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Help I need to clear the runway

by mikej_45
Tags: clear, runway
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mikej_45
#19
Dec2-04, 10:10 PM
P: 19
russ_watters...Again I UNDERSTAND THE MATH, AGAIN, I NO LONGER CARE ABOUT EQUATIONS AND CALCULATIONS. THIS IS NOT A MATHEMATICAL PROBLEM.

I'M LOOKING FOR AN EXPLANATION USING ONLY THE LAWS OF PHYSICS. THERE MUST BE A WAY TO EXPLAIN THIS PROBLEM BY ONLY TALKING ABOUT CONCEPTS.

ALSO, I'M SURE THAT LIFT AND DRAG HAVE SOMETHING TO DO WITH THIS PROBLEM...THEY MUST COME INTO PLAY SOMEHOW BECAUSE THEY ARE SOME OF THE MOST IMPORTANT FACTORS THAT A PILOT MUST TAKE INTO CONSIDERATION WHEN PLANNING HIS/HER TAKE OFF.

ALL I'M TRYING TO FIGURE OUT IS WHY, WHY, WHY, A PLANE NEEDS 70% OF ITS TAKE OFF SPEED AFTER USING 50% OF THE RUNWAY. I KNOW THE LAWS OF PHYSICS APPLY...JUST NOT SURE HOW...

THIS PROBLEM IS DRIVING ME NUTS...I EVEN HAD A DREAM ABOUT IT LAST NIGHT!!

SOMEONE PLEASE HELP...OR I FEAR I MAY END UP IN THE COO COO HOUSE.
Cliff_J
#20
Dec2-04, 11:22 PM
Sci Advisor
P: 789
Mike - Russ has this one down, its acceleration. Lift and drag are not factors in the model (the drag will force this to be even greater) and that's why I used automotive examples for acceleration. Velocity is all the pilot cares about, once he has enough he will have the lift he needs.

The physics are simple, once you start building speed you start covering distance. Sure you can accelerate more but you are covering ground the whole time but more of it as time goes on. From a standing start you don't have any speed and are not using any runway. Here's a quick chart, lets assume 6000ft runway and we can accelerate at 10ft/s^2.

Pos.......Vel..........Time
0.............0...........0
1000.....141..........14.1
2000.....200...........5.9
3000.....245...........4.5
4000.....283...........3.8
5000.....317...........3.3
6000.....347...........3.0

Ok, so look at the graph and think about this. For the first 1000 ft the plane takes a long time, 14.1 seconds just to get to 141 ft/s. Within that 1000ft the chart would look similar, at first it takes a long time to build any speed. But then as you build speed you cover the distance quickly!

So at the end of the first 1000ft you are going 207MPH. If you cut power and just coasted at that speed you'd cover the next 1000ft in 7.1 seconds. But you don't cut power and instead accelerate at the same rate and now cover that 1000ft in 5.9 seconds instead. Plus you're now at 200ft/s or 293MPH. This continues on and on.

So the difference between coasting and full throttle is 1.2 seconds for that 2nd thousand foot section. As you can see (hopefully) from it being illustrated out with points along the way when I say the velocity is using up the distance I mean it is USING up the distance. And since you're accelerating the velocity is increasing more which means its using up even more distance. Imagine how quickly those dotted lines are flying by at 207MPH and you only have 40% of your needed takeoff speed.

Oh, and as a side note, from what I remember from my aviator roommates the official runway distance includes its own buffer. But as can be seen from simple models that don't include drag's nasty affects, that buffer could quickly be used up and now you'd be in trouble.

Cliff
hitssquad
#21
Dec2-04, 11:43 PM
hitssquad's Avatar
P: 1,382
The simplest answer (which is a distillation of Russ's and Cliff's answers) is that acceleration is an increase of speed over time, not over distance. If the plane has accelerated to any speed at all by the time it gets to the halfway point, it will have less time to accelerate during its transit of the second half of the runway than it had during its transit of the first half of the runway. Therefore, assuming the acceleration of the plane is linear, and not increasing logarithmically, the plane necessarily would have to be going much faster than 50% of takeoff speed at the halfway point of the runway.



Quote Quote by mikej_45
I KNOW THE LAWS OF PHYSICS APPLY
Physics has nothing to do with this problem. This is a math problem.
russ_watters
#22
Dec3-04, 01:03 AM
Mentor
P: 22,294
Mike: constant force (thrust) means constant acceleration. Constant acceleration means a linear increase in speed. A linear increase in speed means distance increases as a square function of time. That's all there is to this problem. You asked "why," but its simply a consequence of the math describing Newtonian physics.
I'M LOOKING FOR AN EXPLANATION USING ONLY THE LAWS OF PHYSICS. THERE MUST BE A WAY TO EXPLAIN THIS PROBLEM BY ONLY TALKING ABOUT CONCEPTS.
The explanation using the laws of physics is the math. That's why they call them "Newton's laws." The first is f=ma.
ALSO, I'M SURE THAT LIFT AND DRAG HAVE SOMETHING TO DO WITH THIS PROBLEM...THEY MUST COME INTO PLAY SOMEHOW BECAUSE THEY ARE SOME OF THE MOST IMPORTANT FACTORS THAT A PILOT MUST TAKE INTO CONSIDERATION WHEN PLANNING HIS/HER TAKE OFF.
Lift and drag determine what the takeoff speed is, but have nothing to do the actual acceleration down the runway: that's all determined by the thrust of the engines. At low speed, drag is negligible, and the thrust of the engines is roughly constant (in fact, thrust and drag both increase with speed and for some types of engines the increases cancel each other nearly exactly). This is strictly a Newtonian physics - 1st law problem.
ALL I'M TRYING TO FIGURE OUT IS WHY, WHY, WHY, A PLANE NEEDS 70% OF ITS TAKE OFF SPEED AFTER USING 50% OF THE RUNWAY. I KNOW THE LAWS OF PHYSICS APPLY...JUST NOT SURE HOW...
You already know the answer and shouting isn't going to help you accept that it really is that simple.
mikej_45
#23
Dec3-04, 10:15 AM
P: 19
Is it all really this simple...????!!! I understand everything you've all been saying...it obviously all makes sense...I just can't believe this is so simple...I guess I'm just stressing because these extra credit points are all that separate me from getting an "A" in class.

Well, I'm going to take my research along with all of your info and turn it into my physics teacher...so thanks everyone for the help...I'll let you know what he says.

Thanks again to everyone that helped. Greatly appreciated

M45
Astronuc
#24
Dec4-04, 04:21 PM
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P: 21,869
A lot of good points have been made in this thread and the parallel one - http://www.physicsforums.com/showthread.php?t=54925 , so I won't repeat. But, I did a little research on this matter - which is found under the keywords "takeoff abort" or "takeoff abort point".

I did find on most websites related to aviation, and the FAA (www.faa.gov) as well, that this is a "rule of thumb" of which ostensibly every pilot knows. In fact, one site mentioned 75% of takeoff speed rather than 70% mentioned by all the rest.

But what struck me was that nowhere did I find the technical basis, although there was an inference to Boeing (representative of the aircraft design and manufacturing industry). As an engineer, I am disappointed that something so fundamental is not discussed, not even at the FAA. On the other hand, perhaps I have not have found the proper webpage or e-document.

There were some caveats mentioned - the condition applies to the situation when engine failure occurs during takeoff (i.e. one cannot take credit for reverse thrust - which IIRC does not applied on fixed prop anyway) and it applies to runways in good condition. There were a lot of anecdotal articles on aircraft (private and commercial) that crashed during takeoff, even when this rule was apparently applied.

From my experience, planes decelerate faster than accelerating - even just using flaps and brakes - and certainly with reverse thrust. Then there is the margin of safety to be considered.
Gonzolo
#25
Dec4-04, 06:08 PM
P: n/a
[cough]...post 6...[cough]... I totally agree.

(still waiting for someone to calculate the 70% or demonstrate y=10*sqrt(x) from known physics...)
russ_watters
#26
Dec4-04, 06:33 PM
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P: 22,294
Quote Quote by Gonzolo
[cough]...post 6...[cough]...
Since acceleration is constant, you won't get 70% of takeoff speed at 70% of the runway.
kuba
#27
Dec8-04, 05:00 PM
P: n/a
Generally,
[tex]s=\int v(t) {\mathrm d}t=\int\int a(t) {\mathrm d}t{\mathrm d}t[/tex]

Assuming a constant acceleration [tex]a[/tex], the integration gives

[tex]s=a\cdot t^2/2, v=a\cdot t[/tex]

Let's say you need your takeoff speed that at [tex]s=L[/tex].

Then [tex]t(s=\frac{L}{2})=\sqrt\frac{L}{a},[/tex] and
[tex]t(s=L)=\sqrt\frac{L}{a}\sqrt(2)=\sqrt\frac{L}{a}\cdot1.414[/tex]

Thus [tex]t(s=L/2)=0.707\cdot t(s=L)[/tex]

So, halfway down the road you'll be at 70% of the time it takes to go full way. Now, since [tex]v\sim t[/tex] ([tex]v[/tex] proportional to [tex]t[/tex]), it's obvious that you'll also have 70% of the lift-off velocity.

It's hardly hardcore science nor engineering. I had such problems in elementary school in Poland in the eighties. No wonder the FAA didn't feel like explaining itself

It's a rule of thumb, though. Some jet engines might actually provide bigger net thrust (i.e. engine thrush-drag) when they go faster. So the constant acceleration assumption may not generally hold.

But remember that the 70% above is a very non-conservative figure. You typically can't use the full runway length, probably more like 95% of it in ideal conditions. Also, if you have face wind your intial velocity in relation to air is non-zero, etc. A nice homework problem would be to ask highschoolers to draw a chart of how much shorter would the run-up to take-off be as a function of face wind speed expressed as a percenteage of say [tex]V_2[/tex].
kuba
#28
Dec8-04, 05:09 PM
P: n/a
Oh, BTW lift/drag are ignored in the problem because even though your sense deceives you that they are the most important things related to flying, the actual problem has NOTHING to do with flying. It's a trivial linear motion problem and you could be as well talking about say a rolling drum accelerating down a slope, or ball from a ball bearing falling down in an evacuated tube.

As far as drag goes, in this problem it actually "fixes" non-constant thrust at the engines. Typically, a jet engine becomes more efficient as the intake air velocity increases. This relation is sorta-kinda quadratic. But the aerodynamic drag will grow with square of the velocity too. So the net thrust increase is negligible to the first order I guess.

And as far as lift goes, who cares? We're talking with everything scaled in relation to take-off velocity required in given conditions. So if during flight planning stage the pilots have computed that given the face wind, mass of the aircraft, height above sea level etc. they need a certain take-off speed, they'd better have 70% of that speed when they're halfway throught their run-up, otherwise they simply won't have the runway left to get to 100% needed. That take-off speed where lift=weight at a certain angle of attack is sure not trivial to compute and indeed involves all the lift/drag knowledge you referred to, otherwise called aerodynamics. But we're talking speeds relative to some target speed, not absolute speeds. That's why I used symbolic math above. Because apart from that "elusive" 70%, everything else would only obscure things if it were numeric, not symbolic.
Cliff_J
#29
Dec8-04, 10:57 PM
Sci Advisor
P: 789
From what I remember from roomates in college and pics of runways, there is the official runway and then the extra for a tiny margin of safety. But as we talked about in these two threads, the plane has a lot of velocity near the end and would quickly use up any margin that was not substantial.

And when thought of from a prop plane perspective and the issues of thrust/drag at speed, this rule of thumb leaves very little margin for a real life scenario. I know from experience in a Cessna 142 in college that the last 1/3 of that runway can be pretty white knuckled when you're looking out the window and you're not sure if the little plane couldn't have used a little more acceleration...

Cliff
Gonzolo
#30
Dec19-04, 01:10 PM
P: n/a
Quote Quote by russ_watters
Since acceleration is constant, you won't get 70% of takeoff speed at 70% of the runway.
I mixed up my imaginary v(x) and v(t) graphs.

You cleared things up kuba.


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