Rubbers 5/28/19

By Richard Bleil

No, not the kind that you are thinking about, perv.

I thought tonight we might do another WTH calculation, just for fun. So, I have a very simple question tonight; when the typical person drives, how many rubber molecules do we leave per square inch on the road?

Okay, so how do we start this one? Well, we know that our tires wear out, and typically need to be replaced every 50,000 miles, according to an online search, so we’ll use this as our starting point. But how much rubber is lost in this distance?

I’m thinking…ohhhhh…let’s say geometry. Apparently, a new tire has a typical diameter of 20.832 inches. Let’s save ourselves some heartburn here, and convert to metric immediately. There are 2.54 centimeters in one inch, so the diameter is (20.832*2.54=)52.913 cm. We will need radius, so (52.913/2=)26.457 cm is the radius.

What I’m working towards here is the area of a new tire, which is the circumference of a tire times the width. From geometry, the formula for circumference is C=2*pi*r, where pi is, of course, 3.14159. So, the circumference of a tire is (2*pi*26.457=)166.234 cm.

We will need to repeat the calculation (I hope…kinda flying off the cuff here) for a worn tire. Apparently, new tires have a depth of 10.5/32″, or 0.328 in, or (0.328*2.54=)0.833 cm. Tires are considered to be legally worn out at 2/32″, or 0.0625″, which is 0.159 cm. The difference in tire tread, then, is (0.833-0.159=)0.674 cm. Therefore, the radius of a worn tire would be (26.457-0.674=)25.783 cm. The circumference of a worn tire is (2*pi*25.783=)161.999 cm. This means that the mean (average to most people) circumference of a tire is ((161.999+166.234)/2=)164.117 cm.

Ok, the mean circumference is what I anticipate needing for this calculation. I love doing stuff like this; I’m hoping the reader is seeing how my mind works which, frankly, is probably more frightening than my vampire book, Vampire Genetics by Richard Bleil.

This commercial plug has been brought to you by Vampre Genetics, by Richard Bleil, available on a major online shopping site.

Sorry, couldn’t resist. Okay, we need the average area of the tire. The average width of a tire is, apparently, 21.5 cm (which is 8.46 inches for those who prefer the English units which, ironically enough, is the length that most men like to call twelve inches for some inexplicable reason that I will never understand). Okay, so on average, the area of a tire is (164.77cm*21.5cm=)3,540 cm^2. BUT, and this is a pretty big but, we don’t typically drive on one tire unless we are really REALLY cool uni cyclists on unrealistically large tires. Since we drive on four tires, the mean area of our tires are (3,540*4=)14,200 cm^2.

NOW, this is working out well. What I am really looking for is the volume of the tire that is lost as we drive. This is quite a simple matter now. We take the mean (or average) area of the tire, times the depth of lost tire tread. So, the volume of tire lost in 50,000 miles is (14,200 cm^2 * 0.674 cm=)9,570 cm^3.

Okay, we’re closer to finishing this than you might imagine now. The reason we needed the volume of the tire is because we need the mass of rubber that we lose. To do this, we need the typical density of tire rubber, which is 1.522 g/cm^3. (These numbers are all from online engine searches, which is sufficient for so-called “back of the envelope” calculations.) The mass of the lost rubber is the volume of the lost rubber times the density, or (9,570 cm^3 * 1.522 g/cm^3=)1,460 g. There are 454 g in one pound, so in 50,000 miles, we lose about 32.1 pounds of rubber.

But, we don’t want mass of rubber, we want the number of molecules. To get this, we first need to convert from grams of rubber to moles of rubber, and from moles of rubber, we can find rubber molecules. To get moles of rubber, we divide the mass of the rubber by the molecular mass of rubber. Rubber atomic mass can run from 50,000 to 3,000,000 g/mol. We could use the average here, but I’m going to use the largest value, because that will give us the smallest estimate of molecules left behind. So, we have (1,460g/3,000,000g/mol=)0.000487 moles of rubber. This doesn’t seem like much, especially spread over 50,000 miles, so this post may be a bomb.

But what the heck, let’s forge ahead.

Okay, we know (Avogadro’s number) that there are 602,000,000,000,000,000,000,000 molecules per mole. We know the moles of rubber that are lost, so we actually will lose (0.000487mol*602,000,000,000,000,000,000,000molecules/mol=)293,000,000,000,000,000,000, or 293 hundred thousand trillion molecules.

This is what we lose per 50,000 miles, but we don’t want to know that. We want to know how much we lose per inch. There are 63,360 inches in one mile, so the distance is actually (50,000mi*63,360in/mi=)3,170,000,000 in. But, we don’t want distance; we want area (square inch). To do this, we multiply the distance by the width of the tires. Above, we learned the average width of one tire is 8.46 in, but we are driving on four tires, so our total width is (4*8.46in=)33.8 inches. In 50,000 miles, the total area driven by our tires is (3,170,000,000in*33.8in=)107,000,000,000 in^2 or square inch.

So we have one last step, just divide the number of molecules by the area. We lose (293,000,000,000,000,000,000molecules/107,000,000,000in^2)=2,740,000,000 rubber molecules per square inch.

That’s nearly (ready for this?) 3 BILLION rubber molecules per square inch left on the road by our tires. One could argue that the area of the tires is grooved, so if we assume that only half of the tire tread touches the road, that is still, conservatively, one and a half billion rubber molecules per square inch of road.

Is your mind blown? My mind is blown. I think your mind is blown. My mind isn’t blown. YOUR mind is blown.

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