7) Science of Bloons

It is all trajectory.

Physics of trajectories

A familiar example of a trajectory is the path of a projectile such as a thrown ball or rock. In a greatly simplified model the object moves only under the influence of a uniform homogenous gravitational. This can be a good approximation for a rock that is thrown for short distances for example, at the surface of the moon. In this simple approximation the trajectory takes the shape of a parabola. Generally, when determining trajectories it may be necessary to account for nonuniform gravitational forces, air resistance. This is the focus of the discipline of ballistics.

One of the remarkable achievements of Newtonian mechanics was the derivation of the laws of Kepler, in the case of the gravitational field of a single point mass (representing the Sun). The trajectory is a conic section, like an ellipse or a parabola This agrees with the observed orbits of planets and comets, to a reasonably good approximation. Although if a comet passes close to the Sun, then it is also influenced by other forces, such as the solar wind and radiation pressure, which modify the orbit, and cause the comet to eject material into space.

Newton's theory later developed into the branch of theoretical physics known as classical mechanics. It employs the mathematics of differential calculus (which was, in fact, also initiated by Newton, in his youth). Over the centuries, countless scientists contributed to the development of these two disciplines. Classical mechanics became a most prominent demonstration of the power of rational thought, i.e. reason, in science as well as technology. It helps to understand and predict an enormous range of phenomena. Trajectories are but one example.

Consider a particle of mass

*m*, moving in a potential field

*V*. Physically speaking, mass represents inertia, and the field

*V* represents external forces, of a particular kind known as "conservative". That is, given

*V* at every relevant position, there is a way to infer the associated force that would act at that position, say from gravity. Not all forces can be expressed in this way, however.

The motion of the particle is described by the second-order differential equation

with

On the right-hand side, the force is given in terms of

, the gradient of the potential, taken at positions along the trajectory. This is the mathematical form of Newton's second law of motion: mass times acceleration equals force, for such situations.

**Examples****Uniform gravity, no drag or wind**Trajectories of two objects thrown at the same angle. The blue object doesn't experience any

The case of uniform gravity, disregarding and wind, yields a trajectory which is a parabola. To model this, one chooses

*V* =

*m**g**z*, where

*g* is the acceleration of gravity. This gives the equations of motion

The present example is one of those originally investigated by Galileo Galilei. To neglect the action of the atmosphere, in shaping a trajectory, would (at best) have been considered a futile hypothesis by practical minded investigators, all through the Middle Ages in Europe. Nevertheless, by anticipating the existence of the vacuum, later to be demonstrated on Earth by his collaborator Evangelista Torricelli Galileo was able to initiate the future science of mechanics. And in a near vacuum, as it turns out for instance on the Moon, his simplified parabolic trajectory proves essentially correct.

Relative to a flat terrain, let the initial horizontal speed be

, and the initial vertical speed be

. It will be shown that, the range is

, and the maximum altitude is

. The maximum range, for a given total initial speed

*v*, is obtained when

, i.e. the initial angle is 45 degrees. This range is

, and the maximum altitude at the maximum range is a quarter of that.

**Derivation**The equations of motion may be used to calculate the characteristics of the trajectory.

Let

be the position of the projectile, expressed as a vector

be the time into the flight of the projectile,

be the initial horizontal velocity (which is constant)

be the initial vertical velocity upwards.

The path of the projectile is known to be a parabola so

where

are parameters to be found. The first and second derivatives of

*p* are:

At

*t* = 0

so

.

This yields the formula for a parabolic trajectory:

(Equation I: trajectory of parabola).

**Range and height**The

**range** *R* of the projectile is found when the

*z*-component of

*p* is zero, that is when

which has solutions at

*t* = 0 and

*t* = 2

*v*_{v} /

*g* (the

**hang-time of the projectile**). The range is then

From the symmetry of the parabola the

**maximum height** occurs at the halfway point

*t* =

*v*_{v} /

*g* at position

This can also be derived by finding when the

*z*-component of

*p*' is zero.

**Angle of elevation**In terms of angle of elevation θ and initial speed

*v*:

giving the range as

This equation can be rearranged to find the angle for a required range

(Equation II: angle of projectile launch)

Note that the sine function is such that there are two solutions for θ for a given range

*d*_{h}. Physically, this corresponds to a direct shot versus a

mortar shot up and over obstacles to the target. The angle θ giving the maximum range can be found by considering the derivative or

*R* with respect to θ and setting it to zero.

which has a non trivial solutions at

. The maximum range is then

. At this angle

*s**i**n*(π / 2) = 1 so the maximum height obtained is

.

To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height

*H* =

*v*^{2}*s**i**n*(θ) / (2

*g*) with respect to θ, that is

which is zero when

. So the maximum height

is obtain when the projectile is fired straight up. The equation of the trajectory of a projectile fired in uniform gravity in a vacuum on Earth in Cartesian coordinates is

,

where

*v*_{0} is the initial speed,

*h* is the height the projectile is fired from, and

*g* is the acceleration due to gravity).

**Uphill/downhill in uniform gravity in a vacuum**Given a hill angle α and launch angle θ as before, it can be shown that the range along the hill

*R*_{s} forms a ratio with the original range

*R* along the imaginary horizontal, such that:

(Equation 11)

In this equation, downhill occurs when α is between 0 and -90 degrees. For this range of α we know: tan( − α) = − tanα and sec( − α) = secα. Thus for this range of α,

*R*_{s} /

*R* = (1 + tanθtanα)secα. Thus

*R*_{s} /

*R* is a positive value meaning the range downhill is always further than along level terrain. The lower level of terrain causes the projectile to remain in the air longer, allowing it to travel further horizontally before hitting the ground.

While the same equation applies to projectiles fired uphill, the interpretation is more complex as sometimes the uphill range may be shorter or longer than the equivalent range along level terrain. Equation 11 may be set to

*R*_{s} /

*R* = 1 (i.e. the slant range is equal to the level terrain range) and solving for the "critical angle" θ

_{cr}:

Equation 11 may also be used to develop the "rifleman's rule" for small values of α and θ (i.e. close to horizontal firing, which is the case for many firearm situations). For small values, both tanα and tanθ have a small value and thus when multiplied together (as in equation 11), the result is almost zero. Thus equation 11 may be approximated as:

And solving for level terrain range,

*R* "Rifleman's rule"

Thus if the shooter attempts to hit the level distance R, s/he will actually hit the slant target. "In other words, pretend that the inclined target is at a horizontal distance equal to the slant range distance multiplied by the cosine of the inclination angle, and aim as if the target were really at that horizontal position

**Derivation based on equations of a parabola**The intersect of the projectile trajectory with a hill may most easily be derived using the trajectory in parabolic form in Cartesian coordinates (Equation 10) intersecting the hill of slope

*m* in standard linear form at coordinates (

*x*,

*y*):

(Equation 12) where in this case,

*y* =

*d*_{v},

*x* =

*d*_{h} and

*b* = 0

Substituting the value of

*d*_{v} =

*m**d*_{h} into Equation 10:

(Solving above x)

This value of x may be substituted back into the linear equation 12 to get the corresponding y coordinate at the intercept:

Now the slant range

*R*_{s} is the distance of the intercept from the origin, which is just the hypotenuse of x and y:

Now α is defined as the angle of the hill, so by definition of

tangent,

*m* = tanα. This can be substituted into the equation for

*R*_{s}:

Now this can be refactored and the trigonometric identity for

may be used:

Now the flat range

*R* =

*v*^{2}sin2θ /

*g* = 2

*v*^{2}sinθcosθ /

*g* by the previously used trigonometric identity and sinθ / cosθ =

*t**a**n*θ so:

Bloons Guide V1.2