App Development with the Matlab App Designer - Part 1: A Simple Mechanics Calculator

In this tutorial, I am going to walk you through the basics of the app development with the relatively new Matlab App Designer platform. Mathworks has introduced the App designer in the 2016 Matlab edition, which helps you to build your own apps with the default graphical user interface (GUI) and programming environment. I have come to know this handy tool lately, and I am very eager to share my knowledge with you, especially if you are interested in object-oriented programming. 

Let's see, first how we can access the App Designer option. Open the Matlab and click on the "APPS" from the top menus.

Matlab > APPS > Design App

Next, after clicking the "Design App" icon, the following window will appear. As you can see, there are more options to choose. But, to begin with, we should select a "Blank App".

Once, you select the "Blank App", you will see the following window. On the left hand, you have the default 'drag and drop' items for your app that you can simply select and bring it to the middle window "Design View", where the development works are carried out. On your right, you have the options to customize the default app items of the left side. For example, if you want to increase the font size of your display, you can do it from the right side menu bar.

Now, if you click on the "Code View", you will see default codes already done by Matlab for the basic interface. You can edit codes here or enter your own codes that you would like to incorporate with the simulator.

Now, let's try a very simple example, let's make the very first app by the Matlab App Designer. We will make a calculator for basic calculation in mechanics where it will calculate the force based on the two input data: mass and acceleration. As we know from Newton's second law of motion that the force is a product of the mass and acceleration, this formula is simply implemented in the development of our very first app or simulator by Matlab. Let's follow the following screenshots closely.

In the above screenshot, as you can see, from the left side menu bar, objects are dragged and placed in the middle window or the design view section. To make a calculator with two inputs, I chose a simple display for the mass input and for the acceleration, I selected a knob, which can be rotated to fix a value. To show the output that is "Force", I dragged the text objects similarly from the left side. You can also format the style of the texts, sizes, etc. from the right side menu bar.

Now, it's time to code the objects that we just placed in the design window a while ago. This is the part that we may call object-oriented programming. So, what are the objects that have been selected here - an execution button, text displays, and a knob for the acceleration parameter. If we press the execution button after providing the inputs, we should have the output displayed on the right side. This is the goal for our first app development.

We will now add codes to the "Execute" button for our goal achievement. In the following image, we see that how we can access the "Callbacks", which will eventually take us to the code environment that is shown in the next screenshot.

We need to add the formula "Force = Mass × Acceleration". To do so, as you see below, we need to copy the values of the functions under each object (e.g., "Object Mass (kg)", "Choose Acceleration", and "Force (N)") that are selected and write or organize them as per the given formula. For example, the output display is the "Force (N)" and its function value is "app.ForceNEditField.Value", which is the left side of the equation. Then, the right side is the mass times acceleration that is represented by the "app.ObjectMasskgEditField.Value" and "app.ChooseAccelerationKnob.Value" subsequently under the respective functions' values.

So, that is all! We have achieved our goal - made a very simple app that calculates the force given the mass and acceleration inputs.

#Matlab   #MatlabAppDesigner   #MechanicsCalculator  #ObjectOrientedProgramming

Development of a Simple Configurable Continuous Track Robot with SolidWorks

In this tutorial, I am going to show how to develop a continuous track robot that is flexible in delivering goods in industries. The robot consists a continuous flexible belt or chain, in which three rollers are constrained so that the robot can achieve adjustable motion when it is actuated. This is a very basic configuration that may be made more advanced according to your desired tasks to be accomplished by the robot.

The continuous tracking robot has the following configurable parts:

• Belt or chain drive

• Wheels or rollers

• Motors or actuators

• Couplings or bearings used as joints between parts

The following images shows the configuration for the tracking robot.

For convenience, in SolidWorks, you may have different visualization options. The following image depicts the wireframe type visualization of the assembly, which shows you the hidden lines, corners, and so forth.

The following design shows the engineering or technical drawing of the robot system. All dimensions shown are in millimetres (mm).

The following video shows the motion simulation of the continuous track robot. The animation is made by the SolidWorks Motion Simulation manager. This is a very intuitive tool if you would like to visualize the movement of your structure prior to manufacturing.

#SolidWorks #CADDesign #MotionSimulation #TrackingRobot #ModellingSimulation

A Python Program for Application of the Newton-Raphson Method

In this tutorial, we will develop a simple Python program to implement the famous Newton-Raphson algorithm. Let us consider the following function:

f(x) = exp (-0.5x) (4 - x ) - 2

We will find the roots of the above function by the Newton-Raphson approach and code them using Python. We will start with an initial guess for the solution and then change it to check for other initial guesses to see whether the solution converges or diverges. This is a very basic and fundamental application of the Newton-Raphson method to find the roots of an unknown function. In the following, Python codes are written by Anaconda to find the roots of the unknown function by the Newton-Raphson method. We begin with, x = 2 for our first initial guess.

Python Script for Newton-Raphson Method

import sympy as sp  # Importing symbolic mathematics library

import pandas as pd  # Importing pandas to work with arrays

import math # Importing mathematics library

def f(x):

    return (math.e**(-0.5*x)*(4-x))-2  # Given function 

x = sp.symbols('x')

fp = f(x).diff(x) # Derivative of the function

x_value = [2] # Initial guess

error = [1]

tol = 0.0001

MaxIter = 100

i = 0

while i <= MaxIter and abs(error[i]) > tol: # Setting convergence criteria

    x_new = x_value[i] - (f(x_value[i])/fp.subs(x,x_value[i])) # Newton-Raphson Algorithm

    x_value.append(x_new)

    ernew = (x_new - x_value[i])/x_new

    error.append(ernew)

    i+=1

solution = [[i, x_value[i], error[i]] for i in range(len(x_value))];

solution_new = pd.DataFrame(solution,columns=["No of Iterations", "x_value", "Error"])

print(solution_new)

Results:

When x = 2,

   No of Iterations            x_value                Error

0                 0                  2                    1

1                 1  0.281718171540955    -6.09929355660769

2                 2  0.776886845045375    0.637375541447737

3                 3  0.881707878928567    0.118884084386961

4                 4  0.885703241166645  0.00451094909940144

5                 5  0.885708801994023  6.27839236320089e-6

So, the algorithm converges for the initial guess, x = 2.

When x = 6,

The algorithm diverges, relative error is NAN or infinity.

When x = 8,

The algorithm again diverges, relative error is close to infinity.

As we see from the results that the algorithm converges only when x = 2, but it diverges for the other two cases. The reason behind this is the derivative of the given function is zero when x = 6 and close to zero for x = 8. Since the slope of the function reaches to zero, the algorithm fails to provide a converged solution. From this example, it is evident that for any numerical method to work for a solution, the initial condition is a very important parameter, which needs to have an educated guess.

#Python #PythonScripting #NewtonRaphson #NumericalMethod

Understanding the Role of the Friction Force in Dynamics with an Example

In this blog post, I would like to discuss an example from the web, which may help understand better about the presence and significance of the friction force in dynamics. Friction plays a very vital role in our daily activities, for example walking, driving, etc. are all possible for us because of friction. We mostly know that the friction is a resistive force (opposes the motion), but it is also a driving force (aids the motion). The following example problem enlightens both the facts of the friction force. It is explained below.

Example Problem:

A 0.5 kg wooden block is placed on top of a 1.0 kg wooden block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower and the level table is 0.20. What is the maximum horizontal force that can be applied to the lower block without the upper block slipping?

Here, the lower block has twice the weight than the upper block. In order to keep the upper block from sliding while a horizontal force is applied to the lower block, the static friction between two blocks has to be the key factor. The maximum static force is calculated as follows,

Ff = 𝛍ₛ FN

According to the Coulomb’s friction model, this static friction is solely dependent on the weight of the body. The more the weight is the more static friction is generated. Say, if we apply a horizontal force on the lower block to move it to the right on the table as in the figure, the upper block would slide backward to the left on the table. And, this is dependent on the friction force between the block surfaces, which can resist the upper block’s sliding motion to the opposite to the lower block. For this problem, to satisfy the above mentioned condition, the friction force acting between the block surfaces must act to the right, which is acting same direction as the applied horizontal force on the lower block. This friction force will resist the upper block sliding backward and maintain the same acceleration for both of the blocks as well. Again, this friction force might be termed driving force in a sense to keep the upper block in its position while the actual driving force to create acceleration is done by the horizontal force applied on the lower block.

The following free body diagram for the upper block will show the direction of the forces and calculation of the system acceleration. Also, another free body diagram is shown later for the combined blocks where we apply the Newton’s second law to determine the required horizontal force on the lower block to keep preventing the  sliding of the upper block.

From the above calculation, we see that for the upper block, the direction of acceleration and maximum friction force is the same. So, for the upper block, we may imply that the friction works as a driving force. The magnitude of the friction force is 1.7 N. Next, we are going to calculate the acceleration based on this frictional force and consider the free body diagram of the two blocks together as well.

Here, we see from the combined free body diagram of the problem that the friction force acts opposite to the applied horizontal force. The applied horizontal force to the lower block is the driving force and friction is the resistive force in this case. Now, we will calculate the net maximum horizontal force that can be applied to the lower block without the upper block slipping using the Newton’s second law of motion again.

By this example, we see that the friction is a resistive force when we consider the overall system (two blocks together); however, it is a driving force for the upper block as per the fundamental laws of motion. The compliance or stiffness (inverse of compliance) is a tendency to resist tension or compression in a system, which is a conservative force (restores energy). But friction is a non-conservative force (dissipates energy), and due to its obvious presence in physical systems, we have some losses from the system in terms of heat, wear and tear, and so on. This may serve as an intuitive example, in regards to realizing the friction force in a conspicuous manner.

#FreeBodyDiagram #Friction #CoulombFriction #ResistiveForce #NewtonsLaw #Blog #Blogger

How to Convert Graphics to Solid Body in CATIA

This tutorial demonstrates how to convert a Stereolithography Mesh (.stl) graphics object to solid editable body in CATIA V5. This is useful when we require to modify an STL object file in any CAD processing tool, such as, AutoCAD, Autodesk Inventor, SolidWorks, CATIA, and so forth. It is also important when we plan to do a Finite Element (FE) analysis of an object that is available only in the STL format. For the demonstration purpose, this tutorial uses an STL image of an anonymous patient having Scoliosis, which is available online. You can download the file from this site.

Processing an Image in CATIA

At first, open CATIA V5, then go to 'Start' and select ‘Shape’, then choose ‘Digitized Shape Editor’. See the screenshot below for the process.

After that, in the following window, just click ‘ok’ to create a new part file.

We will now import the (.stl) by clicking the ‘Import’ option as highlighted below.

Next, click ‘Apply’ and then select ‘ok’ to import the (.stl) file, which looks like below:

Since, we have imported an STL file from third-party source, we need to check the quality of the image, and clean it up. There is an option ‘Mesh Cleaner’, click it, and the following window will appear.

A pop-up small window will appear. Select the object, and then click ‘Analyze’. Click on the ‘Isolated Triangles’ option, and change the color, so that you can visualize it clearly. In this case, we see it is yellow.

As you notice that we are in the ‘Deletion’ mode of the ‘Mesh Cleaner’ window. So, we need to delete those unwanted or corrupted triangles to make it a valid solid body. Otherwise, the conversion either might fail or proceed, but in that case with lots of corrupt meshes that will hamper your analysis.

As we see, there are 60999 corrupt triangles, which need to be discarded. This may also remove some parts from the image, but if that part is not significant, then it may be alright to proceed depending on our requirement. We just need to make sure that we have our concerned region unaffected for the analysis. Next, click on the ‘Long Edges’ and continue the same process for removing the corrupt regions. The last option is the ‘Small Angles’, which we need to select and click ‘Apply’. 

After the process is done, the following window will appear. We see that there are three ‘Non-manifold Vertices’ that need to be removed as well in a similar way.

We need to continue the mesh cleaning process at least twice to make sure that we have removed all the corrupt segments. After doing so, we will have the following image without the unwanted meshes.

Next, we need to click on the ‘Structure’ tab and subsequently, click on the ‘Orientation’, and ‘Split in Connected Zones’ options to check the followings. If the color for the ‘Orientation’ stays same, as in this case, ‘Yellow’, then we are good to go. For the ‘Split in Connected Zones’, there should be nothing specified in the window as highlighted by the following two screenshots below.

After the above process, we need to click Start > Shape > Quick Surface Reconstruction. The steps are depicted below.

Next, click Insert > Surface Creation > Automatic Surface. Just follow the following instructions shown by the image.

The following window will appear. Try to keep the parameters same as shown in the tiny popped-up window. Then, hit ‘ok’.

We are almost done in creating the Solid Body. After the automatic surface creation, we need to close the surface, which is depicted below. Click Start > Mechanical Engineering > Part. 

Then follow this sequence: Insert > Surface-Based Features > Close Surface.

We are all done. Finally, we need to save it as a 'CATIA part' file and also export it to (.igs) format in case if somebody wants to perform FE analysis.

The following image shows the closed view of the processed solid 3D Scoliosis skeleton that may be used for any analysis that needs geometry manipulation. 

#Catia #ReverseEngineering #3Drendering #3Dmodel #ImageProcessing #Blog #Blogger

How to Solve a System of Partial Differential Equations in MATLAB?

In this tutorial, we are going to discuss a MATLAB solver 'pdepe' that is used to solve partial differential equations (PDEs). Let us consider the following two PDEs that may represent some physical phenomena. Sometimes, it is quite challenging to get even a numerical solution for a system of coupled nonlinear PDEs with mixed boundary conditions. This example demonstrates how we may solve a system of two PDEs simultaneously by formulating it according to the MATLAB solver format and then, plotting the results. This method is relatively easier and saves time while coding.

∂y₁/∂t = 0.375 ∂²y₁/∂x² + A(y₁ - y₂)          (1)

∂y₂/∂t = 0.299 ∂²y₂/∂x² - A(y₁ - y₂)          (2)

Where, A is a function of sinusoidal x, meaning that A = sin (x). The above two equations have two derivative terms. The time derivative, generally represents parabolic equation, and the spatial derivative defines elliptic equation. So, we have both forms of PDEs and the highest order is two in the space. There is one limitation of the 'pdepe' that you need to have the parabolic term in the PDEs in order to solve by the 'pdepe'. Now, let's assume, we have the following initial conditions:

y₁(x,0) = 1

y₂(x,0) = 0

The boundary conditions are,

∂/∂x y₁(0,t) = 0

y₂(0,t) = 0

∂/∂x y₂(1,t) = 0

y₁(1,t) = 1

The initial and boundary conditions are true if 0 ⩽ x ⩽ 1 and t ⩾ 0. Before we move forward with the coding, we need to understand first how the 'pdepe' solver accepts the PDEs in MATLAB, which form it recognizes. The general form of PDEs that the solver understands is of the following form:

Here,

x is the independent spatial variable.

t is the independent time variable.

y is the dependent variable being differentiated with respect to x and t. It is a two-element vector where y(1) is y₁(x,t) and y(2) is y₂(x,t).

m is the symmetry constant. For cartesian coordinate, the value is 0; for cylindrical coordinate, the value is 1; and for spherical coordinate, it is 2. For our problem, it is 0 as our coordinate system is simply cartesian.

The functions c, f, and s refer to the coefficients in the above two PDE equations (1) and (2), which are required in a form that is usually expected by 'pdepe' solver.

So, in the above mentioned form, the coefficients of the PDEs may be arranged in a matrix and the equations become as,

Now, we are ready to begin the coding. In the MATLAB, we may create three functions, for example, the first function is for the equations, the second function is for the initial conditions, and the third function is for the boundary conditions. Also, we may incorporate all these functions inside another global function that is convenient as we would have everything in a single file. As the MATLAB solvers use the finite difference approach, the time integration is done with the MATLAB ‘ode15s’ solver. So, the ‘pdepe’ takes advantage of the capabilities of the stiff ‘ode15s’ solver for solving the differential-algebraic equations (DAE), which may arise when the PDEs contain elliptic equations. It is also used for handling the Jacobians with a certain sparsity pattern. Now, if there is an error during the solution process, you may try to refine the mesh size as the initial conditions are sensitive and sometimes are not consistent with the mesh size and the solver.

MATLAB Program:

function system_of_PDEs

close all

clear

clc

% Discretization of the simulation domain

x = linspace(0,10,100);

t = linspace(0,10,100);

% Application of the Matlab partial differential equation solver 'pdepe'

m = 0;

sol = pdepe(m,@pde_func,@pde_ics,@pde_bcs,x,t);

% The solution matrices

y1 = sol(:,:,1);

y2 = sol(:,:,2);

figure(1)

surf(x,t,y1)

title('y_1(x,t)')

xlabel('Distance x')

ylabel('Time t')

figure(2)

surf(x,t,y2)

title('y_2(x,t)')

xlabel('Distance x')

ylabel('Time t')

% Equations to solve

function [c,f,s] = pde_func(x,t,y,dydx) 

% Equations arranged for the 'pdepe' solver

c = [1; 1];

f = [0.375; 0.299].*dydx;

A = sin(x);

s = [A; -A];

end

% Initial Conditions

function u0 = pde_ics(x) 

u0 = [1; 0];

end

% Boundary Conditions

function [pl,ql,pr,qr] = pde_bcs(xl,ul,xr,ur,t) 

pl = [0; ul(2)];

ql = [1; 0];

pr = [ur(1)-1; 0];

qr = [0; 1];

end

end

Results:

The following results show the variations of the two dependent variables, y₁, and y₂, with respect to both space (x) and time (t). We may use the MATLAB 'Surface Plot' feature to do that.

#PDE #Matlab #FiniteDifference #HyperbolicEquation #PDEPE #Blog #Blogger