# Aerospace Help I need to clear the runway

1. Nov 29, 2004

### mikej_45

Help I need to clear the runway!!

Ok, so 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 asked my physics teacher why this is, and he assigned in told our class that if we could find out the answer to this question that he would give us extra credit.

My best guess so far is that as the plane builds up speed the air resistance on the surface of the plane gets bigger, so it needs to have 70 percent of take off speed by the time it gets halfway... any help would be great

2. Nov 30, 2004

### GENIERE

Not near enough information. Think about a 50 mile runway or a Harrier military jet..

3. Nov 30, 2004

### mikej_45

True, the distance of the runway would vary for different types of planes but no matter the distance a particular plane needs for clearance every plane needs to have 70 percent of its take off speed by the time it reaches 50 percent of the distance that plane will need to take off.

If a Harrier jet would really need 50 miles to take off than for some reason that jet must have 70 percent of its take off speed by the time it reaches the 25 mile mark.

Likewise if a fighter plane only needs 100 yards to take off it must have 70 percent of its take off speed by the time it reaches the 50 yard mark.

The runway distance will vary from plane to plane, but the percent of take off speed at the halfway mark is, according to my physics teacher and that article I read, always constant.

I just can't figure out why this is...

4. Nov 30, 2004

### brewnog

Do some research, look into the formula which relates velocity over an aerofoil with lift. That might get you started!

5. Nov 30, 2004

### Cliff_J

And think about the rate of acceleration and velocity. If you have constant acceleration, what is the velocity at the half point if you know the velocity at the end?

Here is a readout of a timeslip for drag racing, look at the readout for MPH for 1/8 and 1/4 mile.
http://sco.dragracing.com/drag_racing/101/timeslip.asp

Cliff

6. Nov 30, 2004

### Gonzolo

I think it's an empirical rule of thumb with a safety margin.

Acceleration is constant as far as I know, so in principle, 70% of T/O speed at 70% of the runway would do it, but no one is stupid enough to try that.

If you're at 70% T/O at 50% of the runway, and you start braking you will stop before the runway ends. If you're at 70% T/O speed at 51% of the runway, you might not be able to stop completely before the end of the runway. Consider 70% T/O at 50% runway a point of no return. The number is probably based on common ratios of T/O distance vs landing (braking) distance + a fudge factor. The ratios are probably similar for most planes.

Harriers don't count as conventionnal planes, they can take of vertically. This kind of take off has nothing to do with your problem.

Last edited by a moderator: Nov 30, 2004
7. Dec 1, 2004

### Cliff_J

Gonzolo - rember that position is the second derivative of acceleration. Its a linear relationship only with a constant velocity and the distance traveled is a combination of acceleration and velocity.

Look up the stopping distance of most cars from 60MPH compared to 80MPH. It might vary slightly from a constant acceleration based on test conditions but it does show the importance of the combined effects of acceleration and velocity to determine position.

Cliff

8. Dec 1, 2004

### Gonzolo

Do you think the 70% at halfway can be calculated?

9. Dec 1, 2004

### enigma

Staff Emeritus
Just use the formulas for uniform acceleration. Make the velocity at the end 1 "unit/second", the length of the runway 1 "unit". You can then go through and calculate the acceleration to obtain that speed.

If you now know the acceleration, go through and find out what the velocity would be at a length of 1/2 "unit".

10. Dec 1, 2004

### mikej_45

Ok, I understand the math behind this question: 10*the square root of the percentage of liftoff distance required is equal to the percentage of lift-off speed that should be attained in that distance. EXAMPLE:
√50=7.07=
10*√50=70.7

So since I want to know what speed a plane needs to achieve at the halfway mark on a runway I take the square root of 50 and multiply it by 10...

So I'm not really trying solve this problem mathematically...What I really want to know is WHY you need 70% of your total take off speed by the time you've used 50% of your runway. I don't care about the #'s I just want the concept.

11. Dec 1, 2004

### mikej_45

Ok, I understand the math behind this question: 10*the square root of the percentage of liftoff distance required is equal to the percentage of lift-off speed that should be attained in that distance. EXAMPLE:
√50=7.07=
10*√50=70.7

So since I want to know what speed a plane needs to achieve at the halfway mark on a runway I take the square root of 50 and multiply it by 10...

So I'm not really trying solve this problem mathematically...What I really want to know is WHY you need 70% of your total take off speed by the time you've used 50% of your runway. I don't care about the #'s I just want the concept.

12. Dec 2, 2004

### Cliff_J

Ok, lets simplify and use time instead of distance.

You need 150 units to take off. You accelerate at 5 units/sec^2 and it will take 30 seconds to get to your desired speed.

After 15 seconds you have 75 units of speed and you're halfway there in time. But you are eating up runway very quickly traveling at 75 units of speed, correct?

The other 75 units of speed will take the same amount of time and will need more runway simply because of your existing speed and the distance that will add for the next 15 seconds.

And obviously you know the mathematical relationship so hopefully switching to a time view helps.

Cliff

13. Dec 2, 2004

### mikej_45

Yes, cliff that does help a bit with the overall picture of the problem, but the time, velocity, acceleration, and distance all correlate in one way or another. So they can all be accounted for mathematically. Yes it always helps to view a problem in three dimensions...I really appreciate the help thus far, and again let me state that the help to this point has been quite useful.

I think that the problem here is me. I'm not being clear...or you've been answering my question and I'm not getting it...haha.

Maybe this will further clear things up...I'm looking for an explanation using concepts.

So far I think that Newtons third law (the force exerted by object 1 on object 2 is equal in magnitude but opposite in direction to the force exerted by object 1 on object 2) so if you push on a brick wall with x ammount of force...the wall pushes back against you with an equal force -x. This applies to an airplane because the plane pushes against the air of x density (ρ) the air will then push back on the plane with equal force in the opposite direction...

So do the the equal reactionary force of the air on the plane, the air velocity(I think the air should now have a velocity of its own, as a result of pushing back on the plane) is a major contributer to lift and drag. This can be proven through these equations for lift and drag.

L=1/2ρV^2SCl and D=1/2ρV^2SCd

Where: L= Lift, D=Drag, Cd= drag coefficent, Cl= Lift coefficent, S= wing area,
V^2= velocity squared, ρ= air density

I'm also pretty sure that it has something to do with Bernoulli's equation(the sum of the pressure, the kinetic energy per unit volume, and the potential energy per unity volume has the same value at all points along a streamline). So if we think of the air as a fluid, an object(the plane) moving through a fluid experiences a net upward force resulting from any effect(planes movement) that causes the fluid(air) to change its direction as it flows past the object(plane)

So to simplify...airplane wings are designed so that the air velocity above the wing is greater than that below, resulting in the air pressure above the wing to be less than the pressure below, and there is a net upward force on the wing(lift)

So, I understand the math, I understand why planes are able to fly...I just can't figure out why (according to physics concepts) they need to have 70% take off speed by the time they've used 50% of the runway.

Keep the help coming! thanks

M45

14. Dec 2, 2004

### Gonzolo

As I understand, you are wondering where the y=sqrt(x) relation comes from.

(that is what I am now wondering anyway)

15. Dec 2, 2004

### mikej_45

I'm trying to figure out how the LAWS of physics, like the ones I talked about above(and theres probably several others) effect a plane so that it would require 70% takeoff speed by the time it gets halfway down the runway...I know there has to be a reason somewhere...all the flight info I've read advises pilots to abort their take offs if they have not achieved 70% of their take off speed by the time their at the halfway mark of the runway....so theres a reason...and if theres a reason...there has to be an explanation using the laws of physics.

I'm not sure what your equation represents...

16. Dec 2, 2004

### ceptimus

17. Dec 2, 2004

### Staff: Mentor

Cliff J's and Enigma's posts are probably the most help: First off, ignore lift and drag - they have nothing to do with this calculation. This is entirely an acceleration/distance calculation. You can use some arbitrary values if it helps to picture it, but I'd imagine your physics teacher wants you to solve the equations. So here's what you do (I'm not going to do it for you since it is for school):

1. There is an equation for distance, which is the double-integral of acceleration - find it in your textbook (hint: acceleration is a constant, distance and time are the variables).

2. Use 1 for acceleration, plug in .5 for the distance and solve for time.

3. Use the time you just calculated to find the speed using the equation for acceleration vs time.

edit: btw, your first post doesn't ask a question, but I assume the question is to verify that rule of thumb. My process will give you speed: you're looking for s=.5 or something like that.

Last edited: Dec 2, 2004
18. Dec 2, 2004

### Gonzolo

The equation y = 10*sqrt(x) is what you are using in post 10 which gives rise to 70% at halfway. I'm not sure whether it is empirical or whether it actually comes from theory.

If it's from theory, the first thing I'd go after is finding out whether on the take-off roll, acceleration is considered constant or not.

Assuming acceleration decreases considerably because of air friction, it is perhaps not as much as the sqrt function. So the sqrt would be a worse case scenario for the acceleration curve (it gives 100% at full runway), and thus used a safe rule of thumb.

19. Dec 2, 2004

### mikej_45

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!!

20. Dec 2, 2004

### Cliff_J

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

21. Dec 2, 2004

Acceleration is an increase of speed over time, not over distance

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.

Physics has nothing to do with this problem. This is a math problem.

Last edited: Dec 2, 2004
22. Dec 3, 2004

### Staff: Mentor

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.
The explanation using the laws of physics is the math. That's why they call them "Newton's laws." The first is f=ma.
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.

Last edited: Dec 3, 2004
23. Dec 3, 2004

### mikej_45

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

24. Dec 4, 2004

### Astronuc

Staff Emeritus
Takeoff abort point

A lot of good points have been made in this thread and the parallel one - https://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.

25. Dec 4, 2004

### Gonzolo

[cough]...post 6...[cough]... I totally agree.

(still waiting for someone to calculate the 70% or demonstrate y=10*sqrt(x) from known physics...)