... equations in (26) into (25) and after operating and simplifying we have: ( ) ( ) 0 0 1 1 1 1 1 2 1 3 1 0 ( ) ( ) 0 2 2 2 22 3 2 1 2 0 0 ( ) ( ) 2 3 3 3 1 3 3 3 3 m m m P s s x y z x m m m P ... 2 1 3 1 12 2 2 32 1 3 23 3 3 s s s s s s When 0 1 2i i or 1,2,3i or when 0 3i or 1,2,3i (30 ) a) b)...
Ngày tải lên: 21/06/2014, 11:20
... using Eq. (54) and the state feedback control law 1 , B B T EF x v b (55) into a linear system 3 3 3 3 3 3 3 3 3 3 3 3 x x x x x x 0 ... using Eq. (54) and the state feedback control law 1 , B B T EF x v b (55) into a linear system 3 3 3 3 3 3 3 3...
Ngày tải lên: 21/06/2014, 11:20
Mechatronic Systems, Simulation, Modeling and Control 2012 Part 11 docx
... constitutive relations for the strain S 11 and electric displacement D 3 , as functions stress T 11 and electric field E 3 , take the form: 11 11 31 11 3 31 33 3 S s d T D d ε E ... d and the stress-free electric Mechatronic Systems, Simulation,Modelling and Control2 90 permittivity matrix ε are each represented by a single coefficient, d 31 an...
Ngày tải lên: 21/06/2014, 11:20
Tribology Lubricants and Lubrication 2012 Part 3 docx
... of Figures 11 and 12 in section 3. 3, reaches – 230 MPa. Fig. 32 . SEM-SE image of (a) the damaged raceway of the inner ring of a CRB after rig testing under engine vibrations and (b) an original ... mechanical rolling contact fatigue and shakedown cold working at constant reference level of b/B≈0.71 in Figures 18, 23 and 27 to 31 , completed by Figures 20, 21 and 24, s...
Ngày tải lên: 19/06/2014, 15:20
Mechatronic Systems, Simulation, Modeling and Control 2012 Part 2 pdf
... max1 max1 1 f f N f kf N k kfM s s (33 ) Also, we have: . 1 1| max1 1 max11 f kkfM ff (34 ) Considering Equations (33 ) and (34 ), the following equation is resulted: max1 NM f f s (35 ) The value ... supporter Tip Mechatronic Systems, Simulation,Modelling and Control1 12 frequency was 31 . 93 kHz, admittance phase coincided with 0 at the resonance frequency...
Ngày tải lên: 21/06/2014, 11:20
Mechatronic Systems, Simulation, Modeling and Control 2012 Part 4 pptx
... x p3 x p4 + p 10 p 4 cos x p3 ( ¨ e 1 − r 1 (x) ) p 5 cos x p3 + p 6 sin x p3 + p 7 x p4 + p 10 p 4 cos x p3 x p4 − p 10 p 4 ( p 3 − p 9 ) tan x p3 e (3) 1 + p 10 p 4 tan x p3 { ( p 3 − ... x p3 x p4 + p 10 p 4 cos x p3 ( ¨ e 1 − r 1 (x) ) p 5 cos x p3 + p 6 sin x p3 + p 7 x p4 + p 10 p 4 cos x p3 x p4 − p 10 p 4 ( p 3 − p 9 ) tan x p3 e (3) 1 + p 10 p 4 tan x p3 { (...
Ngày tải lên: 21/06/2014, 11:20
Mechatronic Systems, Simulation, Modeling and Control 2012 Part 6 pdf
... which makes controller’s tuning simpler. The dynamic part of the control law from (26) has the following form: 3 2 1 3 2 ,2 ,1 ,0 3 2 1 3 2 2 0 3 3 3 3 d d d k ... differential equations: 3 (3) 2 (2) 2 (1) 0 3 3 (34 ) 3 (3) 2 (2) 2 (1) 0 3 3 (35...
Ngày tải lên: 21/06/2014, 11:20
Mechatronic Systems, Simulation, Modeling and Control 2012 Part 8 potx
... dependent parts and independent parts. And, this paper proposes that separates platform dependent parts of the RT control framework. 2 .3. 1 Separation of platform dependent parts of RT control ... platform, and individually creation of a simulation program and an RT control program is needed, this paper proposes the Mechatronic Systems, Simulation,Modelling and Control2...
Ngày tải lên: 21/06/2014, 11:20
Mechatronic Systems, Simulation, Modeling and Control 2012 Part 12 potx
... on Robotics and Automation, pp. 32 13- 3218, Taipei, Taiwan, September 20 03 Lim K. B., Gawronski W. (19 93) , Actuators and sensor placement for control of exible structures, Control and Dynamics ... on Robotics and Automation, pp. 32 13- 3218, Taipei, Taiwan, September 20 03 Lim K. B., Gawronski W. (19 93) , Actuators and sensor placement for control of exible struct...
Ngày tải lên: 21/06/2014, 11:20
Tài liệu File and Registry Operations part 3 docx
... { InitializeComponent(); menuFileOpen.Click += new EventHandler(OnFileOpen); chooseOpenFileDialog.FileOk += new CancelEventHandler(OnOpenFileDialogOK); } void OnFileOpen(object Sender, ... FileAccess.Read); long nBytesToRead = inStream.Length; if (nBytesToRead > 65 536 /4) nBytesToRead = 65 536 /4; int nLines = (int)(nBytesToRead/nCols) + 1; string [] lines = new stri...
Ngày tải lên: 24/12/2013, 11:15