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CEE 212--Dynamics

© Keith D. Hjelmstad 2013

Numerical Methodsintegrating the equations of motion

CEE 212—Dynamics

The Mechanics Project

Arizona State University

Keith D. Hjelmstad

Contents.

1. Doing differential equations.2. Differential and integral relations

3. Trapezoidal rule.

4. Numerical integration.

5. Dynamics.

6. Projectile motion (without drag).

7. Projectile motion (with drag, explicit).

8. MATLAB code for explicit method.

9. Implicit computation (1,2).

10. Computing the gradient.

11. The gradient of the residual (1,2).

12. Reflection of the directional derivative.

13. Derivative of the length of a vector (1,2,3).

14. Derivative of a constant times vector.

15. Projectile motion (with drag, implicit).

16. MATLAB code for the Newton “while”loop.

17. Example. MATLAB code.

18. Summary.



CEE 212--Dynamics


Numerical MethodsDoing differential equations

Doing differential equations. The equations of motion

that result from Newton’s law F = ma are ordinary

differential equations (ordinary in the parlance of

differential equations means that the only derivative is

with respect to time). We face two issues in the study ofdynamics: (1) the formulation of the equations of motion

and (2) the solution of those equations. It is the second of

these that this present set of notes is about.

When we think about “solving” a differential equation

the first thing that comes to mind is finding certain

functions that, when substituted back into the original

equations satisfy it. Indeed, that is a pretty good way tothink about what it means to have solved a differential

equation (perhaps not completely or uniquely, but a

function that “works” is a very good thing).

ox

ov

)(t x

)(t v

( )t a

That perspective, however, is far too limiting to get us very far in mechanics. It is actually pretty rare to

be able to find those special functions that satisfy differential equations, especially if those equations are

nonlinear. While we will not abandon that vision of solving a differential equation (which we generally

refer to as a classical solution to the differential equation) we will augment our toolbox with an

alternative approach— numerical integration— as a means to getting solutions to problems we need to

solve in the service of doing engineering. What we will find is that we can get some very simple

algorithms to give us results that are virtually impossible to get by any other means. The price of

admission to these methods is that we must understand their limitations and we must also understand

that we need the help of the computer to do the many repetitive calculations that come up.



CEE 212--Dynamics


( )

( ) ( )

( ) ( ) ( )

t

t t

t t t

=

= =

x

v x

a v x

Numerical MethodsDifferential and integral relations

Differential and integral relations. What we know

(see the notes on Particles I: Kinematics and Kinetics).

is that velocity is the time rate of change of position

and acceleration is the time rate of change of velocity:

( ) ( ) ( )

( ) ( ) ( )

b

a

b

a

t

b at

t

b at

t t d

t t d

τ τ

τ τ

= +

= +

∫

∫

v v a

x x v

We also know that for every differential relation there

must be an associated integral relationship. For the

above equations we can also write

The integral relationships are illustrated in the figure above. In essence, the area under the acceleration

curve between any two time points is equal to the difference in velocity at those two points; the areaunder the velocity curve between any two time points is equal to the difference in position at those two

points. So if we could estimate the area under those curves then we could carry out the integration of the

equations. That is what numerical integration is all about—approximating the area under the curves.

Students of beam theory may recognize the similarity between these relationships and the relationships

between moment, shear, slope, and deflection of beams subjected to transverse loading.

x

v

a

bt

( ) ( )b at t −v v

( ) ( )b at t −x x

at



CEE 212--Dynamics


Numerical MethodsTrapezoidal Rule

Trapezoidal rule. Let us first consider the simple

problem of the integral of a scalar function of a scalar

variable. The geometric interpretation of the integral is

the area under the curve between the two limits of

integration.

Observe in the figure at left that the area under the curve

is approximately equal to the area of the trapezoid

defined by connecting the two points on the curve f (t a)

and f (t b) with a straight line. The area of the trapezoid is

[ ]12 ( ) ( ) ( )b a a b

Area t t f t f t = − +

)(t f

t at bt

( )a f t

( )b f t

( )b

a

t

t f d τ τ ∫

If the points are close enough together this area is agood representation of the actual integral so we say

[ ]( ) ( ) ( ) (1 ) ( )b

a

t

b a a bt

f d t t f t f t τ τ β β ≈ − + −∫

This approximation is called the trapezoidal rule for integration. What is does, essentially, is replace the

task of integration with the evaluation of the function at to places (which is always easy to do). This

simple observation will create a powerful method for integrating differential equations. We cangeneralize the method by introducing the parameter β (which can take values between 0 and 1) to give

the generalized trapezoidal rule

[ ]12

( ) ( ) ( ) ( )b

a

t

b a a bt

f d t t f t f t τ τ ≈ − +∫



CEE 212--Dynamics


Numerical MethodsNumerical integration

Numerical integration. Now let us take this idea

one step further. Suppose that the curve shown in red

at left is the solution to some differential equation

(with its associated integral form):

)(t f

t nt 1nt +

( )n f t

1( )n f t +

1n nt t t +∆ = −

with (known) initial condition x(0) = xo. If I take the

two points to be t a = t n and t b = t n+1 then I can write the

integral form of the equation as

( ) ( ) ( ) ( )b

a

t

b at

x f t x t x t f d τ τ = = + ∫

To solve the differential equation let us break up the region into pieces of width ∆t = t n+1 − t n and

compute the response recursively with the generalized trapezoidal rule (probably with β = 0.5) as

1

1 ( )

n

n

t

n n t x x f d τ τ

+

+

= +

∫ ( )

n n x x t =

where we use a subscript n to represent the value of x at

the time t n (and similarly for t n+1). Also let ∆t = t n+1 − t ndenote the size of the time step.

[ ]1 1 0( ) (1 ) ( ) (0) (given)n n n n o x x t f t f t x x x β β + += + ∆ + − = =

Knowing x0 = xo we can compute x1; knowing x1 we can compute x2. We can work our way forward

until we reach some place that we want to stop. All initial value problems can be approached this way.

This algorithm produces a sequence { x0 , x1 , x2 , x3 ,…, x N } which approximates the solution x(t ) that

we seek as the solution to the differential equation. The quality of the approximation depends on ∆t .



CEE 212--Dynamics


Numerical MethodsDynamics

Dynamics. Let us finally apply numerical integration to

the equations of kinematics for particle dynamics. Recall

the integral equations and initial conditions

1 1

1 1

(1 )

(1 )

n nn n

n nn n

t

t

β β

β β

+ +

+ +

= + ∆ + −

= + ∆ + −

v v a a

x x v v

where an = a(t n), vn = v(t n), and xn = x(t n), and similarly

for the values at time t n+1.

x

v

a

1nt +

1n n+ −v v

1n n+ −x x

nt t

1

1

1

1

( ) ; (0)

( ) ; (0)

n

n

n

n

t

n n ot

t

n n ot

d

d

τ τ

τ τ

+

+

+

+

= + =

= + =∫∫

v v a v v

x x v x x

Applying the generalized trapezoidal rule gives:

Again, this algorithm gives us a means for computing the sequences:

{ } { }0 1 2 0 1 2, , , ..., and , , , ..., N N x x x x v v v v

when the acceleration is known. In applications we will get the acceleration from the equation of

motion. Hence, the boxed equations will apply to all systems. The difference from one application to

another will be the equations of motion. Note that because the integration variable is time, the

equations apply directly to vectors (e.g., x and v) that are functions of time.



CEE 212--Dynamics


Numerical MethodsProjectile motion (without drag)

Projectile without drag. This algorithm

computes the entire evolution of the motion of

the pendulum subjected to gravity forces. It

has the five key steps that all of our dynamics

codes will need.

This is the simplest possible application of

numerical integration of the equations of

motion of dynamics. The equations are linear

so it is also possible to get an exact (classical)

solution to this problem. The exact solution

(which can be used to verify the code) is

The algorithm has been implemented in

MATLAB and is called projectile.m.

212

( ) o ot t g t = + −x x v n

nF mgg −=ox

ov

)(t x

1 x

3 x

n

Read given physical problem data:

for n = 0 to N − 1 Repeat the following:

,o ox v

Read given initial conditions:

Compute initial acceleration from the

equation of motion, using initial conditions:

0 g= −a n

, , ,m c g n

Set numerical analysis parameters:

, , , , ( ) /o f f ot t t N t t t β ∆ = − ∆

1

1 1

1 1

(1 )

(1 )

n

n n n n

n n n n

g

t

t

β β

β β

+

+ +

+ +

= −

= + ∆ + −

= + ∆ + −

a n

v v a a

x x v v

end (Loop over time steps)

Compute the complete state at all points

in the time history:



CEE 212--Dynamics


Numerical MethodsProjectile motion (with drag, explicit)

Explicit version with drag. This version

adds a term to the acceleration computation

to include the drag force. Ideally, we would

like to satisfy the equation of motion (which

gives the acceleration) at n + 1. But we do not

know the new velocity at the moment we

need to compute the new acceleration. So as

an approximation we use the old velocity at

that time

1 /n n ng c v m+ = − −a n v

This explicit version avoids the solution of

the nonlinear equation of motion. For small

enough time steps it generates reasonably

accurate response.

Read given physical problem data:

for n = 0 to N − 1 Repeat the following:

,o ox v

Read given initial conditions:

Compute initial acceleration from the

equation of motion, using initial conditions:

0 /o og c v m= − −a n v

, , ,m c g n

Set numerical analysis parameters:

, , , , ( ) /o f f ot t t N t t t β ∆ = − ∆

1

1 1

1 1

/

(1 )

(1 )

n n n

n n n n

n n n n

g c v m

t

t

β β

β β

+

+ +

+ +

= − −

= + ∆ + −

= + ∆ + −

a n v

v v a a

x x v v

end (Loop over time steps)

Compute the complete state at all points

in the time history:

d c v= −F v

nF mgg −=ox

ov

)(t x

1 x

3 x

n



CEE 212--Dynamics


Numerical MethodsMATLAB code for explicit method

MATLAB code. The core time-stepping loop of

the MATLAB code for the explicit method is

%... Loop over time steps

for i=1:nsteps

%..... Update time

t = t + dt;

%..... Compute new acceleration from equation of motion

speed = norm(vold);anew = -gravity*up - c*speed*vold/mass;

%..... Compute new velocity and position by trapezoidal rule

vnew = vold + dt*(beta*aold + (1-beta)*anew);

xnew = xold + dt*(beta*vold + (1-beta)*vnew);

%..... Put current state (new) into old slot to get ready for

% new step. Note that this approach allows us to only

% store two states as we carry out the calculation

aold = anew;

vold = vnew;

xold = xnew;

end % loop over time steps

Note that we will need some additional code (see projectile.m) to take care of input of physical parameter,

numerical analysis parameters, initialization, keeping track of the history, and output graphics, but this

code is the core engine of the numerical calculation. It forms the basis of all of our dynamics codes.

1 1

1 1

(1 )

(1 )

n n n n

n n n n

t

t

β β

β β

+ +

+ +

= + ∆ + −

= + ∆ + −

v v a a

x x v v

g → gravity

m → mass

t → time

dt → ∆

t β → beta

N → nsteps

Initialized before entering loop

1 /n n ng c v m+ = − −a n v

MATLAB variable

names might differ

from pseudo-code



CEE 212--Dynamics


Numerical MethodsImplicit computation

Implicit version. The nonlinear equations of motion for the

particle with drag along with the trapezoidal integration

equations are

x

v

a

1nt +

1n n+ −v v

1n n+ −x x

nt t

We want to solve these equations at each time step (as

we have done before, except that they were linear

equations before). Before we set up Newton’s method let

us do a little algebraic manipulation of those equations to

put them into a slightly different form. To do so werecognize that everything with an n subscript is known

(from the calculation at the previous time step). So let’s

rewrite the trapezoidal equations in the following form:

The two equations on the left are the trapezoidal equations and the other four equations (to the right)

are simply definitions of terms in the equations at left. The vectors cn and bn really are nothing more

than the part of vn+1 and xn+1 that can be computed from things that are known at time step n. This

form of the equations shows that both vn+1 and xn+1 are linear functions of an+1. That means it will be

easy to substitute them back into the nonlinear equation of motion. We do that next.

1 1 1 1

1 1

1 1

)(

(1 )

(1 )

n n n n

n nn n

n nn n

m mg c v

t

t

β β

β β

+ + + +

+ +

+ +

= + =

+

= + ∆ + −= + ∆ + −

0g a a n v

v v a a

x x v v

1

1

1

1

n

n

nn

nn η

γ +

+

+

+

= +

= +

a

a

x b

v c

2(1 )n n

n

n n

n n

t t

t

β β

β

+ − ∆= + ∆

= + ∆

a

a

b x v

c v

2 2(1 )

)(1

t

t

γ β

η β

− ∆=

= − ∆

where



CEE 212--Dynamics


Numerical MethodsImplicit computation

Implicit version. In the equation of motion we need the speed

and the velocity at the new time step

Since cn and η are constant with respect to finding the next new

state, we will simply think of vn+1 and vn+1 as functions of an+1 in

the following equations. Hence the residual function for

Newton’s method can be written as

1 1 1 1 1 1( ) ( ) ( )n n n n n nm m g c v+ + + + + += + +g a a n a v a

11

1 1 1

nnn

n n nv

η ++

+ + +⋅

= +

=

av c

v v

We need to do some algebra so let us suppress the subscript

n+1 from all the terms that have it while we do the algebra.

With this abbreviated notation then

( ) ( ) ( ),

( ) , ( ) ( ) ( )n

m mg c v

vη

= + +

= + = ⋅

g a a n a v a

v a c a a v a v a

We can substitute the expressions for the velocity and speed directly into the residual function g to get

( ) ( ) ( )( ) n n nm m g c η η η = + + + ⋅ + +g a a n c a c a c a

showing that what we need to do to satisfy the equation of motion is to solve the nonlinear algebraic

equation g(a) = 0 to find the new acceleration a = an+1. This task is what Newton’s method was born to

do. The only thing we need for Newton’s method is the gradient of g. Let’s compute that next.

EE D



CEE 212--Dynamics


Numerical MethodsComputing the gradient

Now augment a by adding ε h

where h = [h1, h2, h3]T . Now take the derivative with

respect to ε using the chain rule for differentiation

because

Computing the gradient. To find the new

acceleration we need to solve a nonlinearalgebraic system of equations of the form

( ) =g a 0

To do that we need to compute the gradient of

g with respect to its argument a. We can

accomplish this task with a formula called the

directional derivative

[ ]0

( ) ( ) ( )d

Dd ε

ε ε =

⋅ = + = ∇

g a h g a h g a h

Let us first show how this is the same as the

partial derivative formula for the gradient of g

that we had used in the notes Newton’s

Method . Write out the equations in long form

1 1 2 3

2 1 2 3

3 1 2 3

( , , ) 0

( , , ) 0

( , , ) 0

g a a a

g a a a

g a a a

=

=

=

where g = [g1, g2, g3]T and a = [a1, a2, a3]

T .

1 1 1 2 2 3 3

2 1 1 2 2 3 3

3 1 1 2 2 3 3

( , , )

( , , )

( , , )

g a h a h a h

g a h a h a h

g a h a h a h

ε ε ε

ε ε ε

ε ε ε

+ + +

+ + +

+ + +

1 2 31 2 3

1,2,3i i i idg g g gh h h i

d a a aε

∂ ∂ ∂= + + =

∂ ∂ ∂

( )

1,2,3i i

i

d a h

h i

ε

ε

+

= =∂

Putting this result into matrix form we have

1 1 11

1 2 31

2 2 2 22

1 2 33

3 3 3 3

1 2 3

g g gdg

a a ad h

dg g g gh

d a a a hdg g g g

d a a a

ε

ε

ε

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

=

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂

[ ]0

( ) ( )d

d ε

ε ε =

+ = ∇

g a h g a h

So

CEE 1 D



CEE 212--Dynamics


The second term does not involve a, therefore the

derivative is zero (the derivative of a constant is

zero): D(mgn) = 0.

The third term is a product of functions that arefunctions of a. Therefore, we need the chain rule

and product rule for differentiation. Those work the

same for the directional derivative as for the

ordinary derivative.

Numerical MethodsThe gradient of the residual

The gradient of the residual. Recall the

residual function of the equation of motionis defined as

( )0n

d D

d ε η ε η

ε = = + + = v c a h h

( )2

2

2 2

D v D D

v Dv

v Dv

Dvv

η η

η

η

= ⋅ + ⋅

= ⋅ + ⋅

= ⋅

⋅=

v v v v

h v v h

v h

v h

The function has three terms added together.

The derivative of the sum is the sum of the

derivatives. So let us compute them each

individually. The first term is:

Now let us compute the directional derivative of the

speed v(a). A simple way to do this computation is to

realize that the square of the speed is the dot product

of the velocity with itself:2

v = ⋅v v

To get set up let us first compute the directional

derivatives of the velocity ( ) n η = + av a c

( ) ( ) ( )m mg c v= + +g a a n a v a

( )0

( ) d

D Dd ε

ε ε =

= ⋅ = + g g a h g a h

We need to compute the gradient of this

function. We will do it with the directional

derivative. For short let us use Dg to

represent the directional derivative of g

( ) ( )0

d D m m m

d ε ε

ε = = + = a a h h

In other words, the directional derivative of

ma in the direction h is mh.

CEE 212 D



CEE 212--Dynamics


Numerical MethodsThe gradient of the residual

The gradient of the residual (cont’d).

Putting the pieces together for the thirdterm we get

where m = v/v is the unit vector pointing inthe direction of the velocity and I is the

identity matrix. Note that the transpose of a

column matrix (vector) is a row vector so

Using the relationships we have computed

previously we get

[ ] ( ) T m cv cvη η = ∇ = + +A g I m m

Now we can finish computing the gradient of the

residual:

[ ]

( )

T

T

T

D m cv

m cv

m cv cv

η

η

η η

= + +

= + +

= + +

g h m m I h

I h m m I h

I m m h

We can identify the gradient of the residual. It is

given in the box below.

This is the gradient matrix that we need for

implementing Newton’s method to solve our nonlinear

system of equations. There is a fair amount of work to

get it, but it is essential for the algorithm.

( )

( ) ( ) ( )

D D m mg c v

D m D mg D cv

= + +

= + +

g a n v

a n v

( ) ( ) ( )

( )

( )

( )T

T

D cv c Dv cv D

c cvv

cv cvv v

cv

cv

cv

η η

η

η

η

η

η

= +

⋅ = +

⋅

= +

= ⋅ +

= +

= +

v v v

v hv h

v h v

h

m h m h

mm h h

mm I h

( ) ( ) ( )T T ⋅ = ⋅ = = m h m m m h m m h mm h

CEE 212 D i



CEE 212--Dynamics


The directional derivative is computed asNumerical MethodsReflection on the directional derivative

Reflection on the directional derivative.

The gradient of a vector valued function ofa vector variable is simply the matrix of

partial derivatives of the function with

respect to the arguments

is just a very happy result of calculus. This

relationship gives us the route to computing the

gradient through the directional derivative (and note

that we never need to assign a value to the arbitrary

vector h).

The beauty of the directional derivative approach is

that it allows a very simple way to deal with functions

whose arguments are vectors (which is very often the

case in mechanics).

which, if you get right down to it, is simply an

ordinary scalar derivative with respect to ε . For the

purposes of that derivative all of the other terms are

treated as constants (so it does not matter so much if

they are vectors or scalars—just carry them along!).

The fact that

( )0

( ) d

D Dd ε

ε ε =

= ⋅ = + g g a h g a h

In order to use this form of the computation

we have to write out in long form how each

of the functions gi depends upon all of the

independent variables ai. That is generally possible but often a huge mess of algebra.

What the directional derivative does for us

is give us a different way of organizing the

calculation.

1 1

1

1

( )

n

n n

n

g g

a a

g g

a a

∂ ∂ ∂ ∂

∇ = ∂ ∂

∂ ∂

g a

( ) [ ]0

d d ε

ε ε =

+ = ∇ g a h g h

CEE 212 D i



CEE 212--Dynamics


Numerical MethodsExample: derivative of the length of a vector

[ ]

( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

12

12

12

12

12

0

0

0

12

0

1

20

12

12

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2

d Dg g

d

d

d

d

d

d

d

ε

ε

ε

ε

ε

ε ε

ε ε ε

ε ε ε

ε ε ε ε ε

ε ε ε ε

=

=

=

−

=

−

=

−

−

⋅ = +

= + ⋅ +

= + ⋅ +

= + ⋅ + + ⋅ +

= + ⋅ + ⋅ + + + ⋅

= ⋅ ⋅ + ⋅

= ⋅ ⋅

⋅=

⋅

v h v h

v h v h

v h v h

v h v h v h v h

v h v h h v h v h h

v v h v v h

v v v h

v h

v v

chain rule

product rule

The computation of the directional derivative

of the length of a vector goes as follows

( )g = ⋅v v v

The length of a vector v

is the square root of the

vector dotted with itself

( ) Dg ⋅

⋅ =⋅

v hv h

v v

Note that there are three basic

tasks: (1) augmenting v by ε h,

(2) taking ordinary derivative

with respect to ε , (which here

involves chain rule and productrule) and (3) evaluating at ε =0.

The final result is

CEE 212 D i



CEE 212--Dynamics


Numerical MethodsDerivative of the length of a vector

T

v v

⋅ =

v vh h

[ ] [ ]T

•

= • • • • = • = ⋅ •

a b a b

[ ]( ) Dg g⋅ = ∇v h h

It is often useful to note how the dot product

notation and the matrix “transpose” notation can be interchanged. We think of a vector as being

represented by a 3 by 1 column matrix. Hence the

transpose of a vector would be a 1 by 3 row

matrix. So the dot product also looks like this

( ) Dg ⋅

⋅ = ⋅

v hv h

v v

Recall that, by the rules of matrix multiplication, a

1 by 3 multiplying a 3 by 1 gives a 1 by 1 matrix

(scalar). The dot product of vectors is a scalar. We

can also note that

Our example. Remember that we are really

trying to compute the gradient of g for use in

Newton’s method. What we got was

[ ]T • • • • = • • • • = • • •

• • • •

a b

This is called the outer product of vectors. The

order of operations is important for vectors.

What we need to have is an expression in a

form that we can factor out the gradient of g

because

[ ]T

gv

∇ =

v

Noting the matrix multiply equivalents wecan write

Therefore the gradient matrix for the length

of a vector can be written as

In other words, it is a row matrix whose

contents are the vector components

normalized by the length of the vector

CEE 212 D ami s



CEE 212--Dynamics


Numerical MethodsDerivative of the length of a vector

( )12

2 2 2

1 2 2

2 2 21 2 2 1 2 3

( )

( , , )

g

v v v

v v v g v v v

= ⋅

= + +

= + + =

v v v

1 2 3

, ,g g g

gv v v

∂ ∂ ∂∇ =

∂ ∂ ∂

( ) ( )

( )( ) ( )

( )

( ) ( )( )

12

12

12

12

12

12

2 2 2 1 111 2 2 12

2 2 21

1 2 2

2 2 2 2 211 2 2 22

2 2 22

1 2 2

2 2 2 3 311 2 2 32

2 2 23

1 2 2

2

2

2

v vgv v v v

v vv v v

v vgv v v v

v vv v v

v vgv v v v

v vv v v

−

−

−

∂= + + = =

∂ + +

∂= + + = =

∂ + +

∂= + + = =

∂ + +2 2 2

1 2 2v v v v= + +

If you are wondering how the computation of

the gradient matrix for the length of a vector looks if we just put it into component form we

can do that. The function can be written out as

We can compute the partial derivatives of g with

respect to each of the arguments as

The gradient matrix is defined as the matrix

of partial derivatives (a row matrix)

1 2 3

31 2

, ,

, ,T

g g gg

v v v

vv v

v v v v

∂ ∂ ∂∇ =

∂ ∂ ∂

= =

v

Putting in the partial derivative expressions

we just computed we get

which is exactly the same as the result we got

using the directional derivative. The choice

of which method to use depends upon the

specific function g. It is not always

immediately obvious at the start which onewill yield the cleaner derivation.

CEE 212 Dynamics



CEE 212--Dynamics


Numerical MethodsExample: derivative of a constant times a vector

[ ]

[ ]

0

0

( ) ( )d

D cd

d c

d

c

ε

ε

ε ε

ε ε

=

=

⋅ = +

= +

=

g v h v h

v h

h

[ ]c c=h I h

[ ] [ ]c∇ =g h I h

Let’s take a look at what happens when we

compute the directional derivative of thefunction

( ) c=g v v

where c is a constant scalar. We want to take the

derivative of g with respect to v. The derivation

goes as follows

That is a pretty simple result, but remember thatwe are looking for the gradient of g to use in

Newton’s method

[ ]( ) Dg ⋅ = ∇v h g h

The gradient should be a 3 by 3 matrix.

multiplying the arbitrary vector h. It looks

like what we got was only the scalar c

multiplying h. We can fix this situation by

noting that any vector that multiplies theidentity matrix is the same vector. Hence,

we can write

Now we can write the directional derivative

in a form that allows us to identify the

gradient matrix

[ ] [ ]

0 0

0 00 0

c

g c cc

∇ = =

I

Hence,

Note that in MATLAB the 3 by 3 identitycan be create as I = eye(3) .

CEE 212--Dynamics



CEE 212--Dynamics


Numerical MethodsProjectile motion (with drag, implicit)

Set parameters:

, , ,

( ) /

(1 )

itmax 10

tol 1. 8

o f

f o

t t t

N nsteps t t t

t

e

β

η β

∆

= = − ∆

= − ∆=

= −

Algorithm. For the implicit calculation we

need to implement Newton’s method. Thatrequires a few additional parameters

Initialize to prepare for Newton loop

( )

1

1

11 1

1

; /

n n

n

T

n n

its its

v v

m m g c v

m cv cv

err

η

η η

+

+

−+ +

= +

= +

= ⋅ =

= + +

= + +

= −

= ⋅

v c a

v v m v

g a n v

A I mm

a a A g

g g

end (Newton “while” loop)

At the current time step compute the state:

1

1; 0;

n n n

n n

err its

t β

+

= =

= + ∆

=

c v a

a a

while (err > tol) and (its < itmax)

Update position and velocity

1 1

1 1

(1 )

(1 )

n n n n

n n n n

t

t

β β

β β

+ +

+ +

= + ∆ + −

= + ∆ + −

v v a a

x x v v

Newton “while” loop

The time-stepping setup is exactly the same

as the explicit calculation. The computation

of the state is a bit more complex. Theexplicit method estimated an+1 as

1 /n n ng c v m+ = − −a n v

This single line gets replaced by the

Newton “while” loop at right.

CEE 212--Dynamics



CEE 212--Dynamics


%... Set initial values for Newton loop

err = 1; its = 0;

cn = vold + beta*dt*aold;

anew = aold;

%... Newton iteration to compute the acceleration anew

while (err > tol) && (its < itmax)

its = its + 1;

vnew = cn + eta*anew;

speed = norm(vnew);

coeff = c*speed*eta;

g = mass*(anew + gravity*up) + c*speed*vnew;

m = vnew/speed;

A = (mass + coeff)*eye(3) + coeff*m*m';

anew = anew - A\g;err = norm(g);

end % while

vnew = vold + dt*(beta*aold + (1-beta)*anew);

xnew = xold + dt*(beta*vold + (1-beta)*vnew);

Numerical MethodsMALAB code for the Newton “while” Loop Variables defined earlier in

the code include:

Previous state:xol d, vol d, aol d

Numerical integration

parametersbet a, et a, dt

Newton iteration parameters

t ol , i t max

Physical constantsc, gr avi t y, mass, up

Arrays:

3 by 1 matricesxol d, vol d, aol d, cnxnew, vnew, anew, g

3 by 3 matrix:A

MATLAB Coding. To help to fix the ideas relative to how those

algorithms fit in the context of a MATLAB code the following

segment has been “clipped” from the projectile motion code that

treats drag with the implicit approach.

CEE 212--Dynamics



CEE 212 Dynamics


Numerical MethodsExample (MATLAB code)

The algorithm gives the following result for

the given conditions. Note the effect of thefluid resistance (red) on the projectile. The

result w/o fluid (blue) is computed by

integrating the equations of motion directly.

*-------------------------------------------*

| PROJECTILE MOTION PROGRAM |

| (Implicit Version) |

| K. D. Hjelmstad |

| |

| Original Version: 07.29.2012 |

| Latest Update: 09.01.2013 |

*-------------------------------------------*

PHYSICAL PROPERTIES

Acceleration of gravity 9.8100

Drag coefficient (rho.Cd.Ap/2) 0.0020

Mass 2.0000

INTEGRATION OF EQUATIONS

Initial time (to) 0.000Final time (tf) 12.000

Time increment (dt) 0.020

Numerical integration (beta) 0.500

Number of time steps 600

Number of steps to report 600

Newton iteration tolerance 1.00e-07

Maximum Newton iterations 10

INITIAL CONDITIONS

Initial position (x) 0.000

Initial position (y) 0.000

Initial position (z) 55.000

Initial velocity (x) 25.000

Initial velocity (y) 15.000

Initial velocity (z) 50.0000 50 100 150 200 250 300

0

20

40

60

80

100

120

140

160

180

200

x

z

Trajectory z vs. x

CEE 212--Dynamics



CEE 212 Dynamics

© K h D H l d 0 3

Numerical Methods

SummarySummary. The classical solution of differential equations is

great for linear systems and is often difficult if not impossible

for nonlinear differential equations. The equations of dynamics

are, most of the time, nonlinear.

In this set of notes we introduce the idea of numerical

integration of the equations of motion using a technique called

the generalized trapezoidal rule to take care of the derivative

relationships between position, velocity, and acceleration.

We observe that when we use the trapezoidal rule in

conjunction with the equation of motion, the set of equations

boils down to a nonlinear algebraic equation that needs to be

solved at each time step. We can use Newton’s Method (see the

previous set of notes) to take care of this task.

When we set up the algorithm we need to solve the problem is

involves a loop over time steps and inside of that loop a Newton

“while” loop to converge on the solution to the nonlinear

algebraic equation for that particular time step.

ox

ov

)(t x

)(t v

( )t a

One of the first things that we realize when we take this approach is that numerical methods exchangecalculus computations for algebra computations. The cost of making that favorable switch is that there

are lots and lots of computations. For this reason, we always use the computer to carry out these

computations. The framework set up in these notes will hold for all of the Computing Projects in this

course. The only thing that will change is the equation of motion.

6 dynamics numerical methods

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