= 106 min @ 148 kt = 261 NM — Page 264, Lesson 267

I want to walk you through the Point of Safe Return, or PSR, and specifically how we handle it when fuel flows are variable — not constant — between the outbound and home legs. Let’s start with the core idea. The Point of Safe Return is the furthest point along a route from which the aircraft can still return to its departure aerodrome within the available fuel, taking into account reserves. Earlier examples in this chapter treated endurance in hours and minutes. But sometimes, the safe endurance is given as an amount of fuel — a fuel quantity — rather than a time. When that happens, we need a formula that accounts for different fuel consumption rates on the outbound leg versus the home leg. Those differences come from changes in altitude, temperature, wind components, and engine configuration. Here’s the formula. Let: - d = distance to the Point of Safe Return, in nautical miles. - F = fuel available for the PSR calculation, which is total fuel less any reserves, in kilograms. - CO = fuel consumption OUT to the PSR, in kilograms per nautical mile. - CH = fuel consumption HOME from the PSR, in kilograms per nautical mile. Now, how do we get CO and CH? Consumption in kilograms per nautical mile is usually obtained by dividing fuel flow — in kilograms per hour — by ground speed — in knots. Alternatively, you can use sector fuel divided by sector distance. Both give you the same unit: kg/NM. The logic is simple: the fuel used to reach the PSR plus the fuel used to return from the PSR must equal the fuel available, F. So we write: d × CO + d × CH = F Factor out d: d × (CO + CH) = F Then solve for d: d = F divided by (CO + CH) That’s the distance to the Point of Safe Return when fuel flows vary. Let’s work through the example in the book to see it in action. We have a TAS of 310 knots. The wind component out to the PSR is +30 knots, meaning a tailwind outbound. Total fuel available less reserves is 39,500 kilograms. Fuel flow out to the PSR at FL270 is 6,250 kilograms per hour. Fuel flow home from the PSR at FL310 is 5,300 kilograms per hour. First, we need ground speeds. Outbound ground speed is TAS plus the wind component: 310 + 30 = 340 knots. Homebound, the wind component reverses sign — it becomes a headwind of 30 knots — so ground speed home is 310 minus 30 = 280 knots. Now calculate CO: fuel flow out divided by ground speed out. That’s 6,250 kg/h divided by 340 kt, which gives 18.38 kilograms per nautical mile. Calculate CH: fuel flow home divided by ground speed home. That’s 5,300 kg/h divided by 280 kt, which gives 18.93 kilograms per nautical mile. Now apply the formula: d = F divided by (CO + CH). So d = 39,500 kg divided by (18.38 + 18.93). That’s 39,500 divided by 37.31, which gives approximately 1,059 nautical miles. To find the time to the PSR, we take that distance and divide by the outbound ground speed. 1,059 nautical miles at 340 knots gives 3 hours and 7 minutes. So the answer is: distance to PSR = 1,059 NM, time to PSR = 3 hours 7 minutes. Now, the book also includes a set of practice questions. These are the book’s practice questions — let’s try them one at a time.