How to change non scalar variable to scalar variable?
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Good day, im tryin to solve a problem in matlab following steps which were given to us. I am to find the integral of an ode with the lower limit being 0 and the upper limit being x, with x having multiple values. The solution being s or (distance) which will also have multiple values. Im trying to correct the script but I am getting an error because the x limit is not scalar. This is step 3 of 8 and im ultimately trying to complete up to step 8.
This is what I have so far thanks to the help I received:
% k = air resistance coefficient
% 𝜃 (theta) = initial angle of elevation of the ball when it was kicked
% x and y are respectively the horizontal and vertical spatial coordinates
% t is the time measured from the moment the ball is kicked
% |v| = mag_v = is the magnitude of the velocity (speed, changes with time)
global g
m = 0.45 % in kg
g = 9.81 % gravitational acceleration in m/s^2
v0 = 108 % in km/h
normv = sqrt((v0*cos(theta).^2) + (v0*sin(theta).^2))
t = 0: 0.1: 2;
theta = 30*pi/180; %initial guess for theta
k = 3 %initial guess for air resistance in N
ICs = [0 0 v0*cos(theta) v0*sin(theta)]
% v = 10
% dx/dt = x1
% dx1 / dt = x2
% mx2 = (-k)|v|(x1)
% dy/dt = y1
% dy1 / dt = y2
% my2 = (-mg)(-k)|v|(y1)
% mx2 = @(t,x)[x(1); x(2); -k*mag_v*x(1)]
% my2 = @(t,y)[y(1); y(2); -m*g -k*mag_v*y(2)]
% y(3) = dx/ dt, y(4) = dy/ dt
[T,Y] = ode45(@(t,y)fun(t,y,g,k,m),t,ICs);
%plot(T,Y(:,1:2))
Y(:,1) % X(t)
Y(:,2) % Y(t)
x = [Y(:, 1)]'
y = [Y(:, 2)]'
interp1(Y(:,1), Y(:,2), 0.7)
f = @ (x) sqrt( 1 + (y(4) ./ y(3)).^2 )
quad (f, 0, x)
%s =
function dy = fun(t,y,g,k,m)
mag_v = sqrt(y(3)^2+y(4)^2);
dy = zeros(4,1);
dy(1) = y(3);
dy(2) = y(4);
dy(3) = -k*mag_v*y(3)/m;
dy(4) = (-m*g-k*mag_v*y(4))/m;
end
12 Commenti
Harsh
il 19 Mag 2023
Hi,
Yes, a least squares optimization can be implemented using a while loop.
In the following points, I explain the steps that can be leveraged to achieve the goal.
- Setting while loop stopping condition: I suggest setting the while loop stop condition according to an error tolerance value instead of stopping the while loop at a zero value. The below provides a sample code for the same:
tol = 0.00001; % Define a tolerance value for the error to stop optimization
while (dE_dtheta > tol || dE_dk > tol)
% Implement the algorithm and variable correction here.
end
- Implementing the least square optimization algorithm: The algorithm is already specified in the attached PDF document under the “LEAST SQUARES METHOD” heading. If you face specific issue in the implementation, feel free to post here.
- Correcting the variable: The correction in initial guess variables (“k” and “theta” in this case) with each "while" loop iteration is an important step of optimization. You can use gradient descent in the following way to perform the same:
alpha = 0.001; % Defines the learning rate or step size for gradient descent.
% After calculating error values in current iteration, update the optimization variables as given below.
theta = theta - alpha*dE_dtheta;
k = k - alpha*dE_dk;
Walter Roberson
il 19 Mag 2023
while V_interpolated >= 0
E = cumsum(V_interpolated - V_var).^2
end
Your while loop must update at least one of the variables that are being tested in the condition, or else you have an infinite loop.
Exception: it is valid to have a loop similar to
while true
do some calculation
if condition is met
break;
end
end
That is, updating the variable or value being tested in a while loop is not strictly necessary if you have a test and a break statement.
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