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Subject: "The four travellers problem"     Previous Topic | Next Topic
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medita8
Member since Jun-28-03
Jun-28-03, 01:07 PM (EST)
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"The four travellers problem"
 
   There was a terse little note to the effect that I should direct my questions here and not through email. So... my quesion is about the four travellers problem. I don't get the solution. It'says to draw a line perpendicular to the plane... What plane? When was a plane mentioned before this? It said that to draw a graph I need a line for the time and a line for the body to move along. So I drew a 2d plane with a y axis and a time axis. Then it told me to draw a time axis perpendicular to this! This is one of those explanations that assumes that what is in the mind of the author is also in the mind of the listener (the definition of autism, incidentally). Please may I have the unobfuscated version of this solution. Many many thanks.


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  Subject     Author     Message Date     ID  
  RE: The four travellers problem alexbadmin Jun-28-03 1
     RE: The four travellers problem medita8 Jun-28-03 2
         RE: The four travellers problem alexbadmin Jun-28-03 3
             RE: The four travellers problem medita8 Jun-29-03 4
                 RE: The four travellers problem medita8 Jun-30-03 5
                     RE: The four travellers problem alexbadmin Jul-10-03 6

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alexbadmin
Charter Member
998 posts
Jun-28-03, 01:12 PM (EST)
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1. "RE: The four travellers problem"
In response to message #0
 
   LAST EDITED ON Jun-28-03 AT 01:12 PM (EST)
 
>There was a terse little note to the effect that I should
>direct my questions here and not through email.

Thank you for complying.

>So... my
>quesion is about the four travellers problem. I don't get
>the solution. It'says to draw a line perpendicular to the
>plane... What plane?

The plane in which the four roads are located.

>When was a plane mentioned before this?

In the very first sentence: "Four roads on a plane, ..."

>It said that to draw a graph I need a line for the time and
>a line for the body to move along. So I drew a 2d plane with
>a y axis and a time axis.

Nope. You have to draw an xy-plane in which the roads are located and a t-axis perpendicular to that plane.

>Then it told me to draw a time
>axis perpendicular to this! This is one of those
>explanations that assumes that what is in the mind of the
>author is also in the mind of the listener (the definition
>of autism, incidentally). Please may I have the unobfuscated
>version of this solution. Many many thanks.


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medita8
Member since Jun-28-03
Jun-28-03, 02:03 PM (EST)
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2. "RE: The four travellers problem"
In response to message #1
 
   Ok, I was going back to look at it and I see what tripped me up. I was reading, "m of 1 and m of 2 intersect" and I thought, they're straight parallel lines, how can they intersect? But it is their projection, from a point in time, out onto the road plane that intersect. Yes?

The intersecting lines are said to define a plane, which I take to mean a plane hovering parallel over the road plane. Now 2 and 3 also meet, but if they're meetings intersect on the same plane, doesn't that mean they all met at the same time?

I'll have another look.

And my god that was a quick reply. Many thanks.


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alexbadmin
Charter Member
998 posts
Jun-28-03, 02:06 PM (EST)
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3. "RE: The four travellers problem"
In response to message #2
 
   >But it is their projection, from a point in time,
>out onto the road plane that intersect. Yes?

Yes. But not only that. They themselves intersect too.

>
>The intersecting lines are said to define a plane, which I
>take to mean a plane hovering parallel over the road plane.

It's defintely not parallel to the plane of the roads. Each of the graphs intersects its corresponding road at time 0. So are the planes.

>Now 2 and 3 also meet, but if they're meetings intersect on
>the same plane, doesn't that mean they all met at the same
>time?

No, why?

>I'll have another look.

A good idea.

>And my god that was a quick reply. Many thanks.

You were lucky.


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medita8
Member since Jun-28-03
Jun-29-03, 06:12 AM (EST)
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4. "RE: The four travellers problem"
In response to message #3
 
   >>But it is their projection, from a point in time,
>>out onto the road plane that intersect. Yes?
>
>Yes. But not only that. They themselves intersect too.

Then I must be seeing this entirely incorrectly. I'm thinking of first one 2d graph that represents speed, y vs t (time). Uniform motion means a line that has no slope, right? So now I'm thinking of myself standing on the road plane. Off in the corner is another axis going up for time and along one horizon are the four mof i lines for uniform motion going straight up parallel with the time axis. How from here can they intersect without having a slope and representing something other than one particular speed?

>>The intersecting lines are said to define a plane, which I
>>take to mean a plane hovering parallel over the road plane.
>
>It's defintely not parallel to the plane of the roads. Each
>of the graphs intersects its corresponding road at time 0.
>So are the planes.

Uh, ok, I must 100% not understand. There's an x and y graph that represents the location of the people. Perpendicular to that is a graph of their uniform speed. The speed graphs start at time 0, don't they? So everyone is walking, the x and y readout displays are spinning and the speed graphs are moving up and at some point two people meet. So we have a dot on the xy graph and then above their heads is a point that corresponds to their time, an xyt coordinate dot. Where is this intersection at time 0?

Many thanks.


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medita8
Member since Jun-28-03
Jun-30-03, 01:22 PM (EST)
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5. "RE: The four travellers problem"
In response to message #4
 
   This morning I decided to imagine the road plane rising up, with the travellers tracing a line behind them which represented the (xyt) points. Is this how it is and so does this mean that the way speed graphs can intersect is if you take them as traced by the travellers into 3d space? In this way, then, 1,2,3, and 4 must be on the same plane, otherwise there couldn't be a meeting between 1 and the rest and 2 and the rest... The thing I have trouble seeing now is how the roads themselves could possibly be set up to allow this...!


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alexbadmin
Charter Member
998 posts
Jul-10-03, 11:33 PM (EST)
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6. "RE: The four travellers problem"
In response to message #5
 
   There's a plane a, which we take to be horizontal and on which four straight lines are drawn. The lines are assumed to be in general position so that no three pass through the same point and no two are parallel.

The time axis is perpendicular to that plane. When a fellow moves at a constant speed along a straight line, the graph of his motion is a line in space that is orthogonally projected onto the road of motion.

There is no need to draw x- and y- axes in the plane a. Any pair will do. But, yes, I think of the plane as an xy-plane with the t-axis sticking out.


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