Strangest Tire Question EVERRR...
Has anyone ever calculated or estimated the area, in square inches/square cm, of the *surface of the inside* of a tire?
Does at least an average exist out there, for broad categories: passenger car, SUV-light truck, and commercial truck?
Just curious, Thanks!
Tire manufacturers would have this as it would be the same as the inner
build die they use when laying up a tire. However the number would vary
a lot because of the myriad of sizes, Take a 205 75R 15, it would have a
larger surface area than a 205 60R 15 simply due to the aspect ratio change.
What tools do we have to solve the problem?
o Let's start with shape & terminology:
That gives us these perhaps relevant terms:
o Tread width
o Section height
o Aspect ratio
o Rolling circumference
Where for now I'll ignore there's a difference between
o free (i.e., unloaded), and,
Then, using this tire-size calculator:
ignoring sig fig approximations (so I can just cut & paste numbers)...
Assume a P200/50R15 for easy numbers & using only the 1st URL above:
o Diameter = 582mm (22.9")
o Width = 200mm (7.9")
o Sidewall = 99mm (3.9")
o Circum. = 1,824mm (71.8")
o Revs/Mile = 882
o Rim diameter = 381mm (15")
Assuming the carcass inside & outside surface is "similar" in both shape &
surface area, we easily arrive at a gross _approximation_ of:
o Tread area = ~200mm wide times ~1,824mm around = ~364,800?sq mm
o Sidewall area = ~99mm times ~1,824mm around times 2 = ~361,152sq mm
Which calculates to...
o Total U shaped carcass = ~364,800?sq mm + ~361,152sq mm
Which comes to roughly ~725,952?sq mm
Given the sidewall area is probably more accurately done with donuts:
Where the terms appear to be:
o Diameter of the Whole Area = 582mm
o Diameter of Inside Area = 381mm
Which calculates out to:
o Area = 152,023.9sq mm
o Outer Perimeter = 1,828.4mm
o Inner Perimeter = 1,196.9mm
o Total Perimeter = 1,196.9mm
Using these numbers for sidewall area in the previous calculation:
o Tread area = ~200mm wide times ~1,824mm around = ~364,800?sq mm
o Sidewall area = ~152,023.9sq mm times 2 = ~304,047.8?sq mm
o Total U shaped carcass = ~364,800?sq mm + ~304,047.8?sq mm
Which comes to roughly ~668,847.8?sq mm
In summary, and knowing the inside dimensions must be smaller than the
outside dimensions, I'd guess it's around 600,000sq mm to 700,000sq mm.
But that's only a gross approximation.
o How much accuracy does the OP need?
Arlen Holder - wow! You went to town on the math!
How precise? Ohh, just an average. It's something I'm working on, so an average
sized automobile tire is all I'm asking about. 215/6R16ish.
Thanks for the hard work though.
So 725,000 sq mm approximates 1,224 sq in. internal area of your example tire.
If a particular vehicle requires 32psi cold air pressure in that tire, that equates to 32 * 1,224sq. in = 39,168 total
pounds of air pressure against the interior wall of that tire! Wow!
If we increase that 32psi figure by 10%, that's 35.2psi.
35.2 x 1,224 = 39,564 total pounds of pressure against the wall of that same
tire. An increase of only 1.01% in the total pressure against it.
So much for my theory that each additional pound of air pressure exponentiates the total pressure imposed on
the inside of that tire.
That's not how physics works. Pneumatic pressure acts
equally on all surfaces. For 35psi, it's 35psi on each part
of the surface of your closed figure- 35psi on the tire,
35psi on the rim - but 35psi only.
I just noticed something interesting when it comes to 'averages'.
o All you need is the tread width & the aspect ratio & rim size.
o For a 40 series tire, it's 2x40, i.e., tread area times 180%
o For a 50 series tire, it's 2x50, i.e., tread area times 200%
o For a 60 series tire, it's 2x60, i.e., tread area times 220%
Notice an aspect ratio of, oh, say, 50 is 50% of the tread width.
o So the surface area is a function of the tread width & aspect ratio.
If the tire is a 50-series tire, the sidewall area is approximately 100%
more than the tread area (i.e., 50% times 2 sidewalls).
For a 60 series tire, the sidewall area is approximately 120% more than the
tread area (i.e., 60% times 2 sidewalls).
In short, for a quick mental calculation all you need is the tread area and
the sidewall-to-tread ratio, where the tread area is simply the width times
the circumference of the tire.
So, for your P215/60R16, this calculator tells me:
Diameter = 26.2"
o Width = 8.5"
o Sidewall = 5.1"
o Circumference = 82.1"
o Revs/Mile = 771
The area of the tread would be its width time the circumference:
o Tread area = 8.5" times 26.2" = 222.7 sq inches
Since it's a 60-series tire, the sidewall area is 120% of that.
o Sidewall area = 222.7sq inches time 120% = 267.24 sq inches
In summary, you just take the tread area and multiply it by the percent:
o For a 40 series tire, surface area is tread area times 180%
o For a 50 series tire, surface area is tread area times 200%
o For a 60 series tire, surface area is tread area times 220%
So rounded off we'll just say 270 sq. inch of area inside the avg. passenger
tire. I think something was wrong with that online calculator I was using last
night to convert square mm to square inches. For a similar tire I was getting,
if I recall,1,220-something square inches? I suspected something was off!
So the air inside the tire inflated to 35psi is pressing down on each one of
those 270 square inches with 35 lbs of force. So a more sane 9,450lbs of
total pressure on the inside wall of that tire. Probably closer to 7,000ish if
you consider that side of the rim which completes the inside of that total air
cavity at each corner of our daily vehicles.
I don't know but it cannot be 270 square inches ( which is
roughly 10" x 27". I can see a ten inch width bead to bead
but the circumference is way beyond 27 inches.
Regardless of arithmetic, the pneumatic (or hydraulic,
formulae are the same for these purposes) pressure at every
place within a closed figure is equal and all of it will be
Entering footprint into the argument.
My car's GVW is 3800 lbs so each tire supports 950 lbs (ideally).
If a my tire has a footprint of 6"x6" = 36" sq, and if each sq in has a
pressure of 36 psi, then the psi total of the footprint is 1296 lbs.
Oddly enough my car tire's max load rating is 1295 lbs. Push on it with
an area and it will push back equally.
The rest of the tire area is at 36 psi and is irrelevant since it does
not support anything.
Oh really? Last time I checked, a tire is a complete unit, with all of
its parts working together to support the vehicle, help maintain
traction, and help bring it to a stop.
That 36psi is pressing against every square inch of the inside of the
tire and the outer circumferential face of the wheel or rim.
Of course the tire sidewalls will slowly buckle! Common sense! They're the vertical
component of a tire. But what you seem to want to believe is that the only part of a
tire that matters is that which touches the ground. In your world, sidewalls don't exist.
In your world, a vehicle drives by and all you see are the treads and the rims.
It doesn't work that way. A tire - all of it - and the wheel or rim it's attached to, are
together a SYSTEM: of support, of absorption, of compliance, and of traction during
a variety of vehicle dynamics.
And whether a fully mounted, inflated, and balanced wheel/tire combo is sitting, waiting
to be put on a car, or is between that car and the road already, the air inside places
pressure on ALL interior surfaces of the cavity created by tire plus rim. Science!
You have me 100% wrong. I was not disagreeing with you. My point was
that the sidewalls also have pressure acting on them and that helps
keeps them where they need to be. As well, the poster to whom you
responded omitted to mention, in his detailing of pressure acting at the
contact patch, is that it also acts in the upwards direction on the
inner rim surface opposite the inner rim surface. Tyres are not designed
by just looking at the contact patch to determine the inflation pressure
hence the load carrying capacity.
Having been a mechanic for over 50 years, my view of a tyre is very
different from yours. I look at it as a sub-component of the vehicle,
its relationship to steering and suspension and, most importantly,
Two primary factors determine the load carrying capacity of a tyre, the
load and the inflation pressure. If you want to increase the load, you
need to increase the pressure. This is well known and is included in
vehicle owners manuals. The rub here is that there are many other forces
acting on tyres, lateral and centrifugal forces being just two, and each
needs to be considered, especially in relation to speed. Therefore the
service use of the tyre becomes important as well.
Well aware of that. Not my point. BTW, a tyre waiting to be put on a car
is *different* to a tyre on a car at 100 mph. Think heat and what that
does to a tyre and the pressure inside it.
I have never been a mechanic (I'm an engineer & scientist), but I do agree
with Xeno, that heat is (almost) everything when it comes to rubber tires.
o Speed is heat
o Load is heat
o Inflation is heat
My postulation is more a matter of wording:
We do not put "33 pounds of air" in a given tire. We put 33 pounds PER SQUARE INCH in
that tire. Ergo, that volume of air is applying 33lbs of force to EVERY SQUARE INCH of
the inside of that tire & rim face combination.
Correct, it applies 33 PSI of force to every square inch. However that
is the total force per square inch that the air applies. You cannot add
each inch up and say that number is the total pressure applied.
It is no different than atmospheric pressure which is around 14.696 PSI.
Using your version of physics every human being would be crushed to
death by that pressure if you consider the average human male has
22.173 square feet of surface area.