Renamed and re-organized files in Scripts/ directory

14 files changed, 17 insertions, 396 deletions

diff --git a/Functions/Heun/Heun.m b/Functions/Heun/Heun.m
deleted file mode 100644
index 5812dfe..0000000
--- a/Functions/Heun/Heun.m
+++ /dev/null
@@ -1,44 +0,0 @@
-function [t,y] = Heun(dydt,tspan,y0,h,es,maxit)
-% [t,y] = Heun(dydt,tspan,y0,h): uses the heun method to integrate an ODE
-% inputs:
-%	dydt = the differential equation of interest (must be anonymous
-%		function)
-%	tspan = [ti,tf] the initial and final values of the independent
-%		variable
-%	y0 = the initial value of the dependent variable
-%	h = step size
-%	es = stopping criterion (%), optional (default = 0.001)
-%	maxit = maximum iterations of corrector, optional (default = 50)
-% outputs:
-%	t = vector of independent variable values
-%	y = vector of solution for dependent variable  
-
-%Error checking
-if nargin<4, error('At least 4 aruments required.');end
-if nargin<5|isempty(es), es = 0.001;end
-if nargin<6|isempty(maxit), maxit = 50;end
-if length(tspan)~=2, error('Argument tspan must have a length of 2');end 
-
-%Trims tspan according to step size (h)
-t=tspan(1):h:tspan(2);
-%if rem(tspan(2),h)~=0
-if t(length(t))~=tspan(2)
-    t=[t,tspan(end)];
-end
-
-y=ones(1,length(t))*y0;
-h=diff(t);
-
-for i=1:length(t)-1
-	S=dydt(t(i),y(i));
-	ynew=y(i)+S*h;
-	for j=1:maxit
-		yold=ynew;
-		ynew=y(i)+h(i)/2*(S+dydt(t(i+1),yold));
-		ea = abs((ynew-yold)/ynew)*100;
-		if ea <= es, break, end
-	end
-	y(i+1)=ynew;
-end
-plot(t,y)
-end
\ No newline at end of file
diff --git a/Functions/days/README.md b/Functions/days/README.md
deleted file mode 100644
index 8a11db2..0000000
--- a/Functions/days/README.md
+++ /dev/null
@@ -1,11 +0,0 @@
-#days.m
-Function to count the total days elapsed in a year according to a given date.
-The syntax of the function is `days(<months>, <days>, <leap>)` where `months` is a integer (1-12). `days` is an integer (1-31) and `leap` 
-
-## Input
-months - month number (1-12). Example: `8` represents August.
-days - day number of the month.
-leap - indicates if the year is a leapyear or a regular year. `0` for regular and `1` for leap year.
-
-## Output
-`nd` - number of days elapsed in the year.
\ No newline at end of file
diff --git a/Functions/days/days.m b/Functions/days/days.m
deleted file mode 100644
index 87163b7..0000000
--- a/Functions/days/days.m
+++ /dev/null
@@ -1,22 +0,0 @@
-function nd = days(mo, da, leap)
-%days - Counts days elapsed in the year.
-%   
-%   days(<months>, <days>, <leap>) determines the days elapsed in a year
-%   based on the current date. This will include the current day. 
-%   Months (1-12),
-%   Example: August 28, no leap year
-%           days(8,28,0)
-
-daysPeM=[0 31 59 90 120 151 181 212 243 273 304 334]; % cummulative days prior each month.
-
-nd = daysPeM(mo) + da;
-
-if leap == 1 & mo >= 3
-    nd = nd + 1;
-end
-end
-
-
-% MECH 105: Homework 4 - Part 1
-% Author: Christian Kolset
-% Date: 5.21.21
\ No newline at end of file
diff --git a/Functions/falsePosition/falsePosition.m b/Functions/falsePosition/falsePosition.m
deleted file mode 100644
index d2cb776..0000000
--- a/Functions/falsePosition/falsePosition.m
+++ /dev/null
@@ -1,45 +0,0 @@
-function [root, fx, ea, iter] = falsePosition(func, xl, xu, es, maxit, varargin)
-%falsePosition finds the root of a function using false position method
-%   Syntax:
-%	[root, fx, ea, iter] = falsePosition(func, xl, xu, es, maxit, varargin)
-%
-%   Input:
-%	func = function being evaluated
-%	xl = lower guess
-%	xu = upper guess
-%	es = desired relative error (default 0.0001%)
-%	maxit = maximum number of iterations (default 200)
-%	varargin, ...= any additional parameters used by the function
-%   Output:
-%	root = estimated root location
-%	fx = function evaluated at root location
-%	ea = approximated relative error (%)
-%	iter = number of iterations performed
-
-if nargin < 3, error('Too few arguments: at least 3 needed'), end
-if nargin < 4 | isempty(es), es = 0.0001; end
-if nargin < 5 | isempty(maxit), maxit = 200; end
-
-iter=0;
-ea=100;
-xr= xl;
-
-while(1)
-	xrold= xr;
-	iter = iter + 1;
-	xr=xu-func(xu)*(xl-xu)/(func(xl)-func(xu));
-	if xr~=0
-		ea = abs((xr-xrold)/xr)*100;
-	end
-	if func(xl)*func(xr)<0
-		xl=xr;
-	else
-		xl=xr;
-	end
-	if iter == maxit | ea <= es, break, end
-end
-root = xr;
-fx=func(xr,varargin{:});
-ea=ea;
-iter=iter;
-end
\ No newline at end of file
diff --git a/Functions/luFactor/luFactor.m b/Functions/luFactor/luFactor.m
deleted file mode 100644
index 4c853dd..0000000
--- a/Functions/luFactor/luFactor.m
+++ /dev/null
@@ -1,87 +0,0 @@
-function [L, U, P] = luFactor(A)
-% luFactor(A)
-%	LU decomposition with pivoting
-% inputs:
-%	A = coefficient matrix
-% outputs:
-%	L = lower triangular matrix
-%	U = upper triangular matrix
-%       P = the permutation matrix
-
-
-[m,n] = size(A);
-if m~=n, error('Enter a square matrix.'); end
-nb = n + 1;
-L=zeros(m,m);
-U=zeros(m,m);
-P=eye(m);
-
-%Partial pivoting
-for j = 1:n - 1
-	for p = j+1:m
-		if (abs(A(j,j)) < abs(A(p,j)))
-			A([j p],:) = A([p j],:);
-			P([j p],:) = P([p j],:);
-		end
-	end
-end
-%	[big,i] = max(abs(A(j:n,j)));
-%	ipr = i + j - 1;
-%	if ipr ~= j
-%		A([j,ipr],:) = A([ipr,j],:);
-%	end
-%end
-
-for i=1:m
-	% Computing L
-	for k=1:i-1
-		L(i,k)=A(i,k);
-		for j=1:k-1
-			L(i,k)= L(i,k)-L(i,j)*U(j,k);
-		end
-		L(i,k) = L(i,k)/U(k,k);
-	end
-
-	% Computing U
-	for k=i:m
-		U(i,k) = A(i,k);
-		for j=1:i-1
-			U(i,k)= U(i,k)-L(i,j)*U(j,k);
-		end
-	end
-end
-
-for i=1:m
-	L(i,i)=1;
-end
-
-%% Checks
-if P*A ~= L*U, error('LU factorization failed to be computed correctly.'); end
-
-% Checks output size
-if size(L) ~= size(A) | size(U) ~= size(A) | size(P) ~= size(A)
-	error('Incorrect output size')
-end
-
-% Checks if matrices are triangular
-%for i=1:m
-%	for j=1:n		
-%	error('Incorrect output: LU decomposition are not triangular')
-%end
-
-% Checks the permutation matrix P is valid
-%P_str = string(P)
-%for k = m
-%	num1=count(P(:,1:n),1)
-%	if num1 ~=1
-%		error('Incorrect P matrix output. Too many 1's on the same line')
-%	end
-%	num0=count(P(:,1:n),0)
-%	if num0 ~= n-1
-%		error('Incorrect P matric output. Wrong number of 0's')
-%	end
-%end
-
-%}
-
-end
\ No newline at end of file
diff --git a/Functions/simpson/README.md b/Functions/simpson/README.md
deleted file mode 100644
index e69de29..0000000
--- a/Functions/simpson/README.md
+++ /dev/null
diff --git a/Functions/simpson/simpson.m b/Functions/simpson/simpson.m
deleted file mode 100644
index acd4a76..0000000
--- a/Functions/simpson/simpson.m
+++ /dev/null
@@ -1,39 +0,0 @@
-function [I] = Simpson(x, y)
-% Numerical evaluation of integral by Simpson's 1/3 Rule
-% Inputs
-%   x = the vector of equally spaced independent variable
-%   y = the vector of function values with respect to x
-% Outputs:
-%   I = the numerical integral calculated
-
-%Error cheching
-if length(x)~=length(y), error('The vectors need to be of equal lengths'),end
-if length(x)==1, error('Vectors must be at least 2 data points long'),end
-if range(x(2:end)-x(1:end-1))~=0, error('The x vector needs to be equally spaced'), end
-%Error  Errorchecking END
-
-lower_bound=min(x);
-interval=length(x)-1;
-if rem(interval,2)==0
-	trap_rule=0;
-	upper_bound=max(x);
-else 
-	warning('Odd number of intervals. Applying Trapezoidal rule on last interval')
-	trap_rule=1;
-	upper_bound=x(end-1);
-end
-
-if length(x)==2
-	I=(upper_bound-lower_bound)*(y(1)+y(2))/2;
-else
-	h=(upper_bound-lower_bound)/interval;
-	SumOdd=sum(y(1:2:end))-y(end-1);
-	SumEven=sum(y(2:2:end))-y(end-2);
-	I=(upper_bound-lower_bound)*(y(1)+4*SumOdd+2*SumEven+y(end))/(3*interval);
-end
-
-if trap_rule==1 & length(x)~=2
-	I_trap=(max(x)-upper_bound)*(y(upper_bound)+y(max(x)))/2;
-	I=I+I_trap;
-end
-end
\ No newline at end of file
diff --git a/Functions/specialMatrix/specialMatrix.m b/Functions/specialMatrix/specialMatrix.m
deleted file mode 100644
index d6c31f8..0000000
--- a/Functions/specialMatrix/specialMatrix.m
+++ /dev/null
@@ -1,51 +0,0 @@
-function [A] = specialMatrix(n,m)
-% This function should return a matrix A as described in the problem statement
-% Inputs n is the number of rows, and m the number of columns
-% It is recomended to first create the matrxix A of the correct size, filling it with zeros to start with is not a bad choice
-
-%--------------------------------------------
-
-if nargin ~= 2
-    error('Error: Please enter two arguments.')
-end
-if (n|m) <= 0
-    error('Error: Index error. Arguments must be positive integers or logical values.')
-end
-
-A = zeros(n,m);     %Creates a "blank" n x m matrix full of zeros
-
-% Now the real challenge is to fill in the correct values of A
-
-A(1,:) = 1:m;       %Lables the first row with column numbers
-A(:,1) = 1:n;       %Lables the first column with row numbers
-
-for j = 2:n
-	for k = 2:m    
-		A(j,k)=A(j-1,k)+A(j,k-1);  % fills each element of the matrix with the sum of the left and above element.
-	end
-end
-end
-
-%---------------------------------------------
-
-%{
-if nargin(specialMatrix) ~= 2
-    error('Error: Please enter two arguments.')
-elseif nargin(specialMatrix) <= 0
-    error('Error: Index errer. Argument must be positive integers or logical values.')
-end
-
-if n<2
-    A = [1:m];		%If matrix is 1 x m then make a matrix with 1 row.
-else
-    A = [1:m;1:m];	%Create row 1
-end
-
-for j = 3:n			%Loop to make 
-    A(j,:) = sum(A);	%
-end				%
-A(:,1)=(1:n);
-
-end
-% Things beyond here are outside of your functions
-%}
\ No newline at end of file
diff --git a/Scripts/README.md b/Scripts/README.md
new file mode 100644
index 0000000..2e8e64f
--- /dev/null
+++ b/Scripts/README.md
@@ -0,0 +1,17 @@
+# [Scripts](Scripts)
+This directory contains the scripts to solve the homeworks assignments below:
+
+## hw02.m
+Both part 1 and part 2 on the excercise are writen in the script. When the script is run part 2 will be skipped. To run part 2, put part 1 in comments and uncomment part 2. To uncomment remove the `%{` and `%}` (line 43 and 70) to prevent the MATLAB interpreter to skip the code.
+
+### Part 1
+Creates a supplots of two graphs. The first plot shows the charge on a capacitor in an electrical circuit. First graph ranges from t = 0 to 0.8 witj, q0=10, R=60, L=9, and C=0.00005. A second plot shows a capacitor with 10 times greater charge (C=0.0005)
+
+### Part 2
+Creates a plot displaying the measured data (red diamond shapes) and the expected function (green dashed line). Ranges from 0 to 70 minutes using a a step size of 30 seconds.
+
+# hw03.m
+This script solves the volume in a tank with the shape of a cylinder with a tuncated cone on top. This script should be turned into a function to solve the volume of a tank with any given the height of cylinder and truncated cone, and diameter of both the cylinder and top of the truncated cone.
+
+## hw11.m
+Script to solve the angle of which to pull a 25 kg block with 150 N using the bisection method of finding roots.
diff --git a/Scripts/hw02/hw2.m b/Scripts/hw02.m
index 6736240..6736240 100644
--- a/Scripts/hw02/hw2.m
+++ b/Scripts/hw02.m
diff --git a/Scripts/hw03/hw3.m b/Scripts/hw03.m
index 7caf82c..7caf82c 100644
--- a/Scripts/hw03/hw3.m
+++ b/Scripts/hw03.m
diff --git a/Scripts/hw11/hw11.m b/Scripts/hw11.m
index 312e816..312e816 100644
--- a/Scripts/hw11/hw11.m
+++ b/Scripts/hw11.m
diff --git a/Scripts/hw2/hw2.m b/Scripts/hw2/hw2.m
deleted file mode 100644
index 6736240..0000000
--- a/Scripts/hw2/hw2.m
+++ /dev/null
@@ -1,60 +0,0 @@
-% Homework:2 
-% Author: Christian Kolset
-clear
-
-%% Part 1
-
-% Function parameters
-q0 = 10;
-R = 60;
-L = 9;
-C = 0.00005;
-
-% Use linspace to create an array of 100 points between 0 and 0.8
-t = linspace(0,0.8);
-
-% Calculate the values of q
-q = q0.*2.718.^(-R.*t/(2*L)).*cos(sqrt((1/(L*C))-(R/(2*L))^2).*t);
-
-% Plot q vs t
-hold on
-subplot(2,1,1)
-plot (t,q,'b--*')
-title('Capacity vs Time Graph')
-xlabel('Time')
-ylabel('Charge')
-%legend('Charge','Time')
-hold off
-
-% Make the capacitor 10x bigger
-q2 = q0.*2.718.^(-R.*t/(2.*L)).*cos(sqrt((1/(L.*10.*C))-(R/(2.*L))^2).*t);
-
-% Plot q2 vs t
-hold on
-subplot(2,1,2)
-plot(t,q2,'rs:')
-title('10x Capacity vs Time Graph')
-xlabel('Time')
-ylabel('Charge')
-%legend('Charge','Time')
-hold off
-
-%% Part 2
-%{
-% Given experimental data
-t_exp = 10:10:60;
-c_exp = [3.4 2.6 1.6 1.3 1.0 0.5];
-
-% Expected function
-t_func = 0:0.5:70;
-c_func = 4.84*2.718.^(-0.034*t_func);
-
-% Plot
-hold on
-plot(t_exp,c_exp,'rd:')
-plot(t_func,c_func,'g--')
-xlabel('Time [minutes]')
-ylabel('Concentration [ppm]')
-legend('Experimental','Predicted')
-hold off
-%}
\ No newline at end of file
diff --git a/Scripts/hw3/hw3.m b/Scripts/hw3/hw3.m
deleted file mode 100644
index 7caf82c..0000000
--- a/Scripts/hw3/hw3.m
+++ /dev/null
@@ -1,37 +0,0 @@
-% HW3
-% Author: Christian Kolset
-% Date: 20.5.21
-
-%% HW3
-% Specify the variables needed to solve this problem (ie. height of each section, diameter, radiaus, ...)
-%   It is alwasy easier to work with variables (diameter_cyl = 25) than to use numbers everywhere, since a 
-%   diameter indicates something specific but the number 25 could mean anything
-diameter_Bot = 25;
-r_bot = diameter_Bot/2;
-h_cone = h-19;
-
-% Specify the height of the water
-h = 20
-% You can comment / uncomment lines below for testing. This will overwrite the previous line for h = 20.
-% For submission, make sure all of the following lines are commented out and h = 20! (OR IT IS MARKED AS WRONG)
-%h = 5
-%h = 19
-%h = 47
-%h = -1
-
-% Now compute the volume. Using conditional statments you will want to first check the height makes sense,
-% and then solve the volume depending on what portion of the tank has been filled.
-% Make sure that your volume is stored in the variable v! (OR IT WILL BE MARKED AS WRONG)
-% You may find it more convenient to move v around in you code, it is only given here to indicate what variable to use.
-%v =
-
-v_cyl = @(h,r_bot) pi*r_bot^2*h;
-v_truncone = @(h_cone,r_bot) 1/3*pi*h_cone*(r_bot^2+r_bot*((h_cone+16.625)/1.33)+((h_cone+16.625)/1.33)^2);
-
-if h > 19
-    v = v_cyl(19,r_bot)+v_truncone(h_cone,r_bot);
-else
-    v = v_cyl(h,r_bot);
-end
-
-fprintf(1,'Volume of Tank: %6.2f\n',v)
\ No newline at end of file