Envelope Equation: Reachable Points and Safe Regions

1. The Reachability Condition

The envelope equation gives us a powerful and direct way to test if a point $(X,Y)$ can be hit. Since the envelope represents the maximum possible height at any horizontal distance, a point is reachable if and only if its height $Y$ is less than or equal to the height of the envelope at $X$.

The Condition for Reachability: A point $(X,Y)$ is reachable by a projectile launched from the origin with speed $v_0$ if:
$Y\le \frac{v_0^2}{2g}-\frac{g{X}^2}{2v_0^2}$

• If $Y<y_{env}(X)$, the point is inside the envelope and can be hit using two different launch angles.
• If $Y=y_{env}(X)$, the point is on the envelope and can be hit with exactly one launch angle.
• If $Y>y_{env}(X)$, the point is outside the envelope and is unreachable.

This inequality is the key to solving all problems related to safe and unsafe regions.

2. Application: Defining Safe and Unsafe Zones

Let's apply this condition to a practical scenario. Imagine an artillery gun at the origin with a fixed muzzle velocity $v_0$. We want to determine the "danger zone" on a horizontal battlefield.

Problem: An enemy tank is moving along the x-axis. What is the maximum horizontal distance the gun can target?

Solution:
The tank is at a height $Y=0$. We apply the reachability condition:
$0\le \frac{v_0^2}{2g}-\frac{g{X}^2}{2v_0^2}$
Rearranging to solve for $X$:
$\frac{g{X}^2}{2v_0^2}\le \frac{v_0^2}{2g}$
$X^2\le \frac{v_0^4}{g^2}$
$|X|\le \frac{v_0^2}{g}$
This result confirms that the maximum horizontal range on level ground is $R_{max}=\frac{v_0^2}{g}$. The "danger zone" on the ground extends from $-\frac{v_0^2}{g}$ to $+\frac{v_0^2}{g}$. This aligns with the x-intercepts we identified in the last lesson.

The concept has significant real-world implications, as highlighted in the article we reviewed.

3. Advanced Application: Deriving a Non-Standard Envelope

Now for a problem that truly tests your understanding. What if the projectiles are not all launched from the same point? In this case, our standard formula for the parabola of safety is insufficient. We must return to the fundamental method of finding an envelope, typically the discriminant method, which you appreciate for its algebraic rigor.

Consider the following complex scenario, which is a classic advanced problem.

As you've seen, the motion of a particle leaving the disk at an angle $\theta$ is described by:
$x(t,\theta )=v_0\cos \theta \cdot t-R\sin \theta$
$y(t,\theta )=v_0\sin \theta \cdot t-\frac{1}{2}g{t}^2-R\cos \theta$
where $v_0=R{\omega }_0$.

Our goal is to find the envelope of this family of trajectories. The family is parameterized by $\theta$. The strategy, as in our discriminant method derivation, is to eliminate the parameter $\theta$ to get a relationship between $x$, $y$, and $t$.

The video shows a clever way to do this. Rearrange the equations:
$x+R\sin \theta =(R{\omega }_0)t\cos \theta$
$y+\frac{1}{2}g{t}^2+R\cos \theta =(R{\omega }_0)t\sin \theta$
To eliminate $\theta$, we can isolate the terms with $\theta$ and use the identity ${\sin }^2\theta +{\cos }^2\theta =1$. This requires some algebraic manipulation. The video takes a slightly different path but arrives at the same result. Let's follow the video's next step.

After eliminating $\theta$, we arrive at an equation that has the form of a quadratic in $t^2$:
$A(t^2{)}^2+B(t^2)+C=0$
where the coefficients A, B, and C depend on $x$, $y$, $g$, $R$, and ${\omega }_0$. Specifically, the video shows:
$\left(\frac{g^2}{4}\right)(t^2{)}^2+(gy-R^2{\omega }_0^2)t^2+(x^2+y^2-R^2)=0$

This is the pivotal moment. This equation must have a real, non-negative solution for $t^2$ for the point $(x,y)$ to be reachable. The boundary of the reachable region—the envelope—corresponds to the case where there is exactly one solution. This is precisely the discriminant condition: $\Delta =0$.

The video applies this exact logic to find the final envelope equation.

By setting the discriminant of the quadratic in $t^2$ to zero, $(gy-R^2{\omega }_0^2{)}^2-4(\frac{g^2}{4})(x^2+y^2-R^2)=0$, and solving for $y$, we get the equation of this new, non-standard envelope. The final result from the video is:
$y_{env}=\frac{R^2{\omega }_0^2}{2g}+\frac{g}{2{\omega }_0^2}-\frac{g}{2R^2{\omega }_0^2}{x}^2$
This is, once again, a downward-opening parabola, but its coefficients are quite different from the standard parabola of safety, reflecting the more complex physical situation.

This exercise demonstrates that the true power lies not in memorizing the formula for the envelope, but in understanding the fundamental principle—be it the derivative or the discriminant method—that allows you to derive it in any situation.

Conclusion

Today we have successfully bridged the gap between the theory of the envelope and its practical application. We have seen how it provides a definitive test for reachability and allows us to map out safe and unsafe regions.

Key Takeaways:

• A point $(X,Y)$ is reachable if it satisfies the inequality $Y\le \frac{v_0^2}{2g}-\frac{g{X}^2}{2v_0^2}$. This inequality is the basis for solving problems involving safe/unsafe zones.
• The most powerful skill is not the application of the formula, but the application of the method used to derive it.
• For complex, non-standard launch conditions, we can derive a unique envelope equation by setting up the system's parametric equations and applying the discriminant method (or the partial derivative method) to find the boundary of possible solutions.

Preview of the Next Lesson:
We have established the spatial boundary of where a projectile can go. The next question is, when does it get there? In our next lesson, we will tackle problems with an added layer of complexity: "Solve complex projectile problems involving constraints, such as hitting a target at a specific time and location." This will require us to synthesize our understanding of the trajectory equation with specific time constraints.