Will it take off?

[PhilMays]

I've now changed my stance. It depends on how toasted the pilots are!

[DizRotus]

According to the Discovery Channel site at http://mythbusters-wiki.discovery.com/page/Episode+94%3A+Air+Plane+Hour?t=anon, the show will air Jan. 30. Mark your cakendar.

[jacksonbart]

from straight dope:

On first encounter this question, which has been showing up all over the Net, seems inane because the answer seems so obvious. However, as with the infamous Monty-Hall-three-doors-and-one-prize-problem (see The Straight Dope: "On Let's Make a Deal" you pick Door #1, 02-Nov-1990), the obvious answer is wrong, and you, Berj, are right--the plane takes off normally, with no need to specify frictionless wheels or any other such foolishness. You're also right that the question is often worded badly, leading to confusion, arguments, etc. In short, we've got a topic screaming for the Straight Dope.

First the obvious-but-wrong answer. The unwary tend to reason by analogy to a car on a conveyor belt--if the conveyor moves backward at the same rate that the car's wheels rotate forward, the net result is that the car remains stationary. An aircraft in the same situation, they figure, would stay planted on the ground, since there'd be no air rushing over the wings to give it lift. But of course cars and planes don't work the same way. A car's wheels are its means of propulsion--they push the road backwards (relatively speaking), and the car moves forward. In contrast, a plane's wheels aren't motorized; their purpose is to reduce friction during takeoff (and add it, by braking, when landing). What gets a plane moving are its propellers or jet turbines, which shove the air backward and thereby impel the plane forward. What the wheels, conveyor belt, etc, are up to is largely irrelevant. Let me repeat: Once the pilot fires up the engines, the plane moves forward at pretty much the usual speed relative to the ground--and more importantly the air--regardless of how fast the conveyor belt is moving backward. This generates lift on the wings, and the plane takes off. All the conveyor belt does is, as you correctly conclude, make the plane's wheels spin madly.

A thought experiment commonly cited in discussions of this question is to imagine you're standing on a health-club treadmill in rollerblades while holding a rope attached to the wall in front of you. The treadmill starts; simultaneously you begin to haul in the rope. Although you'll have to overcome some initial friction tugging you backward, in short order you'll be able to pull yourself forward easily.

As you point out, one problem here is the wording of the question. Your version straightforwardly states that the conveyor moves backward at the same rate that the plane moves forward. If the plane's forward speed is 100 miles per hour, the conveyor rolls 100 MPH backward, and the wheels rotate at 200 MPH. Assuming you've got Indy-car-quality tires and wheel bearings, no problem. However, some versions put matters this way: "The conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation." This language leads to a paradox: If the plane moves forward at 5 MPH, then its wheels will do likewise, and the treadmill will go 5 MPH backward. But if the treadmill is going 5 MPH backward, then the wheels are really turning 10 MPH forward. But if the wheels are going 10 MPH forward . . . Soon the foolish have persuaded themselves that the treadmill must operate at infinite speed. Nonsense. The question thus stated asks the impossible -- simply put, that A = A + 5 -- and so cannot be framed in this way. Everything clear now? Maybe not. But believe this: The plane takes off.

--CECIL ADAMS

[oldtimer]

I like Kari on mythbusters.

[pauln]

What happens on the TV show will not determine what is true, math and physics already indicate that it will take off... the wheels, conveyor, and their speeds have nothing to do with it. I thought this had already been made clear.

[^o)]

The
wheels still seem to be causing some confusion. That's because the role of the
whole thing about the wheels is to provide confusion to make analyzing the
problem challenging and fun, so let's clarify about the wheels and their
speeds...

What the hell, let's analyze all the speeds:

The
fundamental reference is the ground. The speed of the ground under the conveyor
is zero. This is not required for the plane to take off; it could do so during
an Earthquake comprising a single back-wards (or forwards) longitudinal lurch
aligned to the plane's direction of motion that matched the plane's speed (or
didn't match, doesn't matter).

The
speed of the still air (wind speed) is zero with reference to the ground. This
is not required for the plane to take off but it makes all the other speeds less
complicated to understand. The still air speed is equivalent to the still
ground speed.

The
speed of the plane relative to the still air - let's have the plane point West.

The
speed of the conveyor relative to the still air - let's have the top of the
conveyor go East ( the bottom part goes West at the same speed)

Note:
The plane's wheels parts have a variety of speeds depending on the part of the
wheel one is thinking about (because the angular speed of the wheel varies with
radii with reference to it's hub).

The
middle part of the wheels (the hub) has the same speed as the plane with
reference to the still air (with reference to the plane itself the hubs have
zero speed).

The
top part of the wheel is moving West in the direction of the plane's motion
faster than the speed of the plane with reference to the still air. How much
faster varies with how big the radius makes the wheel's circumference, the
bigger the faster, but in all cases where the wheel is of non zero radius the
top is going faster than the plane.

Likewise,
the bottom part of the wheel where it contacts the conveyor is going East with
reference to the plane at slower speed than which the plane is going West with
reference to the still air. The speeds of the top and bottom of the wheel are
equal and opposite in direction with reference to the plane.

Also,
the bottom part of the wheel is dead still with reference to the conveyor -
even though the wheel is turning and rolling on the conveyor. If you will
examine a rolling wheel you can convince yourself that this is true for all
non-skidding rolling wheels (of course all rolling wheels have to skid a tiny
bit - that is how the ground imparts
force to make them turn beyond the angular momentum the wheel already has and
the losses to friction (rolling friction)).

Notice
something interesting and very important here... the contact patch (bottom of
the wheel) is going East with reference to the plane, West with reference to
the still air, and is not moving with reference to the conveyor.

Before
figuring out more about all these speeds, lets summarize and give them formal
names. The subscripts will help keep track - they are just like the ones for
speaker perameters:

Vg Ground speed

Vw Wind speed relative to the ground

Vpg Plane speed relative to ground

Vpc
Plane speed relative to the conveyor

Vcg Conveyor speed relative to the ground

Vwtp
Speed of wheel top relative to the plane

Vwtc
Speed of wheel top relative to the conveyor

Vwhp
Speed of wheel hub relative to the plane

Vwhc
Speed of wheel hub relative to the conveyor

Vwbp
Speed of wheel bottom relative to the plane

Vwbc
Speed of wheel bottom relative to the conveyor

So,
what can we know about all these speeds? Let’s do the easy ones first…

First,
let’s assume the ground is not moving so

Vg=0

And
the wind is not blowing so

Vw=0

We
know that the wheel bottom is motionless with reference to the conveyor so

Vwbc=0
(see the calculations below for proof if you don't believe this after just
thinking about it)

And
we know that the speeds of the top and bottom of the wheel are opposite and
equal with reference to the plane so

Vwtp
= (-Vwbp)

And
we know that the wheel hub and plane have the same speed in reference to
anything

Vpg
= Vwhg ...equal with reference to the ground

And

Vpc
= Vwhc ...equal with reference to the conveyor

For
Vwtp and Vwbp, (wheel top and bottom with reference to the plane) we may find their
speeds by dividing the distance traveled around the wheel in one revolution
divided by the time it takes to make one revolution.

We don’t know the radii of the wheels, but we
do know that the circumference of the wheel is 2 pi times the radius; (2pir).
So this is the arc length of one rotation of the wheel.

With
Vwhc (the speed of the wheel hub with reference to the conveyor in feet per
second) we know that the number or revolutions of the wheel will be

Vwhc/2pir

The time for each revolution during that
second at Vwhc therefore is also Vwhc/2pir

Note,
we will use the plane's speed with reference to the conveyor to calculate the
wheel's top and bottom speed with reference to the plane, ground, and the
conveyor.

For
Vwtp (the top of the wheel speed with reference to the plane) we take the
distance traveled around the wheel with one revolution (2pir) and divide it by
the time it takes to do so (1/(Vwhc/2pir))

Therefore
the angular speed of the edge of the wheel with respect to the plane is
(2pir)/(1/(Vwhc/2pir)) which simplifies to just Vwhc, so

Vwtp
= Vwhc

For
Vwbp (the bottom of the wheel) we get the same magnitude of opposite sign
(opposite direction).

Vwbp
= -(Vwhc)

Both
of these speeds are added to the speed of the conveyor to get the wheel’s top and bottom speeds
with reference to the conveyor, Vwtc and Vwbc.

Vwtc
= Vwhc + Vcg

Vwbc
= Vwhc - Vcg ... = 0 I'll show below in an example calculation

Let’s
put in some numbers to see how this works. Let the plane’s speed (same as the
hub’s speed) in relation to the conveyor be 30 feet per second.

Note:
this would be the case when the speed of the plane with reference to the ground
is 15 feet per second and the conveyor is going in reverse at 15 feet per
second.

And
let the wheels be 18 inches in diameter (radii of 9 inches or ¾ foot).

Wheel
circumference is 2pir which is 2*3.14*.75 feet=4.71 feet

The
plane/hub speed is 30feet/second compared to the conveyor.

The
number of revolutions in 30 feet of travel is 30/4.71=6.37 and since the 30
feet of travel took 1 second

The
revolutions per second is 6.37

The
time period for one revolution is 1/6.37 (one second divided by the number of
revolutions during that second) =0.157 seconds

For
Vwtp (the top of the wheel speed with reference to the plane) we take the
distance traveled around the wheel with one revolution (4.71 feet) and divide
it by the time it takes to do so (0.157 seconds) and get 30 feet per second.
Notice that this is with reference to the plane. Since the plan is going 15
feet per second with reference to the ground, the top of the wheel is going 45
feet per second with respect to the ground because you add the speeds.

Likewise,
the same calculation for the bottom of the wheel with respect to the plane will
be –30 feet per second, but since the plane is going 15 feet per second with
respect to the ground you add them and get the bottom of the wheel going –15
feet per second with reference to the ground. Both the bottom of the wheel and
the conveyor are both going –15 feet per second so the bottom of the wheel
really is stationary with reference to the conveyor.

So
we have verified that

Vwbc=0
the contact point at the bottom of the wheel in reference to the conveyor is stationary.

[theplummer]

Pauln, you math equations I'm sure are correct, but NONE of the equations have anything to do with wether the plane will take off or not. Hell, the converbelt has absolutely nothing to do with the plane taking off.

The only equation that is in play here is the THRUST value that the engine imparts on the plane. If the plane can take off on a normal runway, it can take off anywhere (presuming the tires have the ability to roll across the resting surface long enough for the plane to gain forward momentum so the wings can create enough lift to overcome gravity's effect on the airplane, freeing it from the earth!!!

It doesn't matter what the tire speed is it's all about airspeed over the wings, and the engines thrust WILL make that happen, just as it always does on a normal runway.

[germerikan]

A plane is standing on runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in opposite direction).

The question is:

Will the plane take off or not? Will it be able to run up and take off?

Take another example. A sea plane trying to take off going upstream against a current. Can it? Depends on how fast the current is flowing. The plane must generate enough force to move its pontoons through the water at whatever velocity is required to take off - around 65 mph in a Cessna 152, for example. Let's say the engines can exert enough force to move the plane through the water at, uh, 80 mph. If the river is flowing at less than 15 mhp against us, we can take off. If the river is flowing at faster than 15 mph against us, we can't take off. The pontoons exert too much drag and we can't reach the required 65 mph airspeed. Wheels on a conveyor belt have, for all practicle purposes, no drag. The belt can move beneath us as fast as it wants, we'll still take off.

Why does this work with water and not with a belt???

heheheheheheheheheh

[IndyKlipschFan]

http://dsc.discovery.com/video/?playerId=203711706&categoryId=210013704&lineupId=229524134&titleId=1344511100

woo hooooo a preview of the event!!!

[Jay481985]

Pauln, you math equations I'm sure are correct, but NONE of the equations have anything to do with wether the plane will take off or not. Hell, the converbelt has absolutely nothing to do with the plane taking off.

The only equation that is in play here is the THRUST value that the engine imparts on the plane. If the plane can take off on a normal runway, it can take off anywhere (presuming the tires have the ability to roll across the resting surface long enough for the plane to gain forward momentum so the wings can create enough lift to overcome gravity's effect on the airplane, freeing it from the earth!!!

It doesn't matter what the tire speed is it's all about airspeed over the wings, and the engines thrust WILL make that happen, just as it always does on a normal runway.

On planes thrust is only one part of the equation of becoming airborne..... The answer still holds that the plane will take off but the idea that thrust is the only equation is not fully correct. If I had enough thrust anything will become airborne. I can have a brick fly or even hover or float if I wanted to but it becomes a point of efficency. Bullets have initial force enough that it will project in a near straight line until the intial force is overcome by gravity and air resistance. Same with most missles and rockets although they have a constant force or a force until the fuel is used up, but airplanes add a different equations of wings. Without the wings the plane would need to have produce more thrust then the weight of the aircraft. But in most cases, the plane engine jet or propeller do not produce more thrust then weight. The only one I believe again that has more thrust then weight is the F-15 which can climb vertically without stalling. Airplanes relie on the wings to offset some of the weight as the wing as air moves over it produces lift which offset the weight of the plane to make it lighter than the available thrust so you can have a plane that lets say weighs 10 tons but the thrust is only 5 tons. The wings offset the weight to make it take off that is why planes need to accelerate to a certain speed to have enough lift to take off or stay airborne. Go below 100 mph or 70 mph in cesna's I believe and you will experience stalling or not take off. I have yet to see an airplane just take off like birds with no runway other than the harrier or the jsf35 which work on different principle. Even aircraft carrier airplanes need a steam catapult to take off the short runway. The steam catapult speeds the plane quick enough that it has enough lift to fly.

[theplummer]

A plane is standing on runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in opposite direction).

The question is:

Will the plane take off or not? Will it be able to run up and take off?

Take another example. A sea plane trying to take off going upstream against a current. Can it? Depends on how fast the current is flowing. The plane must generate enough force to move its pontoons through the water at whatever velocity is required to take off - around 65 mph in a Cessna 152, for example. Let's say the engines can exert enough force to move the plane through the water at, uh, 80 mph. If the river is flowing at less than 15 mhp against us, we can take off. If the river is flowing at faster than 15 mph against us, we can't take off. The pontoons exert too much drag and we can't reach the required 65 mph airspeed. Wheels on a conveyor belt have, for all practicle purposes, no drag. The belt can move beneath us as fast as it wants, we'll still take off.

Why does this work with water and not with a belt???

I'll take a stab at this one, since I own a boat that travels 100mph.

Water creats sliding friction on a pontoon (or Hull). The more of the pontoon that is in the water, the greater the friction, the more thrust is needed to propell the pontoon in a forward motion. if your facing a 15+ mph current, the airplane's propeller cannot generate enough velocity to raise the pontoon's out of the water, therefore reducing friction. Although, If the engine was able to produce more power to handle a steeper pitch propeller, this may be overcome, because once the wetted area on the pontoon's is reduced, the water friction on the pontoon's becomes less therefore allowing the plane's available thrust to be used in increasing velocity, as apposed to just maintaining opposing force of greater friction.

There are other things that can be done to help a pontoon break free. Installing notches in the pontoons hull, adding lifting strakes, and finally Sanding the wetted area of the hull with sandpaper. (I know this doesn't make since, because a smooth surface slides across another surface easier right? Not always. When dealing with water, a rougher surface will introduce and hold microbubbles in the scratches in the hull, these microbubbles act like ball bearings, releasing the adhesive properties of the water, therefore loosening the pontoon from the water's hold, therefore it takes less thrust to decrease the wetted area of the hull and reducing friction, again transfering the thrust to forward momentum as apposed to just fighting friction.

When dealing with wheels, tires, bearings, and air friction, these types of friction are much less than water friction. They have less of an effect on the net power needed to overcome them. Rolling friction is the least friction there is, then air friction, followed by sliding friction. Water friction is more than sliding, because it has a adhesive type property to it. That's why water wants to be together all the time. That's why raindrop's form, instead of humidity, or mist.

[theplummer]

Pauln, you math equations I'm sure are correct, but NONE of the equations have anything to do with wether the plane will take off or not. Hell, the converbelt has absolutely nothing to do with the plane taking off.

The only equation that is in play here is the THRUST value that the engine imparts on the plane. If the plane can take off on a normal runway, it can take off anywhere (presuming the tires have the ability to roll across the resting surface long enough for the plane to gain forward momentum so the wings can create enough lift to overcome gravity's effect on the airplane, freeing it from the earth!!!

It doesn't matter what the tire speed is it's all about airspeed over the wings, and the engines thrust WILL make that happen, just as it always does on a normal runway.

On planes thrust is only one part of the equation of becoming airborne..... The answer still holds that the plane will take off but the idea that thrust is the only equation is not fully correct. If I had enough thrust anything will become airborne. I can have a brick fly or even hover or float if I wanted to but it becomes a point of efficency. Bullets have initial force enough that it will project in a near straight line until the intial force is overcome by gravity and air resistance. Same with most missles and rockets, but airplanes add a different equations of wings. Without the wings the plane would need to have produce more thrust then the weight of the aircraft. But in most cases, the plane engine jet or propeller do not produce more thrust then weight. The only one I believe again that has more thrust then weight is the F-15 which can climb vertically without stalling. Airplanes relie on the wings to offset some of the weight as the wing as air moves over it produces lift which offset the weight of the plane to make it lighter than the available thrust so you can have a plane that lets say weighs 10 tons but the thrust is only 5 tons. The wings offset the weight to make it take off.

read the line about the wings creating enough lift for the plane to overcome gravity. Thrust is the only factor that needs to be looked at here because everything else is constant. The damn plane is going to fly with or without a conveyer belt under, so therefore the lift calculations are moot here also. All that is needed is enough excess thrust from the engine to overcome the increased rolling friction that the wheels are going to impart on the plane as they will be traveling exponentially faster as the planes speed increases. Ultimately, in theory, the wheels could reach terminal velocity and not be able to spin faster, at which time sliding friction will then be imparted, but for this practical experiment, the velocity of the plane at the time this happens gravity will be less on tires, allowing sliding friction to be reduced to a minimal figure also.

quit overthinking this one.

[Jay481985]

i edited my previous post to give more detail

And the last sentence is confusing me. Also I said the airplane still takes off... Its just that you put thrust as the only equation for the plane to become airborne. What a plane does with the thrust is two fold, first it provides a forward motion and second it pushes enough air for the airplane's wing to produce lift. And gravity is a constant dependant only on distance from whatever body. 9.8 m^2 the last time I checked at sealevel.

[Jay481985]

A plane is standing on runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in opposite direction).

The question is:

Will the plane take off or not? Will it be able to run up and take off?

Take another example. A sea plane trying to take off going upstream against a current. Can it? Depends on how fast the current is flowing. The plane must generate enough force to move its pontoons through the water at whatever velocity is required to take off - around 65 mph in a Cessna 152, for example. Let's say the engines can exert enough force to move the plane through the water at, uh, 80 mph. If the river is flowing at less than 15 mhp against us, we can take off. If the river is flowing at faster than 15 mph against us, we can't take off. The pontoons exert too much drag and we can't reach the required 65 mph airspeed. Wheels on a conveyor belt have, for all practicle purposes, no drag. The belt can move beneath us as fast as it wants, we'll still take off.

Why does this work with water and not with a belt???

I'll take a stab at this one, since I own a boat that travels 100mph.

let me take a stab at this one too... The poontoon boat has direct force applied by the water when it moves. A wheel does not apply this since the wheel of an aircraft is disconnected meaning that yes the wheel has a connection or it would not be able to hold the airplane but the wheel also spins freely. A car wheel spins freely if it were not connected to the transmission/engine but the wheels that are not connected (if its a fwd or rwd) its more or less the same idea. Put the car in neutral on a "treadmill" and all I need is a rope that is not that thick to hold the car in place. If the car was put into drive and connected to the engine that rope would snap and I would need a stronger rope since the car and the wheels are directly connected

Water creats sliding friction on a pontoon (or Hull). The more of the pontoon that is in the water, the greater the friction, the more thrust is needed to propell the pontoon in a forward motion. if your facing a 15+ mph current, the airplane's propeller cannot generate enough velocity to raise the pontoon's out of the water, therefore reducing friction. Although, If the engine was able to produce more power to handle a steeper pitch propeller, this may be overcome, because once the wetted area on the pontoon's is reduced, the water friction on the pontoon's becomes less therefore allowing the plane's available thrust to be used in increasing velocity, as apposed to just maintaining opposing force of greater friction.

since the water is going 15 mph against you, you would need to go 15 mph faster to get out of the water. think of the 15 mph as a negative number., You go 65 mph but the water is -15 mph you are really only going 40 mph

There are other things that can be done to help a pontoon break free. Installing notches in the pontoons hull, adding lifting strakes, and finally Sanding the wetted area of the hull with sandpaper. (I know this doesn't make since, because a smooth surface slides across another surface easier right? Not always. When dealing with water, a rougher surface will introduce and hold microbubbles in the scratches in the hull, these microbubbles act like ball bearings, releasing the adhesive properties of the water, therefore loosening the pontoon from the water's hold, therefore it takes less thrust to decrease the wetted area of the hull and reducing friction, again transfering the thrust to forward momentum as apposed to just fighting friction.

That theory is likewise to why a golf ball has dimples, the dimples produce tiny turbulence that allow the air to not stick to the surface. This is the same idea with why a ping pong ball when hit tends to curve even though its smooth. Also when you really get into it, try sliding two pieces of float glass across each other, the surface tension makes it very hard. This is due to the theory where microscopic or nanoscopic theory comes to play where it is believed that the molecules tend to fill each other and prevent sliding. But place one drop of water and the glass will slide. That is why you want to polish the pistons and engineblock since the motoroil is there.

When dealing with wheels, tires, bearings, and air friction, these types of friction are much less than water friction. They have less of an effect on the net power needed to overcome them. Rolling friction is the least friction there is, then air friction, followed by sliding friction. Water friction is more than sliding, because it has a adhesive type property to it. That's why water wants to be together all the time. That's why raindrop's form, instead of humidity, or mist.

I think air friction is the lowest, rolling friction is the lowest in a perfect environment, i.e. ball bearings have no friction but that is not true in reality. Also water friction has more friction due to the mass of the water. Its easier to push air out of the way then water because air < mass of water.

[theplummer]

Jay, I'm not going to re-quote the entire quote, but I want to address your statement about traveling -15mph due to the current. and only traveling 40mph. At first, yes you are correct. But, once all thrust forces are implimented, the 15mph current becomes moot, because thrust is being applied to the air, not the water current. The only factor your dealing with regards to the water, (once the plane reaches a forward momentum) is increased friction on the pontoons, and there needs to be available extra thrust to overcome the increased friction. That's why I tried to explain the part about friction on the wetted area. the less wetted area, the less friction, the less its imparted on the planes overall velocity.

NOW, in the case of my raceboat, the current does play a factor, but Not exactly an opposite reaction. In a 15mph current, the propeller is using the water as its thrust factor, and yes you are starting at a -15mph, and ultimately a 65mph boat would be only going 40mph, but that can be overcome also (to a degree). Once the boat is moving across the water fast enough for the hull to create lift, pressure under the hull is increased over still water, causing the hull to react differently. the hull wants to rise out of the water at a slower speed due to the increased force from the moving water.

I can give a real world example of this. I race my boat on the Missouri river, near St. Louis. The river current is 7mph. This can vary from when the river is low 6mph, to when it's high 8mph, but lets use a constant of 7mph. My boat has a 98mph top speed in calm water. When I run downstream, you'd think the boat would travel 104mph. It does not. It will only travel 98mph. Then when I turn around and head upstream, the assumption is that the boat will travel 91mph. It does not. It actually travels 94mph, with relationship to the ground on the water's edge. Why is this? It's because going down stream, the water pressure on the hull is less at top speed, than it is when in calm water, the boat hull then takes on a different "Angle of Attack" with referance to the horizontal plane of the water, (in effect, the 7mph current is imparting an undesirable force to the final outcome of the speed), but when heading into the current, the same 7mph current imparts a positive effect on the hull, allowing the hull to "fly" at a different "angle of attack", therefore making the hull more efficient at the same amount of thrust input. So, heading upstream, my boat is 4mph more efficient than calm water, and 10mph more efficient than heading downstream. I'm willing to hypothosize that the airplane's pontoons will react in a similar fashion, and once the hulls of the pontoons create enough lift to reduce the wetted area of the hull (I.E. reduce wetted friction) the plane will be able to gain speed, because the force is acting only in increased "usable" friction, not as much as a negative as first thought.

There is not enough information about the sea plane to come to a complete conclusion here, because the factors of the pontoon's hulls coefficient of drag at different speeds is not known yet, and we don't know what the "Stall" speed is of the airplane also. I'm guessing that the general consensus of why a sea plane cannot take off in 15mph current or more is due to the fact that the engine and propeller cannot overcome the forces of friction on the hull at when the plane is at rest. I.e. not enough po power to get going.

[Rockets]

You don't need math to figure this out. Just some common sense and some air bearings will do,

If one were to take the wheels off and replace them with a pair of air bearings, is there any doubt at all that the plane would take off? Basically with the air bearings the plane would be floating on a few mircons of air above the conveyor belt. That should kill any argument in my mind.

[jacksonbart]

Murderer!

[bodcaw boy]

yes.

in Christ, because of God's grace,

roy

[Coytee]

heh heh heh

[6]

[Ray Garrison]

If nothing else, the amount of hot air generated in this thread will lift the plane off irregardless of any other considerations.

[Travis In Austin]

Bullets have initial force enough that it will project in a near straight line until the intial force is overcome by gravity and air resistance.

A bullet, fired perfectly level to the earth, will fall to earth in the same time as if the bullet were dropped from a fix point. In other words, if you fire a bullet from a barrel 10' off the ground, that is level to the ground, and you drop a bullet from 10' at the exact same time, they will hit the ground at the exact same time.

Thrust alone will not keep a bullet "airborne" longer then simply dropping one from a fixed height, it has to have a positive trajectory in order to put gravity in issue.

I think your other points are right on. Trust is one factor, but if it is the main one, like with the Harrier, thrust is converted from verticle to horizontal in a smooth transition to make a runway immaterial. However, there are compromises made with the airfoil used, which limits top speed in favor of more lift, more forgiving stall charistics, etc.

It will take off.

Travis