The International Standard Atmosphere, ISA, is an answer to the need for a single reference atmosphere, reasonably close to the day-to-day actual atmosphere and shared between all aviation stakeholders. It can be used for graduating instruments like altimeters or providing reference airplane performances. It has been defined by ICAO in the 70s. And is based, first, on a set of initial conditions at sea level, a pressure of 1013.25 hectopascal, at a temperature of 15 degrees Celsius. Second, a temperature evolution law with altitude, a decrease of 6.5 degrees per 1,000 meters, up to 11,000 meters about 36,000 feet. And then a constant value at -56 degrees Celsius up to 20,000 meters. What happens at higher altitudes, is outside of the scope of aviation. Those data, when combined to the perfect gas law and to the hydrostatic equilibrium of the atmosphere, completely define the temperature, pressure, and density of air, at each altitude. For those interested, I briefly show how we can solve this set of equations to obtain the exact relationship between pressure and altitude. First, in the hydrostatic equilibrium equation, dp equals -rho gdz. We need to replace the air density of rho by its value coming from the perfect gas equation rho equals ps/rT. And we then replace the absolute temperature T. By the standard atmosphere linear evolution law, T0 minus kz. We then obtain a single differential equation where the left number depends only on the static pressure ps. And the right number depends only on the altitude set, g, k, and r being constant parameters. As you can see on the diagram, dps is the variation of ps due to an increase, dz, in altitude. We can now solve this differential equation from the ground up to the current altitude. So, on the left number, we will integrate dps/ps DPS from p0 to the current pressure ps. And for the right number, we will integrate -kdz over T0 minus kz from 0 to the current altitude z. As you can see, both integrals are in the form of dx/x. So this integrates into a logarithm. We obtain the variation of the log of ps between p0 and ps, on the left, is equal to a constant multiplied by the variation of log of T0 minus kz on the right. As log of ps minus log of p0 is, in fact, equal to log of ps/p0. This gives us log of ps/p0 equals a constant multiplied by log of 1 minus kz over T0. And by exponent setting, we obtained the final single law of the evolution of the pressure ps as a function of altitude, ps=p0(1- kz/T0), and the power g/kr, which is a constant number. When we replace the constant values by their numeric values, we obtained this equation with ps in hectopascal and z in meters. This gives us, on the right, the blue curve, giving the evolution of the pressure with altitude. We can do the same kind of computation with a constant value of temperature. And so, layer of altitude between 11,000 meters and 20,000 meters, and we obtain the second equation. As you see on the curves on the right, those two loads perfectly connect at 11,000 meters. Close to 36,000 feet, if you prefer. We can use these formulas as they are, or we can tabulate them using a spreadsheet like or Excel for an easier use for day-to-day computations. [SOUND]
