This section is optional to being able to use the rules of thumb derived below. If you're not a math person, you can skip this section. But if you are like me, especially when it comes to tire safety, you may want to study this section if you:
-
want to prove to yourself that it is mathematically
and logically sound (there's nothing like the certainty gained by making your own conclusions);
-
have a different climate and want to create your own reference table;
-
want to answer questions not answered in this article; or
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use other pressure units for your tire pressures (e.g. BAR, kPa, kgf/cm²).
If any of these apply, you'll need the physics theory and formulas below. (Feel free to verify this against physics on the topic of the relationship between pressure, volume and temperature of gasses.)
Because the tires we usually use (radials) are wrapped by steel belts, the change in the volume of air inside the tire as the air temperature rises is negligible. In other words: the tire is made so that what stretching occurs is very small, enough to be negligible for our calculations. This allows us to simplify the pressure-volume-temperature formula a bit by asserting that the volume of air will not change. Thus the simplified theory:
When a gas is trapped inside a container which has a fixed size (its volume does not change) and the gas is heated, the pressure increases in proportion to the rise in temperature on the Kelvin temperature scale.
From this we can derive the equation: P1/T1 = P2/T2
where:
T2 = new temperature (°K);
P2 = resulting tire pressure from new temperature.
In English: in a fixed-sized container, the ratio of pressure and temperature (using the Kelvin scale) stay the same. In other words, as the temperature increases, so does the pressure.
To bring this theory all the way to application, let us say we are trying to answer the question asked above: given the 43 and 49 PSI measured on front and rear tires respectively, are one or both of them low?
First, we need to know what they are expected to be after warming up (assuming there are no leaks). So we solve for P2:
P2 = T2 * (P1 / T1).
Since P1 and T1 don't change during a ride, (P1 / T1) can be reduced to a constant. If we say P1 = 42 PSI, and T1 = current daily lows (as of the date of this writing) = 66°F = 292.0°K, then we get:
P2 = T2 * (42 PSI / 292.0°K)
P2 = T2 * 0.1438
Finally, we need to plug in the temperatures for the front and rear tires (measured or estimated) to compute the
expected pressure for each. (I have included a Kelvin-degrees lookup table in
Appendix A to make that easier.) Let's say the temperatures for front and rear tires are estimated to be 130 and 155°F respectively (there is more on
how I estimate temperatures below). That translates to 327.6 and 341.5°K respectively. Thus:
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P2(front)
|
= 327.6°K * 0.1438
= 47.1 PSI
|
|
P2(rear)
|
= 341.5°K * 0.1438
= 49.1 PSI
|
The pressure units (PSI, BAR, etc.) for P2 will be the same as for P1. This works for all pressure units.
Now we can answer the question! Given 43 and 49 PSI respectively, are either of them low? Answer: The front tire is low by about 4 lbs, and the rear tire is right about where it should be.
If you want to get more precise than the Kelvin Degrees Lookup Table in
Appendix A, you will need to convert your measured temperatures to the Kelvin scale before plugging them into the above formula to compute P
2. So you will need one of these formulas to compute Kelvin degrees:
Computing Kelvin from Fahrenheit:
°K = ((°F - 32) / 1.8) + 273.15
Computing Kelvin from Celsius:
°K = °C + 273.15
