Added Functions and Scripts

3 files changed, 126 insertions, 0 deletions

diff --git a/Scripts/hw11/hw11.m b/Scripts/hw11/hw11.m
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+% Define problem constants
+g = 9.81;
+mu = 0.55;
+F = 150;
+m = 25;
+
+% Define the function to be solved for. Is the angle specified in radians or degrees?
+%angle=@(-asin(g/(sqrt(mu^2)))+atand(mu)+((4*n2+1)*pi)/2
+func=@(angle) mu*m*g/(cosd(angle)+mu*sind(angle))-F;
+
+% THINK, what makes range sense for angle?
+lower_bound=-90;
+upper_bound=90;
+
+% Plot your function. Does plotting give you an idea about where the root is?
+hold on
+fplot(func)
+fplot(0)
+
+% Finaly solve for the root using the bisection script given to you, which can be called as:
+%[root, fx, ea, iter] = bisect(func, lower_bound, upper_bound);
+
+angle = bisect(func,lower_bound, upper_bound)
+
+% These variables will be checked for your final answer:
+%angle =  % the angle in degree that solves this problem
+fx = func(angle)    % the function value at your solved angle
+ea = ea    % the bisection error estimate
+iter = iter   % the number of bisection iterations
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diff --git a/Scripts/hw2/hw2.m b/Scripts/hw2/hw2.m
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+% 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
+%}
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diff --git a/Scripts/hw3/hw3.m b/Scripts/hw3/hw3.m
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+% 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)
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