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What influences pigeon droppings

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Ever wondered about the physics behind a bird dropping... well, anything... on a target below? It seems random, but can we model it? This article details a clever approach to defining and calculating an "impact probability" for a projectile dropped from a moving source.

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The Model: From Dot Product to Cosine

Let's imagine a bird flying with a constant horizontal velocity, $u_{b}u\_b$, at a height, $hh$. It drops a projectile which is then only acted upon by gravity, $gg$.

Our goal is to define a "probability of impact," $\psi \backslash psi$. A simple way to think about this is to consider the alignment of the projectile's final velocity vector, $\vec{s}\backslash vec\{s\}$, with the horizontal ground vector, $\vec{p}\backslash vec\{p\}$. The more aligned they are (a shallower angle), the "better" the shot, in a sense.

The initial idea was to use a dot product, but a much more elegant solution is to normalize it. This gives us the cosine of the impact angle, $\theta \backslash theta$.

$$\psi =\cos \theta \backslash psi\; =\; \backslash cos\backslash theta$$

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This provides a perfect "probability" score between 0 (a direct vertical hit) and 1 (a perfectly horizontal graze).

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Breaking Down the Motion: A Kinematics Review

To find $\cos \theta \backslash cos\backslash theta$, we first need to find the angle itself. This requires calculating the projectile's final velocity components at the moment of impact.

1. Horizontal Velocity ( $v_{x}v\_x$ ): Assuming no air resistance, the horizontal velocity remains constant.
   $v_x=u_{b}v\_x\; =\; u\_b$

2. Time of Flight ( $tt$ ): The time it takes to fall a height $hh$ from rest is found using:
   $h=\frac{1}{2}g{t}^2⟹t=\sqrt{\frac{2h}{g}}h\; =\; \backslash frac\{1\}\{2\}gt^2\; \backslash implies\; t\; =\; \backslash sqrt\{\backslash frac\{2h\}\{g\}\}$

3. Final Vertical Velocity ( $v_{y}v\_y$ ): The vertical speed upon impact is:
   $v_y=gt=g\sqrt{\frac{2h}{g}}⟹v_y=\sqrt{2gh}v\_y\; =\; gt\; =\; g\backslash sqrt\{\backslash frac\{2h\}\{g\}\}\; \backslash implies\; v\_y\; =\; \backslash sqrt\{2gh\}$

4. Impact Angle ( $\theta \backslash theta$ ): The angle is the arctangent of the ratio of the velocity components.
   $\theta =\arctan \left(\frac{v_y}{v_x}\right)=\arctan \left(\frac{\sqrt{2gh}}{u_b}\right)\backslash theta\; =\; \backslash arctan\backslash left(\backslash frac\{v\_y\}\{v\_x\}\backslash right)\; =\; \backslash arctan\backslash left(\backslash frac\{\backslash sqrt\{2gh\}\}\{u\_b\}\backslash right)$

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The Elegant Solution: A Trigonometric Identity

Now we have our "probability" defined as $\psi =\cos (\arctan \left(\frac{\sqrt{2gh}}{u_b}\right))\backslash psi\; =\; \backslash cos\backslash left(\backslash arctan\backslash left(\backslash frac\{\backslash sqrt\{2gh\}\}\{u\_b\}\backslash right)\backslash right)$. This can be greatly simplified using a right-angled triangle.

If $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{2gh}}{u_b}\backslash tan\backslash theta\; =\; \backslash frac\{\backslash text\{opposite\}\}\{\backslash text\{adjacent\}\}\; =\; \backslash frac\{\backslash sqrt\{2gh\}\}\{u\_b\}$, we can construct the following triangle:

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Velocity Vector Triangle

From this, we can see that $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}\backslash cos\backslash theta\; =\; \backslash frac\{\backslash text\{adjacent\}\}\{\backslash text\{hypotenuse\}\}$. This gives us our final, simplified formula:

$$\psi =\frac{u_b}{\sqrt{u_b^2+2gh}}\backslash psi\; =\; \backslash frac\{u\_b\}\{\backslash sqrt\{u\_b^2\; +\; 2gh\}\}$$

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Conclusions

This powerful formula tells us exactly how each variable affects our impact "probability":

• Higher Bird Speed ( $u_{b}u\_b$ ) $⟹\backslash implies$ Higher $\psi \backslash psi$: As $u_{b}u\_b$ increases, it dominates the denominator, and $\psi \backslash psi$ approaches 1. A faster bird results in a shallower, more "probable" impact.
• Lower Height ( $hh$ ) $⟹\backslash implies$ Higher $\psi \backslash psi$: As $hh$ decreases, the $2gh2gh$ term shrinks, and $\psi \backslash psi$ approaches 1. Dropping from a lower altitude gives a higher score.
• Lower Gravity ( $gg$ ) $⟹\backslash implies$ Higher $\psi \backslash psi$: On the moon, for instance, the projectile would gain less vertical speed, resulting in a shallower impact and a higher $\psi \backslash psi$.

In conclusion, the highest "probability" in this model is achieved when the horizontal velocity is far greater than the final vertical velocity, leading to a shallow, grazing impact.