The Parabola of Safety – Unveiling the Firework’s Outer Limits (Part 2)

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In Part 1 of our pyrotechnic physics journey, we marveled at how, at any given instant, the fragments of an exploding firework form a perfect, falling sphere. We can also look at a different perspective of the picture. Each of those glittering stars traces its own distinct parabolic path through the night sky. If you could see all these paths traced out simultaneously over their entire flight, there seems to be a dome or a ultimate boundary that contains them all.

This boundary is what we call the “envelope of safety” or the “parabola of safety.” It’s the curve that no projectile from that initial burst (given a fixed initial speed) can cross. Understanding this envelope is not just an academic curiosity; it has real-world implications for ensuring the safety of spectators and infrastructure during a display.

So, welcome back! In this second and concluding part, we’ll dive deep into the mathematics and physics required to define this crucial outer limit.

1. Visualizing the Challenge: The Envelope of Trajectories

a. The Intuitive Image

Imagine a time-lapse photo of a single firework shell bursting, but instead of dots of light, each fragment leaves a continuous glowing trail from the moment of explosion until it fizzles out or hits the ground. You’d see a beautiful, intricate web of overlapping parabolic arcs, all originating from the same point but fanning out in different directions.

While the individual stars form a sphere at any instant, the collection of their entire paths sketches out a different, larger shape. This overall boundary is what we’re after.

b. Conceptualizing the Problem

Our goal is to find a single curve, which we can denote as:
Y_e = G(X)
This curve “envelops” all possible individual projectile trajectories. Each individual trajectory can be described by y = f(x, θ), where θ is the launch angle (or some other parameter related to it) and v₀ is the initial speed. No projectile launched from the origin with this initial speed should be able to reach a point outside this envelope.

2. Mathematical Toolkit: Understanding Envelope Curves

What exactly is an envelope curve in mathematics? Consider a family of curves, where each curve in the family is distinguished by a parameter (like our launch angle θ). An envelope curve has some fascinating properties:

a. Properties of an Envelope Curve Yₑ = G(X)

Let our family of curves be described by an equation involving a parameter θ (which we’ll later relate to our launch angle):
y = f(x, \theta)
The envelope, Yₑ = G(X), has the following characteristics:

  1. Tangency: At every point on the envelope, it is tangent to one of the curves in the family y = f(x, θ). The envelope “just kisses” each curve it encloses, or is made up of these points of tangency.
  2. Extremal Property: For a given x (or y), the corresponding point on the envelope often represents an extreme value (maximum or minimum) of y (or x) for the family of curves at that point. In our firework scenario, for any horizontal distance x from the burst, the envelope will give the maximum possible height Yₑ that any projectile can reach. So,
    G(x) = \max_{t} {f(x, \theta)}
  3. Derivative Condition: This extremal property leads to a powerful mathematical condition. If we’re looking for the maximum y for a given x by varying the parameter θ, we can find this by setting the partial derivative of f(x, θ) with respect to θ to zero. Also because all points on the envelope curve are also on at least one of the original equation for the family of curves. So all points on envelope curve need to satisfy the following system of equations:
    \begin{cases} y - f(x, \theta) &= 0\ \frac{\partial f(x, \theta)}{\partial \theta} &= 0 \end{cases}
    Solving this system of equations, eliminating θ, gives the equation of the envelope. This is the formal form of deriving the envelope of a family of curves y=f(x,θ).
  4. Converging Intersections (Locus of Ultimate Intersections): Another intuitive understanding of the envelope curve is to see it as the convergence of the intersection of two curves differing by a small change in θ.

Two curves in the family for a given θ but differing by a small change Δ θ can be described as below: \begin{cases} y &= f(x, \theta) \ y &= f(x, \theta + \Delta \theta)\end{cases}As Δ θ approaches zero, the intersection points of these neighboring curves approach a point on the envelope. The envelope is the collection of all locus of these ultimate intersection points for all starting angle θ.

These properties give us several avenues to mathematically derive the envelope of our firework trajectories.

3. Back to the Physics: The Projectile’s Path

Let’s recall the trajectory equation for a single projectile launched from the origin (0,0) with initial speed v₀ at an angle θ to the horizontal (ignoring air resistance, and with gravity g acting in the negative y-direction). Let t be the time of flight for the projectile.

  1. Parametric Equations:
    The position components are given by:
    \begin{cases} x(t, \theta) &= v_0t \cos \theta\ y(t, \theta) &= v_0 t \sin \theta - \frac{1}{2} g t^2 \end{cases}
  2. Trajectory Equation y = f(x, θ):
    As we have done in Part 1, we can get the trajectory equation y = f(x, θ). To get the path in terms of x and y, we eliminate t from the x equation:
    t = \frac{x}{v_0 \cos \theta}
    Substituting this into the y equation:
    y = (v_0 \sin \theta) \left(\frac{x}{v_0 \cos \theta}\right) - \frac{1}{2} g \left(\frac{x}{v_0 \cos \theta}\right)^2
    This simplifies to:
    y = x \tan \theta - \frac{g x^2}{2 v_0^2 \cos^2 \theta}
    Here we go one step further, using the following trigonometry identity
    \frac{1}{\cos^2 \theta} = \sec^2 \theta = 1 + \tan^2 \theta
    So, we get the trajectory equation:
    y = x \tan \theta - \frac{g x^2}{2 v_0^2} (1 + \tan^2 \theta)
    This quadratic equation of y(x) describes the parabolic path of a single projectile launched at angle θ.
  3. Substitution with m = \tan θ:
    To simplify differentiation or algebraic manipulation, let the parameter m be:
    m = \tan \theta
    The parameter m can range from −∞ to +∞, covering all possible launch angles (except directly up/down, which are limiting cases). Our family of curves becomes y = f(x, m):
    f(x, m) = y = mx - \frac{g x^2}{2 v_0^2} (1 + m^2)
    This is the equation we will work with to find the envelope, with m as our parameter.

4. Finding the Parabola of Safety: Multiple Paths to One Truth

Now, let’s employ the different mathematical strategies suggested by the properties of envelope curves.

a. The Extrema Method: Maximizing Height for Each x (by Completing the Square)

For any given horizontal distance x from the launch point, we want to find the launch parameter m = \tan θ that maximizes the height y. This maximum y will be a point on the envelope. Our trajectory equation is:
y(x, m) = mx - \frac{g x^2}{2 v_0^2} (1 + m^2)
Let’s rearrange this into a standard quadratic form in terms of m:
y(x, m) = -\left (\frac{g x^2}{2 v_0^2}\right) m^2 + x m - \left(\frac{g x^2}{2 v_0^2}\right)
For a fixed x, this is an equation of the quadratic form y = -P m^2 + Q m - Rwhere the coefficients are:
\begin{cases} P(x) &= \frac{g x^2}{2 v_0^2} \quad (\text{Coefficient of } m^2)\ Q(x) &= x \quad (\text{Coefficient of } m)\ R(x) &= \frac{g x^2}{2 v_0^2} \quad (\text{Coefficient of } m) \end{cases}
Since g, x^2, v₀^2 are positive (assuming x ≠ 0), the coefficient P(x) is always positive. This means that for a fixed x, the graph of y versus m is a parabola opening downwards. The maximum value of y occurs at the vertex of this parabola.

We can find the vertex by completing the square, and we get the following form:
\begin{align} y &= -P \left(m - \frac{Q}{2P}\right)^2 + R - \frac{Q^2}{4P} \ &\le R - \frac{Q^2}{4P} \end{align}
As we discussed above in the property of envelope curve, the maximum value of y gives us the envelope Yₑ , so
\begin{align} Y_e &= R - \frac{Q^2}{4P} \ &= -\frac{g x^2}{2 v_0^2} - \frac{x^2}{-\frac{2g x^2}{v_0^2}} \ &= -\frac{g x^2}{2 v_0^2} + \frac{v_0^2}{2g} \end{align}
The condition when y reaches maximum is when the square term in y(m) is zero, or: \begin{align} m &= \frac{Q}{2P} \ &= \frac{x}{2\frac{gx^2}{2v_0^2}} \ &= \frac{v_0^2}{gx} \end{align}
We will come back to this condition of m in the next subsection using the derivative method.
This algebraic method of completing the square confirms the result we would get using calculus, showing that for any horizontal distance x, this Yₑ is the maximum possible height a projectile can achieve.

b. The Derivative Method (Formal Envelope Condition)

This method uses calculus. The general method for finding an envelope of a family of curves described by f(x,y,m)=0 (or, if y is an explicit function of x and m, y − F(x,m) = 0) is to solve the system of equations:
\begin{cases} y-F(x,m) &= 0 \ \frac{\partial F(x,m)}{\partial m} &= 0 \end{cases}
In our case, we have:
F(x,m) = mx - \frac{g x^2}{2 v_0^2} (1 + m^2)The partial derivative with respect to m is:
\frac{\partial F(x,m)}{\partial m} = x - \frac{g x^2}{2 v_0^2} (2m) = x - \frac{g x^2 m}{v_0^2}
Setting this to zero to find the m that (for a fixed x) corresponds to a point on the envelope:
\begin{align} x - \frac{g x^2 m}{v_0^2} &= 0 \ m = \frac{v_0^2}{gx} \end{align}
This is exactly the same condition as what we derived above to reach yₘₐₓ . Substituting this back into y = F(x,m) yields the same envelope:
Y_e = \frac{v_0^2}{2g} - \frac{g}{2 v_0^2} x^2

c. The Discriminant Method: Where Solutions Merge

Rearrange the trajectory equation y = mx − \frac{g x^2}{2 v₀^2} (1 + m^2) into a quadratic equation in terms of m:
\left(\frac{g x^2}{2 v_0^2}\right) m^2 - x m + \left(y + \frac{g x^2}{2 v_0^2}\right) = 0
This is a quadratic equation of the form
Am^2 + Bm + C = 0with coefficient:
\begin{cases} A &= \frac{g x^2}{2 v_0^2} \ B &= -x \ C &= y + \frac{g x^2}{2 v_0^2} \end{cases}
For a given point (x, y):

  • If this quadratic equation has two distinct real solutions for m, it means there are two different launch angles θ (corresponding to m = \tanθ) that can reach the point (x, y). This point is inside the envelope.
  • If it has no real solutions for m, the point (x, y) is unreachable. This point is outside the envelope.
  • If it has exactly one real solution for m (a repeated root), the point (x, y) lies on the envelope. This is the boundary condition.

The condition for a quadratic equation to have exactly one real solution is that its discriminant Δ must be zero. So,
\begin{align} \Delta = B^2 - 4AC &= 0 \ (-x)^2 - 4 \left(\frac{g x^2}{2 v_0^2}\right) \left(y + \frac{g x^2}{2 v_0^2}\right) &= 0 \ x^2 - \frac{2g x^2}{v_0^2} \left(y + \frac{g x^2}{2 v_0^2}\right) &= 0\end{align}
Assuming x ≠ 0 (the launch point x=0, y=0 is a special case on the envelope; if x=0, then y=0 is the only solution), we can divide by x^2:
\begin{align} 1 - \frac{2g}{v_0^2} \left(y + \frac{g x^2}{2 v_0^2}\right) &= 0 \ 1 - \frac{2gy}{v_0^2} - \frac{2g^2x^2}{v_0^4} &= 0\end{align}
Multiply by v₀^4:
v_0^4 = 2gy v_0^2 + g^2 x^2
Now, solve for y (which will be Y_E on the envelope):
y = \frac{v_0^4}{2g v_0^2} - \frac{g^2 x^2}{2g v_0^2}
Or:
Y_e = \frac{v_0^2}{2g} - \frac{g}{2 v_0^2} x^2
Once again, we arrive at the same parabolic equation for the envelope of safety!

d. The Intersection of Neighboring Curves Method

Now one last method! As illustrated above, the envelope curve is the convergence of non-origin intersection of two infinitesimally close members of the family of projectile curves.

In this case, we consider two trajectories with slightly different parameters, m and m + δ m. Their equations are:
\begin{cases} y = mx - \frac{g x^2}{2 v_0^2} (1 + m^2) \ y = (m+\delta m)x - \frac{g x^2}{2 v_0^2} (1 + (m+\delta m)^2) \end{cases}
We find their intersection by solving this system of equations:
mx - \frac{g x^2}{2 v_0^2} (1 + m^2) = (m+\delta m)x - \frac{g x^2}{2 v_0^2} (1 + m^2 + 2m\delta m + (\delta m)^2)
Simplifying this gives:
\begin{align} (\delta m)x - \frac{g x^2}{2 v_0^2} (2m\delta m + (\delta m)^2) &= 0\ x - \frac{g x^2}{2 v_0^2} (2m + \delta m) &= 0 \end{align}
Now, let δ m → 0. The intersection point approaches the envelope, and the equation becomes:
\begin{align} 0 = x - \frac{g x^2}{2 v_0^2} (2m) &= 0 \ m = \frac{v_0^2}{gx} \end{align}
This is the same condition to get the boundary as we saw over and over again. Substituting this value of m back into the original trajectory equation will, as we’ve seen, yield the envelope equation:
\begin{align} Y_e &= mx - \frac{g x^2}{2 v_0^2} (1 + m^2) \ &= \frac{v_0^2}{2g} - \frac{g}{2 v_0^2} x^2 \end{align}
This method highlights the envelope as the locus of points where infinitesimally close members of the family of curves meet.

5. The Grand Unveiling: Summary and Reflections

It’s quite remarkable! Four different mathematical perspectives on the problem of finding the outer boundary of firework trajectories all converge on the exact same solution:
Y_e = \frac{v_0^2}{2g} - \frac{g}{2 v_0^2} x^2
This is the equation of a parabola, often called Torricelli’s parabola or the parabola of safety.

a. The Beauty of Convergent Paths

More important than memorizing the formula is appreciating the underlying concepts. The fact that we can approach this problem by maximizing height (algebraically or with calculus), by looking for the boundary of solvability for launch angles, or by examining the meeting points of infinitesimally different trajectories—and all methods agree—is a testament to the consistency and elegance of mathematical physics. Each method provides a different shade of understanding to the nature of the envelope.

b. Significance of the Parabola of Safety

This isn’t just a neat mathematical result.

  • Firework Safety: For pyrotechnicians, knowing this boundary (and adding appropriate safety factors for things like wind, dud shells, and projectile spread) is critical for determining safe viewing distances for the audience and protecting nearby structures.
  • Military Applications: Historically, understanding projectile envelopes was crucial for artillery. The “parabola of safety” defines the maximum range and the boundary of the area that can be targeted by a gun with a given muzzle velocity. Any point outside this 3D paraboloid of revolution (if we rotate our 2D parabola around the vertical axis, assuming launches in all azimuthal directions) is safe from direct fire.
  • Sports: Think of a baseball player trying to throw a ball as far as possible, or a shot putter. The optimal launch angle (around 45^∘ for maximum range on level ground, which is related to this envelope) comes from this same physics.

c. Viewing Fireworks with New Eyes (Again!)

In Part 1, we saw the instantaneous sphere of light. Now, as you watch those beautiful arcs, try to visualize not just the individual paths, but also that invisible, overarching parabolic cloak – the parabola of safety – that contains them all. It’s a silent testament to the predictable laws governing motion, even amidst the seemingly chaotic beauty of an explosion.

This concludes our deep dive into the physics of firework projectiles. Hopefully, you’ve gained a new appreciation for the intricate dance of math and physics that lights up our celebrations!

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