# How do I calculate the point of no return for a flight with multiple legs?

In every textbook where Points of No Return (PNR) are discussed (or at least, those I've read), the method of calculating them is always between point A to point B as a direct route. This is simple enough to do with the appropriate formula.

What I've never come across is an application of this in a more realistic scenario, where an aircraft will have at least one waypoint along the way, or a change in altitude/flight level, where the wind conditions or a change in TAS result in a change of ground speed.

Take the following example:

Leg Track Ground Speed Wind Distance
A -> B 000 100 090/50 50
B -> C 270 150 090/50 25

What's the PNR for A -> C?

• I spent several minutes wondering what it was I forgot when I submitted the question. That was embarrassing. Edited. Mar 22, 2022 at 10:32
• On the return do you need to backtrack the original legs or can you go straight?
– Jim
Mar 23, 2022 at 9:06

The point of no return is defined as the point where the travel time to the destination (C) is equal to the travel time when returning to the departure (A). On each leg, calculating travel times is trivial with the known ground speed and distance ($$\text{time} = \text{distance} / \text{speed}$$). In our example we would get the following for the time to get to point C as a function of distance $$d$$:

$$\text{time to C} = \begin{cases} (50 - d) / 100 + 25 / 150, & d \leq 50 \\ (75 - d) / 150, & d > 50 \end{cases}$$

When assuming a ground speed of 50 for C -> B (tailwind turned into headwind), a similar equation can be created for the return case:

$$\text{time to A} = \begin{cases} d / 100, & d \leq 50 \\ (d - 50) / 50 + 50 / 100, & d > 50 \end{cases}$$

Plotting the two functions, we get this picture:

The distance where the two time curves intersect, is the point of no return. It can be calculated by equating the two straight line equations for the blue and red curve:

$$\frac{50 - d}{100} + \frac{25}{150} \stackrel{!}{=} \frac{d}{100} \quad \Leftrightarrow \quad d = \frac{100}{3} \approx 33.3$$

• Would graphs of fuel required to reach points A and C give the same final result? Mar 22, 2022 at 12:13
• (Maybe grounds for a new ASE question?) Mar 22, 2022 at 12:13
• @quietflyer Sure, fuel flow per time is roughly constant (slightly decreasing due to the lowering weight), but it does not depend on head- or tailwind. If you fly at different altitude to avoid the headwind, things start to get more interesting... Mar 22, 2022 at 12:17
• Is there a formula you use to calculate that intersection? I'm looking to see if I can build a spreadsheet that calculates a lot of my flight planning. Got everything made, except this. Mar 22, 2022 at 21:44
• @hiigaran I added the formulas for the lines and how the PNR was calculated. They should be easy to generalize for more legs. Mar 23, 2022 at 10:16

The Point of no Return (PNR) is the furthest distance an aircraft can travel from it's departure point and return to this same departure point with the consumption of all usable fuel except fuel reserves ('Safe Endurance(SE)').

There are many practical applications where the journey out and the journey home have different true air speed values and fuel consumption rates (for example, one engine inoperative on the return journey). Since your question doesn't reference fuel consumption rates I am going to go ahead and assume the same fuel consumption throughout for the purpose of the PNR calculation.

Since we need to return home, the distance out ($$d_{1}$$) is equal to the distance home ($$d_{2}$$) so lets just call this $$d$$ $$d = d_{1} = d_{2}$$

The safe endurance is the total time spent on the journey, that is the time out plus the time home. $$SE = t_{1} + t_{2}$$

Furthermore, by rearranging the basic speed equation $$s = \frac{d}{t}$$ for time $$t = \frac{d}{s}$$ we can say: $$SE = \frac{d_{1}}{O} + \frac{d_{2}}{H}$$

Where $$O$$ is the ground speed for the journey out and $$H$$ is the ground speed for the journey home. We already know that $$d = d_{1} = d_{2}$$ so we can factor out $$d$$

$$SE = d \left( \frac{1}{O} + \frac{1}{H} \right)$$

And with some basic algebra...

$$SE = d \left( \frac{O+H} {O\cdot H} \right)$$

and multiplying each side by the inverse of the fraction to isolate distance to the PNR.

$$d_{toPNR} = SE \left( \frac{O\cdot H}{O+H} \right)$$

Now lets consider a multi-leg journey

Firstly, consider the endurance to travel leg one (A to B) and return to the departure point (A)

$$Endurance_{A-B-A} = E_{1}= d_{1} \left( \frac{O_{1}+H_{1}} {O_{1}\cdot H_{1}} \right)$$

The remaining SE avaible for the second leg (B-C) is the original (total) SE minus the endurance if the PNR was just leg 1 $$E_{1}$$ (that is, the endurance for the out and home journey of leg 1).

$$SE_{2} = \left( SE \right)_{Total} - E_{1}$$

Note, If $$E_{1} > \left( SE \right)_{Total}$$ then the PNR lies within the first leg. If $$E_{1} = \left( SE \right)_{Total}$$ then B is the PRN point.

$$d_{toPNR from B} = SE_{2} \left( \frac{O_{2}\cdot H_{2}}{O_{2}+H_{2}} \right)$$

You could continue this process for any consecutive legs, by determining at which leg the the PNR endurance elapses the safe endurance.

To convert this into a time to the PNR from the departure, calculate and sum the times for each leg up to the leg on which the PNR occurs. Then calculate time to PNR for that leg (as a single leg question). Again, rearranging the speed equation $$s = \frac{d}{t}$$ thus, $$t = \frac{d}{s}$$

$$t_{toPNR} =\frac{d_{toPNR}}{O}$$

Substituting in the $$d_{toPNR}$$ equation from earlier:

$$t_{toPNR} = SE \left( \frac{H}{O+H} \right)$$