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        riddles >> medium >> U-boat
        
(Message started by: ThudanBlunder on Sep 5th, 2009, 9:07am)
Title: U-boat
Post by ThudanBlunder on Sep 5th, 2009, 9:07am
An Allied destroyer is at the origin of the Cartesian plane and moving with constant speed V(a). A stationary U-boat, carrying a full complement of torpedoes, is lying in wait somewhere along the positive x-axis at (p,0). The speed of its torpedoes is kV(a), (where k < 1). The U-boat captain knows that he will get only one chance to sink his quarry. What is the heading of the destroyer?
Title: Re: U-boat
Post by Ronno on Sep 6th, 2009, 12:36am
<=arctan(k) with the x axis?
Title: Re: U-boat
Post by ThudanBlunder on Sep 6th, 2009, 3:02am

on 09/06/09 at 00:36:08, Ronno wrote:
<=arctan(k) with the x axis?

Nope. There are two possible answers.
But you seem to be on the right track, so to speak.
Anyway, I was hoping for a solution, not an answer.  :P
Title: Re: U-boat
Post by Eigenray on Sep 8th, 2009, 2:40pm
Let the velocity of the destroyer be v = (vx, vy), and the torpedo be w = (wx, wy).  If vy = 0, there is no chance if vx < 0, or many chances if vx > 0.  So WLOG suppose vy > 0.

First, by time considerations, it's necessary that
(1) [hide]wy http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif vy[/hide],
and by space considerations,
(2) [hide]c = vx wy - vy wx > 0, i.e., w is counterclockwise from v[/hide].

Conversely, these are sufficient because, if t0 is the launch time, and t is the intersection time, we can solve
[hideb]
(vx t, vy t) = (p + wx (t-t0), wy (t-t0) )
t0 = p(wy - vy) / c http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 0
t-t0 = p vy / c > 0
[/hideb]

Let v = |V| (cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif, sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif), w = k|V| (cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif, sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif), where WLOG http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif are both in (0,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif).  Then the conditions become
(1) k sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif,
(2) sin(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif- http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif) > 0, i.e., http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif > http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif, if both are in (0,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif).
Condition (1) implies that sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif k < 1, and that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif lies in a neighborhood of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2 that doesn't contain http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif.  Therefore given condition (1), condition (2) is equivalent to vx > 0.  Therefore, with v fixed, there are:
-no solutions if sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif > k or cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif < 0.
-one solution if sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif = k, cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif > 0
-many solutions if sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif < k, cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif > 0.

So the destroyer makes an angle of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifarcsin(k) with the positive x-axis, where 0 < arcsin(k) < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2.  (If k=0 there is either no chance or many chances.)
Title: Re: U-boat
Post by ThudanBlunder on Sep 9th, 2009, 6:29am
A nice generic approach, Eigenray.

Given the heading of the destroyer and the initial positions and the relative speeds of the destroyer and torpedo, we can generate the set of points where interception is possible. This turns out to be an Apollonius circle.

Let the destroyer be at (0,0), the torpedo be at (p,0), and the speed of the torpedo be k times the speed of the destroyer.

Then
when k > 1 the centre the circle is at (-p/(k2 - 1), 0) and its radius is kp/(k2 - 1)
and
when k < 1 the centre the circle is at (p/(k2 - 1), 0) and its radius is kp/(1 - k2)

Also,
when k = 1 we have a straight line (circle of infinite radius), equidistant from destroyer and torpedo
and
when k  > 1 the circle encloses the destroyer and this implies that, for a given heading of the destroyer, there is only one interception point
and
when k < 1 the circle encloses the torpedo and this implies that, for a given heading of the destroyer, there are (in general) two interception points! (I wonder if this fact was ever used in practice.) Of course, in this puzzle (which I have adapted) there is only one interception point in each quadrant because the heading of the destroyer is a tangent to the circle.


on 09/08/09 at 14:40:04, Eigenray wrote:
(If k=0 there is either no chance or many chances.)

How can there be many chances? Do you mean the destroyer can ram the sub at different speeds?  ::)
Title: Re: U-boat
Post by Eigenray on Sep 9th, 2009, 12:31pm

on 09/09/09 at 06:29:49, ThudanBlunder wrote:
How can there be many chances? Do you mean the destroyer can ram the sub at different speeds?  ::)

I was just pointing out that there is no solution for k = 0 : if the destroyer rams the sub the sub can 'launch' its torpedo at any time before the collision; otherwise, it can't hit it at all.  But for 0 < k < 1 there are two solutions.


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