1. Alpha to Bravo, where Bravo requires alternate fuel to Charlie, and Charlie has an INTER.... day VFR private SE piston.
Start with a review of CAAP 234-1(2.1) at
www.casa.gov.au/sites/default/files/caap...or-aircraft-fuel.pdf
4.1.1 requires, for a small aircraft on Day VFR PVT operations, 30 mins fixed and no variable reserves.
4.2.1 requires you to carry holding fuel if that be required for your flight -
was the hold fuel unnecessary? If holding is required, then you require the fuel to be able to do that hold (should it turn out to be necessary when you get there).
I did taxi + 30 min FR + trip fuel + alt fuel + 30 min hold.
Sounds about right to me ... although this conservative airman would probably take a bit more, depending on how I interpreted the wx forecast. One must keep in mind that the rules give the minimum solution, but not always the sensible solution. That's where training, experience and common sense play their role in operational management.
2. Effects of wind on the CP
As often is the case, it's a bit difficult without seeing the actual question.
However, if we look at the basic CP formula
Dcp = (total distance x GS home) / (GS home + GS on)
and the CP position moves into wind from the half way point between departure and destination.
A problem occurs in real winds (and one
which we tend to ignore) where the magnitude of the crosswind can muddy the waters a little, so let's assume, for this discussion, that we are talking pure HW or TW so that drift considerations don't come into play and we get a basic sort of indication as to what's going on. This will be adequate until wind speeds and drifts become significant and will cover the questions you are going to come up against in the CASA exams.
With this caveat, the equation simplifies to
Dcp = (D x (TAS - W)) / ((TAS - W) + (TAS + W))
If we play around with that, just a little bit, we get
Dcp = (D x (TAS - W)) / (2 x TAS)
Now, if we look at the difference between the nil wind and wind cases, we can come up with
delta Dcp = Dcp nil wind - Dcp wind = (D x TAS) / (2 x TAS) - (D x (TAS - W)) / (2 x TAS)
which simplifies to
delta Dcp = (total distance x wind component) / (2 x TAS)
Now, we can play around with this a bit further and come up with
delta Dcp = ((total distance / 2 ) / TAS ) x wind component
It might make more sense to put this into words, as - in a wind, compared to nil wind (or, if the wind strength increases) -
the distance to the CP moves into wind from the half way point by (half the total distance x wind component/TAS)
and you will see versions of this in various texts about the place. It then is a simple matter to figure out the time interval home or on from the CP.
3. Rate of Climb required to achieve 6%.
We covered this to some extent in a recent thread at
bobtait.com.au/forum/performance/6784-mi...mb-rate?limitstart=0
I was given a weight, DH and headwind, but not the TODA or slope.
Weight and PH/OAT (DH should be OK) are necessary. HW may be or may not be, as discussed in the other thread. TODA and slope are irrelevant.
I ... just used the VTOSS for the given weight.
What "sort" of speed is Vtoss ? What sort of speed do you need to run the gradient calculations ?
Is this correct, or should I have assumed a particular runway length so I could get around the p-chart?
Not necessary, as the only interest you have in the P-chart for this question is to get the takeoff safety speed.
Now, just get back on the horse, back into the books, and knock over the exam next time around ...