Question

Can you explain how to code for homework 2 #3b. How do I write a code for a changing alpha1 and alpha2 and get qe = [xe, ye] out?

Answer 1

In problem 2a, you were able to figure out qe based on alpha1 and alpha2. In 3b, you just need to generalize this such that you just update the alphas and the qe is automatically computed for you. From 3a, you should have the equation that updates each alpha as a function of t. So essentially you need to plug in these alphas into your equation for calculating xe and ye and then graph each versus time. Does this help? Let me know if it's still unclear.

Answer 2

I no longer think I have done 3a correct. I think I ploted alphadot1(xdot1) and alphadot2 (xdot2).

function xdot = f(t, x)

xdot = zeros(2,1);
xdot(1) = .5.*cos(t);
xdot(2) = 1/3;

end

How do I use the code stubs provided to plot alpha1 and alpha2?

Answer 3

What you pasted there looks right...you just need to integrate it just like you did in HW1 using ode45. You could also just integrate it yourself. alpha1 would just be the integral of .5*cos(t) and alpha2 is the integral of 1/3. Keep in mind that both are subject to the initial conditions. That should give you alpha1 and alpha2 as functions of t and you should be able to graph them using the same technique you used in homework 1.

Answer 4

I see now that the code I above gives me alpha1 and alpha2. I need to write code to calculate qe. Something like

xe = cos(alpha1) + 0.5*cos(alpha1 + alpha2);
ye = sin(alpha1) + 0.5*sin(alpha1 + alpha2);

but I am not clear as how I pass alpha1 and alpha2 to a function get xe and ye out and plot xe and ye versus time. Two vectors into a function (alpha1, alpha2) three vectors out (xe, ye, time) and have xe and ye connected to time.

Answer 5

You will have to define the time vector to calculate the alpha1 and alpha2. It doesn't need to be returned by the function because you already have it. All you need to do is pass alpha1 and alpha2 into the function and get xe and ye. Then plot each against the time vector which you already have.

Answer 6

I think I follow. I will try it and get back to you within 15 min. Thanks.

Answer 7

Sounds good. Let me know

Answer 8

Something is going wrong.

case 'a'
tspan = [0 , 40];
x0 = [pi/4 0];
[t, x] = ode45( @hw2_3a , tspan , x0 );

case 'b'
[qe] = fkin2R(x(:,1), x(:,2));


function fkin2R = f(qe)

fkin2R(1) = cos(x(1)) + .5.*cos(x(1) + x(2));
fkin2R(2) = sin(x(1)) + .5.*sin(x(1) + x(2));
qe = [fkin2R(1) fkin2R(2)];

end

I am getting an error message 'Index exceeds matrix dimensions.'.

fkin2R(1) is xe and fkin2R(2) is ye
x(:,1) is alpha1 and x(:,2) is alpha2

I want to get qe as a 2 column vector but it is only coming out as a 1 column vector.

Answer 9

For one thing you defined your function as having one input and you gave it 2 inputs. If you want the function to be input the vector of alphas you should pass it [x(:,1); x(:,2)]. Also your function does not use its input, it uses a variable which isn't available to it. Remember the function can only use inputs you pass it. In your case, this would be qe. So inside the function it should be using qe[1] or qe[2] instaed of x(1), x(2). On another note, qe is the output, not the input. The proper name for the function would be function qe = fkin2R(alpha)

Answer 10

I think I follow. I am working on it now. I will be back with results. Thanks

Answer 11

Just to be a little more clear the function should look like this:

function qe = fkin2R(alpha)

qe(1) = function of alpha
qe(2) = function of alpha
qe = [qe(1); qe(2)];

end

Answer 12

this line may not even be necessary: qe = [qe(1); qe(2)];

Answer 13

working on it

Answer 14

Yeah you're fine. I'm not waiting on you or anything

Answer 15

I get to graphs of horizontal dots. I do not think that is correct.

doPart = 'b';

switch doPart

case 'a'
tspan = [0 , 40];
x0 = [pi/4 0];
[t, x] = ode45( @hw2_3a , tspan , x0 );


case 'b'

alpha = [x(:,1) x(:,2)];
[qe] = fkin2R(alpha);



figure(1);
plot(t, qe(1));


figure(2);
plot(t, qe(2));

end

function qe = fkin2R(alpha)

qe(1) = cos(alpha(1)) + .5.*cos(alpha(1) + alpha(2));
qe(2) = sin(alpha(1)) + .5.*sin(alpha(1) + alpha(2));

end

Answer 16

can you paste your code from @hw2_3a?

Answer 17

Oh I htink i see why...you are only doing part "b" You really need to do everything. Remove the switch...case part of the code.

Answer 18

think*

Answer 19

I still get horizontal ldots.

Answer 20

post you hw23a code

Answer 21

doPart = 'b'; % Set to 'a' or 'b' as desired.

switch doPart

case 'a'
tspan = [0 , 40];
x0 = [pi/4 0]; % Should be a column vector.
[t, x] = ode45( @hw2_3a , tspan , x0 );

figure(1);
plot(t, x(:,1)* 180/pi);
title('\alpha1 vs Time');
xlabel('time (s)');
ylabel('\theta (degrees)');
grid on;

figure(2);
plot(t, x(:,2) * 180/pi); % Factor is to convert to degrees from radians.
title('\alpha2 vs Time');
xlabel('time (s)');
ylabel('\theta (degrees)');

case 'b'
tspan = [0 , 40];
x0 = [pi/4 0]; % Should be a column vector.
[t, x] = ode45( @hw2_3a , tspan , x0 );

alpha = [x(:,1) x(:,2)];
[qe] = fkin2R(alpha);

figure(1);
plot(t, qe(1));
title('Xe vs Time');
xlabel('time (s)');
ylabel('Xe');
grid on;

figure(2);
plot(t, qe(2));
title('Ye vs Time');
xlabel('time (s)');
ylabel('Ye');


end

part a

function xdot = f(t, x)

xdot = zeros(2,1); % Force to be a 2x1 (column) vector.
xdot(1) = .5.*cos(t); % Add in differential equations.
xdot(2) = 1/3;

end

part b

function qe = fkin2R(alpha)

for i = 1 : 81

qe(:,1) = cos(alpha(:,1)) + .5.*cos(alpha(:,1) + alpha(:,2));
qe(:,2) = sin(alpha(:,1)) + .5.*sin(alpha(:,1) + alpha(:,2));

end

end

Answer 22

None of your code actually happens because the doPart variable is never set. You need to remove the case 'a' and case 'b' and switch doPart lines

Answer 23

Or put back the line where you set the doPart variables if you like

Answer 24

Oh I'm sorry I see now you're doing it fine

Answer 25

does not the

doPart = 'b'

at the top of the code make the program do case 'b' ?

Answer 26

Yes. Never mind on that. Does the part a work?

Answer 27

yes. I get two plots and no errors.

Answer 28

One is a sinusoid and the other an upward sloping line yes?

Answer 29

I did get that

Answer 30

Ok good. In your fkin2R function you dont need to have a loop. The way you're figuring out qe is a vector operation and all the operations already occur.

Answer 31

I changed the fkin2R code

function qe = fkin2R(alpha)

qe(1) = cos(alpha(1)) + .5.*cos(alpha(1) + alpha(2));
qe(2) = sin(alpha(1)) + .5.*sin(alpha(1) + alpha(2));

end

Answer 32

I do still get horizontal dots.

Answer 33

alpha = [x(:,1) x(:,2)]; => define this with a semicolon: alpha = [x(:,1); x(:,2)];

Answer 34

Ok done

Answer 35

Did that help?

Answer 36

I still get horizontal dots for my plots.

Answer 37

Ok see what output you're getting for qe. Just take the semicolon away from the end of the line where you fall fkin2R and see what kind of numbers you get. Are they all the same or different?

Answer 38

call*

Answer 39

qe returns

qe =

0.7071 1.2071

Answer 40

so it's only returning 2 values? It should be a whole matrix right

Answer 41

yes

Answer 42

Take away that semicolon I told you to put in and try it again

Answer 43

I removed the semicolon and qe returns

qe =

0.7071 1.2071

Answer 44

Hmm ok no difference. Ok take away the semicolon at the end of this line: alpha = [x(:,1) x(:,2)]; and we can see what the alphas look like

Answer 45

I did. I get two colomns of numbers for alpha.

Answer 46

Ok that's good. somewhere we are losing the rest of the data...

Answer 47

whats the output of this line? cos(alpha(:,1))

Answer 48

a column of numbers between 0.2 to 0.9

Answer 49

function qe = fkin2R(alpha)

qe(:,1) = cos(alpha(:,1)) + .5.*cos(alpha(:,1) + alpha(:,2));
qe(:,2) = sin(alpha(:,1)) + .5.*sin(alpha(:,1) + alpha(:,2));

end

That should be your fkin2R function right

Answer 50

yes

Answer 51

The most recent thing you posted, you removed the colons

Answer 52

I just put them back and ran the program and got horizontal dots

Answer 53

I'm out of battery. I'll look at it some more during class

Answer 54

thanks

Answer 55

Still no luck?

Answer 56

Oh I got it. When you plot, you not specifying (;,1) (;,2). It should look like this: plot(t, qe(:,1));

Answer 57

You see you're getting the right values but only plotting the first x and first y value

Answer 58

I got two graphs that are sinusoidal.

Answer 59

Well they're not purely sinusoidal right... I mean they have peaks and valleys but the amplitude is not the same or anything. It definitely has a sinusoidal aspect to it.

Answer 60

They are diffierent from eachother but they both follow the pattern of repeating three different peak amplitudes.

Answer 61

Well what do you think? Does it make sense?

Answer 62

I have been at this one problem for so long I no longer understand what it should look like.

Answer 63

I now need to figure out part c, can you help?

Answer 64

part c is just plotting x vs y. A parametric plot just plots both x and y as a position of time. Since you've defined the x and y according to the same time vector, it's just plotting x vs. y. This will basically show you the path the end effector would take.

Answer 65

I think I have that. Can you explain how I can understand the results of part b. How would I know if my plots are in the ballpark or not?

Answer 66

Well one thing you could do is calculate a couple points by hand and see if they match the graph. the first point is easy enough. Also if you think about what the second joint is doing...it's going in a constant speed but since it's revolute it goes around in a circle. So you would expect there to be some repetition in the end effector position.

Answer 67

Look at the first point for instance. alpha1 is 45 and alpha2 is 0 so the whole manipulator would be a straight line at 45 degrees right. This would imply that the x and y should be the same for the first point

Answer 68

My part c looks like an infinity symbol. I think that is correct. Thank you for your help.

Answer 69

No problem happy to help