At what point is the faster aircraft cheaper even though it is more expensive?
Hello,
Yesterday I had one of those nights again where my head thought, "No, you're not going to sleep today"…
And the question has been on my mind for a long time, but I can't seem to find a solution. It's a mathematical problem:
You have two planes:
Aircraft A flies 110 knots and costs 240 EUR per flight hour
Aircraft B flies 180 knots and costs 390 EUR per flight hour
Is there a distance beyond which it's cheaper (over the entire flight) to fly with aircraft B (because it then takes less time to complete the journey and thus saves flight hours)? Or how much faster or cheaper would aircraft B have to be for it to eventually become "worthwhile" to fly with aircraft B instead of aircraft A?
Thanks!
(No Ratings Yet)
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In this consideration: Never. Why? Because you do not take into account any costs independent of the route.
The kilometre costs € 1.17 with the smart plane and € 1.18 with the slow plane. So if you only take the route into account no scenario where the slower becomes cheaper.
You need to take into account the cost of starting and landing and variable from the flight path, otherwise no shoe will get out.
Look at flight times from the past. Go back a few years / decades.
You’ll find that we’re not flying faster today than we can. At least not much faster. This is because it is usually more worth flying slower. Otherwise, such a ticket would be much more expensive.
That is why there is no oversonic journey even if we have the planes and have already built them.
We’re getting slower. Civilian oversonic flights have e.g. fallen away.
B is always cheaper.
As you already realized. The ratio of costs to distance covered is linear. Your functional equation is correct. So one is always cheaper than the other.
Because only the track is variable. Is the easiest really to calculate the cost per sea mile.
240 / 110 = 2,18€ per seemeile
390/180 = 2,16€ per seemeile,
Plane B is therefore always cheaper than plane A.
If a simple invoice is made, simply calculate the costs per kilometre.
Set up a function, let the graph draw, then you know it.
Yeah, I’ve already thought about it, but I’m just on a role:
f(x)= x/180*390 or F(x)=x/110*240. Where X stands for the distance. That would then be a linear function which would have its origin at 0 in both formulas. Ergo would almost never “calculate” the faster aircraft. I’m just not sure if my function is correct.
Funfact: the faster aircraft is always counting.