A woman called in to Car Talk to say that she believes the windshield on her parked car was cracked from falling bird poop. Is this possible? Let's find out.

If you would like to listen to this actual segment, here is the link to the Car Talk clip.

What starting info do I have? The bird poop made a 4 inch diameter spot on car. Not sure how thick this spot was. The car was parked.

That is about all I have. So maybe I can make some assumptions: I am going to assume the starting height of the poop was large enough such that the poop was traveling at terminal speed. Falling poop is spherical. Yes, I know rain doesn't fall this way (rain drops aren't shaped like I thought) but you have to start somewhere. The density of poop is the same as water (should be close enough for this estimation). What is the compressive strength of laminate glass (for a windshield)? I couldn't find a definite answer, but it seems 25 MPa to 100 MPa seems plausible (although it could be much higher). For this case, I am going to assume 50 MPa â maybe there was a slight crack previously that made the glass weaker.

What is compressive strength? This is the maximum pressure the material can take before something bad happens. Bad in this case could be a crack, or deformation.Size of Poop

Are you getting tired of me saying "poop"? That is my preferred word for the bird bomb. Ok, so how big would this be? If the poop is spread out in a circle of 4 inches (radius of 0.05 meter), I need to guess at the depth of this circle (really, it is a cylinder). Suppose the poop is 2 mm deep after impact (let me call this *h*). The total volume of poop on the windshield should be the same as the spherical falling poop. Let me call the radius of the poop on the windshield *R* and the radius of the falling sphere, *r*. This means:

Using a value of 2 mm for the height and 5 cm for the radius, I get a radius of the sphere to be about 1.55 cm.

If the density of the poop is 1000 kg/m3 (like water) then it would have a mass of 0.0157 kg.Terminal Speed of Poop

So, the poop is spherical and falling.

Here is a force diagram of some falling poop (at terminal speed):

If the poop is at terminal speed (and thus a constant speed), then the magnitudes of these two forces must be equal.

The gravitational force would be *mg*, where *g* is the local gravitational field.

The magnitude of the air resistance force can be modeled as:

Where Ď is the density of the air (not the poop), *A* is the cross sectional area, and *C* is the drag coefficient (0.47 for a sphere).

Using this, I can solve for the velocity.

I get:

I guess it wouldn't be a bad idea to write this in terms of the radius of the sphere and the density of the poop (I will call Ďp as to not confuse it with the density of air).

I can also get the cross sectional area in terms of the radius.

This gives:

If I use a density of air at 1.2 kg/m3, this gives a terminal speed of 26.8 m/s (60 mph).The Collision With the Glass

This is the difficult part. I need to estimate the force and pressure on the glass during the collision.

Let me make a diagram of the colliding poop (on horizontal glass for simplicity).

So here, the poop is moving down with a speed *v*1 right before it hits the glass. While it is stopping, I will assume the center of mass of the poop moves a distance *r* (I just picked that to make things easier). Also, during this stopping time, there is a force the glass exerts on the poop (and the poop exerts the same force on the glass). I have left off the gravitational force with the assumption that it will be small compared to the glass force. Also, I am just going to be solving for the average force during this time.

To estimate the value of this force, I will use the work-energy principle. It says that the work done on the poop is equal to its change in kinetic energy.

I can write that as:

The work done by the glass is negative since the force is in the opposite direction the poop is moving.

So, using a displacement of *r* and a final speed of zero, this gives:

Using the terminal speed of 26.8 m/s and a radius of 0.0155 meters, I get an average force of 363 Newtons.

However, I am really interested in the maximum force.

Suppose the force as a function of displacement looks something like this:

The red dotted line represents the average force (what I solved for above). The maximum force could be significantly higher. Let me just pretend that the maximum force for this case is 1000 Newtons.

Now what about the pressure? The problem is that the area of contact between the poop and the glass changes as it collides. Also, I really need the contact area when the force is maximum. Suppose that at this time, the poop makes a contact circle with a radius of 0.01 meters?

This would make the pressure on the glass:

In case you didn't notice, 3 MPa is less than 50 MPa for the compressive strength of laminate glass. But all is not lost. This is just an estimate. What if there was a small rock embedded in the poop? This could decrease the contact circle to a radius of perhaps 0.001 meters. Such a small contact area could cause the pressure to go up to 318 MPa.Conclusion

I am leaning towards "possible". If you bird had a poop that big and was dropped from a significant height and perhaps had some debris in the poop, then it could possibly crack a windshield. Should you worry about this threat? I don't think so.

Or just park under tree.

*Photo Credit: Joshua Ganderson / Flickr*

*Rhett Allain is an Associate Professor of Physics at Southeastern Louisiana University. He enjoys teaching and talking about physics. Sometimes he takes things apart and can't put them back together.*

*Follow @rjallain on Twitter.*

*This post was originally published on September 16th, 2011 on Wired Magazine's Science Blog. It has been republished with permission.
*

*If you would like to see your story here, send us an email at tips@jalopnik.com with the subject line "Syndication."*