Question

Hw8 #1, do I integrate by hand and then by MatLab? Where did we talk about integration?

Answer 1

Well integration by MatLab was done in HW1. You can check those solutions if you don't remember. To integrate by hand, just write out the three equations and then integrate them separately.

Answer 2

I see what you mean with Hw#1 but how does doing it by hand work?

Answer 3

Well write out the three equations explicitly for a start

Answer 4

\[\dot g = g * \xi \]

g is the translation matrix, a 3x3
\[\xi\] is the body velocity, a 3x1

the product is a 3x1, in this case 3 numbers

Answer 5

I do not get 3 equations

Answer 6

I meant instead of matrix form write out the equations in regular form.

Answer 7

\[g = \left[\begin{matrix}g11 & g12 & g13\\g21 & g22 & g23 \\ g31 & g32 & g33\end{matrix}\right]\]

\[\xi =\left(\begin{matrix}\xi1 \\ \xi2 \\ \xi3\end{matrix}\right)\]

\[\dot x = g11*\xi1 +g12*\xi2 +g13*\xi3\]
\[\dot y = g21*\xi1 + g22*\xi2 + g23*\xi3\]
\[\dot \theta = g31*\xi1 + g32*\xi2 + g33*\xi3\]

but I have numbers that reduce these equations down to single numbers

Answer 8

Ok just plug in the g11 etc symbolically and get rid of the factors that are 0

Answer 9

You only have numbers that reduce this down to a single numbers in the initial state.

Answer 10

yes, in the initial state.

Answer 11

So, I will always start with these 3 equations and the apply the initial conditions from the point after integration.

Answer 12

Yes just integrate the equations with respect to time and use the initial conditions to figure out constants.

Answer 13

\[\xi \]
is the body velocity. I use this for my equations.

Answer 14

Yep

Answer 15

I integrate form 0 to tau

Answer 16

0 to pi with respect to tau

Answer 17

\[x = \pi \cos \theta +\pi \sin \theta + C\]
This is what I get for x. C is my constant for initial conditions. I know mathematically how \[\pi \] is in front and \[\theta \] has not been evaluated. But, why is pi in front and why has theta not been evaluated?

Answer 18

Does it represent some unknown area under a graph?

Answer 19

I think you forgot the tau in your solution.

Answer 20

Oh I'm sorry you substituted it

Answer 21

Ok so you should be able to get C using the initial condition right?

Answer 22

\[C = -\pi \cos \theta - \pi \sin \theta \]

Answer 23

This means x = 0 always.

Answer 24

Oh I see what you did. You need to figure out C before you substitute that the final time is pi. so\[x = \tau*\cos(\theta)+\tau*\sin(\theta)+C\]

Answer 25

Now say x(0) = 0 and determine C

Answer 26

Is \[x(0), x(\tau) \]?

Answer 27

x(0) is x at tau = 0.

Answer 28

So what is in the ( ) is tau

Answer 29

Yes

Answer 30

I have class

Answer 31

Are you saying C = 0?

Answer 32

for x(0).

Answer 33

If C = 0 for x(0) C=0 all the time. C is just a constant

Answer 34

I get C = - tau.

Answer 35

for x

Answer 36

Ok so C is 0 then right?

Answer 37

for y I get

\[C = \frac{ 0.5 }{ \sin \theta - \cos \theta } - \tau \]

Answer 38

Not correct. What is your equation for y(tau)

Answer 39

I did my indefinite integral incorrectly. My C should absorb any multiples tac on to it. I see that C = 0. I still have problems understanding why I wait to evaluate tau until I do.

Answer 40

So I get a Cx, Cy and Ctheta

Answer 41

Ok do you agree that ydot = sin(theta)-cos(theta)?

Answer 42

yes.

Answer 43

Then do you also agree that y = tau*sin(theta)-tau*cos(theta)+C?

Answer 44

yes

Answer 45

Ok to figure out C, we simply plug in known conditions. In the problem we are given that at tau=0, y = .5. Therefore C = .5.

Answer 46

To figure out y at t=pi, then you would plug in pi for tau.

Answer 47

So at tau=pi y = pi*sin(theta)-pi*cos(theta)+.5. You can figure out theta by integrating thetadot and plugging in pi for tau. I have to go now. I'll check your questions this evening.

Answer 48

do I do the same for x?

Answer 49

Yes of course. x,y,and theta. You have a speed in x, speed in y, and speed in theta. You want to figure out position in x, position in y, and position in theta after certain time.

Answer 50

Thanks.

Answer 51

later

Answer 52

I do not understand the part of #1 that askes if the output makes sense. What does a 'make sense' response look like?

Answer 53

For problem #1, is the MatLab section required or is it there for information reasons?

Answer 54

You are required to plotthe trajectory in MatLab for part 1 yes. As for yourfirst question...you are given an initial positions in x,y,theta and a speed in x,y,theta. Your final answer is your final position in x,y,theta. You should be able to guess what the final configuration will be given the initial and the speed. Compare your guess against the actual results. Also compare the two xis you were given and how they affected the final configuration.

Answer 55

How do we start Problem 2? Do we need to find \[\xi (\hat) \]? And then implement Rodrigues Formula?

Answer 56

I believe there was a formula discussed in class for the closed form solution of the exponential of xi. You can see it here for convenience: http://i.imgur.com/Mcq11.png

Answer 57

I am refreashing myself on the workings of ode45

Answer 58

The problem/solution from homework 1 should be helpful.

Answer 59

I get a line with a positive slope when I plot \[\dot x = \cos \theta + \sin \theta \]

Answer 60

I took the ode45 of
\[\dot x = \cos (5\pi/12) + \sin (5\pi/12)\]
and plotted it. That is where I got a straight line with a positive slope.

Answer 61

So, R(xi3*tau) = [cos(pi/4) -sin(pi/4); sin(pi/4) cos(pi/4]
How do we obtain the Jacobian of {x1; x2} ?

Answer 62

*Jacobian of {xi1, xi2}?

Answer 63

Confused about problem #1 ploting

Answer 64

Will I have an integration m.file for xdot, ydot and thetadot? 3 intergration m.files.

Answer 65

My equations are

\[\dot x = \cos \theta + \sin \theta \]
\[\dot y = \sin \theta - \cos \theta \]
\[\dot \theta = 1/4\]
where does my tau come into play? How do I get the angle theta without solveing the 3 equations?

Answer 66

Ok can we make 2 threads here? billingsley you start another question on the left so I don't get confused. Bebo... I don't understand why you plugged in 5*pi/12 for tau in the differential equation. In this case it just evaluates to a constant and of course the integration will be linear.

Answer 67

That is where I thought theta went. My theta is 5pi/12.

Answer 68

Your final theta may be 5pi/12 but your theta is changing. Not constant.

Answer 69

Ok but how?

Answer 70

How is your theta changing?

Answer 71

Theta changes from 0 to 2pi. This is because you want to see how the body changes in one full rotation.

Answer 72

I'm not following you. Thetadot is given as 1/4 correct?

Answer 73

yes.

Answer 74

This means theta is 1/4*tau correct?

Answer 75

yes

Answer 76

Theta(tau) = (1/4)tau + C

Answer 77

with initial conditions
theta(tau) = (1/4)tau + (pi/6)

Answer 78

Ok well then you know theta is changing in time...right?

Answer 79

I do understand that that is what would happen but I do not see how I can work a changing theta into the equation without changing the equation.

Answer 80

Post your code of how you defined the differential equation in matlab.

Answer 81

function xdot = f(t, x)

xdot = cos(t) + sin(t);

end

function ydot = f(t, y)

ydot = sin(t) - cos(t);

end

function thdot = f(t, x)

thdot = 1/4;

end

I have 3 m.files

Answer 82

Your functions shouldn't input or use t...ode45 takes care of that. You can just have 1 mfile that inputs x. Where x(1)=x x(2)=y x(3)=theta and you should define xdot(1), xdot(2), xdot(3)

Answer 83

function xdot = f(t, x)

xdot = zeros(3, 1);
xdot(1) = cos(t) + sin(t);
xdot(2) = sin(t) - cos(t);
xdot(3) = 1/4;

end

Answer 84

should be cos(theta) not t

Answer 85

ok

Answer 86

Should it be
function xdot = f(theta, x)

xdot = zeros(3, 1);
xdot(1) = cos(theta) + sin(theta);
xdot(2) = sin(theta) - cos(theta);
xdot(3) = 1/4;

end

if so how does tau = pi get in here?

Answer 87

no the inputs are still t,x
theta = x(3)
when you call ode45 you define the timespan as [0 pi] thats how that gets in there

Answer 88

working

Answer 89

I have a plot but it does not look right.