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Aerospace Clearing The Runway Revisited

  1. Dec 1, 2004 #1
    Ok, I asked this question a few days ago. I'm writing this to try to clear up some of the confusion about what it is I'm trying to figure out.

    I read in a flight magazine that in order for a plane to be able to clear the runway during take-off, a plane will need to have 70 percent of its take-off speed by the time it reaches the half-way point on the runway...

    I already know this to be a FACT so I am not trying prove it or trying to solve this question through the use of mathematics(it works)

    The part that has me stumped is the conceptual aspect of this question...

    WHY does a plane need 70% of its total take off speed by the time it reaches the halfway marker of its runway.

    Hopefully i've cleared it up a bit. Thanks for all the help I've been recieving...so far it has led me to research some interesting aspects of aviation physics...but all I'm getting is mathematical solutions, no real concept help. Thanks again.

    M45 :bugeye:
  2. jcsd
  3. Dec 2, 2004 #2


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    Air resistance increases as the square of the speed. It is 4X harder to go from 100 to 200mph than to go from 0 to 100.
  4. Dec 2, 2004 #3

    I missed the origional discussion (and couldn`t find it after a quick check through the previous day or so`s postings), so someone may have already suggested it, but isn`t a simple case of building up the necessary speed??

    For example, in a car is takes less time (and effort) to get from 0mph to 40mph as it would to get from 100mph to 140mph.

    As to exactly why, I guess you would have to look at the factors that will slow the car down, the ones I can think of are air resistance and friction. My guess would be that the air resistance would be greater the higher the speed you are going (more air to pass through?).

    I`m not sure if frictional (as in from the ground) effects would come into effect, as I`d imagine once the plane is airborne (and therefore not in contact with the ground) the same effect occurs, but on the ground there would be an additional frictional effect.

    If the car(or plane) were in space with no frictional surface / air resistance (and for the sake of somplicity no gravity acting on it), then I think it would take the same amount of effort to get from 100-140 as from 0-40.

  5. Dec 2, 2004 #4
    It's not just because of air resistance, or friction. The plane only has a certain power available from its engine(s) and this will result in an ever decreasing acceleration, as the speed builds. Think in how much less time a car can accelerate from 0 to 30 mph, compared to 50 to 80 mph. This would still be the case even with no air resistance.

    Even if the acceleration were constant (which implies an ever increasing power use) then the 70% figure is still about right. If the plane takes 10 seconds to reach 30 metres / second (about 67 mph) from rest then it will travel 150 metres in the process. If it takes only another 10 seconds to double its speed, it will travel an additional 300 metres in doing so.

    The formula for distance travelled with constant acceleration is:

    [tex]s= ut + at^2/2[/tex]

    where s is distance, u is the initial velocity, a is acceleration, and t is time.
    Last edited: Dec 2, 2004
  6. Dec 3, 2004 #5

    Chi Meson

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    All things are true so far. And there is another thing: even if the acdcelration was held constant throughout the take-off, you can see that the airplane will spend much less time on the second half of the runway, since it will be moving much faster over this part. Acceleration is the rate of change of velocity over time, not distance. If the airplane had only half its take-off speed at the halfway point, it would need the same amount of time to gain the same amound of speed. But at the greater speed, the same distance will be covered in much less time.

    Edit: Oh, Ceptimus essentially said this already. Sorry.
  7. Dec 5, 2004 #6


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    Oh yeah. I feel silly. ..

    That is the simplest reason - even in an "ideal" setup without friction or power considerations.

    The second half of the runway is simply traversed *faster*.
  8. Dec 8, 2004 #7
    Bingo :rofl:
  9. Dec 11, 2004 #8


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    The aircraft requires to accelerate up to 70 % of its total take off speed in order to prevent catastrohpy. That's right, it's all very pretty when talking about friction and etc, however when refering to airlines taking off and landing (mainly) things are somewhat different. If one of the engines blows up, the pilots have sufficient time to halt the aircraft before it reaches the end of the runway.
  10. Dec 11, 2004 #9
    Also the faster a plane rises the less noise over houses nearby!
  11. Dec 11, 2004 #10
    What alot of people have neglected to mention is something called the best rate-of-climb airspeed and the best angle-of-climb airspeed. The best rate of climb airspeed means you get to altitude the fastest. The best angle of climb airspeed gives you the greatest gain in altitude for the shortest amount of distance. If there is an obstruction at the end of the runway, you have to clear it after you lift off, or your going to crash into it. In order to clear the tree lines, power lines etc, you want to have your takeoff speed reached as soon as possible. This gives you enought time to clear that obsticle by the time you reach it. The other factors mentioned are important too. You want to have avialable runway to slow down in an emergency. And if your not up to speed very fast, you run out of runway faster as your speed increases.
    Last edited: Dec 11, 2004
  12. Dec 19, 2004 #11
    We must also keep in mind that it wouldn't be very wise to pull the nose up right away at takeoff speed, being dangerously close to stall speed in the possible event of an engine failure. Instead, after rotation the plane accelerates in the ground effect all the way to climb speed.
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