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I am trying to compare trains with airplanes and the distances where taking a train is more viable than an airplane. I guess this would depend on the type of plane, but the Airbus 320 seems like a common enough plane in Europe so I have something along that type in mind.

As far as I know, it would be a good model for airplanes to separate the takeoff and landing phase, where they cover less ground and does take appreciable time on shorter hauls and the cruising phase. I found average speed for the latter but I don't know how much time takeoff takes and how much ground the aircraft can cover during this. Same goes for landing, so my question is, how much time do these take and how far does the aircraft get during these?

I'm also assuming that when looking at city center to city center travel times, taking a plane has about a three hour overhead (airports are far and it takes quite a long time to get in, security check and all), but if anybody actually knows some statistics on this I'd be much obliged :)

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  • $\begingroup$ There is a simple rule of thumb in Europe. For domestic trips, trains always make more sense. For international trips, airplanes make more sense. The reasons are time, pricing and heavy competition in international vs domestic routes. Airliner performance does not play into this. $\endgroup$ – Juan Jimenez Jun 23 at 16:09
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    $\begingroup$ I am right now only interested in time, nothing else. Also, your rule of thumb might apply for Spain, but Vienna - Budapest is already faster and cheaper with a train and there's nothing really fancy going between the two :) $\endgroup$ – fbence Jun 23 at 16:51
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    $\begingroup$ "Takeoff" and "landing" have rather specific meanings in aviation, and are very short in duration. You are probably more interested in the climb and descent portions of the flight. $\endgroup$ – a CVn Jun 23 at 17:23
  • $\begingroup$ When flying, nearly no time is spent below average ground speed of even a high-speed train, so in reality flying speed isn’t the controlling parameter, I believe... nonetheless, good question! $\endgroup$ – Cpt Reynolds Jun 23 at 18:13
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    $\begingroup$ I think this question needs to be restated, the title asks a specific question but the post's body seems confused. $\endgroup$ – zymhan Jun 23 at 18:32
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I answer directly to the title of the question:

What is the average time and distance needed to reach cruising height?

The Takeoff takes very little time, but the climb phase depends on the cruise flight level clearance. Assuming you climb with a vario of 2000ft/min with an average IAS of 200kts. The time to reach the cruise level is equal to:

(Cruise level in ft - takeoff terrain elevation if ft)/2000, the result is in minutes.

So if the departure is at sea level and the cruise altitude is at 30000ft, the climb phase will take : 30000/2000=15min

The descent till arrival could be more complicated because in addition to the pure descent you have to level before intercepting the approach phase which occurs relatively at lower speed. If you are lucky, not having to hold above the airport, the total descent might take 20 to 30 minutes

Routes are not necessarily straight lines, nevertheless you may assume an average cruise True speed of 500 knots.

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  • $\begingroup$ Thanks for answering, but I don't think I actually understand your answer, which is probably due to my lack of knowledge in aviation. I'm guessing that the 500 kts/hour would be true for cruising. You also said climbing and descending will take about say 40 minutes in total. I'm assuming during this 40 minutes the average speed is below 500 kts/hour, so what would be a good estimate for the amount of ground covered in that 40 minutes? $\endgroup$ – fbence Jun 24 at 17:55
  • $\begingroup$ After surfaces retraction the speed goes rapidly to 250 kts indicated, but at sea level no real difference between true speed and indicated airspeed. However at cruise altitude even though being at about 250knots indicated, your true speed may be about 500 knots. So you may assume during climb and descent phases an average true speed of 375 knots. You are right, these phases should be calculated separately, I do apologize. For very long flights roughly we can consider an average true speed of 500 knots, I will correct. $\endgroup$ – user40476 Jun 24 at 18:27
  • $\begingroup$ "500 kts/hr" would be a (weirdly expressed) rate of accelleration, as knots is already nautical miles per hour, thus kts/hr would be nm/h/h. $\endgroup$ – a CVn Jun 24 at 21:11
  • $\begingroup$ Yes indeed, I wanted to write miles/hour instead of kts, to be clearer ....unhappily I did a lapsus $\endgroup$ – user40476 Jun 25 at 3:29
  • $\begingroup$ I thought it was addressed to me, I am sure fbence did a lapsus $\endgroup$ – user40476 Jun 25 at 4:19

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