Tuff math problem with easy intuitive solution.

The following problem can be solved by setting up some appropriate differential equation and solving it. There is an intuitive method of solving the problem without calculations.

Imagine n missiles at the vertices of a regular polygon. 3 at the points of an equilateral triangle; 4 on the corners of a square; 5 on a pentagon; et cetera. Say each polygon has a side 100 miles long, and assume that the missiles travel 500 miles per hour. The missiles pursue each other in a clock wise direction, each pursuing the one on the next vertex, correcting course instantaneously as the quarry moves.

Due to the symmetry of the situation, the missiles will always be on the vertices of a polygon which rotates and contracts as the missiles pursue each other. Very soon the missiles collide at the center of the original polygon.

Assume that the missiles can be treated as points, and courses are corrected continuously to keep each pursuer headed toward his quarry.

For any polygon, can you come up with an algorithm which provides the theoretical distance traveled by each missile before the collision? Is any polygon easier to deal with than the others?

Answer to last question is below. I will post intuitive solution in a few days. If you want to try the differential equations approach, have fun. I would not try that without being promised sex or money for the effort.























I was given the problem for the square in a differential equations course. Almost everybody in the class managed to solve the problem. Years afterwards the intuitive solution occurred to me, and it suggested an algorithm for solving the problem for other polygons. I am not sure the differential equations approach would work anything but the square.

Doesn't the question reduce to:
What is the distance from a vertex to the center of a polygon?

Nope, the thing is that the missiles are heading for the following ones, not the center.
Therefore you calculate the tangential speed + the centripetal speed is the missile speed

polygon corner counts:
P
angle from center between points:
a=2pi/pradius
r=50/sin((1/8-1/(2*P))*PI)
total speed:
500
speed vector component to centre:
v=500*cos(1/4-1/p)*pi
'time until collision:
t=r/v

For the square, the velocity vector of each pursuer and his quarry are always at right angles to each other. As a consequence, the quarry's motion never moves him away from nor toward his pursuer. Therefore, the pursuer must travel the original distance. If the sides of the original square were 100 miles long, each missile travels 100 miles on a spiral course toward the center of the square. It theory, the last part of the journey consists of infinitely many rotations infinitesimally close to the center of infinitesimal squares.

There is a second intuitive solution, which gives the same result. Image the square centered at the origin, with two corners on the Y-Axis, two on the X-Axis. Consider the motion of the missile on the positive Y-Axis. It initially moves an infinitesimal amount diagonally toward the positive X-Axis before correcting its course for the quarry's movement. Now suppose the square is rotated counterclockwise just enough to return the missile to the Y-Axis. It is now on an infinitesimally smaller square, and moves toward its (also) rotated quarry. If this process is kept up, the missile follows a diagonal path on the side of a square which decreases in size, eventually reaching the origin. This description of the motion is equivalent to the original problem, in which the square was viewed as rotating and the missiles were viewed as following a spiral path toward the origin.

The above suggests a construction which will solve the problem for other polygons. Position the polygon with its center at the origin, and one vertex on the positive Y-Axis. From the vertex on the positive Y-Axis, follow the side (extended, except for triangle) to the positive X-Axis. The length of that line to the X-Axis is the distance that must be traveled.

I'm starting to feel very stupid! I got lost by the word "vector"!

<retires to previous corner and mumbles incoherently, safe in own insanity.>

Hey Guv, what do you think of my fast solution above?
Well actually i just had centripetal forces and relative speed excercises a few weeks ago so i had them in mind when i came up with my solution.

1. The missile velocity vector will always be pointed in the same direction relatively to itself and the center.

2. The missile veloctiy vector can be split up into tangential speed and centripetal velocity, where as the centripetal velocity is the part that counts when you calculate the time t=x/v

CyberSurfer, when dealing with moving objects, the direction as well as the speed is important. Going 100 miles per hour south is not the same as going 100 miles per hour west. The concept of speed combined with direction is called velocity. Velocity can be represented as a line with an arrow at one end: -------> . The direction of the line indicates the direction of the motion, while the length corresponds to the speed. The line is called a velocity vector. A lot of problems involving velocity can be visualized and solved by applying geometric methods to the vectors (lines).

Does the above help you understand what a vector is?

Kedaman, I did not understand your equations.

Angle from center between points: a=2pi/pradius (I think this should be a = 2*pi/P for radians, or 360/P for degrees), but since you do not use it anywhere, I guess it does not matter.

r=50/sin((1/8-1/(2*P))*PI)
I just do not understand this equation nor the next. Why not r = 50/sin(a/2)?
Speed vector component to centre: v=500*cos(1/4-1/p)*pi (Why not 500*cos(a/2)?)

At any rate, for the square, the distance traveled has to be 100 miles. I remember this from the formal solution using differential equations. At 500 mph, the time has got to be 12 minutes. Your formulae do not seem to give this result, nor do my variations on your formulae.

Sorry for not being too clear, my formulas are supposed to work with polygons with P corners.

Quote:

---

Angle from center between points: a=2pi/pradius (I think this should be a = 2*pi/P for radians, or 360/P for degrees), but since you do not use it anywhere, I guess it does not matter.

---

I was to use that angle but dunno why i typed that other formula, maybe i had it for another angle that i abandoned later. 50/sin(a/2) is exactly what it should be

Code:

---

__50______
\    |
 \  |
  \  |
 r \ |
    \| <- a/2
    C

---

Also the radius shouldn't be there either, in the formla to get a.

Code:

---

  __
    /|
 Vt/  _____
  /___/    \_____
_/______\_________\
 \b    /
  \ 
  \Vc
  _\
    \
      \
      \C

---

Vt and Vc here are the speed components (should be right angled) in tangential direction and centripetal direction. the angle inside cos, is not a, as you see, it's b = (pi/2-a) So i could as well write the formula:
V=sin(a/2)
I think the confusion is that i used the b angle all the time, which i got another formula for, i