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re: Can this 747 take off?
Posted on 4/11/24 at 11:03 am to Street Hawk
Posted on 4/11/24 at 11:03 am to Street Hawk
quote:
Can this 747 take off?
Not without a 200mph headwind...
![](https://images.tigerdroppings.com/Images/Icons/IconLOL.gif)
Posted on 4/11/24 at 2:31 pm to Street Hawk
Posted on 4/11/24 at 3:44 pm to Street Hawk
This is such a stupid question how it's worded, and is requiring assumptions to be made.
If the plane's engines are on, then it's going to make the plane go forward since the plane's power is simply to push the plane forward, with the wheels similar to a car being in neutral.
WHen they say the wheels and belt are the same speed, i assume what they mean is that same distance is being displaced by both the wheel and the belt as they move simultaneously, thus the plane doesn't actually move horizontally. I guess that's there way of saying ignore other forces like friction and gravity. In actuality, this would require a lot of physics to figure out how both the wheels and belt can be at teh same speed.
This is no different than if you put a toy car on a treadmill and put your finger behind the car and turned the treadmill on. the wheels will turn, and the belt will turn, but the car won't move b/c you're applying the force along the x-axis onto it, same as the planes engines.
If the plane isn't moving along the x-axis, then air isn't passing over and under the wings, and if that isn't happening, then it's impossible to achieve lift.
I'm assuming the way this stupid question is posed, i am too assume the plane will not actually move along the x-axis, even though in reality it woudl be extremely difficult to achieve this.
If the plane's engines are on, then it's going to make the plane go forward since the plane's power is simply to push the plane forward, with the wheels similar to a car being in neutral.
WHen they say the wheels and belt are the same speed, i assume what they mean is that same distance is being displaced by both the wheel and the belt as they move simultaneously, thus the plane doesn't actually move horizontally. I guess that's there way of saying ignore other forces like friction and gravity. In actuality, this would require a lot of physics to figure out how both the wheels and belt can be at teh same speed.
This is no different than if you put a toy car on a treadmill and put your finger behind the car and turned the treadmill on. the wheels will turn, and the belt will turn, but the car won't move b/c you're applying the force along the x-axis onto it, same as the planes engines.
If the plane isn't moving along the x-axis, then air isn't passing over and under the wings, and if that isn't happening, then it's impossible to achieve lift.
I'm assuming the way this stupid question is posed, i am too assume the plane will not actually move along the x-axis, even though in reality it woudl be extremely difficult to achieve this.
Posted on 4/11/24 at 8:33 pm to Street Hawk
Trick question…that 747 identifies as piece of cutlery upon the titanic.
Posted on 4/11/24 at 9:14 pm to Street Hawk
I can’t believe this thread. Let me help out everyone who said no, because they don’t understand the physics or those who said no because of the wording of the question. The answer is yes even within the confines of the question, which is what makes it a trick question.
The question says the conveyor belt is designed to exactly match the speed of the wheels, moving in the opposite direction. The answer is that it’s impossible to achieve this regardless of design. Which is what makes this an engineering brain teaser.
Why is this?
Assume the speed of the conveyor is X, while sitting with no engine thrust the wheels speed can also be described as X, and that formula is X=X (speed of the belt on the left, speed of the wheels on the right) which people said cancels motion because the directional vectors are opposite. Many people who said no stopped here.
Now assume the forward momentum from the engine thrust as Y. When the engines power up the formula changes to X=X+Y. So long as Y isn’t zero this formula becomes mathematically impossible at any value of X. This is as a dependent (x) and independent (y) variable.
So regardless of what speed the conveyor is going it can never ever match the speed of the wheels as the wheels speed is the speed of the belt plus the additional thrust.
So as you approach infinity, assuming nothing would mechanically fail, the plane not only takes off, but it does so as if it was sitting on concrete.
The question is designed to show people studying engineering that some things are mathematically impossible to engineer.
The question says the conveyor belt is designed to exactly match the speed of the wheels, moving in the opposite direction. The answer is that it’s impossible to achieve this regardless of design. Which is what makes this an engineering brain teaser.
Why is this?
Assume the speed of the conveyor is X, while sitting with no engine thrust the wheels speed can also be described as X, and that formula is X=X (speed of the belt on the left, speed of the wheels on the right) which people said cancels motion because the directional vectors are opposite. Many people who said no stopped here.
Now assume the forward momentum from the engine thrust as Y. When the engines power up the formula changes to X=X+Y. So long as Y isn’t zero this formula becomes mathematically impossible at any value of X. This is as a dependent (x) and independent (y) variable.
So regardless of what speed the conveyor is going it can never ever match the speed of the wheels as the wheels speed is the speed of the belt plus the additional thrust.
So as you approach infinity, assuming nothing would mechanically fail, the plane not only takes off, but it does so as if it was sitting on concrete.
The question is designed to show people studying engineering that some things are mathematically impossible to engineer.
This post was edited on 4/11/24 at 9:17 pm
Posted on 4/11/24 at 10:22 pm to Street Hawk
quote:
Prove to us it will work. Please pass along your work. Or do an Einstein thought experiment and just write the prose.
Ok...
Mythbusters actually did this already. This first video is the explanation of why people argue and where they are wrong and you get your answer at the 06:45 mark...
"Yes, the plane will always take off"
Explanation
Video of the experiment...
LINK
Pilot: "I was really surprised it just took off normally. I thought that I would just sit here like a brick."
This post was edited on 4/11/24 at 10:28 pm
Posted on 4/11/24 at 10:47 pm to Street Hawk
In Mythbusters, the plane took off because the wheel speed was twice the normal take off speed. The way this question if written doesn't allow the wheel speed to be higher than the conveyor belt speed, thus the plane can't take off.
Captain Joe explains it here:
PLANE on a CONVEYOR BELT! Will it TAKE-OFF? Explained by CAPTAIN JOE
Captain Joe explains it here:
PLANE on a CONVEYOR BELT! Will it TAKE-OFF? Explained by CAPTAIN JOE
Posted on 4/11/24 at 11:13 pm to Street Hawk
I see we’re now on page 25.
I did go through a few pages, and you’re all completely wrong but can’t yet understand it.
I’m sorry to let you know, but you’re wrong.
I went through this 15 years ago. I took the same stance that you guys did that there was no chance in hell that the plane could possibly takeoff. I was wrong. I didn’t realize it at the time. Eventually it essentially clicked in my head. A lightbulb went off. It’ll happen for you too.
It’s not a trick question or some physics calculation question. The whole idea in concepts is really very simple.
Hopefully you get it soon
I did go through a few pages, and you’re all completely wrong but can’t yet understand it.
I’m sorry to let you know, but you’re wrong.
I went through this 15 years ago. I took the same stance that you guys did that there was no chance in hell that the plane could possibly takeoff. I was wrong. I didn’t realize it at the time. Eventually it essentially clicked in my head. A lightbulb went off. It’ll happen for you too.
It’s not a trick question or some physics calculation question. The whole idea in concepts is really very simple.
Hopefully you get it soon
Posted on 4/13/24 at 9:47 pm to Street Hawk
how is this thing 25 pages long now?
If the belt speed matches the wheel speed 1 for 1 but in the opposite direction then the plane would never move in a forward direction, hence not enough airflow across the wings so it can not take off because it is not moving along the x axis
And adding thrust as a forward vector does not count because the wheels only roll forward as a result of that vector. If the forward vector from engine thrust is combined with the equal vector in the opposite direction from the treadmill motion then the net vector is 0
If the belt speed matches the wheel speed 1 for 1 but in the opposite direction then the plane would never move in a forward direction, hence not enough airflow across the wings so it can not take off because it is not moving along the x axis
And adding thrust as a forward vector does not count because the wheels only roll forward as a result of that vector. If the forward vector from engine thrust is combined with the equal vector in the opposite direction from the treadmill motion then the net vector is 0
This post was edited on 4/13/24 at 9:53 pm
Posted on 4/13/24 at 10:04 pm to Street Hawk
No it can’t. The wheels are the only thing moving in place and there’s no air going over the wings.
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