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.