# Probability of a rock striking a windshield in a particular location

1. Jan 5, 2013

### CatastrophicF

The windshield of a car has a width of 1067mm and a height of 889mm. The windshield also has a crack with a length of 305mm and a height of 1mm. What is the probability of a rock, with a random trajectory and a measurement of 1mmx1mm, hitting the windshield strikes directly in the center of the crack anywhere along its length?

Am I correct to calculate...?
Possible contact area of the windshield = 948,563mm^2
Possible contact area of the crack = 305mm^2
Probabilty of the rock striking the crack directly in the center of the crack anywhere along its length = 1/948,563(1/305) = 1/289,311,715
...or is it simply 1/948,563 and please explain why the former or the latter is correct

Last edited: Jan 6, 2013
2. Jan 6, 2013

### lurflurf

p=Area(crack)/Area(total)=305/948,563

3. Jan 6, 2013

### Stephen Tashi

It isnt clear to me whether the event described is the event that the center of the rock hits the center line of the crack or whether the event is that any point on the rock hits the center line of the crack. (I assume the question isn't whether the center of the rock or some point on it hits the single point that is the "center of the crack".)

4. Jan 6, 2013

### CatastrophicF

It would be the rock hitting the center line of the crack.

5. Jan 6, 2013

### Stephen Tashi

So which alternative is that? - any point on the rock hitting the center line? If so, can't the center of the rock be anywhere in length of 306 mm ? - assuming the crack doesn't extend to the edge of the window.

6. Jan 7, 2013

### jbriggs444

As I read this, it is a homework exercise. A real-world windshield is unlikely to have this kind of aspect ratio, is unlikely to have cracks with a measurable height and one is unlikely to ask about rocks hitting dead-center in a 1 mm air gap that a crack has created.

Assume:

The windshield is rectangular.
The crack is horizontal (implied by the use of the word "height" to characterize it).
The rock is modelled as a square impact area that is 1 mm horizontal by 1 mm vertical.
The distribution of impact points is uniform across the entire area of the windshield, all the way up to the edges.
This distribution specifies where the center of the rock strikes.
The endpoints of the crack are at least 1/2 mm away from the edges of the windshield.
A "hit" is defined as an event where any point on the rock intersects with any point on the centerline of the crack.

[These assumptions mean that the height of the crack is irrelevant. The edges of the crack do not enter into the question. Only its centerline does]

How far above or below the crack centerline can the rock impact and still be considered a hit? How far to the left or right of the crack can the rock impact and still be considered a hit? What is the area of the region containing impact points that will be considered hits? What is the total area of the windshield?

How do things change if the rock is modeled as a circle rather than as a square? How much does this affect the final result?

7. Jan 7, 2013

### Stephen Tashi

I agree. It's the type of problem where you must think about how pedagog's think rather than think about reality. The answer given by lurflurf suits me. I'm just wondering why the problem bothered to give the dimensions of the rock if it only cares about where the center of the rock hit. If it only cared about the center line of the crack being hit by some point on the rock, why did it bother to give a width for the crack? (Maybe part B of the question is to solve the problem for a 2mm x 2mm wide rock and a 3mm wide crack! )