Motion puzzle: Bird flying between a train and a platform

[haruspex]

@anthraxiom , what is the source of this question? Is it something you or a colleague made up?

 

[anthraxiom]

Well, I said "more realistic" as compared with reversing direction instantly. Also, no realistic bird flies for 40 hours at 100 km/h with one-second breaks between velocity reversals.

i really didn't want to find the answer per se i just wanted to see if an expression could be found which is the cause for the bizarre numbers. I really thought a solution could be attained pehaps.

 

[anthraxiom]

@anthraxiom , what is the source of this question? Is it something you or a colleague made up?

yes my friend did make it up on the fly

 

[haruspex]

yes my friend did make it up on the fly

Then we should leave it to your friend to astonish the world with an analytic solution.

 

[MatinSAR]

Then we should leave it to your friend to astonish the world with an analytic solution.

Shouldn't the sequence of touches form a convergent infinite series?

Edit: I mistakenly thought this was like the classic train-and-bird problem with constant speeds. Apologies for the off-topic question!

 

[erobz]

What you should try to do - is take a step back and try to solve the "easier problem". Let the train travel at constant speed, and remove the rest pause of the bird. See if you are able to find a closed form solution for the number of collisions.

 

[Lnewqban]

You lose that rather nice time symmetry for the return trip if you do.

The flight is still symmetrical.
It is only that the bird touches the surface of the train some time after it stops in front of it and starts flying back.
If its stop is delayed, a painfull collision will follow.

I see it as equivalent to the the advancement of the spark jump in any gasoline internal combustion engine.

 

[erobz]

Shouldn't the sequence of touches form a convergent infinite series?

I agree, but the challenge for the accelerating train and pausing bird ,how far do you have to compute until the pattern becomes clear is the problem.

 

[MatinSAR]

I agree, but the challenge for the accelerating train and pausing bird ,how far do you have to compute until the pattern becomes clear is the problem.

I mistakenly thought this was like the classic train-and-bird problem with constant speeds. With acceleration, it definitely becomes more complicated.

 

[erobz]

The flight is still symmetrical.

The bird will be accelerating during the pause on the train, its position will change. It will have a shorter return trip to close out the cycle.

 

[erobz]

I mistakenly thought this was like the classic train-and-bird problem with constant speeds. With acceleration, it definitely becomes more complicated.

The classic problem is "asking how far the bird travels in that case" which is turns out to be trivial if you stop and consider it. Finding the convergent series for the number of collisions seems (at least - I haven't solved it) to be much more work even in that problem (unless I haven't had an "aha" moment yet).

 

[WWGD]

Strange how this question is swinging back and forth aimlessly, just like the bird.

 

[erobz]

Strange how this question is swinging back and forth aimlessly, just like the bird.

I don't understand why you are apparently so frustrated. Is there something wrong with demonstrating the challenges of the problem by asking the OP to solve the "easier problem". Also, this is also how we make progress and learn stuff...by solving an easier problem. Are you really going to try and complain to have this thread shut down? What else is going on that is so pressing we can't be bothered to teach/learn?

 

[WWGD]

I don't understand why you are apparently so frustrated. Is there something wrong with demonstrating the challenges of the problem by asking the OP to solve the "easier problem". Also, this is also how we make progress and learn stuff...by solving an easier problem. Are you really going to try and complain to have this thread shut down? What else is going on that is so pressing we can't be bothered to teach?

Just a joke, guy. I haven't nor will I report anything.Edit: I am, though, ignoring you from now on. Feel free to reciprocate.

 

[erobz]

Just a joke, guy.

No its not, I'm attacked relentlessly on here by the "gang" for this kind of exploration. What you are doing is signaling that this thread needs to be shut down because I'm participating in it and it has exceeded 44 posts. Do me a favor and report it...so I can be banned. I've been asking for it.

 

[Lnewqban]

The bird will be accelerating during the pause on the train, its position will change. It will have a shorter return trip to close out the cycle.

Sorry, erobz, could you clarify the idea?

Not that it changes anything important in your calculations, but it still seems to me that the round flight, from the point of view of the bird, will take the same time between both moments when its velocity becomes zero (resting point and stopping-coming back point).
The following round flight will take a different time.

Where the train is at during those round flights, seems irrelevant to that fact (time forward = time backward), unless the bird overshoots and collides with the machine.

Of course, if the point of contact bird-train is used as point of reference, the first leg of the flight should take longer (more distance) than the second one.

 

[erobz]

Sorry, erobz, could you clarify the idea?

Not that it changes anything important in your calculations, but it still seems to me that the round flight, from the point of view of the bird, will take the same time between both moments when its velocity becomes zero (resting point and stopping-coming back point).
The following round flight will take a different time.

Where the train is at during those round flights, seems irrelevant to that fact (time forward = time backward), unless the bird overshoots and collides with the machine.

Of course, if the point of contact bird-train is used as point of reference, the first leg of the flight should take longer (more distance) than the second one.

In the case where the bird leaves the platform, intercepts the train, instantaneously flips and comes back to the platform the elapsed time to this point is $T = t_1+t_1 = 2 t_1$

I thought that it was suggested-to fudge in some acceleration effects of the bird meeting the train- that the bird will take off from the platform, intercept the train, rest for 1 second on the train and return to the platform. Here the elapsed time from leaving the platform to returning to it is not $T \neq 2t_1$. It is $T = t_1 + t_2$ where $t_2 < t_1$.

So the symmetry I'm talking about losing is that the round trip for the bird between the platform and the platform is no longer $2 t_1$ , I exploited that symmetry in the equations to calculate where the train is when the bird gets back to the platform. If its messy now, its going to be worse.

 

[bob012345]

In the case of constant velocity and instantaneous acceleration to speed of the bird, and the bird being a mathematical point, isn’t the number of contacts going to be infinite even when the total distance flown and total time flown is finite? That’s because no matter how short the remaining distance, the bird is always faster than the train. Maybe it’s Zeno’s bird?

 

[erobz]

In the case of constant velocity and instantaneous acceleration to speed of the bird, and the bird being a mathematical point, isn’t the number of contacts going to be infinite even when the total distance flown and total time flown is finite? That’s because no matter how short the remaining distance, the bird is always faster than the train. Maybe it’s Zeno’s bird?

Do you mean when the bird takes no pause at the platform?

 

[bob012345]

Do you mean when the bird takes no pause at the platform?

Right, it bounces perfectly and instantly.

 

[erobz]

Right, it bounces perfectly and instantly.

I don't know , maybe you have a look at this. I get a formula for $t_n$ that is how long it takes the bird to get to the train and is a constant to the nth power $\propto (1 - 2q)^n$. I find that $q$ ( a constant of speed parameters) is less than $\frac{1}{2}$ so long as the velocity of the bird does not equal that of the train (which would result in a single collision).

Then I sum double all these $t_i$ and I should have a geometric series sum. That has to be finite in $n$ right, because the sum must be equal to $\frac{V_T}{S_o}$ ?

The way I see it if $v_B = v_T$ we get 1 collision. They meet at some point (the middle) and they never separate. I think this proves that if the train is slower than the bird there will be more than 1 collisions, but a finite number.

But I might be goofing up, because I also had the same thought as you until I got half way to a possibly erroneous result. Or my result is fine, and I'm doing a muck up of its interpretation. I'm going to have to work on it.

 

[erobz]

@bob012345 I think you are correct. When I think about what happens when the bird is even just a little bit faster than the train, I see what you are saying. It would always reach a point over half-way of the distance between them at the beginning of the trip out to meet the train, and it seems there will always be a trip out to the train! Its does seem to be Zeno's Bird!

Just a 1 one second pause takes you from infinite to finite number of collisions! How about a $0. \cdots 000001$ pause?

 

[Lnewqban]

In the case of constant velocity and instantaneous acceleration to speed of the bird, and the bird being a mathematical point, isn’t the number of contacts going to be infinite even when the total distance flown and total time flown is finite? That’s because no matter how short the remaining distance, the bird is always faster than the train. Maybe it’s Zeno’s bird?

Just like a bouncing ball on a hard surface, which frequency keeps groing up?

 

[haruspex]

Fwiw, I worked in terms of $v_n$, the velocity of the train at completion of the nth cycle, and obtained a quadratic in $v_n, v_{n+1}$ and constants. The quadratic is quite general, involving the cross product and linear terms with different coefficients.
I should verify it against @erobz' numeric result.

 

[bob012345]

@bob012345 I think you are correct. When I think about what happens when the bird is even just a little bit faster than the train, I see what you are saying. It would always reach a point over half-way of the distance between them at the beginning of the trip out to meet the train, and it seems there will always be a trip out to the train! Its does seem to be Zeno's Bird!

Just a 1 one second pause takes you from infinite to finite number of collisions! How about a $0. \cdots 000001$ pause?

Yes, an infinite series can have a finite sum. It would seem any pause makes the number of collisions finite.

 

[erobz]

It would seem any pause makes the number of collisions finite.

I will never understand mathematics-especially with infinities.

 

[Ibix]

It would seem any pause makes the number of collisions finite.

Consider the case where the bird travels at infinite speed. Then the number of journeys is simply the time to the train's arrival divided by the pause length. That's obviously finite and clearly an upper bound on the number of journeys at any speed.

 

[erobz]

Consider the case where the bird travels at infinite speed. Then the number of journeys is simply the time to the train's arrival divided by the pause length. That's obviously finite and clearly an upper bound on the number of journeys at any speed.

After I just said I will never understand infinities (which is still true), you present this shift in perspective, and one can't help but understand it. Thats why they pay you the big bucks!

 

[bob012345]

Consider the case where the bird travels at infinite speed. Then the number of journeys is simply the time to the train's arrival divided by the pause length. That's obviously finite and clearly an upper bound on the number of journeys at any speed.

But if we consider the case where the train asymptotically approaches the station then there will always be an infinite number of journeys regardless of the speed of the bird or the pause time simply because the train takes an infinite amount of time to get there.

 

[Ibix]

This ma

But if we consider the case where the train asymptotically approaches the station then there will always be an infinite number of journeys regardless of the speed of the bird or the pause time simply because the train takes an infinite amount of time to get there.

If it takes the train an infinite amount of time to reach the station, sure. I didn't realise that case was under discussion.