Help I need to clear the runway!!


by mikej_45
Tags: clear, runway
mikej_45
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#1
Nov29-04, 10:03 PM
P: 19
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.
We are able to consult any source we want, so if anyone knows the answer to this question please help!!

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
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GENIERE
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#2
Nov30-04, 02:33 AM
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Not near enough information. Think about a 50 mile runway or a Harrier military jet..
mikej_45
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#3
Nov30-04, 02:20 PM
P: 19
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...

brewnog
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#4
Nov30-04, 02:50 PM
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Help I need to clear the runway!!


Do some research, look into the formula which relates velocity over an aerofoil with lift. That might get you started!
Cliff_J
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#5
Nov30-04, 03:52 PM
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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
Gonzolo
#6
Nov30-04, 09:20 PM
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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.
Cliff_J
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#7
Dec1-04, 12:36 PM
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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
Gonzolo
#8
Dec1-04, 01:32 PM
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Do you think the 70% at halfway can be calculated?
enigma
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#9
Dec1-04, 03:58 PM
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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".
mikej_45
mikej_45 is offline
#10
Dec1-04, 05:00 PM
P: 19
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.
mikej_45
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#11
Dec1-04, 05:02 PM
P: 19
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.
Cliff_J
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#12
Dec2-04, 08:52 AM
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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
mikej_45
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#13
Dec2-04, 01:24 PM
P: 19
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
Gonzolo
#14
Dec2-04, 01:40 PM
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As I understand, you are wondering where the y=sqrt(x) relation comes from.

(that is what I am now wondering anyway)
mikej_45
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#15
Dec2-04, 02:01 PM
P: 19
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...
ceptimus
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#16
Dec2-04, 02:14 PM
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Did you see my reply in your other runway thread ?

Why have two threads on the same topic anyhow
russ_watters
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#17
Dec2-04, 03:06 PM
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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.
Gonzolo
#18
Dec2-04, 03:12 PM
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Quote Quote by mikej_45
I'm not sure what your equation represents...
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.


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