Physics Puzzle

Stoday said:
You've all missed the significant factor.
Or is that you didn't read the entire topic?

For example said:
...the only answer to be determined is whether or not the motive force of the jet engine is more than double the usual frictional force of the wheels.
So the only relevant part of your entire post is this:

Stoday said:
So the plane will take off, provided the wheels can spin at twice their normal take-off speed.
You are absolutely correct, despite your arrival at the party being only just in time to start clearing up the empty cans and full ashtrays. ;)
 
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LANGLEY MEMORIAL AERONAUTICAL LABORATORY,
NATIONAL ADVISORY CordanmD FOR AERONAUTICS,
LANGLEY FIELD, VA

REPORT No. 583
THE ROLLING FRICTION OF SEVERAL AIRPLANE WHEELS AND TIRES AND THE EFFECT OF ROLLING FRICTION ON TAKE-OFF
...The measurement of the rolling friction was accomplished
by recording the pull between B towing vehicle
and a loaded trailer equipped with the wheels and tires
to be tested. The resistance thus measured included, of
course, that due to the wheel bearings as well as that
of the tires....

CONCLUSIONS

1. The values of rolling-friction coefficient obtained from
the concrete runway varied from 0.009 to 0.035;
the fim turf surface, from 0.023 to 0.054; and on
the soft turf, where onlY the high-pressure ti& Were
tested, from 0.064 to 0.077.
2. The most important factor affecting the rolling friction
coefficient was the character of the ground surface.

3. For comparable conditions, either on a concrete
runway or on firm turf, the standard-type wheels and
tires had the lowest values of rolling-friction coefficient;
the values for the low-pressure tires were only slightly
higher. The highest coefficients were obtained with
the extra low-pressure wheels and tires.
4. In general, the variation in rolling-rfiction coefficient
with either wheel load or tire inflation pressure
Was fairly small.

5. The rolling-friction coefficient was appreciably
greater for wheels equipped with plain bearings than
for the same wheels having roller bearings.
6. The effect on take-off of all the variablea, with the
exception of the ground-surface condition, was generally
quite small; so that, for ordinary calculations of takeoff
performance, the assumption of an average vahe of
rolling-friction coefficient corresponding to a given
ground-surface condition would probably be satisilctory.
Where greater accuracy is desired, however,
the other factors, although of less consequence, should
nevertheless be considered.

Just pulling some ends together ... ref. rolling resistence.
;)
 
The place.
http://www.dfrc.nasa.gov/Education/OnlineEd/NewtonsLaws/
The small Document : 'Takeoff performance'
http://www.dfrc.nasa.gov/Education/OnlineEd/NewtonsLaws/pdf/studtakeoff.pdf

This is a document concerned with the take off of a real Jet trainer aircraft on a normal runway... Aermacchi MB-326 Impala jet trainer. at the National Test Pilot School.
Take off WT 7887 Lbs
Static maximum thrust at Mojave Airport: 2200 Lbs

Take off distance was calculated and then physically measured with the 'real thing'.
Rolling resistence was at first ignored, this lead to a significant error between claculated and real takeoff distance.
With 'rolling resistence' friction taken into account the error was 1.78% or 31.3 ft on a real take off run of 1750 ft.

In my previous post it was noted..
2. The most important factor affecting the rolling friction coefficient was the character of the ground surface.
Speed of tyre to surface made little difference to the measured rolling resistance ... It remained almost constant.

In the calcs for the trainer I ridiculously, in terms of the previous paragraph, doubled the 'rolling friction' from 394.4 Lbs to 788.8 Lbs, this resulted in a reduced acceleration figure from 6.74ft/sec² to 5.133 ft/sec².
Using this figure the run time to take off increased from 22.6 secs to 29.3 secs an increase of 29.5 %.
The take off calculated run distance increased from 1719 ft to 2253 ft an increase of 31%... ( 0.43 miles total length.)

I think this just about bottoms the whole thing out .. with some real data.
I have seen no calcs attempted anywhere else - google didn't find them .. But a search on NASA web site turned up the goods.... theory backed with practice.

;)
 
But again, you have not answered the question based on the data given, you have simply made assumptions. This is not a scientific approach.
 
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empip said:
LANGLEY MEMORIAL AERONAUTICAL LABORATORY,
NATIONAL ADVISORY CordanmD FOR AERONAUTICS,
LANGLEY FIELD, VA

REPORT No. 583
THE ROLLING FRICTION OF SEVERAL AIRPLANE WHEELS AND TIRES l.[/b]
5. The rolling-friction coefficient was appreciably
greater for wheels equipped with plain bearings than
for the same wheels having roller bearings.
.
;)
WOW..I`d never have guessed that .I bet they`ve got CORGI, Oftec......... :rolleyes: :LOL: :LOL:Good job you lot weren`t "The Few" of the Battle of Britain...you`d never have got off the bleedin runway :rolleyes:
 
Eddie M said:
do not assume anything.
.
.
.
As you have no other information, you have to assume that the co-efficient of friction is one.
.
.
.
Bearing friction, etc, not given, therefore assume is zero
An excellent lesson in how not to make assumptions. Not. :rolleyes:
 
take assume to mean regard. Shouting doesn't work with me.


You still haven't answered the question.
 
Eddie M said:
take assume to mean regard. Shouting doesn't work with me.
FYI, THIS IS SHOUTING, but the large text certainly worked with you, since it got your attention.

Eddie M said:
You still haven't answered the question.
Oh really?

So what was I doing at 8:48 am on June 6th said:
...the jet force is hugely greater than the frictional force, therefore the plane will reach lift-off velocity.
:?:
 
Nige F said:
empip said:
LANGLEY MEMORIAL AERONAUTICAL LABORATORY,
NATIONAL ADVISORY COMMITEE FOR AERONAUTICS,
LANGLEY FIELD, VA

REPORT No. 583
THE ROLLING FRICTION OF SEVERAL AIRPLANE WHEELS AND TIRES l.[/b]
5. The rolling-friction coefficient was appreciably
greater for wheels equipped with plain bearings than
for the same wheels having roller bearings.
.
;)
WOW..I`d never have guessed that .I bet they`ve got CORGI, Oftec......... :rolleyes: :LOL: :LOL:Good job you lot weren`t "The Few" of the Battle of Britain...you`d never have got off the bleedin runway :rolleyes:

I wonder how they separated the effect on rolling resistance caused by the wheel bearing friction from that of the differing tyres tested?
Perhaps they used bearings with distinctly different characteristics to force a difference ... Ball / roller against plain?
This was a very practical series of tests and experiments... lots of data.

Only the best ...

....The United States trailed Europe in its accomplishments, its lack of organized research, and also in the amount of funds allocated to military aviation. To help resolve these problems, in 1917, the NACA established the Langley Memorial Aeronautical Laboratory in Virginia. This laboratory would become the most advanced aeronautical test and experimentation facility in the world....

Great times in the development of aircraft.
About NACA

:D :D
 
more crap from softus which we'll just ignore.

Bas I'm still not convinced by the examples given, you mention the thrust pushes the plane forwards and i'll agree to that......initially, the plane starts to move forward but the runway sensors pick up the movement of the wheels and the runway compensates in a backwards motion.

you keep saying the wheels are not important and i still disagree, one they support the weight of the plane on the runway,two they do spin as the plane is moving at a speed governed by the thrust, yes you are correct the plane is moving! the spinning wheels prove this.

Only trouble is the plane isn't moving in airspace it is stationary as the runway is keeping the plane static by compensating for the thrust.

going back to your conveyor example, i do influence the car by mving forward but i'm a force with reference to solid ground, if i climb onto the conveyor i will never be able to move the car forwards bar initial starting delays.

in saying the wheels are not significant you then remove their function ie the obvious support , also the more important function to tell the runway sensors what speed to run the runway at.

because the plane is attached to the wheels and the wheels never gain any ground with respect to the surroundings only to the amount of ground gained on the runway which is cancelling out any movement then the plane is static in the airspace.

put a model plane on a conveyor fix it in place by a stick attached to the conveyor, switch on conveyor, what happens? the plane moves forward!
through airspace but look at the wheels, they are not turning, in this example the equation is still balanced but to get forward motion at the speed of the conveyor the wheels must be static, if they move backwards or forwards they affect the formula, in the example above the wheels are the constant they never move, the variables are the speed of conveyor and the speed of the plane through airspace which funnily enough are the same(due to being attached) so balance in a forward motion.

for the plane to move forward we must upset the balance somehow, turn off the runway sensors so that the runway is static and off goes the plane.
 
kendor said:
more rubbish from softus which we'll just ignore.
So, you're ignoring the answer then. Namely this one:

Back on page 4 said:
...the jet force is hugely greater than the frictional force, therefore the plane will reach lift-off velocity.

kendor said:
Only trouble is the plane isn't moving in airspace it is stationary as the runway is keeping the plane static by compensating for the thrust.
rofl5.gif


put a model plane on a conveyor fix it in place by a stick attached to the conveyor, switch on conveyor, what happens? the plane moves forward!
through airspace but look at the wheels, they are not turning, in this example the equation is still balanced but to get forward motion at the speed of the conveyor the wheels must be static, if they move backwards or forwards they affect the formula, in the example above the wheels are the constant they never move, the variables are the speed of conveyor and the speed of the plane through airspace which funnily enough are the same(due to being attached) so balance in a forward motion.

for the plane to move forward we must upset the balance somehow, turn off the runway sensors so that the runway is static and off goes the plane.
So many words; so little meaning. :)
 
kendor said:
going back to your conveyor example, i do influence the car by mving forward but i'm a force with reference to solid ground,
ALLELUJAH!

All you have to do now is to realise that you applying force to the car without any reference to the conveyor belt IS EXACTLY THE SAME AS the engines thrusting the plane forwards without any reference to the runway.

in saying the wheels are not significant you then remove their function ie the obvious support , also the more important function to tell the runway sensors what speed to run the runway at.

because the plane is attached to the wheels and the wheels never gain any ground with respect to the surroundings only to the amount of ground gained on the runway which is cancelling out any movement then the plane is static in the airspace.
I'll ask you again - in which direction do you think the runway will move, and at what speed, and why?

put a model plane on a conveyor fix it in place by a stick attached to the conveyor, switch on conveyor, what happens? the plane moves forward!
through airspace but look at the wheels, they are not turning, in this example the equation is still balanced but to get forward motion at the speed of the conveyor the wheels must be static, if they move backwards or forwards they affect the formula, in the example above the wheels are the constant they never move, the variables are the speed of conveyor and the speed of the plane through airspace which funnily enough are the same(due to being attached) so balance in a forward motion.
OK - take your model plane, and hold the stick. Then turn on the conveyor.

Oh look - the wheels go round but the plane does not move.

How many combinations of wheels and planes moving and not moving do you need to consider before you realise that they have nothing to do with each other when the forces on the wheels and the forces on the plane are not in the same frame of reference?

for the plane to move forward we must upset the balance somehow, turn off the runway sensors so that the runway is static and off goes the plane.
No - we can have the runway moving either backwards so the the wheels spin faster, or forwards so that they spin slower, or not at all, or even go into reverse.

We could have the runway moving backwards and forwards so that the wheels spin at all sorts of different speeds and directions.

We could have a runway conveyor split into three lengthways with each strip doing different things to the left, right and nosewheels.

But nothing we do will stop the plane from moving forwards through the air and taking off.
 
I'm going to have one last try....

The wheels are there to provide a low-friction barrier between plane and ground, for when the plane is on the ground and moves forward. They provide no traction.

We are told that the conveyor moves backwards at the same speed as the wheels move forwards.

Imagine instead that the wheels are replaced with ice skates, and the conveyor is also made of ice.

Apply thrust to the plane, the plane and, resultingly, the skates move forward. Move the conveyor backwards by the same ammount that the skates have moved forward. Does that put the plane back where it started ?

No. It keeps moving forward and then takes off.

Backwards motion of the conveyor does not convert into backwards motion of the plane.
 
Physics puzzle :rolleyes: ....just when I thought I was out, they pull me back in :D

When you guys all agreed, can someone tell me which came first, the chicken or the egg? ;)
 
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