How does a nail puncture a tire?

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Date: Thu Jan 9 20:09:45 2003
Posted By: Aurelio Ramos, Grad student, Computer Engineering
Area of science: Engineering
ID: 1018555438.Eg

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Message:

It is a bit surprising that very little work has been done in the field 
of tire failure analysis to understand the nail puncture failure 
mechanism.

I believe most experts have concluded from the empirical evidence the 
following things:
1. It is obviously possible
2. Whenever anything is possible and the conditions for it arise often, 
it is bound to happen.

So as long as there are nails on the road (often near construction sites) 
and cars driving over them, it is only a statistical matter of time 
before someone gets a flat.

What this reasoning fails to address is trully understanding the 
mechanism by which it happens.

One way this could be done would be by mounting some high speed cameras 
on a car near each of its tires and driving at various speed ranges over 
pavement sparsely covered with nails, then analyzing the footage as soon 
as a few failures have been recorded to see if they have anything in 
common.

Alternatively, a model could be constructed (or purchased at a toy store) 
and scale models of pavement and nails would be required.

I typically try to do experiments (or find published results) to answer 
scientific questions, however, it's easy to see the high cost of carying 
out this experiment.

For this reason, I will attempt to answer in a speculative way. Keep in 
mind that for this answer to be "science" it is still missing the 
experimental aspect, and I would rather leave that up to those with a 
budget.

When a nail is driven into a tire, it is safe to assume that it happens 
in the first contact. The reason for this assumption is that the surface 
of the tire is compliant (bouncy like a spring) so as soon as contact has 
been made and the nail is lodged in the thread, if no puncture exists, 
repeatedly pushing on the nail will not make any progress, rather it will 
push against the compliant surface of the tire (which makes a "tent" 
around the nail") with no energy being transfered to the tire (no 
puncture). The part of the nail that sticks out will simply be polished 
flush over time against the surface of the thread. Remember that pushing 
on an elastic object (like a spring) causes energy to be absorbed by the 
spring, only to be released once the force is relieved. And energy would 
have be spent in the puncture. Also, the hardness of the tire is unlikely 
to suddenly be reduced to allow the nail to penetrate in a subsequent 
rotation. Finally, the force driving the nail can be assumed to be about 
the same on each rotation (a fraction of the car's weight, as distributed 
according to its dimensions, acceleration rate and grade of pavement, all 
of these are parameters that change slowly from one tire rotation to 
another during normal driving, and can be assumed to be about the same)

This is easy to test: Push on a Ballon with a nail with a certain force. 
If it is not broken in the first attempt, push again with the *same* 
amount of force. It is not likely to suddenly break if pushed again with 
the same force a second time. The force will have to break thru 
a "threshold" for the puncture to happen.

So the nail will either fall out, stick sideways on the thread, or 
penetrate immediately, upon contact.

With this understanding, I would speculate that the nail must be standing 
in a position that allows penetration upon first contact.

The only way I can see this happening is by realizing that asphalt is 
rarely flat. At dimensions similar to a nail's size (about 1 1/2 inch), 
the pavement's surface is highly uneven. Thus, the conditions necessary 
for a puncture would be easy to imagine: A nail must have fallen on the 
pavement in such a way that the head was in a valley, the tip was raised 
at an angle just high enough for puncture, and the head was supported so 
it does not slide away. Finally, the tire must have aproached the nail so 
that its tip met a gap between the thread blocks. This way, the surface 
making contact with the tip is approximately smaller circle than the 
surface tracing the road (the thread). This makes the minimum angle 
required for puncture even lower than if the nail met the surface of the 
thread.

You can easily ilustrate this to yourself: Find a round object with a 
smaller concentric circle drawn on its own surface, and a ruler. Perhaps 
an LP record? Pretend the outer edge of the record is the tire's thread 
surface, and that the label represents the inner surface in a thread gap. 
This is highly exagerated, but illustrates the point by increasing the 
observed effect.

Hold the ruler at an angle (you could try about 30 degrees) If the pencil 
met the outer edge of the record, asuming it stuck, there is a certain 
ratio of forward travel to downward travel for the contact point, You 
can "eyeball" this ratio and remember it. This ratio varies with the 
angle at which the ruler is held initially. For example, if the ruler is 
held flat, there is no forward motion or downward motion and the ratio 
cannot be calculated. As the ruller aproaches flatness, there is more 
down motion than forward motion of the contact point. 

If, instead, the ruler is imagined to stick to the "label's" edge, the 
ratio of forward motion versus downward motion at the contact point is 
different; in fact, higher!. My assumption is that for a nail that is 
sitting at a shallow angle, the higher this ratio, the greater the 
likelyhood for a puncture, because the tire's contact point is traveling 
a little more in the direction the nail is capable of penetrating (along 
its axis)

Of course, in the case of a tire this effect is much more reduced because 
both circles are close in diameter. Also, the contact patch is elastic, a 
significant difference from the LP record and ruler model.

Elasticity would lower the tire so that the distance from the contact 
patch to the center is lower than both the outer diameter and the thread 
gap diameter. I will leave it up to you to come up with an experiment to 
model elastic rotation.

Experimentation would be needed to find out the minimum angle required 
for puncture.

Notice how easy it is for nails to land with a slightly raised tip by 
throwing some nails on ordinary pavement. Try this several times since, 
this being a random event, all you need to observe is that it happens 
with a certain likelyhood. Even 1 in 100 is enough for a puncture.

Another possibility is that someone vandalized your car, the nail was 
driven in when you drove away, and the tire held its pressure for a 
while, just long enough for you to park the car with the nail hidden 
under the tire's contact patch, before you actually found it (thus making 
it seem that vandalism was unlikely.)

But accidents like this do happen and are quite likely under the right 
circumstances.

As far as the likelyhood of 2 nails been driven at once, I can only say 
that if they landed on the same area of the pavement, they could have 
been subject to similar conditions of that area (a valley at their head) 
and that they fell from a container that held many of them (perhaps a 
small spill)

Hope this answers your question,

Your mad scientist,

-Aurelio

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