TIRE DESIGN

How tires Support a weight of a Car ?

You may have wondered how a car tire with 30 pounds per square
inch (psi) of pressure can support a car. This is an interesting
question, and it is related to several other issues, such as how much
force it takes to push a tire down the road and why tires get hot
when you drive (and how this can lead to problems).
The next time you get in your car, take a close look at the tires. You
will notice that they are not really round. There is a flat spot on the
bottom where the tire meets the road. They are squished at the
bottom. The flat portion on the bottom where the tire meets the road
is called the Contact patch OR Footprint. If you were looking up
at a car through a glass road, you could measure the size of the
contact patch. You would multiply the length of the contact patchby
its width to get the area , then add up area for all four tires to get the
total area of the contact patch. For your 2-ton (4,000 lb) car, you
will find that the area of the contact patch is about equal to the
weight of the car divided by the tire pressure. In this case, 4,000
pounds divided by 30 pounds per square inch equals 133 square
inches. That may seem like a lot, but your car's tires are Probably
about 7 inches wide. That means that the contact patch for each tire
Will be about 4.75 inches long. (4.75*7* 4 tires=133 sq.inch)

A tire showing the side and bottom of the contact patch
If you were looking up at a car through a glass road, you could
measure the size of the contact patch. You could also make a pretty
good estimate of the weight of your car, if you measured the area of
the contact patches of each tire, added them together and then
multiplied the sum by the tire pressure.

Inflation pressure x Footprint area x No of tires =Vehicle
weight

Since there is a certain amount of pressure per square inch in the
tire, say 30 psi, then you need quite a few square inches of contact
patch to carry the weight of the car. If you add more weight or
decrease the pressure, then you need even more square inches of
contact patch, so the flat spot gets bigger.

You can see that the underinflated/overloaded tire is less round than
the properly inflated, properly loaded tire. When the tire is
spinning, the contact patch must move around the tire to stay in
contact with the road. At the spot where the tire meets the road, the
rubber is bent out. It takes force to bend that tire, and the more it
has to bend, the more force it takes. The tire is not perfectly elastic,
so when it returns to its original shape, it does not return all of the
force that it took to bend it. Some of that force is converted to heat
in the tire by the friction and work of bending all of the rubber and
steel in the tire. Since an underinflated or overloaded tire needs to
bend more, it takes more force to push it down the road, so it
generates more heat.

Tire manufacturers sometimes publish a coefficient of rolling
friction (CRF) for their tires. You can use this number to calculate
how much force it takes to push a tire down the road. The CRF has
nothing to do with how much traction the tire has; it is used to
calculate the amount of drag or rolling resistance caused by the
tires. The CRF is just like any other coefficient of friction: The
force required to overcome the friction is equal to the CRF
multiplied by the weight on the tire. This table lists typical CRFs
for several different types of wheels.

Tire Type Coefficient of Rolling Friction
Low rolling resistance car tire 0.006 - 0.01
Ordinary car tire 0.015
Truck tire 0.006 - 0.01
Train wheel 0.001

Let's figure out how much force a typical car might use to push its
tires down the road. Let's say our car weighs 4,000 pounds
(1814.369 kg), and the tires have a CRF of 0.015. The force is
equal to 4,000 x 0.015, which equals 60 pounds (27.215 kg). Now
let's figure out how much power that is. You know that power is
equal to force times speed. Therefore, the amount of power used by
the tires depends on how fast the car is going. At 75 mph (120.7
kph), the tires are using 12 horsepower, and at 55 mph (88.513 kph)
they use 8.8 horsepower. All of that power is turning into heat. Most
of it goes into the tires, but some of it goes into the road (the road
actually bends a little when the car drives over it).


From these calculations you can see that the three things that affect
how much force it takes to push the tire down the road (and
therefore how much heat builds up in the tires) are the weight on
the tires, the speed you drive and the CRF (which increases if
pressure is decreased).

If you drive on softer surfaces, such as sand, more of the heat goes
into the ground, and less goes into the tires, but the CRF goes way
up.

Question:] All the wheels of most auto-vehicles are equal in
size, per vehicle. But WHY the wheels of a TRACTOR used
by farmers not the same size?


Historically, tractors used the rear wheels to propel themselves.
Farmers need large tires to avoid compressing the earth, and to
avoid digging in. THE LARGER WHEELS,WHICH ARE USED
AS DRIVE WHEELS, COUPLED WITH LOW AIR PRESSURE
SPREAD THE WEIGHT OF THE TRACTOR OVER A LARGE
AREA ,(IE CONTACT AREA OF THE TIRE WITH GROUND
WILL BE BIGGER). OTHERWISE ,IF GREATER THE FORCE
AND SMALLER THE AREA OF CONTACT THE MORE
LIKELY A WHEEL TO SINK INTO THE MUD.

On the other hand, it is more difficult for the steering mechanism to
turn the large wheels making it more desirable to to keep them
smaller. Thus only the rear tires really need to be large and the front
tires can be small and smooth. Large four-wheel drive farm tractors
have equal size front and rear tires. Some small tractors are four-
wheel drive, and these have medium size lugged tires in front for
extra pulling power.