Will the Plane Take-Off - Merged with MythBusters Show Thread
#641
Senior Moderator
Originally Posted by spooky3ce
Forgot to mention about planes that take off from water...
If a plane is at a stand still but the water is rushing under the plane... the plane will be pushed backwards... If the speed of the water pushing the plane backwards is equal to the speed of the plane moving forward the plane will be at a stand still in the water... Once the plane exceeds the speed of the water pushing it backwards and pushes agains the wind to creat lift, then the plane can take off... This is why planes that take off from water try to find a way to go towards the curent and not against the curent... because the speed of the water plus the speed of the plane plus the wind pushing againts the plane equals a shorter run and easier lift...
If a plane is at a stand still but the water is rushing under the plane... the plane will be pushed backwards... If the speed of the water pushing the plane backwards is equal to the speed of the plane moving forward the plane will be at a stand still in the water... Once the plane exceeds the speed of the water pushing it backwards and pushes agains the wind to creat lift, then the plane can take off... This is why planes that take off from water try to find a way to go towards the curent and not against the curent... because the speed of the water plus the speed of the plane plus the wind pushing againts the plane equals a shorter run and easier lift...
We're not arguing about speed, we are arguing about force. Obviously the plane needs to exceed the speed of the water....it does need to move forward..noone is arguing it doesn't. But The plane will easily be able to exceed the speed of the water, as the force required to do so is not proportional to the speed of the water coming the other way.
#642
Someone stole "My Garage"
Originally Posted by spooky3ce
Forgot to mention about planes that take off from water...
If a plane is at a stand still but the water is rushing under the plane... the plane will be pushed backwards... If the speed of the water pushing the plane backwards is equal to the speed of the plane moving forward the plane will be at a stand still in the water... Once the plane exceeds the speed of the water pushing it backwards and pushes agains the wind to creat lift, then the plane can take off... This is why planes that take off from water try to find a way to go towards the curent and not against the curent... because the speed of the water plus the speed of the plane plus the wind pushing againts the plane equals a shorter run and easier lift...
If a plane is at a stand still but the water is rushing under the plane... the plane will be pushed backwards... If the speed of the water pushing the plane backwards is equal to the speed of the plane moving forward the plane will be at a stand still in the water... Once the plane exceeds the speed of the water pushing it backwards and pushes agains the wind to creat lift, then the plane can take off... This is why planes that take off from water try to find a way to go towards the curent and not against the curent... because the speed of the water plus the speed of the plane plus the wind pushing againts the plane equals a shorter run and easier lift...
#643
Big Block go VROOOM!
Originally Posted by spooky3ce
Correct... The forward thrust is reliant on it pushing against/off the ground... but so that the tires obtain the maximum grip at high speeds there is something called down force. Down force is produced from the air traveling around the car pushing the car down to the ground at high speeds. That is what canards in front of the car are for and the wing on the rear. Without wind; which is produced when a vehicle is moving forward, the car will not create/obtain any down force. Analogy: Down force to a car is like Lift to a Plane... The more lift the higher it flies and the more down force more grip for quicker speed... (The only difference is too much down force can also create a lot of drag holding your maximum speed back).
#644
Big White Chocolate
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Originally Posted by spooky3ce
Forgot to mention about planes that take off from water...
If a plane is at a stand still but the water is rushing under the plane... the plane will be pushed backwards... If the speed of the water pushing the plane backwards is equal to the speed of the plane moving forward the plane will be at a stand still in the water... Once the plane exceeds the speed of the water pushing it backwards and pushes agains the wind to creat lift, then the plane can take off... This is why planes that take off from water try to find a way to go towards the curent and not against the curent... because the speed of the water plus the speed of the plane plus the wind pushing againts the plane equals a shorter run and easier lift...
If a plane is at a stand still but the water is rushing under the plane... the plane will be pushed backwards... If the speed of the water pushing the plane backwards is equal to the speed of the plane moving forward the plane will be at a stand still in the water... Once the plane exceeds the speed of the water pushing it backwards and pushes agains the wind to creat lift, then the plane can take off... This is why planes that take off from water try to find a way to go towards the curent and not against the curent... because the speed of the water plus the speed of the plane plus the wind pushing againts the plane equals a shorter run and easier lift...
#645
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Originally Posted by curls
However, a plane pushes on the air to move, and simply rolls along the runway, skiis along the snow, or floats along the water as a means of keeping it as isolated as possible from the ground.
Think of it this way, if you will: Stand on a treadmill with rollerblades on. Set the treadmill to 5mph. Skate at an effort of 5mph. Net result: You are skating at 5mph but going nowhere relative to the wall next to you. This is how a car, or anything propelled by contact with the ground, would react -- it would stay stationary relative to a stationary object. Now, new scenario: You are on a treadmill, wearing rollerblades, but also have a rope attached to your waist on one end, and a friend in front of (but not on) the treadmill is holding the other end. Set the treadmill to 5mph. Skate at 5mph so your new speed relative to the wall next to you is 0mph (ie: typical treadmill...). Now, have your friend pull the rope at a speed of 3mph. What happens?
- you'd still only be skating at an effort of 5mph (or you could stop skating and glide with NO change in the result)
and
- the wheels on your rollerblades will be rotating at a speed of (5+3=) 8mph. (or in the case of you stopping the skating and simply gliding, the wheel speed is still 8mph!)
and
- You'd be moving forward, relative to the stationarty wall (and the air around you!) at a speed of 3mph.
Apply this principle to the plane whose thrust is NOT generated from contact with the runway/conveyor, but the interaction with the air around it. Net result? The plane DOES move forward with no extra effort(*). And since the plane moves forward, it DOES have air moving over the wings, and as such, WILL FLY.
* The "no extra effort" ignores the very minute increase in rolling friction and bearing efficiency. Thrust requirement will NOT double with every doubling of the conveyor speed, as rolling resistance is almost constant in most circumstances!
Think of it this way, if you will: Stand on a treadmill with rollerblades on. Set the treadmill to 5mph. Skate at an effort of 5mph. Net result: You are skating at 5mph but going nowhere relative to the wall next to you. This is how a car, or anything propelled by contact with the ground, would react -- it would stay stationary relative to a stationary object. Now, new scenario: You are on a treadmill, wearing rollerblades, but also have a rope attached to your waist on one end, and a friend in front of (but not on) the treadmill is holding the other end. Set the treadmill to 5mph. Skate at 5mph so your new speed relative to the wall next to you is 0mph (ie: typical treadmill...). Now, have your friend pull the rope at a speed of 3mph. What happens?
- you'd still only be skating at an effort of 5mph (or you could stop skating and glide with NO change in the result)
and
- the wheels on your rollerblades will be rotating at a speed of (5+3=) 8mph. (or in the case of you stopping the skating and simply gliding, the wheel speed is still 8mph!)
and
- You'd be moving forward, relative to the stationarty wall (and the air around you!) at a speed of 3mph.
Apply this principle to the plane whose thrust is NOT generated from contact with the runway/conveyor, but the interaction with the air around it. Net result? The plane DOES move forward with no extra effort(*). And since the plane moves forward, it DOES have air moving over the wings, and as such, WILL FLY.
* The "no extra effort" ignores the very minute increase in rolling friction and bearing efficiency. Thrust requirement will NOT double with every doubling of the conveyor speed, as rolling resistance is almost constant in most circumstances!
Hypotheticly speaking:
Plane requires to go 50 MPH on a runway to lift off... Thats 50 MPH wind to create enough lift for the plane to take off... (assuming wind speed on that day is 0 MPH)
Put the plane on the treadmill going 50 mph... The wheels will be spinning at 50mph the plane will be going 50 mph and the wind will be at 0 mph... So that the plane can take off, it will need to create enough lift (from the air) around the wings to take off... so the plane would need to be going 100 mph exceeding the speed of the treadmill to move forward on a 50 mph treadmill to create 50 mph wind...
Still wind is to planes as the road is to cars...
Now get 50 mph winds and push it agains a plane going 0mph like a kite... the plane will take off if it drives against the wind... ever see those tunnels that are created to test planes???
The plane would still need a treadmill long enough to exceed the treadmills speed to create enough air to push agains the wings to creat lift...
#646
Senior Moderator
Originally Posted by spooky3ce
You are 100% correct... And i'll be using your calculation...
Hypotheticly speaking:
Plane requires to go 50 MPH on a runway to lift off... Thats 50 MPH wind to create enough lift for the plane to take off... (assuming wind speed on that day is 0 MPH)
Put the plane on the treadmill going 50 mph... The wheels will be spinning at 50mph the plane will be going 50 mph and the wind will be at 0 mph... So that the plane can take off, it will need to create enough lift (from the air) around the wings to take off... so the plane would need to be going 100 mph exceeding the speed of the treadmill to move forward on a 50 mph treadmill to create 50 mph wind...
Still wind is to planes as the road is to cars...
Now get 50 mph winds and push it agains a plane going 0mph like a kite... the plane will take off if it drives against the wind... ever see those tunnels that are created to test planes???
The plane would still need a treadmill long enough to exceed the treadmills speed to create enough air to push agains the wings to creat lift...
Hypotheticly speaking:
Plane requires to go 50 MPH on a runway to lift off... Thats 50 MPH wind to create enough lift for the plane to take off... (assuming wind speed on that day is 0 MPH)
Put the plane on the treadmill going 50 mph... The wheels will be spinning at 50mph the plane will be going 50 mph and the wind will be at 0 mph... So that the plane can take off, it will need to create enough lift (from the air) around the wings to take off... so the plane would need to be going 100 mph exceeding the speed of the treadmill to move forward on a 50 mph treadmill to create 50 mph wind...
Still wind is to planes as the road is to cars...
Now get 50 mph winds and push it agains a plane going 0mph like a kite... the plane will take off if it drives against the wind... ever see those tunnels that are created to test planes???
The plane would still need a treadmill long enough to exceed the treadmills speed to create enough air to push agains the wings to creat lift...
Again, noone is saying the plane will take off from a standing position. Of course not! Only that it will easily be able to move forward on the treadmill in order to achieve lift. The fact that the treadmill is there will have little consequence. It can easily thrust forward and take off.
#647
Someone stole "My Garage"
Originally Posted by spooky3ce
You are 100% correct... And i'll be using your calculation...
Hypotheticly speaking:
Plane requires to go 50 MPH on a runway to lift off... Thats 50 MPH wind to create enough lift for the plane to take off... (assuming wind speed on that day is 0 MPH)
Hypotheticly speaking:
Plane requires to go 50 MPH on a runway to lift off... Thats 50 MPH wind to create enough lift for the plane to take off... (assuming wind speed on that day is 0 MPH)
Originally Posted by spooky3ce
Put the plane on the treadmill going 50 mph... The wheels will be spinning at 50mph the plane will be going 50 mph and the wind will be at 0 mph...
Originally Posted by spooky3ce
So that the plane can take off, it will need to create enough lift (from the air) around the wings to take off... so the plane would need to be going 100 mph exceeding the speed of the treadmill to move forward on a 50 mph treadmill to create 50 mph wind...
Originally Posted by spooky3ce
Still wind is to planes as the road is to cars...
Originally Posted by spooky3ce
Now get 50 mph winds and push it agains a plane going 0mph like a kite... the plane will take off if it drives against the wind... ever see those tunnels that are created to test planes???
Originally Posted by spooky3ce
The plane would still need a treadmill long enough to exceed the treadmills speed to create enough air to push agains the wings to creat lift...
#648
Go Giants
You guys are so thick....
#649
Big White Chocolate
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Originally Posted by spooky3ce
You are 100% correct... And i'll be using your calculation...
Hypotheticly speaking:
Plane requires to go 50 MPH on a runway to lift off... Thats 50 MPH wind to create enough lift for the plane to take off... (assuming wind speed on that day is 0 MPH)
Put the plane on the treadmill going 50 mph... The wheels will be spinning at 50mph the plane will be going 50 mph and the wind will be at 0 mph... So that the plane can take off, it will need to create enough lift (from the air) around the wings to take off... so the plane would need to be going 100 mph exceeding the speed of the treadmill to move forward on a 50 mph treadmill to create 50 mph wind...
Still wind is to planes as the road is to cars...
Now get 50 mph winds and push it agains a plane going 0mph like a kite... the plane will take off if it drives against the wind... ever see those tunnels that are created to test planes???
The plane would still need a treadmill long enough to exceed the treadmills speed to create enough air to push agains the wings to creat lift...
Hypotheticly speaking:
Plane requires to go 50 MPH on a runway to lift off... Thats 50 MPH wind to create enough lift for the plane to take off... (assuming wind speed on that day is 0 MPH)
Put the plane on the treadmill going 50 mph... The wheels will be spinning at 50mph the plane will be going 50 mph and the wind will be at 0 mph... So that the plane can take off, it will need to create enough lift (from the air) around the wings to take off... so the plane would need to be going 100 mph exceeding the speed of the treadmill to move forward on a 50 mph treadmill to create 50 mph wind...
Still wind is to planes as the road is to cars...
Now get 50 mph winds and push it agains a plane going 0mph like a kite... the plane will take off if it drives against the wind... ever see those tunnels that are created to test planes???
The plane would still need a treadmill long enough to exceed the treadmills speed to create enough air to push agains the wings to creat lift...
For instance, if you put a toy car on a treadmill and used your finger to keep it in place, the force required to keep it in place wouldn't increase in portion to the speed of the treadmill. Also, the force to move the toy car forward would be more or less the same than if the treadmill weren't on. (The force applied with your finger would be equivalent to the engines on a plane since the propulsion of a plane isn't transferred via the wheels.)
#650
Senior Moderator
Originally Posted by NetEditor
I think you're confusing the point that the plane needs to overcome the effects of the treadmill. Yes, if you stick something on a treadmill and turn it on, it will move backward. But that's because of inertia caused by the weight of the object and the resistance due to rolling friction.
For instance, if you put a toy car on a treadmill and used your finger to keep it in place, the force required to keep it in place wouldn't increase in portion to the speed of the treadmill. Also, the force to move the toy car forward would be more or less the same than if the treadmill weren't on. (The force applied with your finger would be equivalent to the engines on a plane since the propulsion of a plane isn't transferred via the wheels.)
For instance, if you put a toy car on a treadmill and used your finger to keep it in place, the force required to keep it in place wouldn't increase in portion to the speed of the treadmill. Also, the force to move the toy car forward would be more or less the same than if the treadmill weren't on. (The force applied with your finger would be equivalent to the engines on a plane since the propulsion of a plane isn't transferred via the wheels.)
Exactly.
And no one is debating the need for the plane to move forward relative to the earth, and the need for the associated airflow to achieve lift. This is scientific fact. The argument is that the force to do this will be independent of how fast the treadmill is going in the opposite direction.
#651
Big Block go VROOOM!
Originally Posted by NetEditor
For instance, if you put a toy car on a treadmill and used your finger to keep it in place, the force required to keep it in place wouldn't increase in portion to the speed of the treadmill.
#652
Someone stole "My Garage"
Copied and pasted... This should pretty much /thread:
"This is somewhat of a "trick" question. Not because it is phrased in a
deliberately tricky way, but because people tend to have trouble
thinking about the operation of other vehicles apart from cars which
they know so well.
The heart of the confusion is simply these two important facts:
* cars propel themselves by pushing against the ground via friction
* airplanes propel themselves by pushing against the air
If you can let go of how cars operate and think about what an airplane
does, you'll be able to see the problem clearly.
One good way of tackling this problem is to find a good analogy. But
the analogy must be a valid one else you'll just get more confused.
For example, someone posted the analogy of running on a treadmill. Why
is that a bad analogy? Because one runs by pushing against the ground
via friction between their shoe and the ground. This is how a car
propels itself! It is not how an airplane propels itself, by pushing
against the air. Bad analogy.
Let's use this analogy. Instead of looking at the airplane, let's back
up and go into the airport. Suppose you're walking down to your gate
and pulling your carry-on bag behind you. It's a nice new bag with low
friction wheels. No problem! Up ahead you see one of those moving
walkways. You don't see anyone coming, so you decide to do a little
experiment. You go over to the walkway that is moving TOWARDS you and
place your bag on it. Meanwhile, you step off to the side of the
walkway, and still holding on to the handle of your bag, you continue
to walk along. In fact, you intentionally walk along at the same speed
that the moving walkway is going, just in the opposite direction.
Question: does the bag move or does it remain stationary as you keep
walking? Obviously it moves with you. So why does your bag move
forward when you are walking at the same speed of the conveyor going
in the opposite direction?
The answer to that question is also the answer to the
airplane-conveyor question. To complete the analogy, the pull of your
arm is analogous to the force of the airplane engines. The bag's
wheels are analogous to the airplane tires. Do the nice low-friction
wheels on your bag on the conveyor pull against you anymore than they
do when you're just pulling your bag along normally? No, they don't.
They are free-wheeling, after all. Meanwhile, you're pulling the bag
with the same force in both cases. So in both cases, the bag keeps
moving forward. Likewise with the airplane, the pull of the engines
doesn't change nor does the force on the airplane imparted by the
tires change no matter what the ground is doing underneath the tires.
You have the same force imbalance in either case, and since Force =
mass x accceleration, you have the same acceleration. Remember, we are
talking airplane engines which push against the AIR, not the ground.
The acceleration is with respect to the AIR, thus the airplane
develops a speed relative to the air and can eventually take off.
That's a long winded analogy. Here's a quicker solution. Engineers
learn to draw Free Body Diagrams to understand such problems. A FBD is
just a block diagram which illustrates the forces acting upon an
object. The net force can be calculated from all the contributing
forces. If that net force is not zero, the object accelerates in the
direction of that force. Let's draw a FBD now. Represent the airplane
by a simple rectangle (the shape doesn't matter). Indicate the force
of the engines pulling on the plane with a large arrow, labeled "F_e"
with e for engines. What force are the tires imparting to the plane?
Remember, they spin freely except for bearing friction and rolling
friction, which produces forces that are quite tiny compared to the
engines. Represent those forces with a small arrow, labeled F_t where
t means tire. What other forces are operating? There is a drag force
too. But remember this drag force is also small compared to the
propulsion of the engines at least at takeoff speeds. Label that F_d.
FREE BODY DIAGRAM
F_d |----------| F_e
<---| Plane |------------->
|----------|
<-- F_t
So what do we have here? We haven't put numerical quantities on these
forces, but instead have just been talking in terms of large and
small. That's okay for the purposes of this illustration. It's enough
to know that F_e is going to be much greater than F_d + F_t pulling in
the opposite direction. If that's the case, we have an unbalanced
force on the plane. Therefore, it accelerates. And it accelerates with
respect to the AIR, since F_e is produces by the engines pushing
against the air.
To sum up: Yes, the airplane takes off. The motion of the surface
underneath the freely-spinning tires is irrelevant to the acceleration
of the aircraft since the tires cannot impart any force to the
aircraft (aside from the aforementioned very small rolling and bearing
friction).
I hope this discussion has helped somewhat and not muddied the waters any further.
John Strong
Ph.D., biochemical engineering
M.S., chemical engineering
B.S., mechanical engineering"
Copied and pasted...
"This is somewhat of a "trick" question. Not because it is phrased in a
deliberately tricky way, but because people tend to have trouble
thinking about the operation of other vehicles apart from cars which
they know so well.
The heart of the confusion is simply these two important facts:
* cars propel themselves by pushing against the ground via friction
* airplanes propel themselves by pushing against the air
If you can let go of how cars operate and think about what an airplane
does, you'll be able to see the problem clearly.
One good way of tackling this problem is to find a good analogy. But
the analogy must be a valid one else you'll just get more confused.
For example, someone posted the analogy of running on a treadmill. Why
is that a bad analogy? Because one runs by pushing against the ground
via friction between their shoe and the ground. This is how a car
propels itself! It is not how an airplane propels itself, by pushing
against the air. Bad analogy.
Let's use this analogy. Instead of looking at the airplane, let's back
up and go into the airport. Suppose you're walking down to your gate
and pulling your carry-on bag behind you. It's a nice new bag with low
friction wheels. No problem! Up ahead you see one of those moving
walkways. You don't see anyone coming, so you decide to do a little
experiment. You go over to the walkway that is moving TOWARDS you and
place your bag on it. Meanwhile, you step off to the side of the
walkway, and still holding on to the handle of your bag, you continue
to walk along. In fact, you intentionally walk along at the same speed
that the moving walkway is going, just in the opposite direction.
Question: does the bag move or does it remain stationary as you keep
walking? Obviously it moves with you. So why does your bag move
forward when you are walking at the same speed of the conveyor going
in the opposite direction?
The answer to that question is also the answer to the
airplane-conveyor question. To complete the analogy, the pull of your
arm is analogous to the force of the airplane engines. The bag's
wheels are analogous to the airplane tires. Do the nice low-friction
wheels on your bag on the conveyor pull against you anymore than they
do when you're just pulling your bag along normally? No, they don't.
They are free-wheeling, after all. Meanwhile, you're pulling the bag
with the same force in both cases. So in both cases, the bag keeps
moving forward. Likewise with the airplane, the pull of the engines
doesn't change nor does the force on the airplane imparted by the
tires change no matter what the ground is doing underneath the tires.
You have the same force imbalance in either case, and since Force =
mass x accceleration, you have the same acceleration. Remember, we are
talking airplane engines which push against the AIR, not the ground.
The acceleration is with respect to the AIR, thus the airplane
develops a speed relative to the air and can eventually take off.
That's a long winded analogy. Here's a quicker solution. Engineers
learn to draw Free Body Diagrams to understand such problems. A FBD is
just a block diagram which illustrates the forces acting upon an
object. The net force can be calculated from all the contributing
forces. If that net force is not zero, the object accelerates in the
direction of that force. Let's draw a FBD now. Represent the airplane
by a simple rectangle (the shape doesn't matter). Indicate the force
of the engines pulling on the plane with a large arrow, labeled "F_e"
with e for engines. What force are the tires imparting to the plane?
Remember, they spin freely except for bearing friction and rolling
friction, which produces forces that are quite tiny compared to the
engines. Represent those forces with a small arrow, labeled F_t where
t means tire. What other forces are operating? There is a drag force
too. But remember this drag force is also small compared to the
propulsion of the engines at least at takeoff speeds. Label that F_d.
FREE BODY DIAGRAM
F_d |----------| F_e
<---| Plane |------------->
|----------|
<-- F_t
So what do we have here? We haven't put numerical quantities on these
forces, but instead have just been talking in terms of large and
small. That's okay for the purposes of this illustration. It's enough
to know that F_e is going to be much greater than F_d + F_t pulling in
the opposite direction. If that's the case, we have an unbalanced
force on the plane. Therefore, it accelerates. And it accelerates with
respect to the AIR, since F_e is produces by the engines pushing
against the air.
To sum up: Yes, the airplane takes off. The motion of the surface
underneath the freely-spinning tires is irrelevant to the acceleration
of the aircraft since the tires cannot impart any force to the
aircraft (aside from the aforementioned very small rolling and bearing
friction).
I hope this discussion has helped somewhat and not muddied the waters any further.
John Strong
Ph.D., biochemical engineering
M.S., chemical engineering
B.S., mechanical engineering"
Copied and pasted...
#655
Senior Moderator
Originally Posted by Billiam
For purposes of this conversation, downforce on a race car should not be compared to lift on plane. Downforce is an enhancement on a reace car. It is not a requirement for it to accelerate. Lift OTOH is absolutely required for a plane to do its primary function of getting off the ground.
![Thumbs Up](https://acurazine.com/forums/images/smilies/thumbsup.gif)
#656
Go Giants
Originally Posted by dom
As you were.
#657
Senior Moderator
Originally Posted by curls
Copied and pasted... This should pretty much /thread:
"This is somewhat of a "trick" question. Not because it is phrased in a
deliberately tricky way, but because people tend to have trouble
thinking about the operation of other vehicles apart from cars which
they know so well.
The heart of the confusion is simply these two important facts:
* cars propel themselves by pushing against the ground via friction
* airplanes propel themselves by pushing against the air
If you can let go of how cars operate and think about what an airplane
does, you'll be able to see the problem clearly.
One good way of tackling this problem is to find a good analogy. But
the analogy must be a valid one else you'll just get more confused.
For example, someone posted the analogy of running on a treadmill. Why
is that a bad analogy? Because one runs by pushing against the ground
via friction between their shoe and the ground. This is how a car
propels itself! It is not how an airplane propels itself, by pushing
against the air. Bad analogy.
Let's use this analogy. Instead of looking at the airplane, let's back
up and go into the airport. Suppose you're walking down to your gate
and pulling your carry-on bag behind you. It's a nice new bag with low
friction wheels. No problem! Up ahead you see one of those moving
walkways. You don't see anyone coming, so you decide to do a little
experiment. You go over to the walkway that is moving TOWARDS you and
place your bag on it. Meanwhile, you step off to the side of the
walkway, and still holding on to the handle of your bag, you continue
to walk along. In fact, you intentionally walk along at the same speed
that the moving walkway is going, just in the opposite direction.
Question: does the bag move or does it remain stationary as you keep
walking? Obviously it moves with you. So why does your bag move
forward when you are walking at the same speed of the conveyor going
in the opposite direction?
The answer to that question is also the answer to the
airplane-conveyor question. To complete the analogy, the pull of your
arm is analogous to the force of the airplane engines. The bag's
wheels are analogous to the airplane tires. Do the nice low-friction
wheels on your bag on the conveyor pull against you anymore than they
do when you're just pulling your bag along normally? No, they don't.
They are free-wheeling, after all. Meanwhile, you're pulling the bag
with the same force in both cases. So in both cases, the bag keeps
moving forward. Likewise with the airplane, the pull of the engines
doesn't change nor does the force on the airplane imparted by the
tires change no matter what the ground is doing underneath the tires.
You have the same force imbalance in either case, and since Force =
mass x accceleration, you have the same acceleration. Remember, we are
talking airplane engines which push against the AIR, not the ground.
The acceleration is with respect to the AIR, thus the airplane
develops a speed relative to the air and can eventually take off.
That's a long winded analogy. Here's a quicker solution. Engineers
learn to draw Free Body Diagrams to understand such problems. A FBD is
just a block diagram which illustrates the forces acting upon an
object. The net force can be calculated from all the contributing
forces. If that net force is not zero, the object accelerates in the
direction of that force. Let's draw a FBD now. Represent the airplane
by a simple rectangle (the shape doesn't matter). Indicate the force
of the engines pulling on the plane with a large arrow, labeled "F_e"
with e for engines. What force are the tires imparting to the plane?
Remember, they spin freely except for bearing friction and rolling
friction, which produces forces that are quite tiny compared to the
engines. Represent those forces with a small arrow, labeled F_t where
t means tire. What other forces are operating? There is a drag force
too. But remember this drag force is also small compared to the
propulsion of the engines at least at takeoff speeds. Label that F_d.
FREE BODY DIAGRAM
F_d |----------| F_e
<---| Plane |------------->
|----------|
<-- F_t
So what do we have here? We haven't put numerical quantities on these
forces, but instead have just been talking in terms of large and
small. That's okay for the purposes of this illustration. It's enough
to know that F_e is going to be much greater than F_d + F_t pulling in
the opposite direction. If that's the case, we have an unbalanced
force on the plane. Therefore, it accelerates. And it accelerates with
respect to the AIR, since F_e is produces by the engines pushing
against the air.
To sum up: Yes, the airplane takes off. The motion of the surface
underneath the freely-spinning tires is irrelevant to the acceleration
of the aircraft since the tires cannot impart any force to the
aircraft (aside from the aforementioned very small rolling and bearing
friction).
I hope this discussion has helped somewhat and not muddied the waters any further.
John Strong
Ph.D., biochemical engineering
M.S., chemical engineering
B.S., mechanical engineering"
Copied and pasted...
"This is somewhat of a "trick" question. Not because it is phrased in a
deliberately tricky way, but because people tend to have trouble
thinking about the operation of other vehicles apart from cars which
they know so well.
The heart of the confusion is simply these two important facts:
* cars propel themselves by pushing against the ground via friction
* airplanes propel themselves by pushing against the air
If you can let go of how cars operate and think about what an airplane
does, you'll be able to see the problem clearly.
One good way of tackling this problem is to find a good analogy. But
the analogy must be a valid one else you'll just get more confused.
For example, someone posted the analogy of running on a treadmill. Why
is that a bad analogy? Because one runs by pushing against the ground
via friction between their shoe and the ground. This is how a car
propels itself! It is not how an airplane propels itself, by pushing
against the air. Bad analogy.
Let's use this analogy. Instead of looking at the airplane, let's back
up and go into the airport. Suppose you're walking down to your gate
and pulling your carry-on bag behind you. It's a nice new bag with low
friction wheels. No problem! Up ahead you see one of those moving
walkways. You don't see anyone coming, so you decide to do a little
experiment. You go over to the walkway that is moving TOWARDS you and
place your bag on it. Meanwhile, you step off to the side of the
walkway, and still holding on to the handle of your bag, you continue
to walk along. In fact, you intentionally walk along at the same speed
that the moving walkway is going, just in the opposite direction.
Question: does the bag move or does it remain stationary as you keep
walking? Obviously it moves with you. So why does your bag move
forward when you are walking at the same speed of the conveyor going
in the opposite direction?
The answer to that question is also the answer to the
airplane-conveyor question. To complete the analogy, the pull of your
arm is analogous to the force of the airplane engines. The bag's
wheels are analogous to the airplane tires. Do the nice low-friction
wheels on your bag on the conveyor pull against you anymore than they
do when you're just pulling your bag along normally? No, they don't.
They are free-wheeling, after all. Meanwhile, you're pulling the bag
with the same force in both cases. So in both cases, the bag keeps
moving forward. Likewise with the airplane, the pull of the engines
doesn't change nor does the force on the airplane imparted by the
tires change no matter what the ground is doing underneath the tires.
You have the same force imbalance in either case, and since Force =
mass x accceleration, you have the same acceleration. Remember, we are
talking airplane engines which push against the AIR, not the ground.
The acceleration is with respect to the AIR, thus the airplane
develops a speed relative to the air and can eventually take off.
That's a long winded analogy. Here's a quicker solution. Engineers
learn to draw Free Body Diagrams to understand such problems. A FBD is
just a block diagram which illustrates the forces acting upon an
object. The net force can be calculated from all the contributing
forces. If that net force is not zero, the object accelerates in the
direction of that force. Let's draw a FBD now. Represent the airplane
by a simple rectangle (the shape doesn't matter). Indicate the force
of the engines pulling on the plane with a large arrow, labeled "F_e"
with e for engines. What force are the tires imparting to the plane?
Remember, they spin freely except for bearing friction and rolling
friction, which produces forces that are quite tiny compared to the
engines. Represent those forces with a small arrow, labeled F_t where
t means tire. What other forces are operating? There is a drag force
too. But remember this drag force is also small compared to the
propulsion of the engines at least at takeoff speeds. Label that F_d.
FREE BODY DIAGRAM
F_d |----------| F_e
<---| Plane |------------->
|----------|
<-- F_t
So what do we have here? We haven't put numerical quantities on these
forces, but instead have just been talking in terms of large and
small. That's okay for the purposes of this illustration. It's enough
to know that F_e is going to be much greater than F_d + F_t pulling in
the opposite direction. If that's the case, we have an unbalanced
force on the plane. Therefore, it accelerates. And it accelerates with
respect to the AIR, since F_e is produces by the engines pushing
against the air.
To sum up: Yes, the airplane takes off. The motion of the surface
underneath the freely-spinning tires is irrelevant to the acceleration
of the aircraft since the tires cannot impart any force to the
aircraft (aside from the aforementioned very small rolling and bearing
friction).
I hope this discussion has helped somewhat and not muddied the waters any further.
John Strong
Ph.D., biochemical engineering
M.S., chemical engineering
B.S., mechanical engineering"
Copied and pasted...
/THREAD
#658
Moderator
Join Date: Oct 2004
Location: Not Las Vegas (SF Bay Area)
Age: 40
Posts: 63,277
Received 2,793 Likes
on
1,988 Posts
Originally Posted by Whiskers
You must have me confused for another Whiskers....
i do miss Wsklar
![Shrug](https://acurazine.com/forums/images/smilies/shrug.gif)
#659
Go Giants
Originally Posted by Mizouse
i do miss Wsklar ![Shrug](https://acurazine.com/forums/images/smilies/shrug.gif)
![Shrug](https://acurazine.com/forums/images/smilies/shrug.gif)
#661
Go Giants
Originally Posted by fdl
They're both fat so they clicked right away.
#662
Insert Sarcasm Here
Originally Posted by Whiskers
You must have me confused for another Whiskers....
ah I remember. It was like a lightswitch.
......[whiskers] ah ok, you're right. [/whiskers]
![rofl](https://acurazine.com/forums/images/smilies/rofl.gif)
#663
Senior Moderator
Here are the facts:
Let's use a 737-400 to keep it simple - not sure if all of these values would be needed:
Takeoff speed: 150 mph
Takeoff run: 8,483 ft
Takeoff weight: 100,000 lbs
Takeoff thrust: 22,000 lbf
Runway length: 13,000 feet (longest runway at O'Hare, for example)
Wouldn't it be possible to figure this out mathematically? I would specifically be interested in the speed the wheels / treadmill are moving at the time of takeoff. I think it would be ridiculous!
Math and Physics people!! GO!!!
edit: added Takeoff Run length
Let's use a 737-400 to keep it simple - not sure if all of these values would be needed:
Takeoff speed: 150 mph
Takeoff run: 8,483 ft
Takeoff weight: 100,000 lbs
Takeoff thrust: 22,000 lbf
Runway length: 13,000 feet (longest runway at O'Hare, for example)
Wouldn't it be possible to figure this out mathematically? I would specifically be interested in the speed the wheels / treadmill are moving at the time of takeoff. I think it would be ridiculous!
Math and Physics people!! GO!!!
edit: added Takeoff Run length
Last edited by srika; 12-07-2007 at 05:42 PM.
#664
threads on this question always to turn into crap, lol.
basically what you need to realize is that on a plane, the wheels just spin freely, they dont do anything. you cant think of it like a car.
basically what you need to realize is that on a plane, the wheels just spin freely, they dont do anything. you cant think of it like a car.
#665
Senior Moderator
Originally Posted by tdog2313
threads on this question always to turn into crap, lol.
basically what you need to realize is that on a plane, the wheels just spin freely, they dont do anything. you cant think of it like a car.
basically what you need to realize is that on a plane, the wheels just spin freely, they dont do anything. you cant think of it like a car.
#667
Advanced
Join Date: Jul 2003
Location: Winnipeg, Manitoba
Age: 39
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Ok, here is my theory.
Lets say there is a massive treadmill a few miles long and the belt is moving at 300mph. Anyone who has taken fluid mechanics/dynamics knows that a moving solid surface will have an effect on the air directly above it. Air is a fluid and the air directly touching the moving surface of the treadmill will move at the same speed of the treadmill (300mph). As you move away from the surface vertically the velocity decreases to zero in the form a parabolic function. This is called the boundary layer effect because there is friction between the air and the surface of the treadmill.
![](https://upload.wikimedia.org/wikipedia/commons/thumb/3/38/Boundarylayer.png/320px-Boundarylayer.png)
The picture (from wikipedia) shows air moving over a stationary plate but the same phenomenon occurs if the plate is moving and the air is stationary.
At the front end of the treadmill the boundary layer will be tiny, maybe only few thousandths of an inch. Since the thickness increases the further you move along the treadmill towards the back, there will be a larger layer of moving air above the treadmill. If the air flow is turbulent, the boundary layer increases in thickness faster as you move further from the leading edge of the treadmill.
So imagine if you cranked up the speed of the treadmill to a ridiculous amount so that a huge amount of air is pushed by the treadmill and a large boundary layer is created. The airplane could be placed near the back end of the treadmill where the boundary layer is the thickest. The engines are run so that the thrust overcomes the forces of friction from the rotating wheels and the drag caused by all the moving air. Now the plane is stationary with respect to the ground but if the air is blowing past it fast enough the wings will generate the lift needed to overcome the weight of the plane and it should take off and float in the air.
Also remember that the boundary layer is not infinitely thick so if the plane pulled up and out of that layer then the air velocity would be zero. If there is no air moving past the wings the plane will simply lose lift and fall.
Does this make sense to anyone? Obviously testing this theory with a real plane would be near impossible. I imagine you could do small scale testing of this scenario but replace air with a viscous fluid. Using a higher viscosity and denser fluid will make the boundary layer larger. I'm gonna wait and see what the Mythbusters can come up with but can already guess that it will leave me dissatisfied.
I should mention that I have not done any research and I just thought this up now so I not saying I'm right. Also, it has been a couple of years since I've done fluid mechanics. I just wanted to share my thoughts, thats all.
Lets say there is a massive treadmill a few miles long and the belt is moving at 300mph. Anyone who has taken fluid mechanics/dynamics knows that a moving solid surface will have an effect on the air directly above it. Air is a fluid and the air directly touching the moving surface of the treadmill will move at the same speed of the treadmill (300mph). As you move away from the surface vertically the velocity decreases to zero in the form a parabolic function. This is called the boundary layer effect because there is friction between the air and the surface of the treadmill.
![](https://upload.wikimedia.org/wikipedia/commons/thumb/3/38/Boundarylayer.png/320px-Boundarylayer.png)
The picture (from wikipedia) shows air moving over a stationary plate but the same phenomenon occurs if the plate is moving and the air is stationary.
At the front end of the treadmill the boundary layer will be tiny, maybe only few thousandths of an inch. Since the thickness increases the further you move along the treadmill towards the back, there will be a larger layer of moving air above the treadmill. If the air flow is turbulent, the boundary layer increases in thickness faster as you move further from the leading edge of the treadmill.
So imagine if you cranked up the speed of the treadmill to a ridiculous amount so that a huge amount of air is pushed by the treadmill and a large boundary layer is created. The airplane could be placed near the back end of the treadmill where the boundary layer is the thickest. The engines are run so that the thrust overcomes the forces of friction from the rotating wheels and the drag caused by all the moving air. Now the plane is stationary with respect to the ground but if the air is blowing past it fast enough the wings will generate the lift needed to overcome the weight of the plane and it should take off and float in the air.
Also remember that the boundary layer is not infinitely thick so if the plane pulled up and out of that layer then the air velocity would be zero. If there is no air moving past the wings the plane will simply lose lift and fall.
Does this make sense to anyone? Obviously testing this theory with a real plane would be near impossible. I imagine you could do small scale testing of this scenario but replace air with a viscous fluid. Using a higher viscosity and denser fluid will make the boundary layer larger. I'm gonna wait and see what the Mythbusters can come up with but can already guess that it will leave me dissatisfied.
I should mention that I have not done any research and I just thought this up now so I not saying I'm right. Also, it has been a couple of years since I've done fluid mechanics. I just wanted to share my thoughts, thats all.
#668
I shoot people
<-------- very
![Confused](https://acurazine.com/forums/images/smilies/confused.gif)
#669
Big Block go VROOOM!
Originally Posted by shabaaz
Does this make sense to anyone?
#670
Go Giants
Originally Posted by Smalls
ah I remember. It was like a lightswitch.
......[whiskers] ah ok, you're right. [/whiskers]![rofl](https://acurazine.com/forums/images/smilies/rofl.gif)
......[whiskers] ah ok, you're right. [/whiskers]
![rofl](https://acurazine.com/forums/images/smilies/rofl.gif)
#671
Senior Moderator
iTrader: (5)
Well, since this thread was about Mythbusters, then went back to the whole "will the plane take off" discussion, I just merged the two together. No need to have a new topic dedicated to the same thing as an existing 23 or so page topic that was already beat to death...
#672
Senior Moderator
read through some of the rest of this thread.. this thread = ridiculous.
basically you got the people who are saying the jet will take off (headed by cibs), the people saying it won't (headed by JJ4short), and the people asking if the jet could take off, why don't airports have shorter runways..![Jesus LOL](https://acurazine.com/forums/images/smilies/jesuslol.gif)
btw I did some checking - originally, this question did not include the condition that the speed of the belt would change to match the speed of the plane...
http://txfx.net/2005/12/08/airplane-on-a-conveyor-belt/
http://www.airliners.net/discussions...in/136068/1/#1
it has been beaten to death all over the net and I can't believe we talked about it AGAIN.
basically you got the people who are saying the jet will take off (headed by cibs), the people saying it won't (headed by JJ4short), and the people asking if the jet could take off, why don't airports have shorter runways..
![Jesus LOL](https://acurazine.com/forums/images/smilies/jesuslol.gif)
btw I did some checking - originally, this question did not include the condition that the speed of the belt would change to match the speed of the plane...
http://txfx.net/2005/12/08/airplane-on-a-conveyor-belt/
http://www.airliners.net/discussions...in/136068/1/#1
it has been beaten to death all over the net and I can't believe we talked about it AGAIN.
#673
Senior Moderator
Originally Posted by srika
basically you got the people who are right saying the jet will take off (headed by cibs), the people who are wrong saying it won't (headed by JJ4short), and the idiots asking if the jet could take off, why don't airports have shorter runways.. ![Jesus LOL](https://acurazine.com/forums/images/smilies/jesuslol.gif)
![Jesus LOL](https://acurazine.com/forums/images/smilies/jesuslol.gif)
Fixed
#674
Go Giants
Originally Posted by srika
basically you got the people who are right are saying the jet will take off (headed by cibs), the people saying it won't (headed by JJ4short, who changed his mind to agree with cibs), and the people asking if the jet could take off, why don't airports have shorter runways.. ![Jesus LOL](https://acurazine.com/forums/images/smilies/jesuslol.gif)
![Jesus LOL](https://acurazine.com/forums/images/smilies/jesuslol.gif)
#675
I shoot people
anyone know how they are going to prove it on Mythbusters? Model airplane? A small (real) plane with a very large treadmill?
#676
Senior Moderator
Originally Posted by curls
Copied and pasted... This should pretty much /thread:
"This is somewhat of a "trick" question. Not because it is phrased in a
deliberately tricky way, but because people tend to have trouble
thinking about the operation of other vehicles apart from cars which
they know so well.
The heart of the confusion is simply these two important facts:
* cars propel themselves by pushing against the ground via friction
* airplanes propel themselves by pushing against the air
If you can let go of how cars operate and think about what an airplane
does, you'll be able to see the problem clearly.
One good way of tackling this problem is to find a good analogy. But
the analogy must be a valid one else you'll just get more confused.
For example, someone posted the analogy of running on a treadmill. Why
is that a bad analogy? Because one runs by pushing against the ground
via friction between their shoe and the ground. This is how a car
propels itself! It is not how an airplane propels itself, by pushing
against the air. Bad analogy.
Let's use this analogy. Instead of looking at the airplane, let's back
up and go into the airport. Suppose you're walking down to your gate
and pulling your carry-on bag behind you. It's a nice new bag with low
friction wheels. No problem! Up ahead you see one of those moving
walkways. You don't see anyone coming, so you decide to do a little
experiment. You go over to the walkway that is moving TOWARDS you and
place your bag on it. Meanwhile, you step off to the side of the
walkway, and still holding on to the handle of your bag, you continue
to walk along. In fact, you intentionally walk along at the same speed
that the moving walkway is going, just in the opposite direction.
Question: does the bag move or does it remain stationary as you keep
walking? Obviously it moves with you. So why does your bag move
forward when you are walking at the same speed of the conveyor going
in the opposite direction?
The answer to that question is also the answer to the
airplane-conveyor question. To complete the analogy, the pull of your
arm is analogous to the force of the airplane engines. The bag's
wheels are analogous to the airplane tires. Do the nice low-friction
wheels on your bag on the conveyor pull against you anymore than they
do when you're just pulling your bag along normally? No, they don't.
They are free-wheeling, after all. Meanwhile, you're pulling the bag
with the same force in both cases. So in both cases, the bag keeps
moving forward. Likewise with the airplane, the pull of the engines
doesn't change nor does the force on the airplane imparted by the
tires change no matter what the ground is doing underneath the tires.
You have the same force imbalance in either case, and since Force =
mass x accceleration, you have the same acceleration. Remember, we are
talking airplane engines which push against the AIR, not the ground.
The acceleration is with respect to the AIR, thus the airplane
develops a speed relative to the air and can eventually take off.
That's a long winded analogy. Here's a quicker solution. Engineers
learn to draw Free Body Diagrams to understand such problems. A FBD is
just a block diagram which illustrates the forces acting upon an
object. The net force can be calculated from all the contributing
forces. If that net force is not zero, the object accelerates in the
direction of that force. Let's draw a FBD now. Represent the airplane
by a simple rectangle (the shape doesn't matter). Indicate the force
of the engines pulling on the plane with a large arrow, labeled "F_e"
with e for engines. What force are the tires imparting to the plane?
Remember, they spin freely except for bearing friction and rolling
friction, which produces forces that are quite tiny compared to the
engines. Represent those forces with a small arrow, labeled F_t where
t means tire. What other forces are operating? There is a drag force
too. But remember this drag force is also small compared to the
propulsion of the engines at least at takeoff speeds. Label that F_d.
FREE BODY DIAGRAM
F_d |----------| F_e
<---| Plane |------------->
|----------|
<-- F_t
So what do we have here? We haven't put numerical quantities on these
forces, but instead have just been talking in terms of large and
small. That's okay for the purposes of this illustration. It's enough
to know that F_e is going to be much greater than F_d + F_t pulling in
the opposite direction. If that's the case, we have an unbalanced
force on the plane. Therefore, it accelerates. And it accelerates with
respect to the AIR, since F_e is produces by the engines pushing
against the air.
To sum up: Yes, the airplane takes off. The motion of the surface
underneath the freely-spinning tires is irrelevant to the acceleration
of the aircraft since the tires cannot impart any force to the
aircraft (aside from the aforementioned very small rolling and bearing
friction).
I hope this discussion has helped somewhat and not muddied the waters any further.
John Strong
Ph.D., biochemical engineering
M.S., chemical engineering
B.S., mechanical engineering"
Copied and pasted...
"This is somewhat of a "trick" question. Not because it is phrased in a
deliberately tricky way, but because people tend to have trouble
thinking about the operation of other vehicles apart from cars which
they know so well.
The heart of the confusion is simply these two important facts:
* cars propel themselves by pushing against the ground via friction
* airplanes propel themselves by pushing against the air
If you can let go of how cars operate and think about what an airplane
does, you'll be able to see the problem clearly.
One good way of tackling this problem is to find a good analogy. But
the analogy must be a valid one else you'll just get more confused.
For example, someone posted the analogy of running on a treadmill. Why
is that a bad analogy? Because one runs by pushing against the ground
via friction between their shoe and the ground. This is how a car
propels itself! It is not how an airplane propels itself, by pushing
against the air. Bad analogy.
Let's use this analogy. Instead of looking at the airplane, let's back
up and go into the airport. Suppose you're walking down to your gate
and pulling your carry-on bag behind you. It's a nice new bag with low
friction wheels. No problem! Up ahead you see one of those moving
walkways. You don't see anyone coming, so you decide to do a little
experiment. You go over to the walkway that is moving TOWARDS you and
place your bag on it. Meanwhile, you step off to the side of the
walkway, and still holding on to the handle of your bag, you continue
to walk along. In fact, you intentionally walk along at the same speed
that the moving walkway is going, just in the opposite direction.
Question: does the bag move or does it remain stationary as you keep
walking? Obviously it moves with you. So why does your bag move
forward when you are walking at the same speed of the conveyor going
in the opposite direction?
The answer to that question is also the answer to the
airplane-conveyor question. To complete the analogy, the pull of your
arm is analogous to the force of the airplane engines. The bag's
wheels are analogous to the airplane tires. Do the nice low-friction
wheels on your bag on the conveyor pull against you anymore than they
do when you're just pulling your bag along normally? No, they don't.
They are free-wheeling, after all. Meanwhile, you're pulling the bag
with the same force in both cases. So in both cases, the bag keeps
moving forward. Likewise with the airplane, the pull of the engines
doesn't change nor does the force on the airplane imparted by the
tires change no matter what the ground is doing underneath the tires.
You have the same force imbalance in either case, and since Force =
mass x accceleration, you have the same acceleration. Remember, we are
talking airplane engines which push against the AIR, not the ground.
The acceleration is with respect to the AIR, thus the airplane
develops a speed relative to the air and can eventually take off.
That's a long winded analogy. Here's a quicker solution. Engineers
learn to draw Free Body Diagrams to understand such problems. A FBD is
just a block diagram which illustrates the forces acting upon an
object. The net force can be calculated from all the contributing
forces. If that net force is not zero, the object accelerates in the
direction of that force. Let's draw a FBD now. Represent the airplane
by a simple rectangle (the shape doesn't matter). Indicate the force
of the engines pulling on the plane with a large arrow, labeled "F_e"
with e for engines. What force are the tires imparting to the plane?
Remember, they spin freely except for bearing friction and rolling
friction, which produces forces that are quite tiny compared to the
engines. Represent those forces with a small arrow, labeled F_t where
t means tire. What other forces are operating? There is a drag force
too. But remember this drag force is also small compared to the
propulsion of the engines at least at takeoff speeds. Label that F_d.
FREE BODY DIAGRAM
F_d |----------| F_e
<---| Plane |------------->
|----------|
<-- F_t
So what do we have here? We haven't put numerical quantities on these
forces, but instead have just been talking in terms of large and
small. That's okay for the purposes of this illustration. It's enough
to know that F_e is going to be much greater than F_d + F_t pulling in
the opposite direction. If that's the case, we have an unbalanced
force on the plane. Therefore, it accelerates. And it accelerates with
respect to the AIR, since F_e is produces by the engines pushing
against the air.
To sum up: Yes, the airplane takes off. The motion of the surface
underneath the freely-spinning tires is irrelevant to the acceleration
of the aircraft since the tires cannot impart any force to the
aircraft (aside from the aforementioned very small rolling and bearing
friction).
I hope this discussion has helped somewhat and not muddied the waters any further.
John Strong
Ph.D., biochemical engineering
M.S., chemical engineering
B.S., mechanical engineering"
Copied and pasted...
![Thumbs Up](https://acurazine.com/forums/images/smilies/thumbsup.gif)
#678
AZ Community Team
Join Date: May 2007
Location: N35°03'16.75", W 080°51'0.9"
Posts: 32,488
Received 7,771 Likes
on
4,342 Posts
Originally Posted by is300eater
anyone know how they are going to prove it on Mythbusters? Model airplane? A small (real) plane with a very large treadmill?
BTW: Love >>> Electric (IMHO.
![Thumbs Up](https://acurazine.com/forums/images/smilies/thumbsup.gif)
#679
Senior Moderator
iTrader: (5)
Originally Posted by is300eater
anyone know how they are going to prove it on Mythbusters? Model airplane? A small (real) plane with a very large treadmill?
Originally Posted by WILLDOGS
Set your TiVo's
MythBusters
Air Plane Hour
TV-PG
Dec 12, 9:00 pm
Jamie and Adam take wing to test if a person with no flight training can safely land a 747 and if a plane can take off from a conveyor belt speeding in the opposite direction. Tory, Grant, and Kari jump on some Hollywood-inspired skydiving myths.
http://dsc.discovery.com/tv-schedule...913.x&start=10
MythBusters
Air Plane Hour
TV-PG
Dec 12, 9:00 pm
Jamie and Adam take wing to test if a person with no flight training can safely land a 747 and if a plane can take off from a conveyor belt speeding in the opposite direction. Tory, Grant, and Kari jump on some Hollywood-inspired skydiving myths.
http://dsc.discovery.com/tv-schedule...913.x&start=10
![Thumbs Up](https://acurazine.com/forums/images/smilies/thumbsup.gif)