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zhannawk [14.2K]
3 years ago
10

What advantage might there be to having the encoder located on the motor side of the gearhead instead of at the output shaft of

the gearhead (in other words so that it would rotate at the speed of the armature rather than at the speed of the output shaft of the gearhead)?
Engineering
1 answer:
pickupchik [31]3 years ago
7 0

The most accurate answer to that process is definitely precision. The Rotary encoder is an electro-mechanical device that converts the angular position or motion of a shaft or axle to analog or digital output signals. The efficiency of these devices is subject to the position and angle of the axis in front of the encoder.

Most cars use reduction systems in their gearboxes that convert a certain signal input into an output. Mechanically for example, a 20: 1 reduction box already infers that if there is a revolution in the input at the output there are 20. That same transferred to the encoder pulses would imply greater precision.

For example a decoder with 50 holes would have to read 1000 pulses (50 * 20) which is basically a degree of accuracy of 0.36 degrees. In this way it is possible to conclude that if the assembly of the encoder is carried out next to the motor and not at the output, it can be provided with greater precision at the time of reading.

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A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?​
dexar [7]

Answer:

8.85 Ω

Explanation:

Resistance of a wire is:

R = ρL/A

where ρ is resistivity of the material,

L is the length of the wire,

and A is the cross sectional area.

For a round wire, A = πr² = ¼πd².

For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.

Given L = 500 ft and d = 0.03 in = 0.0025 ft:

R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)

R = 8.85 Ω

5 0
3 years ago
If the channel-Length modulation effect is neglected, ID in the saturation region is considered to be independent of VDS
djverab [1.8K]
The answer is true because if the effect is neglected, the saturation id region is considered true
4 0
3 years ago
This assignment is designed to test your understanding of graphical user interface components and Event Driven programming in Ja
lys-0071 [83]

Answer:

Java program is given below

Explanation:

JavaPadGUI.java

package javapad.gui;

import java.awt.BorderLayout;

import java.awt.Container;

import java.awt.Dimension;

import java.awt.event.ActionEvent;

import java.awt.event.ActionListener;

import java.awt.event.WindowAdapter;

import java.awt.event.WindowEvent;

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

import javax.swing.JButton;

import javax.swing.JFrame;

import javax.swing.JLabel;

import javax.swing.JOptionPane;

import javax.swing.JPanel;

import javax.swing.JScrollPane;

import javax.swing.JTextArea;

public class JavaPadGUI extends JFrame implements ActionListener{

  //constants

  private static final String TITLE = "Macrosoft JavaPad XP";

  private static final String SLOGAN = "Macrosoft: Resistance is futile";

 

  //button names

  private static final String NEW = "New";

  private static final String LOAD = "Load";

  private static final String SAVE = "Save";

  private static final String QUIT = "Quit";

 

  private static final String FILENAME = "hardcode.txt";

  //messages

 

  private static final String QUIT_MSG = "Quitting, Save?";

  private static final String LOAD_ERR = "Could not access file " + FILENAME;

 

 

  //fields

  private JTextArea txtContent;

  private JButton butNew, butLoad, butSave, butQuit;

 

  public JavaPadGUI()

  {

      super(TITLE);

      createUI();

      setSize(new Dimension(400, 300));

      setDefaultCloseOperation(EXIT_ON_CLOSE);

         

  }

 

  private void createUI()

  {

      Container c = getContentPane();

      c.setLayout(new BorderLayout(30,30));

     

      //create the buttons panel

      JPanel butPanel = new JPanel();

      butPanel.add(butNew = new JButton(NEW));

      butPanel.add(butLoad = new JButton(LOAD));

      butPanel.add(butSave = new JButton(SAVE));

      butPanel.add(butQuit = new JButton(QUIT));

     

      //set command names for the buttons, these will be when button is clicked

      butNew.setActionCommand(NEW);

      butLoad.setActionCommand(LOAD);

      butSave.setActionCommand(SAVE);

      butQuit.setActionCommand(QUIT);

     

      JPanel sloganPanel = new JPanel();

      sloganPanel.add(new JLabel(SLOGAN));

     

      //create textarea with scrollbar

      txtContent = new JTextArea(15, 25);

      txtContent.setLineWrap(true);

         

      JScrollPane scroll = new JScrollPane(txtContent);

     

      //now add all components

      c.add(butPanel, BorderLayout.NORTH);

      c.add(scroll, BorderLayout.CENTER);

      c.add(sloganPanel, BorderLayout.SOUTH);

     

     

      //set the actionlistner to buttons to handle button click event

      butNew.addActionListener(this);

      butLoad.addActionListener(this);

      butSave.addActionListener(this);

      butQuit.addActionListener(this);

     

  }

  public void actionPerformed(ActionEvent e)

  {

      String cmd = e.getActionCommand();

      if(cmd.equals(NEW))

          txtContent.setText("");

      else if (cmd.equals(LOAD))

          load();

      else if(cmd.equals(SAVE))

          save();

      else if (cmd.equals(QUIT))

          quit();

  }

  private void quit()

  {

      int option = JOptionPane.showConfirmDialog(this, QUIT_MSG);

      if(option == JOptionPane.YES_OPTION)

      {

          save();

      }

     

      System.exit(0);

  }

 

  private void save()

  {

      try {

          PrintWriter w = new PrintWriter(new File(FILENAME));

          w.write(txtContent.getText());

          w.close();

      } catch (FileNotFoundException e) {

          JOptionPane.showMessageDialog(this,"I/O Error", LOAD_ERR, JOptionPane.ERROR_MESSAGE);

      }

             

  }

  private void load()

  {

      try {

          Scanner s = new Scanner(new File(FILENAME));

          String str = "";

          while(s.hasNextLine())

              str += s.nextLine();

          txtContent.setText(str);

          s.close();

      } catch (FileNotFoundException e) {

          JOptionPane.showMessageDialog(this, LOAD_ERR, "I/O Error",JOptionPane.ERROR_MESSAGE);

      }

     

  }

}

RunJavaPad.java

package javapad;

import javapad.gui.JavaPadGUI;

public class RunJavaPad {

  public static void main(String[] args) {

      JavaPadGUI gui = new JavaPadGUI();

      gui.setVisible(true);

  }

}

7 0
3 years ago
. Two rods, with masses MA and MB having a coefficient of restitution, e, move
GarryVolchara [31]

Answer:

a) V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}

b) U_A = 3.66 m/s

V_B = 4.32 m/s

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let U_A be the initial velocity of rod A

Let U_B be the initial velocity of rod B

Let V_A be the final velocity of rod A

Let V_B be the final velocity of rod B

Using the principle of conservation of momentum:

M_AU_A + M_BU_B = M_AV_A + M_BV_B............(1)

Coefficient of restitution, e = \frac{V_B - V_A}{U_A - U_B}

V_A = V_B - e(U_A - U_B)........................(2)

Substitute equation (2) into equation (1)

M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B..............(3)

Solving for V_B in equation (3) above:

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}....................(4)

From equation (2):

V_B = V_A + e(U_A -U_B)......(5)

Substitute equation (5) into (1)

M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))..........(6)

Solving for V_A in equation (6) above:

V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}.........(7)

b)

M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?

Rod A is said to be at rest after the impact, V_A = 0 m/s

Substitute these parameters into equation (7)

0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s

To calculate the final velocity, V_B, substitute the given parameters into (4):

V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s

c) Impulse, I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)

I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))

I = 0 kg m/s^2

d) %\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%

%\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%

% \triangle KE = -47.85 \%

7 0
3 years ago
For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha
Anton [14]

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

\sigma=\frac{F}{A}

F=\sigma*A

Convert given values to base units:

275 MPa = 275*10^{6} Pa

325 mm^{2} = 0.000325 m^{2}

Substituting in given values:

F = (275*10^{6})*(0.000325)=89375 N

3 0
3 years ago
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