This point is critical

The critical point (CP,) also called the equi-time point (ETP,) is the point exactly halfway in terms of time.

Let's say you are riding your bike to your friends house 10km away and you ride at about 10km/H. Where is the CP? Commonsense tells you it is at the 5km mark, exactly halfway between your house and your friends house.

Now let's say you are riding in the opposite direction to Grandmas house which is at the top of a big hill, again 10km away. You can ride up the hill at 6km/H, and back at 14km/H. Where is the CP? Not at the 5km mark, because it takes you more time to ride up the hill than down. We need to use the speeds to find a ratio and apply to the distance, thus;

CP = D x Vhome/(Vout+Vhome)

CP = 10 x 14/(6+14)

CP = 7km

To test, we'll see how long it takes us to ride from the CP to our destination, and how long back.

3km / 6kmH = 0.5 * 60 = 30min

7km / 14kmH = 0.5 * 60 = 30min

So here we are, riding to grandmas house, when we remember we have to call mum to let her know we are on our way to grandmas house, do we turn back and let her know, of do we ride on to grandmas house and ring her from there? Up until the 3km/7km mark, if we turn around it will take less then 30min to ride home. After the 3km/7km mark, it will take less then 30min to ride on.

This is similar to the problem that aircraft face in flight when deciding to continue on or return home in response to changed circumstances. In nil wind, the CP is exactly halfway, but if this is any headwind or tailwind (most likely) the CP will move, and it is worth noting it will move into the headwind (uphill.)

We are flying a B727 from Mount Isa to Alice Springs, a distance of 1050NM. Our cruising speed is 450kts TAS, we have a 50kt headwind and we are past the halfway mark with only 500NM to go. A passenger on board has a medical emergency. 500NM ahead of us is Alice Springs hospital and 550NM behind us is Mount Isa hospital and a whole lot of nothing in between. Do we go on or turn back? Alice is closer, but due to the ground speed difference maybe it will take less time to fly back to Isa.

CP = 1050 * 500 / (400+500)

CP = 583NM from Mount Isa (remembering the CP always moves into wind.)
or
CP = 467NM from Alice Springs.

In this case we go back.

To test, let's fly on to Alice from the CP

400/467*60 = 51min
500/583*60 = 51min.

RESULT.

Conventional

Gratz to Lex who just got the ok to fly taildraggers. I'm glad I got my qual before several thousand hours of tricycle experience made it tougher. Still, he did alright I reckon in that he got the nod after only a few weeks of non-intensive training, an hour here and there, and a false start in an underpowered Champ.

Well done mate.

Point of no return

The point of no return (PNR) is not just an expression, it has a very real meaning in aviation. The point of no return is the point at which the aircrafts endurance is enough to either go on to our destination or turn back. If we choose to go on, we no longer have sufficient fuel to return to our destination, we have reached the point of no return. On over water legs this is an important point, if there is any emergency, our choice is already made, however suitable it may be to cater to our situation, if the weather at our destination is less than ideal, a decision should be made here whether to attempt an approach or abort and go home.

Let's say we are going from point A to point C, via point B. Point B is exactly in the middle between points A and C and there is no wind. We have only enough fuel to go from A to C and no more. Where would the PNR be? It would be at point B. Point B is halfway, and at point B we would have used exactly half our fuel. Up until we reach point B we can still change our minds and go home. At point B we can go home or we can continue on to our destination. Once we've passed point B we are committed to going on to point C.

Now let's say we have more than enough fuel to go from point A to point C, but not enough to go from point A to point C and back again. This is more usually the case with commercial aircraft. The PNR is somewhere between point B and point C, but where? Confusing the situation is the wind which is a tailwind from A to C, but would be a headwind were we to turn back to A again.

And so we find ourselves at B, needing to know how much further we could fly before we reach the PNR. We have a certain amount of fuel left;

Fuel on board 14000 Kg.

And we can calculate how much fuel will be required to fly back from where we are (B) to home (A.)

Fuel required to fly back to A from B 4000 Kg

We do this and subtract it from the fuel we have on board, whatever is left over is the amount of fuel we have available to fly out to the PNR and back to B.

14000 - 4000 = 10000 Kg

If the wind was nil, we could divide by the SGR to find how many miles we could fly using this fuel, halve the distance and we would have our answer. Or to put it another way, divide by twice the SGR, the SGR out being the same as the SGR back. This is the heart of the problem - wind and different aircraft configurations mean that the SGR out is almost never the same as the SGR back. But the equation is still the same;

Fuel at B after fuel back to A is subtracted;

10000Kg

Divided by SGRout + SGRhome;

10000 / (8.5 + 10.7) = 520nm

EDIT - It's been pointed out that I didn't explain what SGR means. The SGR is the specific ground ratio, or the amount of fuel used per distance of ground covered, and varies by fuel flow, airspeed and head/tail wind. In this case the SGR is KG of fuel used per ground NM covered. The B727 conveniently averages 10 KG per NM on average in cruise configuration. A headwind increases the SGR, increasing the amount of fuel used per NM, while a tailwind decreases it.

Flight planning - headwind/tailwind

I'm studying for my ATPL and I am stuck on flight planning. In fact I've been stuck on this one subject for almost a year. For my own edification, I'm going to describe some typical AFPA (ATPL Flight Planning Aeroplane) questions, and their solutions, so I can study wherever I happen to have internet access. Perhaps some more erudite and learned ATPL pilots can happen along and give me some pointers, too.

Maximum headwind/minimum tailwind.

When flying from place to place you need to ensure that you land with a minimum amount of fuel necessary to cover emergencies. That fuel is called reserve fuel, and may not be used, except in an emergency. Using your reserves alone constitutes an emergency.

Say that I am flying a B727-200LR from point B to point C, with point A a long way behind me, and point C still a ways to go. I've already burnt up a fair bit of fuel getting here and I need to ensure that I don't touch my reserve fuel getting to point C. Aloft I can have a headwind, a tailwind, or no wind at all (unlikely.) If the headwind is sufficiently strong I may get nowhere at all.

If I am 450nm from my destination and I have 7000Kg of fuel aboard, 3000Kg of which is reserve fuel, how strong a headwind can I accept at point B and be reasonable confident of getting to point C without using my reserves? I will ignore the descent and approach for the pourposes of this problem.

Firstly, let's see how much fuel I can burn getting to point C.

7000-3000 = 4000Kg of fuel. This is my flight fuel or fuel burn off (FBO,) the amount of fuel I can use getting to point C without using my reserves.

SGR = FBO/dist. The distance to go is 450nm. 4000/450 = 8.889 This is my SGR(required) This is the most Kg of fuel per nm I can burn. If my fuel economy is worse than this, no good.

Another way of calculating SGR = Fuel flow per hour/ground speed in kts. I calculate my fuel flow from the B727 flight manual in this configuration as being 3600Kg/H.

8.889 = 3600/Gs , or Ground speed = 3600/8.889 = 404kts.

I calculate my True Air Speed (TAS) in this configuration, which is 436kts.

My Gs(req)-TAS = my allowable head/tailwind.

404 - 436 = -32 Kts. This is the maximum headwind I can accept. A positive number would indicate a tailwind is required.

I'll test this by giving myself another 2000Kg of fuel onboard to use. If my reasoning is correct, then the answer should give me a much greater accepatable headwind.

6000/450 = 13.334 SGR(req)
3600/13.334 = 269kts Gs
269 - 436 = -167 kts allowable headwind.

Let's try it the other way with less fuel. If I have less fuel to cover the same distance, then logic tells me I need less headwind, maybe even a tailwind helping me along, to get me where I am going without eating into reserves.

2000/450 = 4.445 SGR(req)
3600/4.445 = 809kts Gs
809 - 436 = 373kts tailwind required.

RESULT.

Mo Money

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Frightening, hmmm?

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Thank you