修士論文

Size: px
Start display at page:

Download "修士論文"

Transcription

1 The flexible cotrol of the Quadruped Walkig Robot by Positio-Based Impedace Cotrol

2

3

4 HONDA ASIMO SONY SDR-3X McGhee The OSU Hexapod ReCUS 1980 Waldro ASV Whitaker AmblerHalme MECANT 1,000kg Brooks Muybridge 1960 Mosher Walkig Truck McGhee Phoy Poy 1979 PV

5 6 TITANNINJA 7 8 TITAN 8 TITAN 9 TITAN Tekke Patrush 10 AIBO JROB ZMP ZMP ZMP ZMP ZMP Free Gait

6 Fig.1.1 Stability margi 36 Fig ZMP 3 TITAN 4 5 3

7 Float Force sesor Ma Quadruped Walkig Robot Fig.1.1 Quadruped Walkig Robot image 4

8 ZMP ZMP Zero Momet Poit(ZMP) ZMP 37 Fig.2.1 ZMP ZMP ZMP Fig.2.1 Cotrol image of ZMP type ad Pedulum type 5

9 ZMP Fig.2.2 Swig leg Swig leg Ceter of gravity Supportig leg Supportig leg Ceter of gravity Support polygo Fig.2.2 Support polygo 23 1 Fig2.3 T 1 Tg Tg β = (1) T 6

10 0.75 3/4 Leg Support phase Swig phase Fig2.3 Walk cycle 24 ZMP ZMP 7

11 2 ZMP

12 Duty factor Supportig leg Table.1 classificatio of Gait Leg phase Leg phase (Right ad Left)(Frot ad Rear) Gait Velocity of walk or 4 Crawl Slow Trot Pace Bouce Quick 25 Tipover Stability Margi Eergy Stability Margi Dyamic Eergy Stability Margi NE 1440RSM 16 0 Leg.2 Leg.1 Ceter of gravity Stability margi Leg.4 Leg.3 Fig.2.3 Stability margi 9

13 3 4 TITAN Fig.3.1 Y X Z 3 Leg.1 X 100 Y 200 Leg.2 X 100 Y 200 Leg.3 X 100 Y 200 Leg.4 X 100 Y -200 Coordiate system of mai part Y Z X Coordiate system of leg Fig.3.1 Coordiate system 10

14 32 TITAN l l 243 l2 3 1 Z Y X 2 Joit 2 3 Joit 3 Joit 1 L3 1 L1 L2 Fig.3.2 Simplified leg mechaism 33 Joit ( l + l cosθ l θ ) X = cosθ + (2) cos ( l + l cosθ l θ ) Y = siθ + (3) cos 3 3 Z = l + θ (4) 2 siθ 2 l3 si 3 11

15 θ = arcta Y 1 X (5) θ θ 2 3 = arcta Z l l ± arccos 2 2 l X cosθ1 1 2l2 2 3 X ( l ) l2 l3 + cosθ 1 + Z 1 = θ 2 ± π + arccos 2l2l3 X ( ) cosθ l1 + Z 1 ( ) + X 2 2 cosθ l 1 1 Z 2 (7) (6) 34 Fig.3.3 TITAN Leg 1 Leg Leg 3 Leg 4 90 Fig.3.3 Movable agle 12

16 Leg.4 Leg.1 Leg.3 Leg.2 Swig leg Swig leg Gravity move Swig leg Swig leg Gravity move Gravity move Swig leg Swig leg Gravity move Swig leg Swig leg Fig.4.1 Walk sequece 13

17 m/s Y β Liftig Placig Z 100 Y 100 Z 100 Liftig Placig T 1 = T3 = 0.25msec T 2 = 50msec Liftig 100 Placig T1 T2 T3 T Fig.4.2 swig phase 14

18 Ceter of gravity Quadruped Walkig Robot Ky Cy Cx Kx Cz Kz My Mx Mz Fig.5.1 Virtual mechaical impedace 15

19 52 F = M Pg + C Pg + K( Pg Pg0) (8) 0 F Pg Pg Pg Pg M C K Pg Pg Pg 1 = (9) dt Pg Pg Pg 1 = (10) dt dt Pg 1 Pg F Pg Pg 1 Pg Pg 1 1 = m + C + K( Pg Pg 0 dt dt ) (11) Pg 1 Pg 1 Pg 2 = (12) dt Pg

20 Pg = F M + dt ( Pg Pg ) 1 M dt C dt Pg + C dt + K 1 + KPg 0 (13) Pg Pg = Pg Pg 1 (14) Pg Pf i P i Fig.5.2 TITAN Positio Based Impedace Cotrol Robot Pg Robot TITAN Titech Robot Driver Positio Based Impedace Cotrol Positio cotrol system of Robot Pg Pfi Titech Robot Driver Robot F Pi Fig.5.2 System of Quadruped Walkig Robot 17

21 TITAN 54 18

22 Start Force Sesig Walk Plaig orbit of the ceter of gravity Make a foot positio If foot positio i movable area? No Program stop Yes Joit agle plaig by iverse kiematics Out put Fig.6.1 Cotrol flow 19

23 62 IFS-67M25A 50-I40 PC Table.2 Receiverboard F Force ceser PC Fig.6.2Simulatio system Table.2 Specificatio of F/T Sesor IFS-67M25A 50-I40NITTA Measuremet bouds of X ad Y directio 200[N] Measuremet bouds of Z directio 400[N] Measuremet bouds of each axis momet 12.5[Nm] Diameter 67[] Height 25[] Weight 180[g] Data output cycle 8000[Hz] 20

24 63 Leg1 1 Support polygo Y Leg2 2 Y-itercepts Stability margi X Ceter of Gravity Fig.6.3 Relatio betwee a support polygo ad a stability margi S y = mx + 15 S 2 = (16) 2 1+ m y1 y2 m = (17) x x 1 2 Y = y 1 + mx 1 (18) 21

25 S Y X 64 C M K M 1kgfC 100NS/ K 100N/ C MS-DOS 22

26 X 85 Y 70 Z 30 Lifted leg Supportig leg Ceter of gravity Support polygo Before After Fig.7.1 Chage of Support polygo 23

27 72 F Fig.7.2 Orbit chage at X directio G G T Pg G = (19) Pg G Pg G = (20) Fl Fg 24

28 G Fl ( 1 β )T (21) Pf i Fg G β T (22) Pf i X 25 Y 25 Y Fig.7.3 F Fig.7.3 Orbit chage at Y directio 25

29 8 X Fig.8.1,8.2 Y X Fig.8.1 Fig.8.2 Fig.8.3,8.4 X Y Fig.8.5 Fig.8.6 Fig.8.7 Z Fig.8.8 X Fig.8.9 Y Fig XY Fig

30 Fig.8.1 Stability margi at 3 legs support Fig.8.2 Stability margi at 4 legs support 27

31 Fig.8.3 Orbit chage o X directio Fig.8.4 Orbit chage o Y directio 28

32 Fig.8.5 Locus of Ceter of gravity by itermittet crawl gait Fig.8.6 Stability margi of itermittet crawl gait Fig.8.7 Leg positio of itermittet crawl gait o Z directio 29

33 Fig.8.8 Orbit o X directio Fig.8.9 Orbit o Y directio 30

34 Fig8.10 Leg positio of leg.1 ad leg.4 o X directio Fig8.11 Leg positio of leg.2 ad leg.3 o X directio Fig8.12 Leg positio o Y directio 31

35 Fig8.13 Leg positio of leg.1 ad leg.4 o X directio Fig8.14 Leg positio of leg.2 ad leg.3 o X directio Fig8.15 Leg positio o Y directio 32

36 Fig.8.16 Ceter of gravity ad force o X directio Fig.8.16 Ceter of gravity ad force o Y directio Fig.8.16 Ceter of gravity ad force o Z directio 33

37 9 1 34

38 10 35

39 1 "4 " Vol.16No.5pp "YT " C Vol.62No.593pp "YT " C Vol.62No.601pp " C Vol.62No.596pp " " C Vol.60No.575pp " " C Vol.61No.589pp " NINJA-1 " C 57540pp " " Vol.18No.2pp " TITAN " Vol.17No.8pp " Vol.16No.8pp "4 " Vol.3No.4 pp " MARS " Vol.4No.3pp " MARS " Vol.6No.1pp "3 - -" '02 (ROBMEC'02) Ju "3 " 13 Mar "3 " 18 vol.1, pp Sep " 4 36

40 " Vol.9No.7pp "4 " Vol.2No.6 pp "4 " Vol.4 No.4pp "4 " Vol.9No.3pp " 13 pp " Vol.17 No.2pp " 4 " Vol.16 No.3 pp "4 " Vol.18No.6pp " Vol.6No.2pp " 4 " Vol.9No.6pp " 3 " Vol.4No.3pp " 4 TITAN " Vol.9No.4pp "3D GDA " Vol.13No.5pp "3 " 20 Sep "4 " Vol.12No.7pp " Vol.18No.3pp " Vol.14No.7pp " " Vol.13No.7pp

41 " R.B. McGheead A.A. Frak"O the Stability Properties of Quadruped Creepig Gaits"Math. Bioscieces Vol.3pp [] ". " pp " Vol.16 No.8pp "" Vol.16 No.5pp "" Vol.16 No.7pp

Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n

Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n Part2 47 Example 161 93 1 T a a 2 M 1 a 1 T a 2 a Point 1 T L L L T T L L T L L L T T L L T detm a 1 aa 2 a 1 2 + 1 > 0 11 T T x x M λ 12 y y x y λ 2 a + 1λ + a 2 2a + 2 0 13 D D a + 1 2 4a 2 2a + 2 a

More information

18 A Study on the Running and Jumping by a One-leg Robot ( )

18 A Study on the Running and Jumping by a One-leg Robot ( ) 18 A Study on the Running and Jumping by a One-leg Robot ( ) 109535 1...3 1.1...3 1....4 1.3...4...5.1...5....6 3...7 3.1 CAD...7 3.1.1...7 3.1.CAD...7 3.1.3...9 3.1.4...10 3.1.5... 11 3.1.6...1 3....1

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

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

3 m/sec 8.35 39.06 3.22 2.15 13.72 52.78 15.00 2.12 2.69 12.62 27.62 3 m/ 772 79 68 263 410 1,182 3 m/sec 3.87 0.63 8.00 3.12 1.38 12.50 12.50 2.00 2.50 1.00 5.50 5.50 m/ 105 122 20 247 247 3 m/sec 0.23

More information

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx

More information

CG38.PDF

CG38.PDF ............3...3...6....6....8.....8.....4...9 3....9 3.... 3.3...4 3.4...36...39 4....39 4.....39 4.....4 4....49 4.....5 4.....57...64 5....64 5....66 5.3...68 5.4...7 5.5...77...8 6....8 6.....8 6.....83

More information

極限

極限 si θ = ) θ 0 θ cos θ θ 0 θ = ) P T θ H A, 0) θ, 0 < θ < π ) AP, P H A P T PH < AP < AT si θ < θ < ta θ si θ < θ < si θ cos θ θ cos θ < si θ θ < θ < 0 θ = h θ 0 cos θ =, θ 0 si θ θ =. θ 0 cos θ θ θ 0 cos

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

平成14年度

平成14年度 14 Estimation of Ground Reaction Force and Joint Moment by Plantar Pressure Measurement Device 1055063 1. 1 1.1 1 1.2 1 1.3 2 1.4 2 1.4 a) 2 1.4 b) 3 1.5 3 1.6 3 2. 2.1 5 2.2 5 2.2.1 2.2.1.1 7 2.2.1.1.1

More information

1 a b = max{a, b}, a b = mi{a, b} a 1 a 2 a a 1 a = max{a 1,... a }, a 1 a = mi{a 1,... a }. A sup A, if A A A A A sup A sup A = + A if A = ± y = arct

1 a b = max{a, b}, a b = mi{a, b} a 1 a 2 a a 1 a = max{a 1,... a }, a 1 a = mi{a 1,... a }. A sup A, if A A A A A sup A sup A = + A if A = ± y = arct 27 6 2 1 2 2 5 3 8 4 13 5 16 6 19 7 23 8 27 N Z = {, ±1, ±2,... }, R =, R + = [, + ), R = [, ], C =. a b = max{a, b}, a b = mi{a, b}, a a, a a. f : X R [a < f < b] = {x X; a < f(x) < b}. X [f] = [f ],

More information

mt_4.dvi

mt_4.dvi ( ) 2006 1 PI 1 1 1.1................................. 1 1.2................................... 1 2 2 2.1...................................... 2 2.1.1.......................... 2 2.1.2..............................

More information

1 P2 P P3P4 P5P8 P9P10 P11 P12

1 P2 P P3P4 P5P8 P9P10 P11 P12 1 P2 P14 2 3 4 5 1 P3P4 P5P8 P9P10 P11 P12 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 & 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1! 3 2 3! 4 4 3 5 6 I 7 8 P7 P7I P5 9 P5! 10 4!! 11 5 03-5220-8520

More information

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1 ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD

More information

DVIOUT

DVIOUT A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)

More information

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 = #A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/

More information

. (.8.). t + t m ü(t + t) + c u(t + t) + k u(t + t) = f(t + t) () m ü f. () c u k u t + t u Taylor t 3 u(t + t) = u(t) + t! u(t) + ( t)! = u(t) + t u(

. (.8.). t + t m ü(t + t) + c u(t + t) + k u(t + t) = f(t + t) () m ü f. () c u k u t + t u Taylor t 3 u(t + t) = u(t) + t! u(t) + ( t)! = u(t) + t u( 3 8. (.8.)............................................................................................3.............................................4 Nermark β..........................................

More information

untitled

untitled - k k k = y. k = ky. y du dx = ε ux ( ) ux ( ) = ax+ b x u() = ; u( ) = AE u() = b= u () = a= ; a= d x du ε x = = = dx dx N = σ da = E ε da = EA ε A x A x x - σ x σ x = Eε x N = EAε x = EA = N = EA k =

More information

1.1 ft t 2 ft = t 2 ft+ t = t+ t 2 1.1 d t 2 t + t 2 t 2 = lim t 0 t = lim t 0 = lim t 0 t 2 + 2t t + t 2 t 2 t + t 2 t 2t t + t 2 t 2t + t = lim t 0

1.1 ft t 2 ft = t 2 ft+ t = t+ t 2 1.1 d t 2 t + t 2 t 2 = lim t 0 t = lim t 0 = lim t 0 t 2 + 2t t + t 2 t 2 t + t 2 t 2t t + t 2 t 2t + t = lim t 0 A c 2008 by Kuniaki Nakamitsu 1 1.1 t 2 sin t, cos t t ft t t vt t xt t + t xt + t xt + t xt t vt = xt + t xt t t t vt xt + t xt vt = lim t 0 t lim t 0 t 0 vt = dxt ft dft dft ft + t ft = lim t 0 t 1.1

More information

85 4

85 4 85 4 86 Copright c 005 Kumanekosha 4.1 ( ) ( t ) t, t 4.1.1 t Step! (Step 1) (, 0) (Step ) ±V t (, t) I Check! P P V t π 54 t = 0 + V (, t) π θ : = θ : π ) θ = π ± sin ± cos t = 0 (, 0) = sin π V + t +V

More information

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx 4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan

More information

mugensho.dvi

mugensho.dvi 1 1 f (t) lim t a f (t) = 0 f (t) t a 1.1 (1) lim(t 1) 2 = 0 t 1 (t 1) 2 t 1 (2) lim(t 1) 3 = 0 t 1 (t 1) 3 t 1 2 f (t), g(t) t a lim t a f (t) g(t) g(t) f (t) = o(g(t)) (t a) = 0 f (t) (t 1) 3 1.2 lim

More information

(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)

(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10) 2017 12 9 4 1 30 4 10 3 1 30 3 30 2 1 30 2 50 1 1 30 2 10 (1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10) (1) i 23 c 23 0 1 2 3 4 5 6 7 8 9 a b d e f g h i (2) 23 23 (3) 23 ( 23 ) 23 x 1 x 2 23 x

More information

空気の屈折率変調を光学的に検出する超指向性マイクロホン

空気の屈折率変調を光学的に検出する超指向性マイクロホン 23 2 1M36268 2 2 4 5 6 7 8 13 15 2 21 2 23 2 2 3 32 34 38 38 54 57 62 63 1-1 ( 1) ( 2) 1-1 a ( sinθ ) 2J D ( θ ) = 1 (1-1) kaka sinθ ( 3) 1-2 1 Back face hole Amplifier Diaphragm Equiphase wave surface

More information

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) ( 6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b

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

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

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n ( 3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc

More information

1 3 1.1.......................... 3 1............................... 3 1.3....................... 5 1.4.......................... 6 1.5........................ 7 8.1......................... 8..............................

More information

( )

( ) 18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................

More information

1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O

1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O : 2014 4 10 1 2 2 3 2.1...................................... 3 2.2....................................... 4 2.3....................................... 4 2.4................................ 5 2.5 Free-Body

More information

a) Extraction of Similarities and Differences in Human Behavior Using Singular Value Decomposition Kenichi MISHIMA, Sayaka KANATA, Hiroaki NAKANISHI a

a) Extraction of Similarities and Differences in Human Behavior Using Singular Value Decomposition Kenichi MISHIMA, Sayaka KANATA, Hiroaki NAKANISHI a a) Extraction of Similarities and Differences in Human Behavior Using Singular Value Decomposition Kenichi MISHIMA, Sayaka KANATA, Hiroaki NAKANISHI a), Tetsuo SAWARAGI, and Yukio HORIGUCHI 1. Johansson

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

08-Note2-web

08-Note2-web r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2 r(t), v(t), a(t) t dr(t) dt r(t) =(x(t),y(t),z(t)) = d 2 r(t) dt 2 = ( dx(t) dt ( d 2 x(t) dt 2, dy(t), dz(t) dt dt ), d2 y(t) dt 2, d2 z(t)

More information

最新耐震構造解析 ( 第 3 版 ) サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 第 3 版 1 刷発行時のものです.

最新耐震構造解析 ( 第 3 版 ) サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます.   このサンプルページの内容は, 第 3 版 1 刷発行時のものです. 最新耐震構造解析 ( 第 3 版 ) サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/052093 このサンプルページの内容は, 第 3 版 1 刷発行時のものです. i 3 10 3 2000 2007 26 8 2 SI SI 20 1996 2000 SI 15 3 ii 1 56 6

More information

B

B B YES NO 5 7 6 1 4 3 2 BB BB BB AA AA BB 510J B B A 510J B A A A A A A 510J B A 510J B A A A A A 510J M = σ Z Z = M σ AAA π T T = a ZP ZP = a AAA π B M + M 2 +T 2 M T Me = = 1 + 1 + 2 2 M σ Te = M 2 +T

More information

IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................

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

1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0

1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 : 2016 4 1 1 2 1.1......................................... 2 1.2................................... 2 2 2 2.1........................................ 2 2.2......................................... 3 2.3.........................................

More information

(u(x)v(x)) = u (x)v(x) + u(x)v (x) ( ) u(x) = u (x)v(x) u(x)v (x) v(x) v(x) 2 y = g(t), t = f(x) y = g(f(x)) dy dx dy dx = dy dt dt dx., y, f, g y = f (g(x))g (x). ( (f(g(x)). ). [ ] y = e ax+b (a, b )

More information

1. ( ) 1.1 t + t [m]{ü(t + t)} + [c]{ u(t + t)} + [k]{u(t + t)} = {f(t + t)} (1) m ü f c u k u 1.2 Newmark β (1) (2) ( [m] + t ) 2 [c] + β( t)2

1. ( ) 1.1 t + t [m]{ü(t + t)} + [c]{ u(t + t)} + [k]{u(t + t)} = {f(t + t)} (1) m ü f c u k u 1.2 Newmark β (1) (2) ( [m] + t ) 2 [c] + β( t)2 212 1 6 1. (212.8.14) 1 1.1............................................. 1 1.2 Newmark β....................... 1 1.3.................................... 2 1.4 (212.8.19)..................................

More information

9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x

9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x 2009 9 6 16 7 1 7.1 1 1 1 9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x(cos y y sin y) y dy 1 sin

More information

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 ( 1 1.1 (1) (1 + x) + (1 + y) = 0 () x + y = 0 (3) xy = x (4) x(y + 3) + y(y + 3) = 0 (5) (a + y ) = x ax a (6) x y 1 + y x 1 = 0 (7) cos x + sin x cos y = 0 (8) = tan y tan x (9) = (y 1) tan x (10) (1 +

More information

BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B

BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B 2000 8 3.4 p q θ = 80 B E a H F b θ/2 O θ/2 D A B E BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF :

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

D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y

D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y 5 5. 2 D xy D (x, y z = f(x, y f D (2 (x, y, z f R 2 5.. z = x 2 y 2 {(x, y; x 2 +y 2 } x 2 +y 2 +z 2 = z 5.2. (x, y R 2 z = x 2 y + 3 (2,,, (, 3,, 3 (,, 5.3 (. (3 ( (a, b, c A : (x, y, z P : (x, y, x

More information

ac b 0 r = r a 0 b 0 y 0 cy 0 ac b 0 f(, y) = a + by + cy ac b = 0 1 ac b = 0 z = f(, y) f(, y) 1 a, b, c 0 a 0 f(, y) = a ( ( + b ) ) a y ac b + a y

ac b 0 r = r a 0 b 0 y 0 cy 0 ac b 0 f(, y) = a + by + cy ac b = 0 1 ac b = 0 z = f(, y) f(, y) 1 a, b, c 0 a 0 f(, y) = a ( ( + b ) ) a y ac b + a y 01 4 17 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy + r (, y) z = p + qy + r 1 y = + + 1 y = y = + 1 6 + + 1 ( = + 1 ) + 7 4 16 y y y + = O O O y = y

More information

( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +

( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x + (.. C. ( d 5 5 + C ( d d + C + C d ( d + C ( ( + d ( + + + d + + + + C (5 9 + d + d tan + C cos (sin (6 sin d d log sin + C sin + (7 + + d ( + + + + d log( + + + C ( (8 d 7 6 d + 6 + C ( (9 ( d 6 + 8 d

More information

77

77 O r r r, F F r,r r = r r F = F (. ) r = r r 76 77 d r = F d r = F (. ) F + F = 0 d ( ) r + r = 0 (. 3) M = + MR = r + r (. 4) P G P MX = + MY = + MZ = z + z PG / PG = / M d R = 0 (. 5) 78 79 d r = F d

More information

KENZOU

KENZOU KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................

More information

BH BH BH BH Typeset by FoilTEX 2

BH BH BH BH Typeset by FoilTEX 2 GR BH BH 2015.10.10 BH at 2015.09.07 NICT 2015.05.26 Typeset by FoilTEX 1 BH BH BH BH Typeset by FoilTEX 2 1. BH 1.1 1 Typeset by FoilTEX 3 1.2 2 A B A B t = 0 A: m a [kg] B: m b [kg] t = t f star free

More information

9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 =

9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 = 5 5. 5.. A II f() f() F () f() F () = f() C (F () + C) = F () = f() F () + C f() F () G() f() G () = F () 39 G() = F () + C C f() F () f() F () + C C f() f() d f() f() C f() f() F () = f() f() f() d =

More information

熊本県数学問題正解

熊本県数学問題正解 00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (

More information

( ) x y f(x, y) = ax

( ) x y f(x, y) = ax 013 4 16 5 54 (03-5465-7040) nkiyono@mail.ecc.u-okyo.ac.jp hp://lecure.ecc.u-okyo.ac.jp/~nkiyono/inde.hml 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy

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

A

A A05-132 2010 2 11 1 1 3 1.1.......................................... 3 1.2..................................... 3 1.3..................................... 3 2 4 2.1............................... 4 2.2

More information

,, 2. Matlab Simulink 2018 PC Matlab Scilab 2

,, 2. Matlab Simulink 2018 PC Matlab Scilab 2 (2018 ) ( -1) TA Email : ohki@i.kyoto-u.ac.jp, ske.ta@bode.amp.i.kyoto-u.ac.jp : 411 : 10 308 1 1 2 2 2.1............................................ 2 2.2..................................................

More information

128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds = 0 (3.4) S 1, S 2 { B( r) n( r)}ds

128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds = 0 (3.4) S 1, S 2 { B( r) n( r)}ds 127 3 II 3.1 3.1.1 Φ(t) ϕ em = dφ dt (3.1) B( r) Φ = { B( r) n( r)}ds (3.2) S S n( r) Φ 128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds

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

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

I ( ) 2019

I ( ) 2019 I ( ) 2019 i 1 I,, III,, 1,,,, III,,,, (1 ) (,,, ), :...,, : NHK... NHK, (YouTube ),!!, manaba http://pen.envr.tsukuba.ac.jp/lec/physics/,, Richard Feynman Lectures on Physics Addison-Wesley,,,, x χ,

More information

. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n

. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n 003...............................3 Debye................. 3.4................ 3 3 3 3. Larmor Cyclotron... 3 3................ 4 3.3.......... 4 3.3............ 4 3.3...... 4 3.3.3............ 5 3.4.........

More information

limit&derivative

limit&derivative - - 7 )................................................................................ 5.................................. 7.. e ).......................... 9 )..........................................

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

1

1 PalmGauss SC PGSC-5G Instruction Manual PalmGauss SC PGSC-5G Version 1.01 PalmGauss SC PGSC5G 1.... 3 2.... 3 3.... 3 3.1... 3 3.2... 3 3.3 PalmGauss... 4 3.4... 4 3.4.1 (Fig. 4)... 4 3.4.2 (Fig. 5)...

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

Microsoft PowerPoint - 粉体特論2018 [互換モード]

Microsoft PowerPoint - 粉体特論2018 [互換モード] 8 9 40 4 q c q q i 4 4 u[m/s] Q ( /)[m] / [s] / s] u[m/s] A[m ] [m Q π N q N q Q c / s] [m N qc N qc Q c Q + Q i t c V Q c 44 τa ma Re 4 / ) ( u u t u V ma ) ( ) ( u u A 4 Re m ma Fr V m ) ( Fr 4 45 Fr

More information

平成15年度

平成15年度 5 Ivesigaio of a Gof Swig Robo i Cosideaio of he Mass of a A 6578 - - -3 3 - - DD 3 6 3-6 3-7 3-- 8 3--, 8 3--3 3--4, 3--5 3--6 3-3 4 Li 4 4-4- 5 5 9 5-5-- 9 5-- 5-5-- 5-- 5-- 5-- 5--3 6 6-6- 7 5 7 -,,,,,,,,,,,,

More information

untitled

untitled 8- My + Cy + Ky = f () t 8. C f () t ( t) = Ψq( t) () t = Ψq () t () t = Ψq () t = ( q q ) ; = [ ] y y y q Ψ φ φ φ = ( ϕ, ϕ, ϕ,3 ) 8. ψ Ψ MΨq + Ψ CΨq + Ψ KΨq = Ψ f ( t) Ψ MΨ = I; Ψ CΨ = C; Ψ KΨ = Λ; q

More information

M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -

M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 - M3............................................................................................ 3.3................................................... 3 6........................................... 6..........................................

More information

,., 5., ,. 2.2,., x z. y,.,,,. du dt + α p x = 0 dw dt + α p z + g = 0 α dp dt + pγ dα dt = 0 α V dα dt = 0 (2.2.1), γ = c p /c

,., 5., ,. 2.2,., x z. y,.,,,. du dt + α p x = 0 dw dt + α p z + g = 0 α dp dt + pγ dα dt = 0 α V dα dt = 0 (2.2.1), γ = c p /c 29 2 1 2.1 2.1.1.,., 5.,. 2.1.1,. 2.2,., x z. y,.,,,. du dt + α p x = 0 dw dt + α p z + g = 0 α dp dt + pγ dα dt = 0 α V dα dt = 0 (2.2.1), γ = c p /c v., V = (u, w), = ( / x, / z). 30 2.1.1: 31., U p(z),

More information

橡実験IIINMR.PDF

橡実験IIINMR.PDF (NMR) 0 (NMR) 2µH hω ω 1 h 2 1 1-1 NMR NMR h I µ = γµ N 1-2 1 H 19 F Ne µ = Neh 2mc ( 1) N 2 ( ) I =1/2 I =3/2 I z =+1/2 I z = 1/2 γh H>0 2µH H=0 µh I z =+3/2 I z =+1/2 I z = 1/2 I z = 3/2 γh H>0 2µH H=0

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

x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin

x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin 2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ

More information

example2_time.eps

example2_time.eps Google (20/08/2 ) ( ) Random Walk & Google Page Rank Agora on Aug. 20 / 67 Introduction ( ) Random Walk & Google Page Rank Agora on Aug. 20 2 / 67 Introduction Google ( ) Random Walk & Google Page Rank

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

noted by Kazuma MATSUDA (force) (mechanics) (statics) (dynamics) O F OA O (point of application) (line of action) (vector) F F F

noted by Kazuma MATSUDA (force) (mechanics) (statics) (dynamics) O F OA O (point of application) (line of action) (vector) F F F noted by Kazuma MATSUDA 2016 5 29 1. (force) (mechancs) (statcs) (dynamcs) 1.1 1 1 O F OA O (pont of applcaton) (lne of acton) (vector) F F F F 1.2 ( ) 1 (kg-wt) (Internatonal System of Unt, SI) kg SI

More information

f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a

f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a 3 3.1 3.1.1 A f(a + h) f(a) f(x) lim f(x) x = a h 0 h f(x) x = a f 0 (a) f 0 (a) = lim h!0 f(a + h) f(a) h = lim x!a f(x) f(a) x a a + h = x h = x a h 0 x a 3.1 f(x) = x x = 3 f 0 (3) f (3) = lim h 0 (

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

2009 2010 2 23 (MHD ) GFV (Galium Field Visualizer) GFV OpenGL GFV GFV GFV 1 1 2 2 2.1.................... 2 2.2................................. 2 2.3...................... 3 3 6 3.1 GFV....................

More information

4‐E ) キュリー温度を利用した消磁:熱消磁

4‐E ) キュリー温度を利用した消磁:熱消磁 ( ) () x C x = T T c T T c 4D ) ) Fe Ni Fe Fe Ni (Fe Fe Fe Fe Fe 462 Fe76 Ni36 4E ) ) (Fe) 463 4F ) ) ( ) Fe HeNe 17 Fe Fe Fe HeNe 464 Ni Ni Ni HeNe 465 466 (2) Al PtO 2 (liq) 467 4G ) Al 468 Al ( 468

More information

重力方向に基づくコントローラの向き決定方法

重力方向に基づくコントローラの向き決定方法 ( ) 2/Sep 09 1 ( ) ( ) 3 2 X w, Y w, Z w +X w = +Y w = +Z w = 1 X c, Y c, Z c X c, Y c, Z c X w, Y w, Z w Y c Z c X c 1: X c, Y c, Z c Kentaro Yamaguchi@bandainamcogames.co.jp 1 M M v 0, v 1, v 2 v 0 v

More information

有機性産業廃棄物の連続炭化装置の開発

有機性産業廃棄物の連続炭化装置の開発 ( ) Development of the apparatus conveyer type which carbonizes continuously organic industrial waste (About the form of blade in conveyer) 1055047 1 1-1 1 1-2 1-3 2 2 2-1 2-2 2-3 2-4 7 3 3-1 20 3-2 3-3

More information

2 7 V 7 {fx fx 3 } 8 P 3 {fx fx 3 } 9 V 9 {fx fx f x 2fx } V {fx fx f x 2fx + } V {{a n } {a n } a n+2 a n+ + a n n } 2 V 2 {{a n } {a n } a n+2 a n+

2 7 V 7 {fx fx 3 } 8 P 3 {fx fx 3 } 9 V 9 {fx fx f x 2fx } V {fx fx f x 2fx + } V {{a n } {a n } a n+2 a n+ + a n n } 2 V 2 {{a n } {a n } a n+2 a n+ R 3 R n C n V??,?? k, l K x, y, z K n, i x + y + z x + y + z iv x V, x + x o x V v kx + y kx + ky vi k + lx kx + lx vii klx klx viii x x ii x + y y + x, V iii o K n, x K n, x + o x iv x K n, x + x o x

More information

B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:

B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13: B. 41 II: ;; 4 B [] S 1 S S 1 S.1 O S 1 S 1.13 P 3 P 5 7 P.1:.13: 4 4.14 C d A B x l l d C B 1 l.14: AB A 1 B 0 AB 0 O OP = x P l AP BP AB AP BP 1 (.4)(.5) x l x sin = p l + x x l (.4)(.5) m d A x P O

More information

215 11 13 1 2 1.1....................... 2 1.2.................... 2 1.3..................... 2 1.4...................... 3 1.5............... 3 1.6........................... 4 1.7.................. 4

More information

i

i i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,

More information

genron-3

genron-3 " ( K p( pasals! ( kg / m 3 " ( K! v M V! M / V v V / M! 3 ( kg / m v ( v "! v p v # v v pd v ( J / kg p ( $ 3! % S $ ( pv" 3 ( ( 5 pv" pv R" p R!" R " ( K ( 6 ( 7 " pv pv % p % w ' p% S & $ p% v ( J /

More information

3-2 PET ( : CYRIC ) ( 0 ) (3-1 ) PET PET [min] 11 C 13 N 15 O 18 F 68 Ga [MeV] [mm] [MeV]

3-2 PET ( : CYRIC ) ( 0 ) (3-1 ) PET PET [min] 11 C 13 N 15 O 18 F 68 Ga [MeV] [mm] [MeV] 3 PET 3-1 PET 3-1-1 PET PET 1-1 X CT MRI(Magnetic Resonance Imaging) X CT MRI PET 3-1 PET [1] H1 D2 11 C-doxepin 11 C-raclopride PET H1 D2 3-2 PET 0 0 H1 D2 3-1 PET 3-2 PET ( : CYRIC ) ( 0 ) 3-1-2 (3-1

More information

Microsoft Word - CTCWEB講座(4章照査)0419.doc

Microsoft Word - CTCWEB講座(4章照査)0419.doc 1912 1914 3 58 16 1 58 2 16 3 4 62 61 4 16 1 16 1914 ( 3) 1955 (30) 1961 (36) 1965 (40) 1970 (45) 1983 (58) 1992 ( 4) 1999 (11) 2004 (16) 2 1 2 3 4 5 6 7 8 9 1 10 2 11 12 13 14 15 16 17 18 19 20 21 22

More information

LD

LD 989935 1 1 3 3 4 4 LD 6 7 10 1 3 13 13 16 0 4 5 30 31 33 33 35 35 37 38 5 40 FFT 40 40 4 4 4 44 47 48 49 51 51 5 53 54 55 56 Abstract [1] HDD (LaserDopplerVibrometer; LDV) [] HDD IC 1 4 LDV LDV He-Ne Acousto-optic

More information

dvipsj.4852.dvi

dvipsj.4852.dvi NAOSITE: Nagasaki University's Ac Title Author(s) 3D 計測によるコンクリート打継部及び曲面板の振動解析に関する研究 松田, 浩 ; 和田, 眞禎 ; 小嶋, 悟 ; 崎山, 毅 ; 森田, 千尋掲維 ; 仲村, 政彦 ; 山本, 晃 ; 鶴田, 健 Citation 応用力学論文集 Vol.3, pp.5-4, Issue Date URL http://hdl.handle.net/69/648

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