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Why Tyre Pressure is All Importar

7th October 1939, Page 24
7th October 1939
Page 24
Page 25
Page 24, 7th October 1939 — Why Tyre Pressure is All Importar
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Variations in Weight Should Be Automatically Looked After By the Tyre and the Air, Both of Which are Naturally Adaptable.

IF all things be normal—mainly, that is, if inflation pressure and load be correct—a tyre will assume its normal working 'shape. This shape is important, for on it depends, to a large extent, the well-being of the tyre. It is so designed that the correct shape is automatically assumed when the pressure is appropriate to the load.

. A distorted condition, due to underinflation or , overloading, is responsible for the majority of tyre failures. Excessive flexing is the most common form of tyre distortion.

The effective shape of a tyre is not truly round. It is not intended to , retain, the shape to which it was moulded. It is designed to yield in the right places, and is so constructed that this give-and-take, if confined to sensible limits, will do no harm.

There is a slight flattening at the point of " road contact, both in a lengthways and sideways direction, and the wall flexes slightly. The process of adaption to the contour of the road is known as deflection, and the resultant area in contact with the ground is elliptical in shape.

How Weight is Carried.

There is a definite relationship between the air pressure, the load, and the area.of ground contact. The entire weight of the vehicle falls on that part of the tyre which touches the road. The rest is. doing no work whatsoever until the revolution of ' the wheel calls it into use.

Let us assume that we have a fourwheeled vehicle of a laden weight of 2 tons ; let us assume also, in order to keep our example as simpleas possible, that each wheel be carrying an equal share of the total load, i.e., 10:cikt. per tyre. Thus, each tyre, or that part of the tyre which touches the road, is supporting 1,120 lb. This is the downward pressure.

• . If, for example, the area of road contact of each tyre be 20 sq. ins. 11 ten a.. downward pressure of 1,120 lb.

. is carried on this area. This is equivalent to 56 lb. per sq. in.

A.s this load pressure is exerted on the tyre from above, it is obvious that unless the cover has equivalent air pressure inside, exerting an 'upward lift, it will not be able to assume its normal shape. In other words, it will be forced to yield to . the superior pressiire of the vehicle. The same applies in the case of a thin metal gas container. If the internal pressure were not sufficient it would be crushed by the atmospheric pressure, which exerts a weight of 14i lb. on every sq. in.

If a tyre be required to carry-a certain weight, it is designed to present a definite area of tread to the road, which, in relation to the air pressure, is capable of counterbalancing the weight of the vehicle. If you multiply the air pressure per sq. in. by the number of sq. ins, of ground contact, the result (in lb.) is approximately equal to the weight resting on the tyre.

It can be roughly expressed as a formula, where A — area of road contact per tyre (sq. ins.).

P pressure in tyres (lb.).

W = total weight of vehicle (lb.).

— = weight on each tyre (lb.).

4 The formula is : W 4A x P.

In the case of the vehicle previ ously mentioned, it is easy to see how this works out in practice :— 4480 = 4 x 20 x 56 (lb.)

Or

4480 — = 20 x 56 = 1,120 lb. per tyre.

4 Thus the external weight is coon balanced by the internal air press at the point of road contact.

Now, if the pressure in the tyre reduced, it does not upset the cal lotions, because automatically tyre would yield to a greater ext and the increased deflection wo call into use a bigger area of n contact. If, in the case mentioi above, the pressure dropped ft 56 lb. per sq. in. down to 40 lb., ' area of contact would inimedial be increased from 20 sq. ins. 28 sq. ins. The upward pressure the air remains the same-1,120 lh although the pressure per sq. in. been materially reduced. " weight is distributed over a• Wi area. But this state of affairs is for the tyre itself, because it is . supported in its proper shape.

In the same way, if the weight the vehicle be increased by the at tion of passengers or load, the t3 if the pressures be not adjusted, 1 deflect in the same way, spread out the load over a greater area, the weight per sq. in. will rem unaltered.

" Riding on the Rim."

When the pressure drops so 1 that the tyre's capacity for increas the road-contact area is ft absorbed, then the load is not bo by air any longer. This is 'popula known as "riding on the rim."

If the weight of the vehicle decreased, the relationship 1 remain the same, because the red lion in load will reduce the amo of tyre deflection and the area of rx contact. So a decrease in weight 25 per cent, would result in a aim: decrease of tread-contact area, the inflation pressure and wei pressure would remain the same sq. in.

Extremes in this direction are. bad. A high pressure and a Ii load will prevent the tyre fr deflecting sufficiently, which result in rapid wear, due partly excessive vibration and partly to fact that the tyre rides Mainly on centre line of the tread.

It has been said that the wheel a lorry weighing 5 tons would sink any deeper into soft grot than that of a small car if the pressure in the tyres were the sat Although this is rather an eXtre example, it is quite true in princij and the explanation is closely C ected with this business of the ?.lationship between air pressure, md, and road-contact area.

We will assume that, with load, le small car weighs 10 cwt., i.e., 80 lb. on each tyre. This load, nis:Muted over a contact area of, say, 0 sq. ins, per tyre, results in a prosare of 28 lb. per square in. This is le weight of the vehicle on the tyre nd the weight wessure of the tyre n the ground.

The air pressure must be equal to ounterbalance it, otherwise the tyre fill be sandwiched between the car nd the road surface. Alternatively, f course, the pressure could be flowed to drop so that the contact rea is increased, but in this example Je are dealing with a fixed pressure

28 lb.

We will allow the lorry to have six n)eels. This does not affect the alculations in the least, and it helps o bring the figures within more pracical limits. Five tons distributed ,ver six wheels is equal to 1,8664 lb. )er wheel. If the pressure in the yres be the same as in the small ear, .e., 28 lb., the tyre will be deflected

o such an extent that the area of ;round contact is 013R sq. ins:

The tyre pressure is 28 lb. per sq. in., and this allows of such deflection that the weight is spread out an enormous extent, so that the weight pressure is the same. Consequently, each sq. in. of ground that supports the 5-ton lorry is subjected to the same weight as each sq. in. that supports the car ; that is why the lorry does not sink any deeper. If the air pressure in the lorry tyres were increased, down into the soft ground it would go.

Securing Additional Grip.

If a vehicle becomes bogged it is wise to let out a little air from the tyres, so that the area of ground contact is increased. Often this will give that additional grip which will make all the difference.

So that is why it is all a question of air pressure. The air, and its container, the tyre, are very adaptable bodies, and will speedily alter their shape to accommodate alterations in weight. Thus, assuming a constant air pressure, the tyre can increase its tread-to-road contact in order that more sq. ins, of inflation pressure can be called into use to deal with increases of load; the weight per sq. in. that falls on the ground remains the same, no matter how much the load be increased.

A few interesting facts come to light as a result of the close relationship between the three important factors. For instance, to find the area of ground contact of any tyre, when the weight and tyre pressure are known, the correct formula is — P = A.

4 To find the weight when the pressure and area are known, the formula is 4A x P = W, and to find the pressure in the tyres when the area and weight are known it is ÷ A = P.

4 These formul are approximately correct, but not quite exact. This is partly due to the fact that a tyre is not absolutely round in section and, consequently, presents an area of road contact which is slightly out of proportion to the other figures. Also, a certain small proportion of the weight may be carried by the wall of the tyre.'

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