solved by Maple.
In the case of a triangle, it can be noted that the two
flies that were the furthest apart are the ones who meet first!
| next curve | previous curve | 2D curves | 3D curves | surfaces | fractals | polyhedra |
MUTUAL PURSUIT CURVES (or fly curves)
| Problem first posed by Lucas in 1877.
Paper "Polygons of pursuit", Bernhart, 1959 Wikipedia: fr.wikipedia.org/wiki/Probl%C3%A8me_des_souris Very thorough website in German: did.mat.uni-bayreuth.de/material/verfolgung/node5.html Animation by G. Tulloue: www.sciences.univ-nantes.fr/sites/genevieve_tulloue/Meca/Cinematique/4mouches.php |
When n points M1,M2,…,Mn
(traditionally, flies, mice, ladybirds...) chase one another at the same
constant speed, with Mk chasing
Mk+1
(and Mn chasing M1),
the trajectories of these points are mutual pursuit curves.
| To get the figure opposite, we had the differential system
stemming from the n relations
solved by Maple.
In the case of a triangle, it can be noted that the two
flies that were the furthest apart are the ones who meet first!
|
|
When the initial figure is a regular
polygon with the points in their apparition order, then the trajectories
are logarithmic spirals
with asymptotic points the centre of the polygon.
The parameter of the spiral is 
,
and the polar tangential angle is 
where
n is the number of sides of the polygon.
The length of the trajectory of any fly is therefore
equal to 
where R is the radius, and
the length of the side.
|
 |
 |
|
|
|
|
| Demonstration in the case of a square.
The flies constantly form an EFGH square with the
coordinates shown in the figure.
|
Figure : Elisabeth Busser |
 |
 |
 |
See also similar curves in
3D.
 |
Did the anonymous Ivorian artist who designed this engraving think they were tracing pursuit curves? |
| next curve | previous curve | 2D curves | 3D curves | surfaces | fractals | polyhedra |
© Robert FERRÉOL, Alain ESCULIER 2021
,
results in 
, hence the differential equation of the curve:
. By passing in polar coordinates the equation transforms into 
,
hence the solution 
,
with a = 1 into applying the initial condition.
;
with this, the speed of the flies is 
,
it decreases towards 0, and the time course is infinite, for a length 
(side of the starting square
,
the flies have a speed