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How would you plot a graph which a ball then rolls down (say a y=x^2 graph)

David K.
on 27 Aug 2019

I attached a .m file I created that should work pretty well. It uses a numerical solution by calculating the slope at discrete time steps and determining the acceleration. I only considered a point mass so I ignored rotation. It can also take both static and kinetic friction into account (im slightly worried I messed up here but I think it is right). The function takes in a function handle for what you want the slope to be. I made it so that the part of the function you want the ball on needs to be between x = 0 and 10 but that should be easily changed by changing the function itself.

I did not get around to allowing arguments to be dropped from the function or personally checking arguments and throwing errors.

Let me know how it works for you!

Jim Riggs
on 29 Aug 2019

OK. I see what you are doing. This might work for the frictionless case. (I regret to say that, for security reasons, I do not have Matlab on this machine, so I cannot download and run your code. I would like to try it out, but my Matlab is on an isolated network)

On a side note, I would sugest that you get in the habit of using vector forms whenever possible, avoiding trig forms, as vector calculations are much more robust, and do not have issues when changing quadrants. For example, I once worked on a missile program where the guidance law was coded using a lot of trig expressions. A navigation error in early flight tests caused the missile to believe that it went below the ground level while in terminal homing on the target, resulting in a negative altitude calculation. This caused an unexpected sign change in the guidance law, and the missule started flying away from the target, missing by a significant distance. This error would have been avoided using vector equations, and it would have hit the target even with the navigation error. You might argue that it's reasonable that missiles should't be required to work under the ground, but in the real world, these things happen, and it's always best to hit the target when possible.

For the problem at hand, I would define a unit tangent vector for the curve:

Ut = [Utx, Uty] = [dx, dy] / mag[dx, dy] (where dy is obtained from the numerical first derivative formula).

Now the unit normal vector Un = [Unx, Uny] = [ -Uty, Utx]

Use these two unit vectors and the dot product to obtain the tangent and normal quantities. No trig function required, and it works for all directions with no singularities.

This will also allow you to apply energy conservation principles because you are retaining the full velocity vector. In a frictionless environment the system energy remains constant, and the total velocity should be a function of the Y position, as potential energy is converted to kinetic energy, and vice-versa. You can use this relationship to test the accuracy of your solution, but only if you calculate the total velocity, and not just the x component.

So, I assure you that if you use vectors in place if trig functions, you are more likely to hit the target :)

Oh, and for the record, it does make a difference using central vs. fwd/bkwd difference for the slope calculation. Try it and see how it compares. Use conservation of energy as the accuracy metric and see how the step size effects it.

David K.
on 29 Aug 2019

Welp, I went about trying to calculate frictionless case for checking the velocity. My velocity function based on conservation of energy was

mghmax-mgh = 1/2 mv^2

v = sqrt(2*g*(hmax-h));

Then I tried to calculate the velocity in the y direction and it went pretty terribly. I couldn't do the acceleration in the same way so I used the difference in y position. And the velocities were super off the expected values at some places. It seems that xvelocity was pretty close if I use the expected x part of the expected velocity but the y part is very off.

I guess to make this a decent answer I will need to change it at some point. Ug, I really thought modeling each time step as a slope would work.

btw, the only reason I would need to be in the habit of vector forms is for making better answers here since I do not do this stuff for work xD.

Jim Riggs
on 29 Aug 2019

Yes, it's very important to have a way to check yourself. It sounds like you understand the problem and you are on the right track.

Another way to test your ramp assumption/approximation is to actually use a linear function for the curve, and see if it works there. If not, there is some implementation issue.

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Jim Riggs
on 29 Aug 2019

Edited: Jim Riggs
on 29 Aug 2019

Here is a model for the kinematics.

clear data % I use data to save values in the time loop

func = @(x) x.^2; % this is the user function

% set model parameters and initial conditions

dx = 0.001; % step used to compute numerical derivatives

dt = .01; % integration time step

tstart = 0; % starting value of time

tstop = 12; % time to stop

x = 0.1; % starting value of X

y = func(x) % starting value of Y

speed = 40; % initial speed

grav = 9.806; % acceleration due to gravity

G = [0, -grav]; % gravity vector

nstep = (tstop-tstart)/dt; % numer of calculation steps

cnt = 1; % itteration counter

% initial energy state (per unit mass)

Ep = gravity*y; % potential energy

Ek = 0.5*speed^2; % kinetic energy

Etot = Ep + Ek; % total system energy

% initialize saved data table

% (you can modify this to save the parameters that you want)

data = zeros(nstep,8);

data(1,1) = tstart;

data(1,2) = x;

data(1,3) = func(x); % truth value of y

data(1,4) = y; % numerical approximation of y

data(1,5) = speed;

data(1,6) = Ep;

data(1,7) = Ek;

data(1,8) = Etot;

% calculation time loop

cnt = 1;

while cnt <= nstep

time = cnt*dt + tstart;

dy = (func(x+dx/2) - func(x-dx/2)) / dx; % first derivative

deltax = dx; % step change in X value

deltay = dy*dx; % corresponding change in Y value

mag = sqrt(deltax^2+deltay^2); % magnitude of step change

% compute the unit tangent vector

Tx = deltax/mag;

Ty = deltay/mag

T = [Tx, Ty]; % unit tangent vector

% compute accelerations

At = dot(G, T); % acceration in the tangential direction

% update states (numerical integration)

speed = speed + At*dt;

delta = speed*dt; % dstance traveled along curve

x = x + delta*Tx; % updated X position

y = y + delta*Ty; % updated Y position

% update energy states

Ep = gravity*y;

Ek = 0.5*speed^2;

Etot = Ep + Ek;

cnt=cnt+1

% save data

data(cnt,1) = time;

data(cnt,2) = x; % X position

data(cnt,3) = func(x); % truth Y position

data(cnt,4) = y; % calculated Y position

data(cnt,5) = speed;

data(cnt,6) = Ep; % potential energy

data(cnt,7) = Ek; % kinetic energy

data(cnt,8) = Etot; % total energy

end % end of time loop

% extract data vectors from data table

time = data(:,1);

X = data(:,2);

fX = data(:,3);

Y = data(:,4);

speed = data(:,5);

Ep = data(:,6);

Ek = data(:,7);

Etot = data(:,8);

% Plot data

figure;

plot(X,fX,'r'); % truth Y vs. X

hold on;

plot(X,Y,'ob'); % calculated Y vs X

grid on;

legend('fX','Y');

%.... (add additional plots as desired)

Note that for the function Y = X^2, with X starting near zero there must be a nonzero initial velocity or it won't do anything.

Also note the use of unit vectors and dot product (no trig functions).

Speed is a signed quantity. Positive speed causes x to increase. As the point rises along the curve, the speed drops to zero, and then goes negative as it falls backward.

Numerical integration errors will result in errors in the conservation of energy. Etot will not be constant, but will drift due to integration errors. Try different values for dt and you will see different ammounts of error in the energy conservation. (smaller dt should produce smaller errros). The plot of Y vs X will also become closer to the plot of fX vs X as the time step becomes smaller.

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