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Author Topic: Calculation flighttime versus distance and choosing the correct aircraft  (Read 1169 times)

MagicsAvon

  • Former member
A while ago I picked up a formula on the forum that let me calculate the correct aircraft based on flight time and distance. Some how I must have lost it. The only thing I have left is the following formula I did for an ATR. 174 minutes one-road trip time x 4,58 = 801,50. Now divide the distance 614 NM by the 801,50 = 0.766. The only thing I can't remember is where the 4,58 is based on. I know this formula is somewhere in this forum, hidden in some story, but I can't find it. Any reference to the correct url. will be appreciated

Offline Sami

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Um, what? :P

There's absolutely no need for such calculations or formulas (some hardcore players may argue otherwise but that is just rubbish..).

Use your common sense. ATR on a 7 hour sector is not a good idea for example. The differences in flight time between comparable models are usually within 5-10 minutes (if you compare two mid-size jets for example) and thus negligible.

MagicsAvon

  • Former member
Hmm, of course I know what you mean, but I am sure there was a kind of calculation by the mixture of flight-time and distance for choosing the most efficient aircraft. For instance right now I fly F27-200 for short-distance and BAC-11 for 650-1200 NM. Now I am considering to replace my 27's, and thus flying short distance as well with my BAC 's  (world4). I know there is a handy formula to find a better brake even point than just the 650 NM I am using now. Just because also the use of fuel per hour comes into consideration on (very) short flights. It is just I have a hard time to find it again on the forum.

Offline LemonButt

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I consider myself a hardcore player (with a degree in math even) and there is no formula for calculating "optimal" aircraft.  You can, for example, calculate roughly how many miles you can fly with a fixed number of flights per day.  So if you can fly 2000nm with 4 flights/day, you can do 1000-300-300-400 or 500-500-500-500 etc.  The problem with trying to boil it down to a single formula is that flying north-south or east-west results in different flight times because of the jetstream/wind.  Flying east/west can result in a 60+ minute delta compared to north/south.

If you're talking about a fuel calculation, then I'm sure you can concoct something that takes into account takeoff/landing the come up with the delta between flying a prop versus jet, but the game does it for you and again, east/west or north/south matters.

In the end, the correct aircraft is the one that you can operate profitably, whether it is an ATR or an A380.

Sanabas

  • Former member
I know there is a handy formula to find a better brake even point than just the 650 NM I am using now.

There isn't. Whoever told you there was a formula to pick the correct plane for a route was talking crap.

If you want to know which plane's quicker for the round trip, just compare the two round trips in the route planning. If you really feel the urge to have a pointless formula, then 1 leg = distance/speed + turn time. 480 NM on a 240 kt f27 with 45 minute turns = 2:45. 480 NM on a 430? kt BAC with 60 min turns = 2:07. Do a general formula, and you'll find that ~135 NM is where the higher speed of the BAC beats the shorter turns of the F27. Which is basically pointless. Unless the route is really short, the faster plane will be faster overall. Always.

If you start looking at fuel per hour, then fuel burn divided by speed divided by number of pax = how much fuel to take 1 pax 1 NM. That formula is actually somewhat useful for comparing similar models, gives you a better idea than just fuel burn alone will. But the turboprop will be more efficient, almost always.

So use the turboprop unless the route is too big, too long, or just nicely fits the schedule of a bigger plane. No formula required.

Offline Sami

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If you really feel the urge to have a pointless formula, then 1 leg = distance/speed + turn time.

However that too is inaccurate especially on shorter sectors as AWS models climb, cruise and descent segments of the flight separately and each have different speed profiles for each aircraft (and the difference in climb vs cruise speed isn't the same for all models of course). But if one wants to simplify, then that works.
(^just "for info" since most do not know how detailed the flight model is)

Compare a 50nm route with a prop and a jet and the overall duration is the same (+/-5 min perhaps due roundings), even though jet is supposed to be 200 kts+ faster.
« Last Edit: May 05, 2014, 01:04:56 PM by sami »

MagicsAvon

  • Former member
Thanks guys. Great response)...I see there is no real point of no return, a distance where the F27 absolutely needs to be  by replaced  by  bigger aircraft like the BAC-11 in this case. I will make my own calculations, based on speed, pax, time and fuel-consumption and additional (landing)costs. Somewhere in the mixture of facts and personal flavor lies the answer. )

Offline LotusAirways

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Just to add more detail to the above, it is fundamental to understand that speed/distance varies a lot with small and long distances. Let me explain with an example:

1) 240 kts TAS. This means the plane can fly 240 NM in one hour, and since 240 NM divided by 60 minutes equals 4, then the plane flies 4 NM per minute.

2) Now look at a short route, for instance 75 NM distance. You would expect the plane to cover the distance in 19 minutes because 75 NM / 4 NM per minute = 19 minutes of flying. The reality is more like 40 minutes. Both in game and in real life. Now use the 40 minutes trip time and multiply by 4 NM, which equals 160 NM --this is the optimum distance the plane can travel in 40 minutes. Finally, divide the travelled 75 NM by the optimum distance of 160 NM, it equals 0.47. The efficiency is 0.47.

3) Move on to a medium route, for instance 500 NM. You would expect the plane to cover the distance in 500 NM / 4 NM per minute, a 125 minutes flight. The reality is more like 155 minutes. As above, 155 minutes multiplied by 4 NM equals 620 NM. Divide 500 NM by 620 NM, it gives an efficiency of 0.81. 

4) Try a longer route, 700 NM. Again, the expectation is 700 NM / 4 NM per minute, 175 minutes flight. Reality, 190 minutes. Do as above. Efficiency: 0.92.

Hope it is clear.
LA 

MagicsAvon

  • Former member
Thanks Lotus)..by reading your answer I get the idea this is what I was looking for. My guess you have mentioned this before as an answer on a topic, and it was there I started to use the info like a formula)..thx again

MagicsAvon

  • Former member
Game talk - Game World #2 / Re: Changing Small Aircraft Type
on: October 30, 2013, 12:48:23 PM

Yup, this answer of you Lotus, gave me the idea of the formula I am talking about. Thx,..again..

Offline LotusAirways

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Hehehe,
Sorry mate, no formula to find the "right" airplane for a route. The above is just to show that the longer the route is, and the more efficiently the formula Sanabas described (distance in NM / Speed in NM per hour), works. And vice versa. Nothing more.

 

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