After S crosses the critical circle, the max angular speed of C is always greater than the max angular speed of S. In fact, critical circle is defined by this only - the circle below which the max angular velocity of S is greater than that of C (and vice versa above this circle). So, till S reaches this critical circle, C can just sit and take rest. But, after S crosses this circle, no matter what direction S or C choses, the angular separation between S and C will continuously decrease from pi (on the critical circle) to zero (when C finally catches S at the edge) So, just after crossing the critical radius, S starts swimming in some (randomly chosen) direction. Now, C has the choice of going in the same direction (ie. 'chasing', which is inefficient) OR can run in opposite direction (ie. 'meeting head on', which is better). But, as soon as S sees C running head on, he will immediately change his angular direction - following the same trajectory but in opposite angular direction. But, C should not change his direction again because, the angular separation between them has reduced from pi (from the critical circle) in the instantaneous time that all these happened so far. If C changes, he is just going to run for an angle>pi while S has gained radial distance towards the edge in this same instantaneous time. Now, your earlier zig-zag case is the one in which C changes his direction every time S changes his direction (which is in turn because he sees C running head on once again) - this way, C will oscillate at his initial position and S will also oscillate about his initial angular position but will gain radial advantage and will finally escape.Hence, for analytical purposes, we can safely 'assume' that S and C are moving in the same angular direction (both moving either clockwise or anticlockwise) from the time S gets out of his critical circle. From the above paragraph, we can say that C will not change his direction. And, since the relative angular velocity between them is minimum in this case, S will happily continue swimming in this (same) direction and not change his direction. None of these arguments, so far, require that the max speed of one be known to the other Now, set up the equations to derive the min ratio of speeds.Phew, that was long. And, be careful when you say that 'I used so-n-so principle from economics rather than mathematics to solve'. There are a lot of mathematicians lurking in this forum and they will pounce on you

Hi The Theorist,Oh I see there is also a point missing. My answer to this problem is "no definite answer to the min speed of R" as it depends on when R discoversthat he is fooled, i.e when he finds out chasing on course is the best. It also depends on the angle of the ziz-zag pattern S takes.All of these will determine how close S to the edge of the pool before R discovers that he is fooled. Then it determines the min speed of R required to catch S.My answer is not "the ultimate escape for S using the ziz-zag strategy". Of course if R cannot discover, S can escape. Your arguments say that R should not change direction again once the separation angle of decreases. There is no contradiction with what you wrote in bold. Msccube

QuoteIf he knows this important information, of course he knows that staying on course will be the best way as he knows that his angular speed is faster than that of the swimmer. Chaser (C) will definitely know that his angular speed is greater than the swimmer's (S) since S is outside the critical circle. A more relevant question that you can ask is 'How can C know that S has crossed the critical circle?'. As long as S is able to maintain C at an angle pi, C will know that S is within the critical circle. Even if S pretends by swimming at a slower speed, he cannot do that forever (else, he is just going to waste his potential). It is in his best interest to exit the true critical circle with an angular separation of pi. So, as soon as the false critical circle is reached, he will increase his speed till his max so that he exits the true critical circle with max separation. Also, this is how C will know S's max speed (although this is not required).You may ask 'why is it in S's best interest to exit the true critical circle with an angular separation of pi?'. If that is not so, then C will be closer to S on one side and obviously, any strategy that C uses should start by C choosing the closer side to run first. After this, even if S increases his speed to his max speed, it is definitely not as good as having C start from the max angle of pi.QuoteThen, he must bet! He will ask whether he should change direction to run a shorter distance.You should be careful when you say shorter distance. As soon as S exits the critical circle, irrespective of S's and C's direction, the angular separation between S and C will reduce from pi, in the direction in which C is running. So, if C continues to run in this same direction, he will have to cover a shorter (angular) distance, x (<pi). So, if he starts running in the opposite direction, he will actually have to cover a greater (angular) distance, 2pi-x (>pi)!! But then, it is the relative angular velocity that will be more (wc+ws) in the reversed C direction than in the original direction (wc-ws). Anyways, I addressed this just to give you the picture.Now, let me assume that C could not infer S's max speed and chooses to change his direction a few minutes after S exits the critical circle. Now, what prevents S from reversing his direction also? In fact, if S reverses now (and C has reversed already), it is even better for S because, C will try to catch S with the (once again) reduced angular velocity (wc-ws) but with a greater distance, 2pi-x (>pi) to cover. This can continue recursively until S reaches the edge.I was trying to say that not many would accept the statement '..used economics rather than math..' (while they might accept '..used music rather than math..'). It is like saying '..used physics instead of mathematics..'. Anyways, lets not digress here.Edit: I see that you have edited your earlier post. I will read that and reply later.

Quoteit depends on when R discovers that he is fooled, i.e when he finds out chasing on course is the best.Like I said in my previous post, irrespective of whether R (or C) knows S's max speed (and hence knowing when R crosses the critical circle), R should never change his direction outside the critical circle (can even say inside the critical circle, but that doesn't matter because S will anyways maintain R at pi). If he does change once (thereby deciding to cover >pi distance rather than the original <pi distance, but at greater relative velocity), S will also change once thereby making R cover the greater distance at the same relative angular velocity as before changing direction. This is pure wastage of time for R. If R changes twice, S will change twice..... The more number of times R changes, the better it is for S (who will also change the same number of times).

QuoteOriginally posted by: TheTheoristQuoteit depends on when R discovers that he is fooled, i.e when he finds out chasing on course is the best.Like I said in my previous post, irrespective of whether R (or C) knows S's max speed (and hence knowing when R crosses the critical circle), R should never change his direction outside the critical circle (can even say inside the critical circle, but that doesn't matter because S will anyways maintain R at pi). If he does change once (thereby deciding to cover >pi distance rather than the original <pi distance, but at greater relative velocity), S will also change once thereby making R cover the greater distance at the same relative angular velocity as before changing direction. This is pure wastage of time for R. If R changes twice, S will change twice..... The more number of times R changes, the better it is for S (who will also change the same number of times).HiIf the speed of the runner is less than 4.6 times of the speed of the swimmer, can the runner still catch the swimmer once the swimmerchanges his direction outside the critical circle, and the runner does not change his direction?I think that I have explained enough.Msccube

I do not know how you get this number 4.6 (I never got to know since the time it was quoted early on, in this thread)Since my argument has been that R is forced to run 'behind' (ie, same angular direction as) S, and if S changes his direction and comes head-on (which is suicidal), there is no reason why R should change his direction - he is only going to catch S faster!If R starts out by running 'head-on' (ie, opposite angular direction), ONLY THEN will S change his direction so that R comes 'behind' him. After this, S will not change his direction (as per my arguments in first paragraph).And from my arguments in my recent posts, you should be able to see why R will not change his direction even if he is behind S (because, S will change his direction again and have R behind him once again. The cost: R will now end up having to cover 2pi-x which is greater than the original angle of separation, x, and at the same relative angular velocity).If you wish to continue your argument further, please (also) quote specific lines from my recent posts and prove them wrong rather than posting your arguments alone. Also, please give a sketchy calculation of whatever number you post.

QuoteOriginally posted by: TheTheoristI do not know how you get this number 4.6 (I never got to know since the time it was quoted early on, in this thread)Since my argument has been that R is forced to run 'behind' (ie, same angular direction as) S, and if S changes his direction and comes head-on (which is suicidal), there is no reason why R should change his direction - he is only going to catch S faster!If R starts out by running 'head-on' (ie, opposite angular direction), ONLY THEN will S change his direction so that R comes 'behind' him. After this, S will not change his direction (as per my arguments in first paragraph).And from my arguments in my recent posts, you should be able to see why R will not change his direction even if he is behind S (because, S will change his direction again and have R behind him once again. The cost: R will now end up having to cover 2pi-x which is greater than the original angle of separation, x, and at the same relative angular velocity).If you wish to continue your argument further, please (also) quote specific lines from my recent posts and prove them wrong rather than posting your arguments alone. Also, please give a sketchy calculation of whatever number you post.Hi The value of x= 4.6 was obtained at the very early stage of the thread (page one). If you understand how to get x = 4.6, you will understand my arguments.Msccube

umm, I do not see any supporting argument for the post that had introduced 4.6. For your info, I am quoting that post below...QuoteIt is not so difficult to come up with a strategy that would allow the swimmer to get away if the ratio (pursuer's speed) / (swimmer's speed) = 4 (or, say, ).As for finding the minimum ratio ensuring the pursuer catches the swimmer, I think it is a difficult problem to solve, let alone explain here without detailed drawings, etc. I think the answer would be something like the root of the equation.I would say, Subsequent posts that asked for supporting calculations were in vain. Last I knew, the equation and the ratio (4.6) was lifted from some paper that did not give any convincing arguments either. So, could you please explain how YOU arrived at 4.6 and how it is related to your argument (which, as far as I could understand, concludes that there can be no way of calculating the ratio)And again, if you think that my argument or logic is false at any step, please quote that line and present your specific counter argument.

HiI am very busy for the time being. Please give me sometime and I will reply to you later.It will need to write a long article to illustrate the point. Your arguments have some problem. I will explain in my reply later.Msccube

QuoteIt will need to write a long article to illustrate the point.Well, please do that. As for me, I have come to a point where it is nearly impossible to explain my argument in any different way than how it has been explained in my recent posts. So, I would expect any counter-argument to quote and address specific lines from any (or all) of my recent posts and not just the last ones.

Hi The theorist,Sorry for replying late. I was very busy.The argument of x = 4.6 has been posted by me (Msccube) on Aug 02, 06. I repeated part of my post here.What happens if the runner does not change his course at the first time that the swimmer changes his direction. That is the runner not buying the ziz-zag game. The runner will run nearly 270 degrees depending on the angle of the direction of the swimmer to AB. Let us keep the problem simple. We assume that the swimmer swims at 90 degrees to AB to the first order of approximation.The distance that the swimmer must swim to reach the edge is where x = v/s.The distance that the runner must run to catch the swimmer is Equating the time for the swimmer to travel and the runner to travel, , we getSolving this equation, we get x= 4.6.So if the speed of R is less than x= 4.6 times of the swimmer, R cannot catch S if R does not change course and S continues to swim in this direction.QuoteOriginally posted by: TheTheoristChaser (C) will definitely know that his angular speed is greater than the swimmer's (S) since S is outside the critical circle. A more relevant question that you can ask is 'How can C know that S has crossed the critical circle?'. As long as S is able to maintain C at an angle pi, C will know that S is within the critical circle. Even if S pretends by swimming at a slower speed, he cannot do that forever (else, he is just going to waste his potential). It is in his best interest to exit the true critical circle with an angular separation of pi. So, as soon as the false critical circle is reached, he will increase his speed till his max so that he exits the true critical circle with max separation. Also, this is how C will know S's max speed (although this is not required).I am afraid that it is not right. If S is within the critical circle. S can swim at a speed less than his true critical speed. S can start the ziz-zag strategy any time he likes. Therefore, R (C) cannot know whether S has been swimming at the true critical circle. So R cannot know S's true max speed.QuoteOriginally posted by: TheTheoristYou should be careful when you say shorter distance. As soon as S exits the critical circle, irrespective of S's and C's direction, the angular separation between S and C will reduce from pi, in the direction in which C is running. So, if C continues to run in this same direction, he will have to cover a shorter (angular) distance, x (<pi). So, if he starts running in the opposite direction, he will actually have to cover a greater (angular) distance, 2pi-x (>pi)!! But then, it is the relative angular velocity that will be more (wc+ws) in the reversed C direction than in the original direction (wc-ws). Anyways, I addressed this just to give you the picture.Now, let me assume that C could not infer S's max speed and chooses to change his direction a few minutes after S exits the critical circle. Now, what prevents S from reversing his direction also? In fact, if S reverses now (and C has reversed already), it is even better for S because, C will try to catch S with the (once again) reduced angular velocity (wc-ws) but with a greater distance, 2pi-x (>pi) to cover. This can continue recursively until S reaches the edge.I am sorry that this is wrong again! In this case, you cannot think in term of relative angular separation. I have shown why we need x = 4.6 at least!The answer is on the first part of this post. You also miss the point whether R has enough time to decrease the relative angular separation to zero before S reaches the edge. The logic of my argument on the first part of this post should explain why you are wrong.PM me if you have any question. I do not think that we should repeat the argument again and again on the post.QuoteOriginally posted by: TheTheoristLike I said in my previous post, irrespective of whether R (or C) knows S's max speed (and hence knowing when R crosses the critical circle), R should never change his direction outside the critical circle (can even say inside the critical circle, but that doesn't matter because S will anyways maintain R at pi). If he does change once (thereby deciding to cover >pi distance rather than the original <pi distance, but at greater relative velocity), S will also change once thereby making R cover the greater distance at the same relative angular velocity as before changing direction. This is pure wastage of time for R. If R changes twice, S will change twice..... The more number of times R changes, the better it is for S (who will also change the same number of times).It is wrong again! If the speed of R is less than x= 4.6, and R does not change direction, then after S changes direction and S continues to swim in that new direction, R cannot catch S. The first part of this post has shown the point.Therefore, if R does not know S's max speed, R must gamble!I have no intention to offense any mathematician here. It is nothing wrong to apply new concept on any question. To me, knowledge is knowledge. The division into physics, math, musics, philosophy is only by human beings themselves. Applying different areas of knowledge may lead to new insight and new discovery. Therefore, I cannot see why it is a problem to apply economic concept to a math problem. Decades ago, someone applied principles of musics (violation of spring) to physics and math, and they led us to quantum mechanics.