I need to make a plot using spline interpolation?

This is the problem I'm having trouble with. I really don't know where to start with it, so if someone could get me some pointers that would be amazing.

Answer 1

[Star Strider]

Start by creating two vectors, ‘Force’ and ‘Elongation’ out of the data you are provided. The documentation for the interp1 function has all the information you need.

Not to give away the answer, but remember that the rubber specimen has an initial length, and you are to find the force an a specified greater total length, so make allowances for that.

[Star Strider]

You had your independent and dependent variables backwards. You want force as a function of length, not the inverse. With that correction, the code almost writes itself.

I don’t know where ‘yfun’ came from, but you don’t need it for this problem.

This is how I would do it:

F = [0 0.6 0.9 1.16 1.18 1.19 1.24 1.48 1.92 3.12 4.14 5.34 6.22 7.12 7.86 8.42];

E = [0 1.2 2.4 3.6 4.8 6.0 7.2 8.4 9.6 10.8 12.0 13.2 14.4 15.6 16.8 18];

xi=linspace(min(E), max(E), 50);
yispl=interp1(E,F,xi,'spline');

rest_length = 3;
total_length = 11.5;
stretch_length = total_length - rest_length;
ystretch=interp1(E,F,stretch_length,'spline');

subplot(1,3,1)
plot(E,F,'o', xi,yispl,'--')
hold on
plot(stretch_length, ystretch, 'gp', 'MarkerSize',15)
hold off
grid
xlabel('Stretched Length (in)')
ylabel('Force ()')

fprintf(1, '\n\tForce at total length = %.1f" (stretched length = %.1f") = %.2f\n\n', total_length, stretch_length, ystretch)

It’s also best not to use generic variables such as ‘x’ and ‘y’ when descriptive ones such as ‘E’ and ‘F’ will convey more information and decrease the opportunity to confuse them. A small point, but one that becomes easy to appreciate in complicated code.

Answer 2

[Image Analyst]

Here's my spline demo:

% Demo to show spline interpolation.
% Clean up / initialize
clc; 
close all; 
clear all; 
workspace; % Display workspace panel.

% Create the original knot points.
lengthX = 10;
x = 1:lengthX;
y = rand (lengthX,1);
% Plot it and show how the line has sharp bends.
plot(x, y, '-sr', 'LineWidth', 2);
set(gcf, 'Position', get(0,'Screensize')); % Maximize figure.

% Use splines to interpolate a smoother curve,
% with 10 times as many points,
% that goes exactly through the same data points.
samplingRateIncrease = 10;
newXSamplePoints = linspace(1, lengthX, lengthX * samplingRateIncrease);
smoothedY = spline(x, y, newXSamplePoints);

% Plot smoothedY and show how the line is
% smooth, and has no sharp bends.
hold on; % Don't destroy the first curve we plotted.
plot(newXSamplePoints, smoothedY, '-ob');
title('Spline Interpolation Demo', 'FontSize', 20);

legend('Original Points', 'Spline Points');

% Mathworks Demo code from their Help
% x = 0:10;
% y = sin(x);
% xx = 0:.25:10;
% yy = spline(x,y,xx);
% plot(x,y,'o',xx,yy)

[Image Analyst]

It's not the interp1() function like your homework asked for but the concept is pretty similar, and perhaps this demo will help someone.

Answer 3

[Image Analyst]

Your code is very close, but basically you switched x and y. Try it this way - I just switched x and y and fancied up the plot a bit.

% Create sample data training points.
xElongation = [0 1.2 2.4 3.6 4.8 6.0 7.2 8.4 9.6 10.8 12.0 13.2 14.4 15.6 16.8 18];
yForce = [0 0.6 0.9 1.16 1.18 1.19 1.24 1.48 1.92 3.12 4.14 5.34 6.22 7.12 7.86 8.42];

% Plot "training points".
plot(xElongation, yForce, 'ro', 'LineWidth', 2);
grid on;
hold on;
fontSize = 20;
xlabel('Elongation', 'FontSize', fontSize);
ylabel('Force', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);

% Do a fit everywhere.  Use, say, 200 points.
xFit = linspace(xElongation(1), xElongation(end), 200);
yFittedForce = interp1(xElongation, yForce, xFit, 'spline');
% Plot the fit
plot(xFit, yFittedForce, 'b.-', 'MarkerSize', 5);
% Find the force when elongation = 11.5
y115 = interp1(xElongation, yForce, 11.5, 'spline');
message = sprintf('The Force at 11.5 is %f', y115);
uiwait(helpdlg(message));

[Image Analyst]

Because it started at 3 inches, and if elongation is the length beyond 3 inches, then you might need to find the value at an x value of 8.5, because 3 + 8.5 = 11.5. As you can see it's a trivial change.

[Elise M.]

Thank you!! This helped so much.

[Elise M.]

This makes so much more sense now- I really appreciate the help.