Robot Platform Project(RPP) "Spur" "YP-Spur" rev. 4 [ ] Robot Platform Project(RPP) WATANABE Atsushi 1.,,., Fig. 1.,,,,,.,,,..,,..,,..,,,,. "
|
|
- しのぶ とどろき
- 5 years ago
- Views:
Transcription
1 Robot Platform Project(RPP) "Spur" "YP-Spur" ev. 4 [.8.9] Robot Platform Project(RPP) WATANABE Atsushi.,,., Fig..,,,,,.,,,..,,..,,..,,,,. "",,, Spur.,, Robot Platform Project, "YP-Spur".,,, 98 99,. [][3][4].,,, [5]., Robot Platform Project * ). * ) Robot Platform Project Fig.,, YP-Spur.3.., Spur,.. "" "",, (Power Wheeled Steering/PWS). Figure " LR-", PWS. PWS,,,,.,,, PWS, ω r (t), ω l (t), R r, R l, ()T,. ( ) ( Rr = R r T R l R l T ) ( ωr (t) ω l (t), (x, y, θ),,,. θ(t) = x(t) = y(t) = t t t ω(τ)dτ + θ() v(τ)cos[θ(τ)]dτ + x() v(τ)sin[θ(τ)]dτ + y() ) () ()
2 (message queue) ms x, y, x, y, Fig. " LR-" PWS,.,,., PWS, 3, ( ),.,, 3,., "",,.,, "Spur". 3. YP-Spur, Spur,. MICHI 98, (),, Spur.,. MC689CPU., " 9 ". Spur CPU,. L-Spur 99, 9. Spur T85CPU, Liberos., "Spur", Spur L-Spur. T85CPU (T-Loco ). vxv_tools 5,,. Spur, SH. SHSpur ef,ef ef,ef encoded ms ms Fig. 3, l 5 ms x, y,, l encoded, l 5 ms, l, l YP-Spur 5 7., SH.,,. MRspur 8 9,,. SHSpur. YP-Spur,, Robot Platform Project, SHSpur,.,,,., URG SCIP. (VV ),,.
3 4. YP-Spur YP-Spur,,. Figure 3, YP-Spur. ypspur-coordinater, Spur.,, ypspurcoordinater,. ypspur-coordinater,,,.,, 5.. sh_vel, ypspur-coordinater,,, PI. PI,, PWS,. PWS, []. [5].,, PWS. 5. YP-Spur YP-Spur 6,,,. Spur_line Spur_circle Spur_spin Spur_stop_line Spur_stop Spur_vel,,.,,,.,,,.,, Fig. 3,,. v re f (t + t),, v max, a max,. v re f (t + t) < v max (3) vre f (t + t) < a max (4) t, t, t., ω re f (t + t), ω max, α max,. 6. ω re f (t + t) < ω max (5) ωre f (t + t) < α max (6) t YP-Spur, t,, η(t), φ(t), ω di f f (t),. η(t) φ(t) ω di f f (t), K η, K φ, K ω, () ( d dt ω)re f (t),. ( d dt ω)re f (t) = K η η(t) K φ φ(t) K ω ω di f f (t) (7), t,, ω re f (t), ω re f (t + t) = + t ( K η η(t) K φ φ(t) K ω ω di f f (t) ) (8), η(t), η max, (9)., φ(t), (). η(t) < η max (9) π < φ(t) < π () v re f (t), v re f (t),. ( ) v re f (t) = v re f (t) sign v re f (t) (),,,
4 ,., (),,.,, (),,..,,,. α g (α =..3, g ),, v max_by_ω (t) = α g,. 6. q(t) d(t) () Desired Line Spur_line(,,.536 ) Fig. 5 : (, ), (, ) Desired Line Spur_line(,, ) - Fig. 4 YP-Spur, YP-Spur, Fig. 4 d(t), q(t),,, (8). η(t) = d(t) φ(t) = q(t) (3) ω di f f (t) =, " LR-",, Fig. 5, 6., " LR-", YP-Spur..,.3m/s,.5rad/s.,., (), θ(t), v re f,, y(t) = t t v(τ)sin[θ(τ)]dτ v re f θ(τ)dτ, x, d dt η(t) vre f φ(t) Fig. 6 : (, ), (, )., d dt φ(t) = ω di f f (t) =.,, ω re f (t), (7),. d 3 dt 3 φ(t) = K η v re f d φ(t) K φ dt φ(t) K ω d φ(t) (4) dt (4),,,,, K η, K φ, K ω,,.,.,
5 ,. (9, ),,,,.,,. (4), x φ (x), Desired path Spur_circle( 3,, ) d 3, K η dt 3 φ (x) = (v re f ) vre f φ (x) K φ d (v re f ) dt φ (x) K ω v re f d dt φ (x) (5) φ (x) = φ( x v re f ) (5),,,,,.,,.,,.,. (5),., [6],,, Fig. 8 : (, ), (, ).3m Desired path Spur_circle(,, ) Fig. 9 : (, ), (, ) m η(t) = d(t) φ(t) = q(t) (6) ω di f f (t) = ω required 6. Fig. 7 r q(t) d(t) YP-Spur, YP-Spur, Fig. 7, d(t), q(t),,, (8). r,., ω max ( < ω max ) r + d(t) ω required = r+d(t) ( r + d(t) ) < ω max ω max ( > ω max ) r + d(t) ω required,,., " LR-",, Fig. 8,9., " LR-", YP-Spur..,.3m/s,.5rad/s.,.,,
6 Fig. r d(t + T) q(t + T) Spur,,,,, K η, K φ, K ω, (5),. Spur, Fig.,, T d(t + T ), q(t + T ), (8). η(t) = d(t + T ) φ(t) = q(t + T ) ω di f f (t) = (7), T,., T, T. 7. YP-Spur,,,.,., PWS 3,,,., Spur, 3,., YP-Spur, θ(t), θ re f, ω re f (t)., α max, ω re f (t). θ(t) + t+t t ω(t +t ) = dt = θ re f d dt = α max,. ω re f (t + t) = sign(θ(t) θ re f ) α max θ(t) θ re f (8), v re f (t).,,. 7. Fig. q(t) l(t) d(t) YP-Spur, YP-Spur, Fig.,,, ω re f (t), v re f (t).,.,,., v re f (t) (9), d(t), q(t),, l(t), () (8).
7 re f v (t + t) = sign( l(t) ) amax l(t) (9) 構築することが可能になっている. 今後も, 利用形態や他の技 術の進歩にあわせて, 走行制御コマンド系および走行制御系 を整備していくことは, 移動ロボットに関する研究を行って { d(t) vre f (t + t) < vmax otherwise () φ (t) = q(t) ωdi f f (t) = ω (t) 一方, 直線上での停止コマンドの実行中, 停止する直線との 距離が近いときには, ロボットの位置は停止する直線の水平 方向には制御されておらず, 外から力が加わればその位置が 変化する場合がある. ただし, 停止する直線との距離が十分長 ければ, ロボットは初め直線追従と同様に動作するので, 指定 した位置の点に向かって進む. この場合, 停止する直線に近づ いた際に外乱が働かなければ, 最終的には指定した位置の近 傍で停止することが期待できる. η (t) = 8. おわりに 本稿では, 移動ロボット走行制御コマンド系および走行制 御系, "Spur"の考え方と, 年度現在の知能ロボット研究 室の標準走行制御系である, "YP-Spur" の実装について報告 した. 本稿で説明した内容の多くは, 研究室の先人たちが残し た多くの論文 ソースコードなどの財産を参考にしたもので ある. 現在の"Spur"は, 計算機技術の発展に伴って, 走行制御系の 大部分をラップトップコンピュータ上で動作させる構造に変 化した. 実際の使用例でも, 軌跡追従制御の代わりに, 画像処 理や測域センサデータ処理で得られた情報に基づいた制御を いく上できわめて重要である. 参考文献 [] S. Yuta, Y. Kanayama. An implementation of MICHI - A locomotion command system for intelligent mobile robot, in Proc. of International Conference on Advanced Robotics, pp. 7-34, 985 [] S. Iida, S. Yuta, Control of Vehicle with Power Wheeled Steerings Using Feed-forward Dynamics Compensation, in Proc. of Annual Conference on the IEEE Industrial Electronics Society, pp , 99 [3] 飯田重喜, 油田信一. 車輪型移動ロボットのための走行制御コマ ンド系と軌跡制御方式, in Proc. of 第 回日本ロボット学会ロ ボットシンポジウム, pp. 85-9, 99 [4] 飯田重喜. 車輪型移動ロボットの走行制御システムに関する研 究, 筑波大学大学院博士課程工学研究科 学位請求論文, 99 [5] 坪内孝司. 車輪移動体の制御, in Proc. of 日本ロボット学会主催 第 43 回講習会 ロボット工学入門シリーズ 移動技術編 移 動ロボットのやさしい解説, pp , 995 [6] 渡辺敦志. 繰り返しシミュレーションとヒューリスティック最 適化による制御系のパラメータ決定手法 広い速度域に対応し た移動ロボットの直線追従パラメータ生成 In Proc. of the 9 年度第 回山彦シンポジウム 訂正内容 初稿 p.4, 6 安定性の議論を修正 p., Fig. 3 中に制御 通信周期を追加 p., Fig. 3 中の制御 通信周期を修正 実装して動作させるなど, 柔軟に移動ロボットのシステムを Fig. 年度プラットフォーム整備のための作業日, "山彦"シリーズと"ビーゴ"シリーズ, および筑波大学 知能ロボット研究室メンバー
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( ), ( ) Patrol Mobile Robot To Greet Passing People Takemi KIMURA(Univ. of Tsukuba), and Akihisa OHYA(Univ. of Tsukuba) Abstract This research aims a
( ), ( ) Patrol Mobile Robot To Greet Passing People Takemi KIMURA(Univ. of Tsukuba), and Akihisa OHYA(Univ. of Tsukuba) Abstract This research aims at the development of a mobile robot to perform greetings
More information85 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 informationSICE東北支部研究集会資料(2012年)
77 (..3) 77- A study on disturbance compensation control of a wheeled inverted pendulum robot during arm manipulation using Extended State Observer Luis Canete Takuma Sato, Kenta Nagano,Luis Canete,Takayuki
More informationUntitled
http://www.ike-dyn.ritsumei.ac.jp/ hyoo/dynamics.html 1 (i) (ii) 2 (i) (ii) (*) [1] 2 1 3 1.1................................ 3 1.2..................................... 3 2 4 2.1....................................
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 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 information1 3 1.1.......................... 3 1............................... 3 1.3....................... 5 1.4.......................... 6 1.5........................ 7 8.1......................... 8..............................
More informationGmech08.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 informationSICE東北支部研究集会資料(2014年)
計測自動制御学会東北支部第 291 回研究集会 (214.1.23) 資料番号 291- スライディングモード制御による除雪ロボットの軌道追従制御 Line following control for the snow removal robot using the sliding mode control 駒米光太郎 *,Ernesto Rivas*, 中村久一郎 *, 水戸部和久 * Koutarou
More information1, 2, 2, 2, 2 Recovery Motion Learning for Single-Armed Mobile Robot in Drive System s Fault Tauku ITO 1, Hitoshi KONO 2, Yusuke TAMURA 2, Atsushi YAM
1, 2, 2, 2, 2 Recovery Motion Learning for Single-Armed Mobile Robot in Drive System s Fault Tauku ITO 1, Hitoshi KONO 2, Yusuke TAMURA 2, Atsushi YAMASHITA 2 and Hajime ASAMA 2 1 Department of Precision
More informationr d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t
1 1 2 2 2r d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t) V (x, t) I(x, t) V in x t 3 4 1 L R 2 C G L 0 R 0
More information画像工学特論
.? (x i, y i )? (x(t), y(t))? (x(t)) (X(ω)) Wiener-Khintchine 35/97 . : x(t) = X(ω)e jωt dω () π X(ω) = x(t)e jωt dt () X(ω) S(ω) = lim (3) ω S(ω)dω X(ω) : F of x : [X] [ = ] [x t] Power spectral density
More information,, 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 informationhttp://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 informationDecember 28, 2018
e-mail : kigami@i.kyoto-u.ac.jp December 28, 28 Contents 2............................. 3.2......................... 7.3..................... 9.4................ 4.5............. 2.6.... 22 2 36 2..........................
More informationGmech08.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 informationII ( ) (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 informationω 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 information1 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( ) ( )
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 informationUntitled
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(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 informationLLG-R8.Nisus.pdf
d M d t = γ M H + α M d M d t M γ [ 1/ ( Oe sec) ] α γ γ = gµ B h g g µ B h / π γ g = γ = 1.76 10 [ 7 1/ ( Oe sec) ] α α = λ γ λ λ λ α γ α α H α = γ H ω ω H α α H K K H K / M 1 1 > 0 α 1 M > 0 γ α γ =
More information(5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b)
(5) 74 Re, bondar laer (Prandtl) Re z ω z = x (5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b) (5) 76 l V x ) 1/ 1 ( 1 1 1 δ δ = x Re x p V x t V l l (1-1) 1/ 1 δ δ δ δ = x Re p V x t V
More informationKENZOU
KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................
More informationmain.dvi
6 FIR FIR FIR FIR 6.1 FIR 6.1.1 H(e jω ) H(e jω )= H(e jω ) e jθ(ω) = H(e jω ) (cos θ(ω)+jsin θ(ω)) (6.1) H(e jω ) θ(ω) θ(ω) = KωT, K > 0 (6.2) 6.1.2 6.1 6.1 FIR 123 6.1 H(e jω 1, ω
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
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 information64 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振動と波動
Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
More informationLCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t)
338 7 7.3 LCR 2.4.3 e ix LC AM 7.3.1 7.3.1.1 m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x k > 0 k 5.3.1.1 x = xt 7.3 339 m 2 x t 2 = k x 2 x t 2 = ω 2 0 x ω0 = k m ω 0 1.4.4.3 2 +α 14.9.3.1 5.3.2.1 2 x
More information2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................
More informationII 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 informationm(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 information2. 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 informationThe 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 informationKorteweg-de Vries
Korteweg-de Vries 2011 03 29 ,.,.,.,, Korteweg-de Vries,. 1 1 3 1.1 K-dV........................ 3 1.2.............................. 4 2 K-dV 5 2.1............................. 5 2.2..............................
More information9 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 informationgrad 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構造と連続体の力学基礎
II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton
More information振動工学に基礎
Ky Words. ω. ω.3 osω snω.4 ω snω ω osω.5 .6 ω osω snω.7 ω ω ( sn( ω φ.7 ( ω os( ω φ.8 ω ( ω sn( ω φ.9 ω anφ / ω ω φ ω T ω T s π T π. ω Hz ω. T π π rad/s π ω π T. T ω φ 6. 6. 4. 4... -... -. -4. -4. -6.
More information211 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 information5 H Boltzmann Einstein Brown 5.1 Onsager [ ] Tr Tr Tr = dγ (5.1) A(p, q) Â 0 = Tr Âe βĥ0 Tr e βĥ0 = dγ e βh 0(p,q) A(p, q) dγ e βh 0(p,q) (5.2) e βĥ0
5 H Boltzmann Einstein Brown 5.1 Onsager [ ] Tr Tr Tr = dγ (5.1) A(p, q) Â = Tr Âe βĥ Tr e βĥ = dγ e βh (p,q) A(p, q) dγ e βh (p,q) (5.2) e βĥ A(p, q) p q Â(t) = Tr Â(t)e βĥ Tr e βĥ = dγ() e βĥ(p(),q())
More information二軸ブラシレスモータドライバ TF-2MD3-R6 User's Manual
TF-2MD3-R6 User s Manual (WATANABE Atsushi) 1 rev.3 - January 14th, 2014-3.0 http://creativecommons.org/licenses/by-sa/3.0/ Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041,
More informationmt_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 informationt 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 informationchap1.dvi
1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f
More information29
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 informationxia2.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 information006 11 8 0 3 1 5 1.1..................... 5 1......................... 6 1.3.................... 6 1.4.................. 8 1.5................... 8 1.6................... 10 1.6.1......................
More information1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h
IB IIA 1 1 r, θ, φ 1 (r, θ, φ)., r, θ, φ 0 r
More informationarctan 1 arctan arctan arctan π = = ( ) π = 4 = π = π = π = =
arctan arctan arctan arctan 2 2000 π = 3 + 8 = 3.25 ( ) 2 8 650 π = 4 = 3.6049 9 550 π = 3 3 30 π = 3.622 264 π = 3.459 3 + 0 7 = 3.4085 < π < 3 + 7 = 3.4286 380 π = 3 + 77 250 = 3.46 5 3.45926 < π < 3.45927
More informationv_-3_+2_1.eps
I 9-9 (3) 9 9, x, x (t)+a(t)x (t)+b(t)x(t) = f(t) (9), a(t), b(t), f(t),,, f(t),, a(t), b(t),,, x (t)+ax (t)+bx(t) = (9),, x (t)+ax (t)+bx(t) = f(t) (93), b(t),, b(t) 9 x (t), x (t), x (t)+a(t)x (t)+b(t)x(t)
More informationφ s i = m j=1 f x j ξ j s i (1)? φ i = φ s i f j = f x j x ji = ξ j s i (1) φ 1 φ 2. φ n = m j=1 f jx j1 m j=1 f jx j2. m
2009 10 6 23 7.5 7.5.1 7.2.5 φ s i m j1 x j ξ j s i (1)? φ i φ s i f j x j x ji ξ j s i (1) φ 1 φ 2. φ n m j1 f jx j1 m j1 f jx j2. m j1 f jx jn x 11 x 21 x m1 x 12 x 22 x m2...... m j1 x j1f j m j1 x
More information128 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 information25 22 1 2 2 4 2.1........................ 5 2.2.................. 7 3 9 3.1................................. 9 3.2............................. 10 3.3................. 10 3.4....................... 12
More informationI 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 informatione 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 informationx E E E e i ω = t + ikx 0 k λ λ 2π k 2π/λ k ω/v v n v c/n k = nω c c ω/2π λ k 2πn/λ 2π/(λ/n) κ n n κ N n iκ k = Nω c iωt + inωx c iωt + i( n+ iκ ) ωx
x E E E e i ω t + ikx k λ λ π k π/λ k ω/v v n v c/n k nω c c ω/π λ k πn/λ π/(λ/n) κ n n κ N n iκ k Nω c iωt + inωx c iωt + i( n+ iκ ) ωx c κω x c iω ( t nx c) E E e E e E e e κ e ωκx/c e iω(t nx/c) I I
More informationA
A04-164 2008 2 13 1 4 1.1.......................................... 4 1.2..................................... 4 1.3..................................... 4 1.4..................................... 5 2
More information知能科学:ニューラルネットワーク
2 3 4 (Neural Network) (Deep Learning) (Deep Learning) ( x x = ax + b x x x ? x x x w σ b = σ(wx + b) x w b w b .2.8.6 σ(x) = + e x.4.2 -.2 - -5 5 x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x,
More information知能科学:ニューラルネットワーク
2 3 4 (Neural Network) (Deep Learning) (Deep Learning) ( x x = ax + b x x x ? x x x w σ b = σ(wx + b) x w b w b .2.8.6 σ(x) = + e x.4.2 -.2 - -5 5 x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x,
More information6 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 information1 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 informationẍ = 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 informationA = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B
9 7 A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B x x B } B C y C y + x B y C x C C x C y B = A
More information(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1)
1. 1.1...,. 1.1.1 V, V x, y, x y x + y x + y V,, V x α, αx αx V,, (i) (viii) : x, y, z V, α, β C, (i) x + y = y + x. (ii) (x + y) + z = x + (y + z). 1 (iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y
More information1 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 informationA (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 information08-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 information1 (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 information4‐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 informationdynamics-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: ( ) : ( ) :! ( ) / 32
2015 9 30 ( ) 2015 9 30 1 / 32 : ( ) : ( ) :! ( ) 2015 9 30 2 / 32 http://nalab.mind.meiji.ac.jp/koudai2015/ ( ) 2015 9 30 3 / 32 ( ) ( ) ( : ) 1 2 (2014) 1 M2 2 ( ) 2015 9 30 4 / 32 ( ) ( ) ( : ) 1 2
More information( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )
n n (n) (n) (n) (n) n n ( n) n n n n n en1, en ( n) nen1 + nen nen1, nen ( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( ) ( n) Τ n n n ( n) n + n ( n) (n) n + n n n n n n n n
More informationI 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 informationS: E: O: C: V : 5
( ) 2004 1 S: E: O: C: V : 5 1 1 2 2 2.1.................................... 2 2.2........................ 2 2.3........................... 3 3 7 3.1.................................... 7 3.2....................................
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 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 informationMicrosoft PowerPoint - ロボットの運動学forUpload'C5Q [互換モード]
ロボットの運動学 順運動学とは 座標系の回転と並進 同次座標変換行列 Denavit-Hartenberg の表記法 多関節ロボットの順運動学 レポート課題 & 中間試験について 逆運動学とは ヤコビアン行列 運動方程式 ( 微分方程式 ) ロボットの運動学 動力学 Equation of motion f ( ( t), ( t), ( t)) τ( t) 姿勢 ( 関節角の組合せ ) Posture
More informationNote.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核融合…予稿集
9:30 9:45 9:45 10:00 10:05 10:10 10:10 10:40 2 3 10:40 11:10 11:10 11:40 11:40 12:00 6 7 8 9 10 11 13:10 13:20 13:20 14:00 14:00 14:20 14:20 14:40 14:50 15:20 15:20 15:50 15:50 15:55 14 15 16 17 18 19
More information( ) 2002 1 1 1 1.1....................................... 1 1.1.1................................. 1 1.1.2................................. 1 1.1.3................... 3 1.1.4......................................
More informationsikepuri.dvi
2009 2 2 2. 2.. F(s) G(s) H(s) G(s) F(s) H(s) F(s),G(s) H(s) : V (s) Z(s)I(s) I(s) Y (s)v (s) Z(s): Y (s): 2: ( ( V V 2 I I 2 ) ( ) ( Z Z 2 Z 2 Z 22 ) ( ) ( Y Y 2 Y 2 Y 22 ( ) ( ) Z Z 2 Y Y 2 : : Z 2 Z
More informationSICE東北支部研究集会資料(2012年)
77 (..3) 77- Simulation of Disturbance Compensation Control of Dual Manipulator for an Inverted Pendulum Robot Using The Extended State Observer Luis Canete Kenta Nagano, Takuma Sato, Luis Canete,Takayuki
More information1 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 information1. (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 informationC : q i (t) C : q i (t) q i (t) q i(t) q i(t) q i (t)+δq i (t) (2) δq i (t) δq i (t) C, C δq i (t 0 )0, δq i (t 1 ) 0 (3) δs S[C ] S[C] t1 t 0 t1 t 0
1 2003 4 24 ( ) 1 1.1 q i (i 1,,N) N [ ] t t 0 q i (t 0 )q 0 i t 1 q i (t 1 )q 1 i t 0 t t 1 t t 0 q 0 i t 1 q 1 i S[q(t)] t1 t 0 L(q(t), q(t),t)dt (1) S[q(t)] L(q(t), q(t),t) q 1.,q N q 1,, q N t C :
More information, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,
6,,3,4,, 3 4 8 6 6................................. 6.................................. , 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p,
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)
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 informationf(x,y) (x,y) x (x,y), y (x,y) f(x,y) x y f x (x,y),f y (x,y) B p.1/14
B p.1/14 f(x,y) (x,y) x (x,y), y (x,y) f(x,y) x y f x (x,y),f y (x,y) B p.1/14 f(x,y) (x,y) x (x,y), y (x,y) f(x,y) x y f x (x,y),f y (x,y) f(x 1,...,x n ) (x 1 x 0,...,x n 0), (x 1,...,x n ) i x i f xi
More informationCG38.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 informationPowered by TCPDF ( Title 第 11 講 : フィッシャー統計学 II Sub Title Author 石川, 史郎 (Ishikawa, Shiro) Publisher Publication year 2018 Jtitle コペンハーゲン解
Powered by TCPDF (www.tcpdf.org) Title 第 11 講 : フィッシャー統計学 II Sub Title Author 石川, 史郎 (Ishikawa, Shiro) Publisher Publication year 018 Jtitle コペンハーゲン解釈 ; 量子哲学 (018. 3),p.381-390 Abstract Notes 慶應義塾大学理工学部大学院講義ノート
More informationΓ Ec Γ V BIAS THBV3_0401JA THBV3_0402JAa THBV3_0402JAb 1000 800 600 400 50 % 25 % 200 100 80 60 40 20 10 8 6 4 10 % 2.5 % 0.5 % 0.25 % 2 1.0 0.8 0.6 0.4 0.2 0.1 200 300 400 500 600 700 800 1000 1200 14001600
More information1 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高知工科大学電子 光システム工学科
卒業研究報告 題 目 量子力学に基づいた水素分子の分子軌道法的取り扱いと Hamiltonian 近似法 指導教員 山本哲也 報告者 山中昭徳 平成 14 年 月 5 日 高知工科大学電子 光システム工学科. 3. 4.1 4. 4.3 4.5 6.6 8.7 10.8 11.9 1.10 1 3. 13 3.113 3. 13 3.3 13 3.4 14 3.5 15 3.6 15 3.7 17
More informationB 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 informationreport.dvi
998 Technical Report Driftless [9] driftless ( ) driftless, P θ η P dη/ R l, R r W ω l, ω r P θ P dη/ 5-855 -- d/, d/ d = dη d cos θ, = dη sin θ () d = d tan θ () u, u dη/ dθ/ u = dη u = dθ = R lω l +
More informationlim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d
lim 5. 0 A B 5-5- A B lim 0 A B A 5. 5- 0 5-5- 0 0 lim lim 0 0 0 lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d 0 0 5- 5-3 0 5-3 5-3b 5-3c lim lim d 0 0 5-3b 5-3c lim lim lim d 0 0 0 3 3 3 3 3 3
More information