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Merge pull request #17 from rcc02007/master
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{ | ||
"cells": [], | ||
"metadata": {}, | ||
"nbformat": 4, | ||
"nbformat_minor": 2 | ||
} |
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function sigma_z=boussinesq_lookup(q,a,b,z) | ||
% function that determines stress under corner of an a by b rectangular platform | ||
% z-meters below the platform. The calculated solutions are in the fmn data | ||
% m=fmn(:,1) | ||
% in column 2, fmn(:,2), n=1.2 | ||
% in column 3, fmn(:,2), n=1.4 | ||
% in column 4, fmn(:,2), n=1.6 | ||
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fmn= [0.1,0.02926,0.03007,0.03058 | ||
0.2,0.05733,0.05894,0.05994 | ||
0.3,0.08323,0.08561,0.08709 | ||
0.4,0.10631,0.10941,0.11135 | ||
0.5,0.12626,0.13003,0.13241 | ||
0.6,0.14309,0.14749,0.15027 | ||
0.7,0.15703,0.16199,0.16515 | ||
0.8,0.16843,0.17389,0.17739]; | ||
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m=a/z; | ||
n=b/z; | ||
if n < 1.3 | ||
f=fmn(:,2); | ||
elseif n > 1.5 | ||
f=fmn(:,4); | ||
else | ||
f=fmn(:,3); | ||
end | ||
[~,i_fit]=sort(abs(m-fmn(:,1))); | ||
x=fmn(i_fit(1:4),1); | ||
y=f(i_fit(1:4)); | ||
f_out = Newtint(x,y,m); | ||
sigma_z=q*f_out; | ||
end |
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function [t,r] = coriolis(L) | ||
% In class we ran this function using L=41.8084 (the latitude of Storrs, CT). This | ||
% function takes the latitude (L) in degrees and mass (m) and calculates the trajectory | ||
% of a particle with a 100 N load directed North. The initial conditions are set as L | ||
% (in radians) * radius of Earth, 0 m/s initial x-velocity, 0 m E-W position (add to | ||
% -72.261319 degrees for longitude of Storrs), 0 m/s initial E-W velocity, 10 m initial | ||
% altitude, 0 m/s initial z-velocity [L*pi/180*R 0 0 0 10 0], the force is given as 100 | ||
% N North, 0 West and 9.81*m z (neutrally buoyant) [100 0 9.81*m] | ||
% | ||
% the output of myode is ddr=[dx/dt d2x/dt2 dy/dt d2y/dt2 dz/dt d2z/dt2]' and the input | ||
% to myode is r=[x dx/dt y dy/dt z dz/dt]' | ||
% using ode23 solver solves for r as a function of time, here we solve from 0 to 200 s | ||
% r(:,1) = x (the north-south position from 0 to 200 s) | ||
% r(:,3) = x (the West-East position from 0 to 200 s) | ||
% r(:,5) = x (the altitude from 0 to 200 s) | ||
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% define ordinary differential equation in terms of 6 first order ODE's | ||
function ddr = myode(t,r,R,L) | ||
g=9.81; % acceleration due to gravity m/s^2 | ||
l=10; % 10 m long cable | ||
we=2*pi/23.934/3600; % rotation of Earth (each day is 23.934 hours long) | ||
ddr=zeros(4,1); % initialize ddr | ||
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ddr(1) = r(2); % x North(+) South (-) | ||
ddr(2) = 2*we*r(4).*sin(L)-g/l*r(1); % dx/dt | ||
ddr(3) = r(4); % y West (+) East (-) | ||
ddr(4) = -2*we*(r(2).*sin(L))-g/l*r(3); % dy/dt | ||
end | ||
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R=6378.1e3; % radius of Earth in m | ||
L=L*pi/180; | ||
[t,r]=ode45(@(t,r) myode(t,r,R,L),[0 30000], [1 0 0 0 ]); | ||
figure() | ||
z=-sqrt(10^2-r(:,1).^2-r(:,3).^2); | ||
figure(1) | ||
we=2*pi/23.934/3600; % rotation of Earth (each day is 23.934 hours long) | ||
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plot(t,tan(we*sin(L)*t),t,-r(:,3)./r(:,1),'.') | ||
xlabel('time (s)','Fontsize',18) | ||
ylabel('-y/x','Fontsize',18) | ||
% Plot Coriolis effect path | ||
figure(2) | ||
title('Path at 0 hr, 4.1 hrs, 8.3 hrs','Fontsize',24) | ||
N=length(t); | ||
i1=[1:100]; i2=[floor(N/2):floor(N/2)+100]; i3=[N-100:N]; | ||
plot3(r(i1,1),r(i1,3),z(i1)) | ||
hold on | ||
plot3(r(i2,1),r(i2,3),z(i2),'k-') | ||
plot3(r(i3,1),r(i3,3),z(i3),'g-') | ||
xlabel('X (m)','Fontsize',18) | ||
ylabel('Y (m)','Fontsize',18) | ||
zlabel('Z (m)','Fontsize',18) | ||
title('Coriolis acceleration Foucalt Pendulum') | ||
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% figure() | ||
% % Plot Eotvos effect for deviation upwards | ||
% plot(1e-3*(r(:,1)-r(1,1)),r(:,5)) | ||
% xlabel('North (+km)','Fontsize',18) | ||
% ylabel('Altitude (+m)','Fontsize',18) | ||
% title('Eotvos effect with north force') | ||
% | ||
end | ||
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function [t,y] = eulode(dydt,tspan,y0,h,varargin) | ||
% eulode: Euler ODE solver | ||
% [t,y] = eulode(dydt,tspan,y0,h,p1,p2,...): | ||
% uses Euler's method to integrate an ODE | ||
% input: | ||
% dydt = name of the M-file that evaluates the ODE | ||
% tspan = [ti, tf] where ti and tf = initial and | ||
% final values of independent variable | ||
% y0 = initial value of dependent variable | ||
% h = step size | ||
% p1,p2,... = additional parameters used by dydt | ||
% output: | ||
% t = vector of independent variable | ||
% y = vector of solution for dependent variable | ||
if nargin<4,error('at least 4 input arguments required'),end | ||
ti = tspan(1);tf = tspan(2); | ||
if ~(tf>ti),error('upper limit must be greater than lower'),end | ||
t = (ti:h:tf)'; n = length(t); | ||
% if necessary, add an additional value of t | ||
% so that range goes from t = ti to tf | ||
if t(n)<tf | ||
t(n+1) = tf; | ||
n = n+1; | ||
end | ||
y=zeros(n,length(y0)); | ||
y(1,:)=y0; | ||
for i = 1:n-1 %implement Euler's method | ||
y(i+1,:) = y(i,:) + dydt(t(i),y(i,:),varargin{:})'*(t(i+1)-t(i)); | ||
end |
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