The Machmeter — Page 92, Lesson 107

Let’s pick up right where the Machmeter’s behaviour in different atmospheres leaves off, and go straight into the worked problems, because these tie the theory to the numbers you’ll actually compute. First, a quick recap of the key relationship we’ll use over and over: Mach number equals TAS divided by LSS — local speed of sound. And LSS itself equals 38.95 times the square root of the absolute temperature, T in Kelvin. That constant 38.95 gives you knots when T is in Kelvin. Keep that formula handy; every problem here leans on it. Problem 1: What is the speed of sound at FL380 in ISA conditions? In the ISA atmosphere, FL380 is above the tropopause, so the temperature is constant at minus 56.5°C, which is 216.5 Kelvin. Plug that in: LSS = 38.95 × √216.5, which works out to 573 knots. So at that altitude, sound travels at 573 knots. You can also do this on the navigation computer: place the Mach number index arrow against the temperature in °C, locate M 1.0 — that’s the blue 10 on the inner Mach number scale — and read off the TAS on the outer scale. Since Mach 1.0 means TAS equals LSS, that readout gives you the speed of sound directly. Problem 2: Determine the TAS corresponding to M 0.70 at JSA MSL, which is plus 15°C or 288 Kelvin. On the computer, set the Mach number index against plus 15°C in the Airspeed window. Against 7 — for M 0.7 — on the inner scale, read off the answer, 463 knots, on the outer scale. Alternatively, calculate it: TAS = Mach number × LSS = 0.7 × 38.95 × √288. That’s 0.7 × 661, which gives 463 knots. Notice the LSS at MSL in JSA is 661 knots — that’s the standard sea-level speed of sound. Problem 3: Calculate, without a computer, the altitude in the JSA atmosphere at which a TAS of 450 knots corresponds to Mach 0.80. Start from Mach number = TAS / LSS, so LSS = TAS / Mach number = 450 / 0.8 = 562.5 knots. But LSS also equals 38.95 × √T, so √T = LSS / 38.95 = 562.5 / 38.95 = 14.44. Square that: T = 14.44² = 209° absolute, which is minus 64°C. Now, minus 64°C occurs at FL395 in the JSA — and note the JSA has no tropopause, so that temperature lapse continues right up. So the answer is FL395. Problem 4: If a decrease of 0.12 in the Mach number results in a decrease of 80 knots in the TAS, what is the local speed of sound? Again, Mach number = TAS / LSS. A change in Mach number equals the change in TAS divided by LSS, because LSS is constant here. So 0.12 = 80 / LSS, which means LSS = 80 / 0.12 = 666.7 knots. That’s your local speed of sound. So the pattern across all four problems: you’re always juggling Mach number, TAS, and LSS, and LSS always comes from temperature via that 38.95 √T formula. Get comfortable flipping between the formula and the navigation computer, because both routes appear in the exam.