Henon Research

File Name: Real to Fraction File Submitter: Henon Research File Submitted: 21 Feb 2012 File Updated: 22 Feb 2012 File Category: PLC Sample Code Add-On Instruction Float to Fraction FTF this Add-On convert a real to Fraction. Add Documentation inside the file. Good work. Click here to download this file

File Name: RND LINEAR CONGRUENTIAL GENERATOR Casual Numbers File Submitter: Henon Research File Submitted: 11 Feb 2012 File Updated: 12 Feb 2012 File Category: PLC Sample Code A Linear Congruential Generator (LCG) Represents one of the oldest and best-known pseudorandom number generator algorithms. Click here to download this file

File Name: Complex numbers for Allen-Bradley File Submitter: Henon Research File Submitted: 08 Feb 2012 File Updated: 09 Feb 2012 File Category: PLC Sample Code to apply the mathematics of complex numbers,<BR closure_uid_n926ba="1383" Pc="null">also on plc.<BR closure_uid_n926ba="1384" Pc="null">I send you some Add-On Instructions for dealing with complex numbers this for RsLogix5000 and Siemens S7 Click here to download this file

File Name: Solve system whit Gauss-Jordan File Submitter: Henon Research File Submitted: 06 Jun 2011 File Updated: 07 Jun 2011 File Category: PLC Sample Code In the matrix A [ i , j ] put the System of N-Equation. In the vector b put the solutions §§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§ Example 1: Linear System 3 equation (X,Y,Z) 3 X + 2 Y - Z = 10 - X + Y + Z = -2 2 X - Y + 2 Z = -6 | 3 | | 2 | | -1 | | 10 | X | -1 | + Y | 1 | + Z | 1 | = | -2 | | 2 | | -1 | | 2 | | -6 | Matrix A := Matrix[1,1]= 3 ; Matrix[1,2]= 2 ; Matrix[1,3]= -1 Matrix[2,1]= -1 ; Matrix[2,2]= 1 ; Matrix[2,3]= -2 Matrix[3,1]= 2 ; Matrix[3,2]= -1 ; Matrix[3,3]= -6 Vector b:= Vector[1] =10 ; Vector[2] = -2 ; Vector[3] = -6 ; Solution := Solution [1] := 1.0 ; Solution [2] := 2.0 ; Solution [3] := -3.0 ; X = 1 ; Y = 2 ; Z = -3 §§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§ Example 2: Linear System 5 equation for resolve Polynomial 4th grade exampl. Polynomial whit 5 points: P0(-1,-1) ; P1( 1, 3) ; P2( 5, 3.5) ; P3( 6, 4.5) ; P4( 7, 7) ; Write in the Matrix A [ i, j ] Matrix A := Matrix[1,1]= (-1)^4 ; Matrix[1,2]= (-1)^3 ; Matrix[1,3]= (-1)^2 ; Matrix[1,4]= (-1) ; Matrix[1,5]=1; Matrix[2,1]= (1)^4 ; Matrix[2,2]= (1)^3 ; Matrix[2,3]= (1)^2 ; Matrix[2,4]= (1) ; Matrix[2,5]=1; Matrix[3,1]= (5)^4 ; Matrix[3,2]= (5)^3 ; Matrix[3,3]= (5)^2 ; Matrix[3,4]= (5) ; Matrix[3,5]=1; Matrix[4,1]= (6)^4 ; Matrix[4,2]= (6)^3 ; Matrix[4,3]= (6)^2 ; Matrix[4,4]= (6) ; Matrix[4,5]=1; Matrix[5,1]= (7)^4 ; Matrix[5,2]= (7)^3 ; Matrix[5,3]= (7)^2 ; Matrix[5,4]= (7) ; Matrix[5,5]=1; Write in the Vector [ ] Vector b:= Vector[1] = -1 ; Vector[2] = 3 ; Vector[3] = 3.5 ; Vector[4] = 4.5 ; Vector[5] = 7 Solutions := Solution [1] := 3.27380234e-003 ; Solution [2] := 0.03363105; Solution [3] := -0.56577414 ; Solution [4] := 1.966369 ; Solution [5] := 1.5625005 §§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§ Example 3: Linear System 6 equation for resolve Polynomial 5th grade example. Mototion Interpolation whit Polynomial whit 2 Points : P0 (Time0, Position 0) Start point whit (Velocity 0, Acceleration 0) P1 (Time1, Position 1) End point whit (Velocity 1, Acceleration 1) Write in the Matrix A [ i, j ] Matrix A := X0 = time0 ; X1 = time1 Row1 X0 ^5 + X0 ^4 + X0 ^3 + X0 ^2 + X0 + 1 (Position P0) Row2 5 * X0 ^4 + 4 * X0 ^3 + 3 * X0 ^2 + 2 * X0 + 1 + 0 (Velocity P0) Row3 20 * X0 ^3 + 12 * X0 ^2 + 6 * X0 + 2 + 0 + 0 (Acceleration P0) Row4 X1 ^5 + X1 ^4 + X1 ^3 + X1 ^2 + X1 + 1 (Position P1) Row5 5 * X1 ^4 + 4 * X1 ^3 + 3 * X1 ^2 + 2 * X1 + 1 + 0 (Velocity P1) Row6 20 * X1 ^3 + 12 * X1 ^2 + 6 * X1 + 2 + 0 + 0 (Acceleration P1) Vector b:= Vector[1] = Position P0 ; Vector[2] = Velocity P0 ; Vector[3] = Acceleration P0 ; Vector[4] = Position P1 ; Vector[5] = Velocity P1 ; Vector[6] = Acceleration P1 ; Interpolation Polynomial Position := Position := s1* t^5 + s2* t^4 +s3* t^3 + s4* t^2 + s5* t + s6; Click here to download this file

File Name: Solve system whit Gauss-Jordan File Submitter: Henon Research File Submitted: 06 Jun 2011 File Category: PLC Sample Code This Add-On Instruction Solve the equations System whit Gauss-Jordan Reduction In the matrix A [ i , j ] put the System of N-Equation. In the vector b put the solutions §§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§ Example 1: Linear System 3 equation (X,Y,Z) 3 X + 2 Y - Z = 10 - X + Y + Z = -2 2 X - Y + 2 Z = -6 | 3 | | 2 | | -1 | | 10 | X | -1 | + Y | 1 | + Z | 1 | = | -2 | | 2 | | -1 | | 2 | | -6 | Matrix A := Matrix[1,1]= 3 ; Matrix[1,2]= 2 ; Matrix[1,3]= -1 Matrix[2,1]= -1 ; Matrix[2,2]= 1 ; Matrix[2,3]= -2 Matrix[3,1]= 2 ; Matrix[3,2]= -1 ; Matrix[3,3]= -6 Vector b:= Vector[1] =10 ; Vector[2] = -2 ; Vector[3] = -6 ; Solution := Solution [1] := 1.0 ; Solution [2] := 2.0 ; Solution [3] := -3.0 ; X = 1 ; Y = 2 ; Z = -3 §§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§ Example 2: Linear System 5 equation for resolve Polynomial 4th grade exampl. Polynomial whit 5 points: P0(-1,-1) ; P1( 1, 3) ; P2( 5, 3.5) ; P3( 6, 4.5) ; P4( 7, 7) ; Write in the Matrix A [ i, j ] Matrix A := Matrix[1,1]= (-1)^4 ; Matrix[1,2]= (-1)^3 ; Matrix[1,3]= (-1)^2 ; Matrix[1,4]= (-1) ; Matrix[1,5]=1; Matrix[2,1]= (1)^4 ; Matrix[2,2]= (1)^3 ; Matrix[2,3]= (1)^2 ; Matrix[2,4]= (1) ; Matrix[2,5]=1; Matrix[3,1]= (5)^4 ; Matrix[3,2]= (5)^3 ; Matrix[3,3]= (5)^2 ; Matrix[3,4]= (5) ; Matrix[3,5]=1; Matrix[4,1]= (6)^4 ; Matrix[4,2]= (6)^3 ; Matrix[4,3]= (6)^2 ; Matrix[4,4]= (6) ; Matrix[4,5]=1; Matrix[5,1]= (7)^4 ; Matrix[5,2]= (7)^3 ; Matrix[5,3]= (7)^2 ; Matrix[5,4]= (7) ; Matrix[5,5]=1; Write in the Vector [ ] Vector b:= Vector[1] = -1 ; Vector[2] = 3 ; Vector[3] = 3.5 ; Vector[4] = 4.5 ; Vector[5] = 7 Solutions := Solution [1] := 3.27380234e-003 ; Solution [2] := 0.03363105; Solution [3] := -0.56577414 ; Solution [4] := 1.966369 ; Solution [5] := 1.5625005 §§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§ Example 3: Linear System 6 equation for resolve Polynomial 5th grade example. Mototion Interpolation whit Polynomial whit 2 Points : P0 (Time0, Position 0) Start point whit (Velocity 0, Acceleration 0) P1 (Time1, Position 1) End point whit (Velocity 1, Acceleration 1) Write in the Matrix A [ i, j ] Matrix A := X0 = time0 ; X1 = time1 Row1 X0 ^5 + X0 ^4 + X0 ^3 + X0 ^2 + X0 + 1 (Position P0) Row2 5 * X0 ^4 + 4 * X0 ^3 + 3 * X0 ^2 + 2 * X0 + 1 + 0 (Velocity P0) Row3 20 * X0 ^3 + 12 * X0 ^2 + 6 * X0 + 2 + 0 + 0 (Acceleration P0) Row4 X1 ^5 + X1 ^4 + X1 ^3 + X1 ^2 + X1 + 1 (Position P1) Row5 5 * X1 ^4 + 4 * X1 ^3 + 3 * X1 ^2 + 2 * X1 + 1 + 0 (Velocity P1) Row6 20 * X1 ^3 + 12 * X1 ^2 + 6 * X1 + 2 + 0 + 0 (Acceleration P1) Vector b:= Vector[1] = Position P0 ; Vector[2] = Velocity P0 ; Vector[3] = Acceleration P0 ; Vector[4] = Position P1 ; Vector[5] = Velocity P1 ; Vector[6] = Acceleration P1 ; Interpolation Polynomial Position := Position := s1* t^5 + s2* t^4 +s3* t^3 + s4* t^2 + s5* t + s6; Click here to download this file

Hi guys This my opinion: Before you create the master virtual axle, and after lockup your two axis whit gearing instruction “MAG” . Master command Reference to Virtual Axle From Virtual Axle to Real Master (Axle1). From Virtual Axle to Slave (Axle2) Virtual Master = Axle 1 MAG(Slave1,Master,Gear_Ctrl,0,1.0,1,1,Command,Real,Disabled,100,% of Maximum) Virtual Master = Axle 2 MAG(Slave2,Master,Gear_Ctrl,0,1.0,1,1,Command,Real,Disabled,100,% of Maximum) For Axle 1 put on Properties-Axis / Limit set “Position Error Tolerance” to correct value detect from auto-tune For Axle 2 put on Properties-Axis / Limit set “Position Error Tolerance” to high value (Limit Axis1 multiply for 5 ) In this way the Axis 2 (Slave) isn’t possible to go “Error-Position” . After you check the actual Error-Position of Axle2, and compare this error to Trigger-Tolerance position. When the absolute error-position is over the your Trigger-Tolerance position, you must actualized correction whit Motion Axis Move instruction “MAM” over the Gearing instruction MAG+MAM In this way you can possible Repartition Torque reference to slave Axle.

Hi Gentlemen, ask if you realized Connection among PanelView Plus with a PLC Siemens S7 in Ethernet, the KepServer using Ethernet to the PanelView. I premise that Making a will the PC Project in Simulation with the KepServer Enterprise V4.0 works. However then when I transfer the lines compiled to the PanelView, it doesn't work me. The Firmware Installed on the PanelView is the Version V4.06 (I have Installed the Options S7-TCP/IP for the KepServer) and I have adjourned the to Sign. I have Already tried the same configuration in MPI (+Adapter RS232/MPI) with success however the Ethernet doesn't work. Can you help me? Thanks.

Yes you can possible dowload to PanelView only Compile File *.mer. maybe you can also upload File Compiled *.mer. Whit RsLinx or directly to CompactFlash Memory.

There is some people that can help me to find the documentation whit Conecting Drive VZ3000 Reliance Japan and her PSC 5000 of the Reliance Japan. Would interface be been able in net with card 1756-DCM the VZ3000? Thanks to all of you!

To use RsView Studio - Machine Edition V4.00.00 (CPR7). Remembered that you can transfer (UpLoad /UnLoad) only files Compiled and not the Sources (File.mer) The Files Source is saved as File Backup (File.apa)

Sorry the correct Device is ControlLogix Module 1756-DM30 'Drive Module Interfacce' Interfacce whit UDC-Net. For me the most important to find the documentation on the interfacce between VZ3000G and the card PSC5000, (Reliance Japan) Thanks. Char Description '&%ççCapio !! sucoloçç° de on sucoloçç° '

Yes it is possible in Siemens SCL? You made the struct in a DB. The struct contains your data this is your DB example DB10 Before you create UDT example 10 (Create new Source.File) Source1.File // Make Your UDT10 Type UDT10 Structur Limit : REAL; // Limit of your Device Max : INT; // Max Value Time : S5Time; // data Time for your use End_Structur // Make Your UDT11 Type UDT11 Structur Device: INT; // Number of Device Param: Array[1..50] of UDT10; // Structur of your Parameters End_Structur // Make Your DB10 Data_Block DB10 UDT11 Begin Source2.File FUNCTION_BLOCK FCB10 Title = 'Test' // Test for this Example // // Version : '1.0' Author : 'HenonR' Name : Test Family : Examp VAR_INPUT Point1: INT; // Pointer 1st Element Point2: INT; // Pointer 2nd Element Block: UDT11; // data Block (Inser DB10) Param: INT; // Number of Parameter Device: INT; // Number of Device Limit: REAL; // Limit Value END_VAR VAR_OUTPUT MAX: REAL; // Max value END_VAR BEGIN MAX := Block.Param[Point1].Max; Block.Device := Device; Block.Param[10].Limit := Limit; END_FUNCTION_BLOCK This is a Example for Insert Input (Block) you use an other FC or FB in SCL, because in KOP,FUP,AWL is very Hard. FUNCTION FC11: REAL BEGIN FB10,DB20 ( Point1:= 1 ,Point2:= 2 ,Block:= DB10 ,Param:= 3 ,Device:= 1 ,Limit := 0.768 , MD100 := MAX); FC11 := MAX; // RETURN Value of Function FC11 OK := True; END_FUNCTION