jφ e jt cos φ =cos 4 +5cos 3 +3cos 6 =.69 sin φ =sin 4 +5sin 3 +3sin 6 =.9 =.69 +.9 =7.74 [.]

More information

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1 1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2

More information

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)

More information

A 2 3. m S m = {x R m+1 x = 1} U + k = {x S m x k > 0}, U k = {x S m x k < 0}, ϕ ± k (x) = (x 0,..., ˆx k,... x m ) 1. {(U ± k, ϕ± k ) 0 k m} S m 1.2.

A 2 3. m S m = {x R m+1 x = 1} U + k = {x S m x k > 0}, U k = {x S m x k < 0}, ϕ ± k (x) = (x 0,..., ˆx k,... x m ) 1. {(U ± k, ϕ± k ) 0 k m} S m 1.2. A A 1 A 5 A 6 1 2 3 4 5 6 7 1 1.1 1.1 (). Hausdorff M R m M M {U α } U α R m E α ϕ α : U α E α U α U β = ϕ α (ϕ β ϕβ (U α U β )) 1 : ϕ β (U α U β ) ϕ α (U α U β ) C M a m dim M a U α ϕ α {x i, 1 i m} {U,

More information

n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................

More information

1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2

1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2 filename=mathformula58.tex ax + bx + c =, x = b ± b 4ac, (.) a x + x = b a, x x = c a, (.) ax + b x + c =, x = b ± b ac. a (.3). sin(a ± B) = sin A cos B ± cos A sin B, (.) cos(a ± B) = cos A cos B sin

More information

3 filename=quantum-3dim110705a.tex ,2 [1],[2],[3] [3] U(x, y, z; t), p x ˆp x = h i x, p y ˆp y = h i y, p z ˆp z = h

3 filename=quantum-3dim110705a.tex ,2 [1],[2],[3] [3] U(x, y, z; t), p x ˆp x = h i x, p y ˆp y = h i y, p z ˆp z = h filename=quantum-dim110705a.tex 1 1. 1, [1],[],[]. 1980 []..1 U(x, y, z; t), p x ˆp x = h i x, p y ˆp y = h i y, p z ˆp z = h i z (.1) Ĥ ( ) Ĥ = h m x + y + + U(x, y, z; t) (.) z (U(x, y, z; t)) (U(x,

More information

() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y

() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y 5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =

More information

Gmech08.dvi

Gmech08.dvi 145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2

More information

6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4

6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4 35-8585 7 8 1 I I 1 1.1 6kg 1m P σ σ P 1 l l λ λ l 1.m 1 6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m

More information

1. 4cm 16 cm 4cm 20cm 18 cm L λ(x)=ax [kg/m] A x 4cm A 4cm 12 cm h h Y 0 a G 0.38h a b x r(x) x y = 1 h 0.38h G b h X x r(x) 1 S(x) = πr(x) 2 a,b, h,π

1. 4cm 16 cm 4cm 20cm 18 cm L λ(x)=ax [kg/m] A x 4cm A 4cm 12 cm h h Y 0 a G 0.38h a b x r(x) x y = 1 h 0.38h G b h X x r(x) 1 S(x) = πr(x) 2 a,b, h,π . 4cm 6 cm 4cm cm 8 cm λ()=a [kg/m] A 4cm A 4cm cm h h Y a G.38h a b () y = h.38h G b h X () S() = π() a,b, h,π V = ρ M = ρv G = M h S() 3 d a,b, h 4 G = 5 h a b a b = 6 ω() s v m θ() m v () θ() ω() dθ()

More information

さくらの個別指導 ( さくら教育研究所 ) 1 φ = φ 1 : φ [ ] a [ ] 1 a : b a b b(a + b) b a 2 a 2 = b(a + b). b 2 ( a b ) 2 = a b a/b X 2 X 1 = 0 a/b > 0 2 a

さくらの個別指導 ( さくら教育研究所 ) 1 φ = φ 1 : φ [ ] a [ ] 1 a : b a b b(a + b) b a 2 a 2 = b(a + b). b 2 ( a b ) 2 = a b a/b X 2 X 1 = 0 a/b > 0 2 a φ + 5 2 φ : φ [ ] a [ ] a : b a b b(a + b) b a 2 a 2 b(a + b). b 2 ( a b ) 2 a b + a/b X 2 X 0 a/b > 0 2 a b + 5 2 φ φ : 2 5 5 [ ] [ ] x x x : x : x x : x x : x x 2 x 2 x 0 x ± 5 2 x x φ : φ 2 : φ ( )

More information

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k 63 3 Section 3.1 g 3.1 3.1: : 64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () 3 9.8 m/s 2 3.2 3.2: : a) b) 5 15 4 1 1. 1 3 14. 1 3 kg/m 3 2 3.3 1 3 5.8 1 3 kg/m 3 3 2.65 1 3 kg/m 3 4 6 m 3.1. 65 5

More information

l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r

l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r 2 1 (7a)(7b) λ i( w w ) + [ w + w ] 1 + w w l 2 0 Re(γ) α (7a)(7b) 2 γ 0, ( w) 2 1, w 1 γ (1) l µ, λ j γ l 2 0 Re(γ) α, λ w + w i( w w ) 1 + w w γ γ 1 w 1 r [x2 + y 2 + z 2 ] 1/2 ( w) 2 x2 + y 2 + z 2

More information

meiji_resume_1.PDF

meiji_resume_1.PDF β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E

More information

untitled

untitled 0. =. =. (999). 3(983). (980). (985). (966). 3. := :=. A A. A A. := := 4 5 A B A B A B. A = B A B A B B A. A B A B, A B, B. AP { A, P } = { : A, P } = { A P }. A = {0, }, A, {0, }, {0}, {}, A {0}, {}.

More information

1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1

1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V

More information

20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................

More information

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s [ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =

More information

Z: Q: R: C: sin 6 5 ζ a, b

Z: Q: R: C: sin 6 5 ζ a, b Z: Q: R: C: 3 3 7 4 sin 6 5 ζ 9 6 6............................... 6............................... 6.3......................... 4 7 6 8 8 9 3 33 a, b a bc c b a a b 5 3 5 3 5 5 3 a a a a p > p p p, 3,

More information

(e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ,µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R,µ R,τ R (2.1a

(e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ,µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R,µ R,τ R (2.1a 1 2 2.1 (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ,µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R,µ R,τ R (2.1a) L ( ) ) * 2) W Z 1/2 ( - ) d u + e + ν e 1 1 0 0

More information

c 2009 i

c 2009 i I 2009 c 2009 i 0 1 0.0................................... 1 0.1.............................. 3 0.2.............................. 5 1 7 1.1................................. 7 1.2..............................

More information

B 1 B.1.......................... 1 B.1.1................. 1 B.1.2................. 2 B.2........................... 5 B.2.1.......................... 5 B.2.2.................. 6 B.2.3..................

More information

3 m = [n, n1, n 2,..., n r, 2n] p q = [n, n 1, n 2,..., n r ] p 2 mq 2 = ±1 1 1 6 1.1................................. 6 1.2......................... 8 1.3......................... 13 2 15 2.1.............................

More information

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5. A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c

More information

( ) ,

( ) , II 2007 4 0. 0 1 0 2 ( ) 0 3 1 2 3 4, - 5 6 7 1 1 1 1 1) 2) 3) 4) ( ) () H 2.79 10 10 He 2.72 10 9 C 1.01 10 7 N 3.13 10 6 O 2.38 10 7 Ne 3.44 10 6 Mg 1.076 10 6 Si 1 10 6 S 5.15 10 5 Ar 1.01 10 5 Fe 9.00

More information

grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( )

grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( ) 2 9 2 5 2.2.3 grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = g () g () (3) grad φ(p ) p grad φ φ (P, φ(p )) y (, y) = (ξ(t), η(t)) ( ) ξ (t) (t) := η (t) grad f(ξ(t), η(t)) (t) g(t) := f(ξ(t), η(t))

More information

The Physics of Atmospheres CAPTER :

The Physics of Atmospheres CAPTER : The Physics of Atmospheres CAPTER 4 1 4 2 41 : 2 42 14 43 17 44 25 45 27 46 3 47 31 48 32 49 34 41 35 411 36 maintex 23/11/28 The Physics of Atmospheres CAPTER 4 2 4 41 : 2 1 σ 2 (21) (22) k I = I exp(

More information

I 1

I 1 I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg

More information

( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c) yoshioka/education-09.html pdf 1

( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c)   yoshioka/education-09.html pdf 1 2009 1 ( ) ( 40 )+( 60 ) 1 1. 2. Schrödinger 3. (a) (b) (c) http://goofy.phys.nara-wu.ac.jp/ yoshioka/education-09.html pdf 1 1. ( photon) ν λ = c ν (c = 3.0 108 /m : ) ɛ = hν (1) p = hν/c = h/λ (2) h

More information