KORD airport for instance charges domestic vs international arrivals differently.
I could see that this may have something to do with imports/taxes/tariffs etc, but why are the landing fees measured in \$ per 1,000lbs?
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5$\begingroup$ comparative damage to the runway and taxi routes $\endgroup$– ratchet freakMar 19, 2014 at 13:36
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3$\begingroup$ because the costs to handle passengers from a B777 are higher than the costs of handling a C152? Or because the profits on a B777 flights are higher than a C152 flight, so more can be charged. $\endgroup$– DeltaLima ♦Mar 19, 2014 at 13:41
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$\begingroup$ Because a widebody jetliner takes up a lot bigger chunk of airport than an ultralight does? $\endgroup$– VikkiOct 3, 2019 at 22:02
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2 Answers
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From Anne Graham's book, Managing Airports:
This charging mechanism uses the ‘ability to pay’ principles, since airlines using larger aircraft are in a better position to pay higher charges. Some costs such as runway wear and tear do increase with weight and also larger aircraft require vortex separations, which can reduce the number of aircraft movements during a certain period. Overall, however, there is not a strong relationship between aircraft weight and airfield cost. A flat rate landing charge for all aircraft types may be more appropriate, particularly at congested airports. This is because the cost of occupying the congested runway is movement related and independent of aircraft size. Each aircraft movement will consume the same resource.
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5$\begingroup$ I disagree with Ms. Graham's assertion that the costs are independent of aircraft size. You can fit a lot more smaller aircraft into a line on a taxiway, and it's been my experience that smaller aircraft use less runway / take off quicker. Would OSHKOSH function if all aircraft were 777 sized? $\endgroup$ Mar 19, 2014 at 16:25
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4$\begingroup$ @DanPichelman The quote specifically addreses that when it says "Some costs such as runway wear and tear do increase with weight and also larger aircraft require vortex separations, which can reduce the number of aircraft movements during a certain period." and then goes on to say that there is not a strong relationship so I don't know what you are disagreeing with. Also wake turbulence avoidance procedures call for a specific time between aircraft of a similar size, so a similar number of operations will take place. Oshkosh is special and operating on a waiver. $\endgroup$ Mar 19, 2014 at 18:53
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$\begingroup$ Bonus question: does anybody know WHEN is the weight measured? Is fuel part of this measurement or just airframe? $\endgroup$– KORD4meMar 19, 2014 at 23:30
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4$\begingroup$ It is not measured. Most airports have a weight-related landing charge based on maximum takeoff weight (MTOW) which is fixed and constant per aircraft model. It is a design limitation of the aircraft. It therefore does not depend on the actual load the aircraft is carrying (and by load I mean cargo, passengers, baggage, fuel, etc.) $\endgroup$– molgarMar 20, 2014 at 8:09
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$\begingroup$ @molgar: That is not entirely true. Each aircraft model will be certified with a certain MTOW, however over time the production process will be refined to eliminate structural weight, plus the calculations will be tweaked based on actual performance, to enhance the MTOW. It is also possible for an operator to certify the aircraft with a lower MTOW than the manufacturers recommendation (such as if they only perform domestic flights, and choose to close off some of the fuel tanks. It is therefore more accurate to say a particular aircraft is certified for a MTOW at the point of registration. $\endgroup$ May 9, 2014 at 15:26
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The wear and tear of the runway depends very strongly on the weight of aircrafts, so I expect it to be at least taken into account. By discussing the physics of this below, I will show that the effect is stronger than intuition may commonly suggest.
It is certainly not the only factor to determine the fee, and may even be a minor one in most cases, but it could be of practical importance for more uncommon cases, (like when a runway is already in bad shape in some way).
I'll do a rough estimate of the incremental damage caused by a plane, on average, in comparison to planes of different weight.
The wear of the runway and taxiways depend on the load per area that is caused by the wheels. Take note that heavier planes use larger wheels to reduce that load per area.
Regarding the basic physics, the load per area is comparable to the axle load of cars. For the street wear caused by a car, this article [1] states that it is proportional to the axle load raised to the fourth power.
That means a double load per area causes 16 times the wear, or four times load per area 256 times the wear.
Even without an estimate of how much reduction of the load per area is provided by using larger wheels, I think that means that, for all practical purposes, the wear is caused only by the "large" airplanes.
[1]: "Die Kosten der Verkehrsinfrastruktur" Page 12 (German, PDF) refering to work of the american State Highway and Transportation Officials (AASHO) during the 1950ies, see Wikipedia (German) on axle load.
answered May 7, 2014 at 11:17