I have encountered the following system of differential equations in lagrangian mechanics. Can you suggest a numerical method, with relevant links and references on how can I solve it. Also, is there a shorter implementation on Matlab or Mathematica?
mx (y dot)^2 + mgcosy - Mg - (M=m)(x double dot) =0
gsiny + 2(x dot)(y dot + x (y double dot)=0
where (x dot) or (y dot)= dx/dt or dy/dt, and the double dot indicated a double derivative wrt time.
You can create a vector Y = (x y u v)' so that
dx/dt = u
dy/dt = v
du/dt = d²x/dt²
dv/dt = d²y/dt²
It is possible to isolate the second derivatives from the equations, so you get
d²x/dt² = (m*g*cos(y) + m*x*v² - M*g)/(M-m)
d²y/dt² = -(g*sin(y) - 2*u*v)/x
Now, you can try to solve it using standard ODE solvers, such as Runge-Kutta methods. Matlab has a set of solvers, such as ode23. I didn't test he following, but it would be something like it:
function f = F(Y)
x = Y(1); y = Y(2); u = Y(3); v = Y(4);
f = [0,0,0,0];
f(1) = u;
f(2) = v;
f(3) = (m*g*cos(y) + m*x*v*v - M*g)/(M-m);
f(4) = -(g*sin(y) - 2*u*v)/x;
[T,Y] = ode23(F, time_period, Y0);
Related
I'm creating a game for kids. It's creating a triangle using 3 lines. How I approached this is I create two arcs(semi circle) from two end points of a base line. But I couldn't figure how to find the point of intersection of those two arc. I've search about it but only found point of intersection between two straight lines. Is there any method to find this point of intersection? Below is the figure of two arcs drawn from each end of the baseline.
Assume centers of the circle are (x1, y1) and (x2, y2), radii are R1 and R2. Let the ends of the base be A and B and the target point be T. We know that AT = R1 and BT = R2. IMHO the simplest trick to find T is to notice that difference of the squares of the distances is a known constant (R1^2 - R2^2). And it is easy to see that the line the contains points meeting this condition is actually a straight line perpendicular to the base. Circles equations:
(x - x1)^2 + (y-y1)^2 = R1^2
(x - x2)^2 + (y-y2)^2 = R2^2
If we subtract one from another we'll get:
(x2 - x1)(2*x - x1 - x2) + (y2 - y1)(2*y - y1 - y2) = R1^2 - R2^2
Let's x0 = (x1 + x2)/2 and y0 = (y1 + y2)/2 - the coordinates of the center. Let also the length of the base be L and its projections dx = x2 - x1 and dy = y2 - y1 (i.e. L^2 = dx^2 + dy^2). And let's Q = R1^2 - R2^2 So we can see that
2 * (dx * (x-x0) + dy*(y-y0)) = Q
So the line for all (x,y) pairs with R1^2 - R2^2 = Q = const is a straight line orthogonal to the base (because coefficients are exactly dx and dy).
Let's find the point C on the base that is the intersection with that line. It is easy - it splits the base so that difference of the squares of the lengths is Q. It is easy to find out that it is the point on a distance L/2 + Q/(2*L) from A and L/2 - Q/(2*L) from B. So now we can find that
TC^2 = R1^2 - (L/2 + Q/(2*L))^2
Substituting back Q and simplifying a bit we can find that
TC^2 = (2*L^2*R1^2 + 2*L^2*R2^2 + 2*R1^2*R2^2 - L^4 - R1^4 - R2^4) / (4*L^2)
So let's
a = (R1^2 - R2^2)/(2*L)
b = sqrt(2*L^2*R1^2 + 2*L^2*R2^2 + 2*R1^2*R2^2 - L^4 - R1^4 - R2^4) / (2*L)
Note that formula for b can also be written in a different form:
b = sqrt[(R1+R2+L)*(-R1+R2+L)*(R1-R2+L)*(R1+R2-L)] / (2*L)
which looks quite similar to the Heron's formula. And this is not a surprise because b is effectively the length of the height to the base AB from T in the triangle ABT so its length is 2*S/L where S is the area of the triangle. And the triangle ABT obviously has sides of lengths L, R1 and R2 respectively.
To find the target T we need to move a along the base and b in a perpendicular direction. So coordinates of T calculated from the middle of the segment are:
Xt = x0 + a * dx/L ± b * dy / L
Yt = y0 + a * dy/L ± b * dx / L
Here ± means that there are two solutions: one on either side of the base line.
Partial case: if R1 = R2 = R, then a = 0 and b = sqrt(R^2 - (L/2)^2) which makes obvious sense: T lies on the segment bisector on a length of sqrt(R^2 - (L/2)^2) from the middle of the segment.
Hope this helps.
While you have not stated clearly, I assume that you have points with coordinates (A.X, A.Y) and (B.X, B.Y) and lengths of two sides LenA and LenB and need to find coordinates of point C.
So you can make equation system exploiting circle equation:
(C.X - A.X)^2 + (C.Y - A.Y)^2 = LenA^2
(C.X - B.X)^2 + (C.Y - B.Y)^2 = LenB^2
and solve it for unknowns C.X, C.Y.
Not that it is worth to subtract A coordinates from all others, make and solve simpler system (the first equation becomes C'.X^2 + C'.Y^2 = LenA^2), then add A coordinates again
So I actually needed this to design a hopper to lift grapes during the wine harvest. Tried to work it out myself but the algebra is horrible, so I had a look on the web -in the end I did it myself but introduced some intermediate variables (that I calculate in Excel - this should also work for the OP since the goal was a calculated solution). In fairness this is really much the same as previous solutions but hopefully a little clearer.
Problem:
What are the coordinates of a point P(Xp,Yp) distance Lq from point Q(Xq,Yq) and distance Lr from point R(Xr,Yr)?
Let us first map the problem onto to new coordinate system where Lq is the origin, thus Q’ = (0,0), let (x,y) = P’(Xp-Xq,Yp-Yq) and let (a,b) = R’(Xr-Xq,Yr-Yq).
We may now write:
x^2 + y^2 = Lq^2 -(1)
(x-a)^2 + (y-b)^2 = Lr^2 -(2)
Expanding 2:
x^2 – 2ax + a^2 + y^2 -2ay + b^2 =Lr^2
Subtracting 1 and rearranging
2by = -2ax + a2 + b2 - Lr^2+ Lq^2
For convenience, let c = a^2 + b^2 + Lq^2 + Lr^2 (these are all known constants so c may be easily computed), thus we obtain:
y = -ax/b + c/2b
Substituting into 1 we obtain:
x^2 + (-a/b x + c/2b)^2 = Lq^2
Multiply the entire equation by b^2 and gather terms:
(a^2 + b^2) x2 -ac x + c/4 + Lq^2 b^2 = 0
Let A = (a2 + b2), B= -ac ,and C= c/4 + Lq^2 b^2
Use the general solution for a quadratic
x = (-B +-SQRT(B^2-4AC))/2A
Substitute back into 1 to get:
y= SQRT(Lq^2 - x^2 )
(This avoids computational difficulties where b = 0)
Map back to original coordinate system
P = (x+Xq, y + Yq)
Hope this helps, sorry about the formatting, I had this all pretty in Word, but lost it
I have a problen in scilab
How can I plot functions containing if and < like
function y = alpha(t)
if (t < 227.8) then
y = 0.75;
elseif (t < 300) then
y = 2.8 - 0.009 .* t;
else
y = 0.1;
end
endfunction
and
function [r]=minus_alpha(t)
r = 1 - alpha(t)
endfunction
When I use
x = linspace(0,300)
plot(x, alpha(x))
I got the error message
WARNING: Transposing row vector X to get compatible dimensions
plot2d: falsche Größe für Eingangsargument: inkompatible Größen.
Error 999 : in plot2d called by plot
Sorry for german mix. Thank you.
You can avoid explicit loop and be more efficient using the followin code
function y = alpha(t)
y=0.1*ones(t);
y(t<227.8)=0.75;
i=t>=227.8&t<300;
y(i)=2.8 - 0.009 .* t(i);
endfunction
It is really sad to see a great majority of Scilab community is not aware of vectorized operations. You can change your function to:
function y = alpha(t)
y = 0.1;
if t < 227.8 then
y = 0.75;
elseif t < 300 then
y = 2.8 - 0.009 * t;
end
y = 1 - y;
endfunction
and then use feval to broadcast the function over the sequence:
x = linspace(0, 300);
plot2d(x, feval(x, alpha));
which results:
rule of thumb if you are using for loop you need to revise your code and if someone is offering you a code where there is unnecessary for loop you shouldn't probably use it.
All the proposed answers are overcomplicated considering that the function alpha in the original demand is piecewise-affine. In Scilab in can be coded that way:
x = linspace(0,400,1000);
plot(x,linear_interpn(x,[227.8 300],[0.75 0.1]))
i.e. you just have to know the nodes coordinates (here abscissae) and value of the function at nodes. The function linear_interpn does also multilinear interpolation, it is worth knowing it guys...
If you check the output of your alpha(x), you will see that it is just a scalar (not a vector). I guess you wanted something like this, so it's necessary to iterate through t to compute each value of y based on the value of t:
clc;
clear;
function y = alpha(t)
for i=1:size(t,"*")
if t(i) < 227.8 then
y(i) = 0.75;
elseif t(i) < 300 then
y(i) = 2.8 - 0.009 * t(i);
else
y(i) = 0.1;
end
end
endfunction
x = linspace(0,300);
plot2d(x,alpha(x));
If you find the answer useful, please do not forget to accept it, so others will see that your problem is solved.
Before your answers (thank you) my workaround was a combination of indicator functions composed with floor and exp( -t^2):
function y = alpha(t)
y = floor(exp(-(t .* (t-T1)) / (T1*T1))) * 0.75
+ floor(exp(-((t-T2) .* (t- T1) / (2000)))) .* (2.8-0.009 .* t)
+ floor(exp(-((t-T2) .* (t-1000) / (200000))))*0.1
endfunction
I'm following the machine learning course on Coursera, and one of the lectures provides a contour plot of a cost function for linear regression on one variable:
Source: https://www.coursera.org/learn/machine-learning/lecture/nwpe2/cost-function-intuition-ii
I thought it would be useful from an educational standpoint to be able to reproduce this chart. I don't have any octave experience, so I would need step by step instructions that I could paste into the octave command window.
Anyone here able to help with this?
Update:
I ended up with the following:
function cost = calc_cost (theta0, theta1)
x = 1:10;
y = x.*2;
cost = arrayfun( #(t0, t1) ( 1/(length(x)) * sum( ((t0 + t1*x) - y).^2 )), theta0, theta1);
endfunction
[xx, yy] = meshgrid( -3000:50:3000, -3000:50:3000) ;
zz = calc_cost(xx, yy);
contour(xx, yy, zz )
If you are no able to rewrite your cost funktion so that it accepts matrix inputs you can use arrayfun:
function cost = calc_cost (theta0, theta1)
x = 1:10;
y = x.*2;
cost = ( 1/(length(x)) * sum( ((theta0 + theta1*x) - y).^2 ) );
endfunction
[x,y] = meshgrid (linspace(-5000,5000,20), linspace(-500,500,20));
z = arrayfun (#calc_cost, x, y);
contour (x, y, z)
print out.png
gives
I have followed the tutorial on http://www.mit.edu/people/abbe/matlab/ode.html and prepared a function as follows:
function dxy = diffxy(xy)
%
%split xy into variables in our equations
%
x = xy(1);
xdot = xy(2);
y = xy(3);
%
% define the derivatives of these variables from equations
%
xdot = xdot;
ydot = 3*x + 2*y + 5;
xdoubledot = 3 - ydot + 2*xdot;
%
%return the derivatives in dxy in the right order
%
dxy = [xdot; xdoubledot; ydot]
end
When I call it using
[T, XY] = ode45('diffxy',0,10,[0 1 0])
I get an error
??? Error using ==> diffxy
Too many input arguments.
I also tried
XY= ode45(#diffxy,[0 10],[0;1;0])
Anybody have any idea?
haven't read the whole tutorial but aren't you supposed to define your function as
function dxy = diffxy(t, xy)
where t is time vector
I am a new user of MATLAB. I want to find the value that makes f(x) = 0, using the Newton-Raphson method. I have tried to write a code, but it seems that it's difficult to implement Newton-Raphson method. This is what I have so far:
function x = newton(x0, tolerance)
tolerance = 1.e-10;
format short e;
Params = load('saved_data.mat');
theta = pi/2;
zeta = cos(theta);
I = eye(Params.n,Params.n);
Q = zeta*I-Params.p*Params.p';
% T is a matrix(5,5)
Mroot = Params.M.^(1/2); %optimization
T = Mroot*Q*Mroot;
% Find the eigenvalues
E = real(eig(T));
% Find the negative eigenvalues
% Find the smallest negative eigenvalue
gamma = min(E);
% Now solve for lambda
M_inv = inv(Params.M); %optimization
zm = Params.zm;
x = x0;
err = (x - xPrev)/x;
while abs(err) > tolerance
xPrev = x;
x = xPrev - f(xPrev)./dfdx(xPrev);
% stop criterion: (f(x) - 0) < tolerance
err = f(x);
end
% stop criterion: change of x < tolerance % err = x - xPrev;
end
The above function is used like so:
% Calculate the functions
Winv = inv(M_inv+x.*Q);
f = #(x)( zm'*M_inv*Winv*M_inv*zm);
dfdx = #(x)(-zm'*M_inv*Winv*Q*M_inv*zm);
x0 = (-1/gamma)/2;
xRoot = newton(x0,1e-10);
The question isn't particularly clear. However, do you need to implement the root finding yourself? If not then just use Matlab's built in function fzero (not based on Newton-Raphson).
If you do need your own implementation of the Newton-Raphson method then I suggest using one of the answers to Newton Raphsons method in Matlab? as your starting point.
Edit: The following isn't answering your question, but is just a note on coding style.
It is useful to split your program up into reusable chunks. In this case your root finding should be separated from your function construction. I recommend writing your Newton-Raphson method in a separate file and call this from the script where you define your function and its derivative. Your source would then look some thing like:
% Define the function (and its derivative) to perform root finding on:
Params = load('saved_data.mat');
theta = pi/2;
zeta = cos(theta);
I = eye(Params.n,Params.n);
Q = zeta*I-Params.p*Params.p';
Mroot = Params.M.^(1/2);
T = Mroot*Q*Mroot; %T is a matrix(5,5)
E = real(eig(T)); % Find the eigen-values
gamma = min(E); % Find the smallest negative eigen value
% Now solve for lambda (what is lambda?)
M_inv = inv(Params.M);
zm = Params.zm;
Winv = inv(M_inv+x.*Q);
f = #(x)( zm'*M_inv*Winv*M_inv*zm);
dfdx = #(x)(-zm'*M_inv*Winv*Q*M_inv*zm);
x0 = (-1./gamma)/2.;
xRoot = newton(f, dfdx, x0, 1e-10);
In newton.m you would have your implementation of the Newton-Raphson method, which takes as arguments the function handles you define (f and dfdx). Using your code given in the question, this would look something like
function root = newton(f, df, x0, tol)
root = x0; % Initial guess for the root
MAXIT = 20; % Maximum number of iterations
for j = 1:MAXIT;
dx = f(root) / df(root);
root = root - dx
% Stop criterion:
if abs(dx) < tolerance
return
end
end
% Raise error if maximum number of iterations reached.
error('newton: maximum number of allowed iterations exceeded.')
end
Notice that I avoided using an infinite loop.