Bird & Train: Solving the Mystery

[kudoushinichi88]

ok well I'm totally confused then. The first case was that train goes along line x=0 upwards, and the bird flies at x=-1 or something right beside it, but it also has a very slight component in the x direction, so that it eventually gets into the train.

Lemme try answering this question again.

Let ground be our frame of reference.

Let a car of a train start at coordinates (0,-a) where a is a large number and the train is moving at a velocity, v along the the line x=0 (the y-axis) in the positive direction.

At the start, let the bird be at coordinates (1,-a).

Now I am slightly confused here because, if the bird is moving at the same velocity as the train, the bird and the train will never meet. So, I'm assuming that the y-component of the velocity of the bird is the same as the train.

So let the y-component of the velocity of the bird equals the train and the x-component of the velocity of the bird is in the negative direction and has a low magnitude.

Let the bird enters the car of the train through one of the side windows at coordinates (0,0).

The line traced by the moving car of the train is x=0 for y > -a if the train continues move indefinitely.

The line traced by the flying bird is

$$y = -ax \ \mbox{for} \ 0 > x > 1$$

Now, if I'm sitting inside the train and looking out the window, I will see an almost stationary bird that is moving closer and closer to the window, wouldn't I?

Switching back to the original frame of reference (the ground),

Once the bird enters the train, the y-component of the velocity of the bird increases (the bird accelerates) if the bird retains the same flapping as it does outside the train. Since the air inside the train is moving wrt to the ground, it should cause the bird to accelerate compared to the bird flying in still air, outside the train.

The x-component of the velocity of the bird remains the same.

The bird should exit the train through the window a few windows in front of the window opposite the window it entered. (exits through the window closer to the front of the train than the window that it used to enter the train)

So, the line decribed by the bird would be

$$y = -ax \ \mbox{for} \ 0 < x < 1$$

and

$$y = -ax + b \ \mbox{for} \ x < 0$$

Where b is caused by the acceleration of the bird in the train and

assuming that the width of the train is infinitesimally small.

Anyway, in reality, the width of the train wouldn't be infinitesimally small, but, because of the acceleration of the bird inside the train, the line described by the bird while it's flying inside the train in our frame of reference should have a steeper gradient compared to the line it described when it's flying outside the train. That is what I think of this question, for now...

 

[capnahab]

But what if the bird was trying to get into the bar car? He/She would certainly be flapping their his/her wings faster.

 

[kudoushinichi88]

into the what??

 

[capnahab]

The point is that the bird would be applying its air brakes and could make the transition.

 

[kudoushinichi88]

This next one is that the train is again going upwards on line x=0, but now we have a bird flying on the line y=0 in the positive x-direction (it has no component in the y-direction!), and the bird and train meet at (0,0) and the bird flies into the train through the open window. You are telling me that it will not hit the wall in this case? The x and y directions are completely orthogonal to each other. The only way the bird can survive is if there is another window on the other side of the train which it can quickly pop out of before the wall gets there.

Lemme try this one too.

Let the ground be our frame of reference.

An infinitely long train is moving along the line x = 0 (the y-axis) at a velocity, v1 in the positive direction.

Let our bird start at coordinates (-c,0) and is flying along the line y=0 (the x-axis) at velocity, v2 which has the same magnitude as v1 but in the direction perpendicular to v1 and towards the positive direction on the x-axis. (means it's flying towards the train in the direction perpendicular to the train.) The bird has zero velocity in the y-component.

The bird enters the train through one of the side windows at coordinates (0,0).

The bird now is in the train. Because of the air in the train is moving wrt to the ground, it causes the bird to accelerate and have a velocity in the positive direction of the y-axis.

The bird now exits the train through the window opposite the window it enters.

Hence the line our bird describes on our frame of reference is

$$y=0 \ \mbox{for} x<0$$

and

$$y=d \ \mbox{for} x>0$$

where d is caused by the acceleration of the bird while it's flying in the train, assuming the width of the train is infinitesimally small.

The bird is assumed to maintain a constant flapping throughout and all windows are assumed to be open.From these lines of thought, what I can deduce is, if I am in the train, I would observe the bird entering a window and exiting the window right opposite the window through which it entered but it exits from a position slightly closer to the rear of the train.

 

[kudoushinichi88]

Well, I'm assuming the bird didn't do that. If the bird applies 'air brakes' then it's velocity would change and the whole thing would be messed up.

 

[capnahab]

Instead of a ground frame of reference I prefer coordinate or at least cartesian.

 

[Hootenanny]

How on Earth has this thread reached 42 posts?

 

[kudoushinichi88]

I wonder too.

 

[capnahab]

I say neuter all of the cats. Let's see it that gets a rise.

 

[Hootenanny]

I say neuter all of the cats. Let's see it that gets a rise.

I wasn't trying to get a 'rise', I was merely commenting that the OP's question was answered in the second post (and clarified again in post #9). Since then, the thread has just been misinterpreted, taken out of context and over-complicated.

 

[Mephisto]

I wasn't trying to get a 'rise', I was merely commenting that the OP's question was answered in the second post (and clarified again in post #9). Since then, the thread has just been misinterpreted, taken out of context and over-complicated.

that's actually a quite correct insight :)
anyone feels like discussing the cases where the train moves at 0.99c ? :)
just joking, let it die. The questions were answered.

 

[Shooting Star]

Typical physicist answer = first we assume a spherical bird flying in a vacuum !

I disagree. A typical physicist's answer would have been a point mass moving only along the x-axis in vacuum.

(Hey Hootenanny, this is another way how the number of posts increase. If the number still continues to rise, I promise to contribute more. This thread is a gold mine...)

 

[algis.j]

I think Mephisto assumed initial paths are parallel. The bird will enter the train and accelerate toward the front because air resistance is much lower.
If he keeps going sligthly transversally; passing by one other side window and enter another train crossing in the other way; he will go toward the end of this second train - and if he finally goes out by a fourth window he will keep his way as before entering his set of trains.

But if I were to be him; I just stop at the first bar having half bottle of Vodka to cheer my luck :p

 

[kudoushinichi88]

What if the train is going in the opposite direction of the bird but the magnitude of the velocity is about the same? It means the bird decelerates in the train, right

 

[Hootenanny]

What if the train is going in the opposite direction of the bird but the magnitude of the velocity is about the same? It means the bird decelerates in the train, right

It would accelerate toward the rear of the train.

I wonder too.

I thought we'd agreed this thread was done?

 

[Shooting Star]

It would accelerate toward the rear of the train.

I thought we'd agreed this thread was done?

Instead of a ground frame of reference I prefer coordinate or at least cartesian.

Not until we've settled this one, at least.

 

[jVincent]

Well, let's say you have this senario:

Meassuring EVERYTHING in reference to the train.
In the train the airvelocity is zero. outside the airvelocity is negative 80 km/hour. (The speed of the train relative to the tracks).
So outside you see a bird flapping its wings to stay in air, moving with 0 km/hour relative to the train.

Bird enters window, and is now in a medium of 0 km/hour wind, and has a velocity of 0 km/hour. So either it keeps flapping , slowly accelerating again, eventually hitting the front wall in the train( or landing on a seat). Or, it gets confused by the wind change, stops flapping and crashes to the ground.

You seem to be thinking that it will suddenly be traveling with 80km/hour relative to the train because it "swiches frame of reference" which is NOT what will happen.

 

[Shooting Star]

There are some birds which can cruise for long times without flapping. What if it was one of those type? Will it experience a sudden discontinuity or continue to cruise with equanimity?

 

[jVincent]

This is not that important, but first:
Birds that cruise travel at a negative velocity with respect to the train (since they are not flapping they are slowly slowing down), at the same time, they are only staying in the air, due to the friction of the wind, so if it travels at the same speed as the wind, it falls down(Same speed meaning the same direction too).

When the bird enter the train, there is no longer any wind, so they are left at exactly the same scenario as the flapping birds from before. They are at 0 km/h in a 0km/h wind field. So its left with the choice: flap or drop.

 

[Shooting Star]

Oh no, they are not slowing down at all. They're using the fact that the train drags some of the air aound it, and utilising that to cruise at a steady speed, and keeping up with train. A few flaps now and then is permissible.

 

[TVP45]

Oh no! A drafting blue heron.

 

[Cyrus]

Jesus christ. Stop posting in here and read the answers already provided.

 

[Shooting Star]

Oh no! A drafting blue heron.

It's a pleasure discussing with you. You are so perceptive.

 

[Shooting Star]

Jesus christ. Stop posting in here and read the answers already provided.

Oh, all right. Let's end this one, and just remind everyone that the answer was given in the 2nd post.