Maximizing the Range with Newton's Method

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Maximizing the Range of a Ballistic Projectile

Now we return to the problem posed in Section 1: Ballistics, finding the maximum range of the ballistic trajectory. Given that the object was launched at altitude of 10 meters with speed meters/sec, and writing the acceleration due to gravity as

we asked which angle will cause our projectile to travel the greatest horizontal distance. For that, we observed that the correct angle would be that for which

You saw that such a will be a solution to the equation:

Exploration: Maximizing the range with Newton's method

Suppose we approximate the solution to this equation using Newton's method. You should be on the Solving Equations with Newton's Method page. You will find a control panel:

with the equation in x set to . In the future, you may enter any differentiable function of x, number of steps, initial guess and so on to do Newton's method here.

Where did the equation come from? If you go to the symbolic calculator you will find the following script in the pink script window:

readexact;
make g 49/5;
make v 50;
make alt 10;
'Define the range function as a function of the angle, x'
hor(x) := (v^2*sin(2*x)+2*v*cos(x)*sqrt(v^2*sin(x)^2+2*alt*g))/(2*g);
'Now define the derivative of the range function as a function of the angle, x'
dhor(x) := value d(hor(x),x);
readfloat;
precision 15;

We made that definition also in the book script, so it was defined when the book opened. But we placed the script in the script window also so that you can later come back and change the values of speed v and altitude alt in the lines:

make v 50;
make alt 10;

Then you can remake the definition with these parameter values. To do that, just press the button.

In any case, the panel is already set up to use Newton's Method to approximate the maximum range when the speed v is 50 meters and the altitude h is 10 meters. It will do 20 steps, starting with the first guess: . All of that information is in the control panel.

Now, press the button and you will see:

Starting at , the system "converged" to the solution angle in radians 0.766533025682704 (to 15 place precision) after just 5 steps. This is roughly 43.919107229012077 degrees. The yellow graph on the left "flatlines" very quickly as the sequence converges. So the answer for these parameters is less than 45 degrees (0.785398163397448 radians). If you change the altitude and velocity, you will of course get different answers. There is nothing in Newton's Method that will tell us that for sufficiently fast projectiles and positive altitude, it is always less than 45 degrees. We will come to that question with a more theoretical approach in a moment.

Compare this to the answer we obtained earlier in the Ballistics microworld, where we simply asked the symbolic calculator for a solution:

calc roots(dhor(x), 50, [0,1.57]);

The calculator returned 0.76653302568272075 radians or 43.919107229013036 degrees. These answers agree to 13 decimal places. In the calculator, if you type

calc hor(0.76653302);

You will see: 264.91336705606466. Now try it with some neighboring angles.

If you return to the Ballistics Lab and use this angle (get the scroll bar menu, Settings to set the angle to 43.9191072)

you will see 264.9133670 meters. Try some adjacent angles.

Now, you may see this in a more dramatic way using the original Newton's Method page. Just press the button to go to that page. It has a different control panel.

Replace x^2-200 with dhor(x), and replace the initial x = 11 with x = 1. So the control panel should now look like:

And you are ready to go. Press the Graph f(x) button and you will see:

The vertical red line and the message in the MathEdit shows . Not very close to 0. Now press the Next x button a few times and you will see the convergence to 0.766533025682704 radians at the fifth step, just as before.

Now, we would like to answer the question whether for positive altitudes, the maximizing angle will always be less than 45 degrees (0.785398163397448 radians). The solution using Newton's Method above required that the altitude h = 10 and the speed V = 50 be fixed in advance. Suppose we want an "engineering" solution to the problem in which we seek a solution in terms of V and h. Here, we want to leave V and h variable, but would like an approximate angle for maximum range in terms of V and h . We also would like conditions on V and h that guarantee that this approximation is as close as we like to the actual value. We use differential approximation to solve this problem.

First, we make an observation about the relationship between the altitude h and the velocity V for vertical projectiles. This will help us decide how to approximate the answer.

Question 1: Show that if a projectile, fired vertically from ground level is to attain a height of meters, it must be launched with velocity at least

Solution: In this case, the height attained is:

from Galileo's formula, where is the initial velocity. The maximum height is attained when and so this height is

Solving for in terms of we see that

End of Solution

This consequence of conservation of energy tells us that the initial altitude h becomes less and less important as we choose V so large that is very small.

Now our earlier formula for the time the projectile is in the air given angle (derived in Section 1: Ballistics) says

Notice that

To see that, we consider the function

Certainly,

Maximizing the range using differential approximation

By the method of differential approximation, if is small, then (considering constant)

Now

and so

Now let

Substituting, we see that

Thus, we may write an approximate formula for the time the object will remain in the air:

Our earlier formula for the time the projectile is in the air is

Now we just saw that

This approximation is good as long as

is small

And we know from energy considerations that it is reasonable to make this assumption, because it means that the initial altitude h is small compared to the height to which a vertical projectile would rise from the ground with initial velocity V.

Of course, is not exactly equal to , so we are no longer even trying to find an exact solution. But the horizontal distance the object will travel over this interval of time is

and this may be simplified to

Differentiating this, we see that the derivative will be

Now we would like to solve the (simpler) equation

for and to come to some conclusions about the solution. First of all, we see that

entails

This is a quartic equation in . But since it is also quadratic in you may first solve for then take the square root. Next take the arcsine of the answer. This will give an angle that we think maximizes the distance under these circumstances. Remember our initial assumption that

As , and this means that if is the height to which a projectile fired vertically will rise from the ground then

Now, in the equation

let . Then this becomes the equation

We may solve this quadratic equation to get:

Since is very small by assumption, we may also assume that

so that we may take the square root. There are two answers then.

and

The angles will be

and

Now when is small

(or )

The other answer is very close to 0, and does not represent a solution. If you approximate this numerically with, say, we see that is approximately equal to 0.780545641194 radians. So the correct angle in this case is a little

smaller than radians (or ). The advantage of this approach is that, given altitude h, we can conclude that for V large enough, the maximizing angle is less than (but as close as we like to) .

This is not bad at all, but it depends, of course, on the circumstance that we can solve this (quartic) equation exactly.