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Chapter 10 Second-order systems
10.1 Nested functions
In the Section 7.1, we saw an example of an M-file with more than one function:
function res = duck() error = error_func(10) end function res = error_func(h) rho = 0.3; % density in g / cm^3 r = 10; % radius in cm res = ... end
Because the first function ends before the second begins, they are at the same level of indentation. Functions like these are parallel, as opposed to nested. A nested function is defined inside another, like this:
function res = duck() error = error_func(10) function res = error_func(h) rho = 0.3; % density in g / cm^3 r = 10; % radius in cm res = ... end end
The top-level function, duck, is the outer function and error_func is an inner function.
Nesting functions is useful because the variables of the outer function can be accessed from the inner function. This is not possible with parallel functions.
In this example, using a nested function makes it possible to move the parameters rho and r out of error_func.
function res = duck(rho) r = 10; error = error_func(10) function res = error_func(h) res = ... end end
Both rho and r can be accessed from error_func. By making rho an input argument, we made it easier to test duck with different parameter values.
10.2 Newtonian motion
Newton’s second law of motion is often written like this
where F is the net force acting on a object, m is the mass of the object, and a is the resulting acceleration of the object. In a simple case where the object is moving along a straight line, F and a are scalars, but in general they are vectors.
Even more generally, if F and a vary in time, then they can be thought of as functions that return vectors; that is, F is a function and the result of evaluating F(t) is a vector that describes the net force at time t. So a more explicit way to write Newton’s law is
The arrangement of this equation suggests that if you know m and a you can compute force, which is true, but in most physical simulations it is the other way around. Based on a physical model, you know F and m, and compute a.
So if you know acceleration, a, as a function of time, how do you find the position of the object, p? Well, we know that acceleration is the second derivative of position, so we can write a differential equation
Where a and p are functions of time that return vectors, and ptt is the second time derivative of p.
Because this equation includes a second derivative, it is a second-order ODE. ode45 can’t solve this equation in this form, but by introducing a new variable, v, for velocity, we can rewrite it as a system of first-order ODEs.
The first equation says that the first derivative of p is v; the second says that the derivative of v is a.
Let’s start with a simple example, an object in freefall in a vacuum (where there’s no air resistance). Near the surface of the earth, the acceleration of gravity is g = −9.8 m/s2, where the minus sign indicates that gravity pulls down.
If the object falls straight down (in the same direction as gravity), we can describe its position with a scalar value, altitude. So this will be a one-dimensional problem, at least for now.
Here is a rate function we can use with ode45 to solve this problem:
function res = freefall(t, X) p = X(1); % the first element is position v = X(2); % the second element is velocity dpdt = v; dvdt = acceleration(t, p, v); res = [dpdt; dvdt]; % pack the results in a column vector end function res = acceleration(t, p, v) g = -9.8; % acceleration of gravity in m/s^2 res = g; end
The first function is the rate function. It gets t and X as input variables, where the elements of X are understood to be position and velocity. The return value from freefall is a (column) vector that contains the derivatives of position and velocity, which are velocity and acceleration, respectively.
Computing pt is easy because we are given velocity as an element of X. The only thing we have to compute is acceleration, which is what the second function does.
acceleration computes acceleration as a function of time, position and velocity. In this example, the net acceleration is a constant, so we don’t really have to include all this information yet, but we will soon.
Here’s how to run ode45 with this rate function:
>> ode45(@freefall, [0, 30], [4000, 0])
As always, the first argument is the function handle, the second is the time interval (30 seconds) and the third is the initial condition: in this case, the initial altitude is 4000 meters and the initial velocity is 0. So you can think of the “object” a a skydiver jumping out of an airplane at about 12,000 feet.
Here’s what the result looks like:
The bottom line shows velocity starting at zero and dropping linearly. The top line shows position starting at 4000 m and dropping parabolically (but remember that this parabola is a function of time, not a ballistic trajectory).
Notice that ode45 doesn’t know where the ground is, so the skydiver keeps going through zero into negative altitude. We will address this issue later.
10.4 Air resistance
To make this simulation more realistic, we can add air resistance. For large objects moving quickly through air, the force due to air resistance, called “drag,” is proportional to v2:
Where c is a drag constant that depends on the density of air, the cross-sectional area of the object and the surface properties of the object. For purposes of this problem, let’s say that c = 0.2.
To convert from force to acceleration, we have to know mass, so let’s say that the skydiver (with equipment) weighs 75 kg.
Here’s a version of acceleration that takes air resistance into account (you don’t have to make any changes in freefall:
function res = acceleration(t, p, v) a_grav = -9.8; % acceleration of gravity in m/s^2 c = 0.2; % drag constant m = 75; % mass in kg f_drag = c * v^2; % drag force in N a_drag = f_drag / m; % drag acceleration in m/s^2 res = a_grav + a_drag; % total acceleration end
The sign of the drag force (and acceleration) is positive as long as the object is falling, the direction of the drag force is up. The net acceleration is the sum of gravity and drag. Be careful when you are working with forces and accelerations; make sure you only add forces to forces or accelerations to accelerations. In my code, I use comments to remind myself what units the values are in. That helps me avoid nonsense like adding forces to accelerations.
Here’s what the result looks like with air resistance:
Big difference! With air resistance, velocity increases until the drag acceleration equals g; after that, velocity is a constant, known as “terminal velocity,” and position decreases linearly (and much more slowly than it would in a vacuum). To examine the results more closely, we can assign them to variables
>> [T, M] = ode45(@freefall, [0, 30], [4000, 0]);
And then read the terminal position and velocity:
>> M(end,1) ans = 2.4412e+03 % altitude in meters >> M(end,2) ans = -60.6143 % velocity in m/s
Exercise 1 Increase the mass of the skydiver, and confirm that terminal velocity increases. This relationship is the source of the intuition that heavy objects fall faster; in air, they do!
In the previous section, we saw that the terminal velocity of a 75 kg skydiver is about 60 m/s, which is about 130 mph. If you hit the ground at that speed, you would almost certainly be killed. That’s where parachutes come in.
Exercise 2 Modify acceleration so that after 30 seconds of free-fall the skydiver deploys a parachute, which (almost) instantly increases the drag constant to 2.7.
What is the terminal velocity now? How long (after deployment) does it take to reach the ground?
10.6 Two dimensions
So far we have used ode45 for a system of first-order equations and for a single second-order equation. The next logical step is a system of second-order equations, and the next logical example is a projectile. A “projectile” is an object propelled through space, usually toward, and often to the detriment of, a target.
If a projectile stays in a plane, we can think of the system as two-dimensional, with x representing the horizontal distance traveled and y representing the height or altitude. So now instead of a skydiver, think of a circus performer being fired out of a cannon.
According to the Wikipedia1, the record distance for a human cannonball is 56.5 meters (almost 186 feet).
Here is a general framework for computing the trajectory of a projectile in two dimensions using ode45:
function res = projectile(t, W) P = W(1:2); V = W(3:4); dPdt = V; dVdt = acceleration(t, P, V); res = [dPdt; dVdt]; end function res = acceleration(t, P, V) g = -9.8; % acceleration of gravity in m/s^2 res = [0; g]; end
The second argument of the rate function is a vector, W, with four elements. The first two are assigned to P, which represents position; the last two are assigned to V, which represents velocity. P and V are vectors with elements for the x and y components.
The result from acceleration is also a vector; ignoring air resistance (for now), the acceleration in the x direction is 0; in the y direction it’s g. Other than that, this code is similar to what we saw in Section 10.3.
If we launch the human projectile from an initial height of 3 meters, with velocities 40 m/s and 30 m/s in the x and y directions, the ode45 call looks like this:
ode45(@projectile, [0,10], [0, 3, 40, 30]);
And the result looks like this:
You might have to think a little to figure out which line is which. It looks like the flight time is about 6 seconds.
Exercise 3 Extract the x and y components of position, plot the trajectory of the projectile, and estimate the distance traveled.
Exercise 4 Add air resistance to this simulation. In the skydiver scenario, we estimated that the drag constant was 0.2, but that was based on the assumption that the skydiver is falling flat. A human cannonball, flying head-first, probably has a drag constant closer to 0.1. What initial velocity is needed to achieve the record flight distance of 65.6 meters? Hint: what is the optimal launch angle?
10.7 What could go wrong?
What could go wrong? Well, vertcat for one. To explain what that means, I’ll start with catenation, which is the operation of joining two matrices into a larger matrix. “Vertical catenation” joins the matrices by stacking them on top of each other; “horizontal catenation” lays them side by side.
Here’s an example of horizontal catenation with row vectors:
>> x = 1:3 x = 1 2 3 >> y = 4:5 y = 4 5 >> z = [x, y] z = 1 2 3 4 5
Inside brackets, the comma operator performs horizontal catenation. The vertical catenation operator is the semi-colon. Here is an example with matrices:
>> X = zeros(2,3) X = 0 0 0 0 0 0 >> Y = ones(2,3) Y = 1 1 1 1 1 1 >> Z = [X; Y] Z = 0 0 0 0 0 0 1 1 1 1 1 1
These operations only work if the matrices are the same size along the dimension where they are glued together. If not, you get:
>> a = 1:3 a = 1 2 3 >> b = a' b = 1 2 3 >> c = [a, b] ??? Error using ==> horzcat All matrices on a row in the bracketed expression must have the same number of rows. >> c = [a; b] ??? Error using ==> vertcat All rows in the bracketed expression must have the same number of columns.
In this example, a is a row vector and b is a column vector, so they can’t be catenated in either direction.
Reading the error messages, you probably guessed that horzcat is the function that performs horizontal catenation, and likewise with vertcat and vertical catenation.
These operations are relevant to projectile because of the last line, which packs dPdt and dVdt into the output variable:
function res = projectile(t, W) P = W(1:2); V = W(3:4); dPdt = V; dVdt = acceleration(t, P, V); res = [dPdt; dVdt]; end
As long as both dPdt and dVdt are column vectors, the semi-colon performs vertical catenation, and the result is a column vector with four elements. But if either of them is a row vector, that’s trouble.
ode45 expects the result from projectile to be a column vector, so if you are working with ode45, it is probably a good idea to make everything a column vector.
In general, if you run into problems with horzcat and vertcat, use size to display the dimensions of the operands, and make sure you are clear on which way your vectors go.
The flight of a baseball is governed by three forces: gravity, drag due to air resistance, and Magnus force due to spin. If we ignore wind and Magnus force, the path of the baseball stays in a plane, so we can model it as a projectile in two dimensions.
A simple model of the drag of a baseball is:
where Fd is a vector that represents the force on the baseball due to drag, Cd is the drag coefficient (0.3 is a reasonable choice), ρ is the density of air (1.3 kg/m3 at sea level), A is the cross sectional area of the baseball (0.0042 m2), v is the magnitude of the velocity vector, and V is a unit vector in the direction of the velocity vector. The mass of the baseball is 0.145 kg.
For more information about drag, see http://en.wikipedia.org/wiki/Drag_(physics).
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