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* Two slots per five minutes are for prop planes.
* Three slots per five minutes are for regional planes.
* Four slots per five minutes are for narrow-body planes.
* One slot per five minutes are for wide-body planes.
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IMO that would be another case of regulation by rules.
Sure the exact numbers would be subject to discussion, but this way you'd limit the number of options no matter what. If I'd want to operate a hub with 100 747-8i, I'd be penalized. I'd be forced to operate small equipment.
Top priority needs to be a system that allows anyone to operate any equipment at any time from/to anywhere as long as there are no "natural" limitations (nighttime ban, noise regulations, slots)
The most practical way I see is the aforementioned system of dynamic landing fees.
Let's play around with a little example that is certainly be subject to discussion:
Well, take your time, it's a bit lengthily, I'm bored right now... :wacko: :blink: -_-
Say we're operating a Dash8-Q200A (DH2) with 36 seats.
Condition at present
Landing fees (Tempelhof server) range from 0 for a 0-bar airfield to 142$ for a 10-bar airport.
Even if I link two 10-bar airports (say LHR-FRA) it doesn't cost me more than 142$ or 4$ per seat. It doesn't matter what route I'm flying and it doesn't matter how many slots are left.
Target
We need the following factors:
Aircraft size (MTOW), airport size at both origin and dest., slot occupancy at both origin and dest.
This needs to be packed into a formula with slot occupancy being the dynamising element.
Let's assume the following base data for the DH2 (these numbers are subject to discussion):
For basic landing fee (BLF) we take MTOW in kg / (100kg/$) -> 15,960kg/(100kg/$)= 160$
The A380 would consequently have 560,000kg/(100kg/$) = 5600$ for example
Airport size (demand-) factors (ASF): 0 for 0-bar, 0.1 for 1-bar and up to 1 for a 10-bar.
Connecting between a 3-bar and a 6-bar airport would calculate to 0.3*0.6 = 0,18 for example.
Connecting a 1-bar to a 10-bar (the typical feeder flight) would be 0.1.
Connecting between two 6-bar would logically be 0.36.
Slot occupancy factor (SOF)
(Let's take both origin and dest. to have identical values for the return trip - makes it easier to control and understand)
The formula could look like this: SOF = 1+5*(Oo*Od)3 (max SOF = 6)
Oo = Occupancy origin (/100%), Od = Occupancy dest. (/100%)
Example 1: FRA-LHR
Oo = 96%/100% = 0.95
Od = 98%/100% = 0.98
SOF = 1+5*(0.95*0.98)3 = 5.03
Example 2: FRA-DTM
Od = 0.11
SOF = 1.006
The full formula for a DH2 onroute FRA-LHR:
Actual landing fee (slot fee)
SF = BLF*ASF*SOF = 160$*1*5.03 = 809$ (= 22$/seat)
FRA-DTM:
SF = 160$*0.5*1.006 = 80$ (= 2$/seat)
Ok, the A380 on FRA-LHR would cost a fortune:
SF = 5,600$15.03 = 28.168$
A size correcting factor SCF would be in order here. So to say a factor favouring lager aircraft - I think that's what the discussion also was about...
Let's simply assume we want our heaviest airframe (the 388) to cost only half (SCF = 0.5) and the smallest airframe roughly SCF = 1.5
=> SCF = (MTOW388/MTOWxx)0.133 - 0.5
SCF for a 388 would be (560,000/560,000)0.133 - 0.5 = 0.5
SCF for a BNI would be ( 560,000/2990)0.133 -0.5 = 1.51
SCF for our DH2: (560,000/15,960)0.133 -0.5 = 1.11
SCF for a 73J: (560,000/85,130)0.133 -0.5 = 0.78
So, a final formula could be
SF = BLF*ASF*SOF*SCF
Our DH2 would cost us on the FRA-LHR trip at given occupancy rates: 160$*1*5.03*1.11 = 893$
The 388 would cost us 5,600$*1*5.03*0.5 = 14,084$