10 Nov 09
Imagine four ants on a table, with the ants forming the vertices of a square of side length "b". Simultaneously, each ant starts crawling directly toward the ant in front of it (i.e. ant #1 chases ant #2, #2 chases #3, #3 chases #4, and #4 chases #1), all moving at the same speed "k".
(1) How long does it take before the ants meet in the centre?
(2) What shape does the path of each ant take?
11 Nov 09
Originally posted by forkedknightAnd a more general problem, what about ants at X number of vertices, like triangles, pentangles and so forth, does the path differ in geometry or are they similar?
It depends on whether you take the theoretical case where the "ants" are points, or the practical case where they have volume/area relative to the size of the square.
In the latter case, they make slightly fewer revolutions until they meet...
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11 Nov 09
Originally posted by PBE6Is the path a logarithmic spiral?
Imagine four ants on a table, with the ants forming the vertices of a square of side length "b". Simultaneously, each ant starts crawling directly toward the ant in front of it (i.e. ant #1 chases ant #2, #2 chases #3, #3 chases #4, and #4 chases #1), all moving at the same speed "k".
(1) How long does it take before the ants meet in the centre?
(2) What shape does the path of each ant take?
11 Nov 09
1 edit
Originally posted by sonhouseSimilar - it just changes the tightness of the spiral.
And a more general problem, what about ants at X number of vertices, like triangles, pentangles and so forth, does the path differ in geometry or are they similar?
I get the following solution. If they start off a distance a from the centre (not exactly as it was posed, but it's simpler this way), and there are N vertices, then the path in radial coordinates (r, theta) is:
r = a - kt.sin(PI/N)
theta = -cot(PI/N).ln{1 - kt/[a.sin(PI/N)]}
12 Nov 09
Originally posted by mtthwThe last equation, theta =-cot(PI/N) you put a dot after that and before the natural log, I think that's what your ln means. Does the dot symbolize multiplication?
Similar - it just changes the tightness of the spiral.
I get the following solution. If they start off a distance a from the centre (not exactly as it was posed, but it's simpler this way), and there are N vertices, then the path in radial coordinates (r, theta) is:
r = a - kt.sin(PI/N)
theta = -cot(PI/N).ln{1 - kt/[a.sin(PI/N)]}
I see you also use it between the a and sin, same thing? a times the sin of (.....)
I usually see the * symbol for multiplying. Just wondered. I want to put it into my newly downloaded scilab when I figure out how to use it. Ever hear of scilab? A freeware math program, pretty powerful. I just got it today and still working out some details of the download. The manual is over two thousand pages! (pdf file)
It will be fun to put this in and see if I can graph them but I need to make sure exactly what your equation says. Thanks.
12 Nov 09
Originally posted by PBE61. Depends on how fast they are walking!
Imagine four ants on a table, with the ants forming the vertices of a square of side length "b". Simultaneously, each ant starts crawling directly toward the ant in front of it (i.e. ant #1 chases ant #2, #2 chases #3, #3 chases #4, and #4 chases #1), all moving at the same speed "k".
(1) How long does it take before the ants meet in the centre?
(2) What shape does the path of each ant take?
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13 Nov 09
Originally posted by PBE6Actually, im going to say that the shape of the path for 2 ants from opposite corners is kindof like a cubed root function. I think it is not a spiral.
Imagine four ants on a table, with the ants forming the vertices of a square of side length "b". Simultaneously, each ant starts crawling directly toward the ant in front of it (i.e. ant #1 chases ant #2, #2 chases #3, #3 chases #4, and #4 chases #1), all moving at the same speed "k".
(1) How long does it take before the ants meet in the centre?
(2) What shape does the path of each ant take?
And im thinking the meet in the center the same time that it would take one ant to walk from one vertex to the next of the square
so
t=b/k
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13 Nov 09
2 edits
Originally posted by joe shmoJust in case my result is vague, here is what i did (not very mathematical)
Actually, im going to say that the shape of the path for 2 ants from opposite corners is kindof like a cubed root function. I think it is not a spiral.
And im thinking the meet in the center the same time that it would take one ant to walk from one vertex to the next of the square
so
t=b/k
I drew a square and then decided to move each ant an arbitrary distance (d) from its original position. the direction of any ant at any given time can be found by the direction of a line connecting the position of any 2 ants in sequence at that time. Now there are 4 lines at an angle to each side, next the ants travel equal distances on those lines, you connect those lines and repeat as desired...The intersections of the lines give the path traveled...which is like a curvey X
Actually just put the square on to the cartesian coordinates
thinking that the lines I mentioned are tangent to the path of the ants ( I guess they represent the velocity vectors)
then the slope of one of the lines is ( if the length of the square is b) is:
Let C stand for "change in"
slope = Cy/(b-Cx)
so
dy/dx = y/(b-x)]
and doing separation of variables and integrating I get
y = +/-e^(b-x) as the path function (with/without the appropriate constants).
ahhh... let me know if my differential equation solving is incorrect, im not entirely sure, I haven't studied them yet.
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13 Nov 09
1 edit
Originally posted by joe shmonevermind the path function....lol, my assumtions must be wrong or my equation solving I think i just wanted e to be there, because the solution according to my above post yields a line.....haha( maybe both)
Just in case my result is vague, here is what i did (not very mathematical)
I drew a square and then decided to move each ant an arbitrary distance (d) from its original position. the direction of any ant at any given time can be found by the direction of a line connecting the position of any 2 ants in sequence at that time. Now there are 4 lines at an differential equation solving is incorrect, im not entirely sure, I haven't studied them yet.
the path function is more difficult than I thought
😞
13 Nov 09
Here's how I did it (for the 4 corner case).
The key thing to note is that, due to symmetry, if they start off as a square they are always a square. This means:
- the velocity towards the centre is a constant k/sqrt(2)
- the velocity perpendicular to a line drawn from the centre is a constant k/sqrt(2)
Solving the motion towards the centre is easy.
dr/dt = -k/sqrt(2)
r = (b - kt)/sqrt(2)
So they reach the centre at t = b/k (you got that bit right!)
For the radial motion:
r.dtheta/dt = k/sqrt(2)
=> dtheta/t = k/(b - kt)
=> theta = -ln(1 - kt/b)
Now consider what that does as t -> b/k.
Comparing to my n-version, so a = b/sqrt(2), they match. At least, they do if you correct the mistake I've just noticed while writing this: it should have been:
theta = -cot(PI/N).ln[1 - kt.sin(PI/N)/a]