Four heat seeking missiles
 

Four heat seeking missiles are fired similtaneously from the four corners of a horizontal plane square formation of side 100 Km, high up in the air. They are all pointed to adjacent missiles in clockwise cyclic order so that they follow similar curved paths. Neglecting the effect of gravity, determine the distance travelled by each missile before they collide at the centre.

hint: this problem is best done graphically and please display the curved path.

Mally :coffee:

[spoiler]http://mathworld.wolfram.com/MiceProblem.html[/spoiler]

Mice
 

[QUOTE=axn1;94398][spoiler]http://mathworld.wolfram.com/MiceProblem.html[/spoiler][/QUOTE]
:smile:
Thank you axn1. You are spot on. So for 4 what will be the distance ?
A question for the others.
Mally :coffee:

Mally,

You didn't specify that the missles all travel at identical speeds. However this is required for the curved paths to be "similar" in the strictest sense.

The distance that each missle travels is [spoiler] 100 Km [/spoiler]

[QUOTE=Wacky;94421]Mally,

You didn't specify that the missles all travel at identical speeds. However this is required for the curved paths to be "similar" in the strictest sense.

The distance that each missle travels is [spoiler] 100 Km [/spoiler][/QUOTE]
There's an interesting observation that can be made for the 'sqare' problem (not the general problem linked by axn1).

Each target's velocity is perpendicular to its pursuer's velocity, with no component from or towards the pursuer. Therefore, the distance is closed purely by the motion of the pursuer, which is directly toward the target. This explains why the path is exactly as long as the original side of the square.

same speeds
 

[QUOTE=Wacky;94421]Mally,

You didn't specify that the missles all travel at identical speeds. However this is required for the curved paths to be "similar" in the strictest sense.

The distance that each missle travels is [spoiler] 100 Km [/spoiler][/QUOTE]

:smile:
You are quite right Wacky. My fault and omission of identical speeds

Mally :coffee:

orthogonal
 

[QUOTE=drew;94458]There's an interesting observation that can be made for the 'sqare' problem (not the general problem linked by axn1).

Each target's velocity is perpendicular to its pursuer's velocity, with no component from or towards the pursuer. Therefore, the distance is closed purely by the motion of the pursuer, which is directly toward the target. This explains why the path is exactly as long as the original side of the square.[/QUOTE]

:smile:
Fine observation Drew. For the square, the velocity does not matter as it cancels out.
The general problem of axni1 was excellent.
Mally :coffee:

I say that they never intersept each other. In my world the square plane that you describe slices through Mt. Fuji. Once they get close enough to each other, the mountain obscures their view of each other. They wind up,, going around the mountain until they run out of fuel.
:lol:

Just finding a little hole.

Little hole
 

[QUOTE=Uncwilly;94547]I say that they never intersept each other. In my world the square plane that you describe slices through Mt. Fuji. Once they get close enough to each other, the mountain obscures their view of each other. They wind up,, going around the mountain until they run out of fuel.
:lol:

Just finding a little hole.[/QUOTE]
:smile:
In your world Maybe!

I have flown several times over Mt. Fuji every time we took off from Tokyo and it never obscured our view as we flew well above it, unlike Mt Blanc, where our Boeing 707 in 1966, crashed into it, and not a trace of it is left.

I was flying then and was happy not to have been on that ill fated flight.

Well as I understand you, yours is a terrestial oriented world, and I can well understand your viewpoint. Try shifting it higher than mountains and view the stars above. They are very much there too.! :sad:

Mally :coffee:

Interesting observation
 

[QUOTE=drew;94458]There's an interesting observation that can be made for the 'sqare' problem (not the general problem linked by axn1).

Each target's velocity is perpendicular to its pursuer's velocity, with no component from or towards the pursuer. Therefore, the distance is closed purely by the motion of the pursuer, which is directly toward the target. This explains why the path is exactly as long as the original side of the square.[/QUOTE]
:smile:
Your observation is correct.

Now Drew Im interested in solving it non graphically. Could you help ?

Mally :coffee:

let x be the distance between two adjacent ants (missiles or whatever)

dx/dt = 0 throughout.

So x is the same as it started.

David