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# An "Oldie" with a twist of lemon

Best Answer Prime, 22 February 2013 - 02:08 AM

I have a cold, my head is stuffed, I feel like an Ogre. I am not figuring out the differential equations to deny hungry Ogre his meal. However, here is some food for thought for those Brain Denizens who took the maiden's side. Indeed, she could go slower. Mayhap, someone will try carrying out the required calculations, make an error and give the maiden a wrong advice.

Spoiler for maiden voyage

To make the contest fair, the maiden should have only so much time for rowing. Having exhausted it, she falls asleep. Whereafter, the Ogre should start blowing at the boat pushing it towards the shore...

Go to the full post

49 replies to this topic

### #41 bonanova

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Posted 24 February 2013 - 09:35 AM

Consider  my diagram for straight line tangential escape.

From the red dot [landing point] move clockwise a small angle DELTA a and draw a blue dot.

Call the shore length between the dots h.

Draw an arc centered on the rower's starting point clockwise from the red dot [radius is y, the rower's distance to the red dot.]

Call this arc s.

Now draw a line from the blue dot back to the rower's starting position.

The distance from the blue dot on the shore to the intersection with arc s is the change DELTA y of the rower's distance to the blue dot instead of the red dot.

The red dot, blue dot and intersection point form a right triangle whose hypotenuse is h = R.DELTA a and short side is DELTA y and included angle is a.

Thus cos(a) = DELTA y / R DELTA a

Letting DELTA a go to zero, with have the rate of increase path for the rower:

dy/da = R cos(a)

Call the ogre's path O. His path increases by h, the shore length between the red and blue dots: DELTA O = R DELTA a.

Again, in the limit we have

dO/da = R

So with respect to the angle a, the rower's and ogre's paths increase in a ratio of cos(a).

Recall that cos(a) = 1/f = the speed ratio of ogre and rower.

Thus, the difference in arrival time for the two racers with respect to angle is zero at the red dot.

This means the rower's advantage over the ogre is either a maximum or a minimum at a.

Note that the triangle we drew actually has curved sides, so the equality is only strict as DELTA a goes to zero.

They curve away from each other, so that DELTA y increases super linearly, while DELTA O increases linearly with DELTA a.

Thus, for angles greater than a, the ogre gains an advantage.

The rower's advantage is thus a maximum.

Conclusion: the red dot is the optimal landing point, and a straight line is the shortest distance to it.

Q.E.D.

• 0

Vidi vici veni.

### #42 CaptainEd

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Posted 26 February 2013 - 04:34 PM

This is a good clear thorough analysis of the tangential escape--a better linear escape than the straight-ahead escape or my wait-until-Ogre-has-committed-22-degrees escape.

I'm sorry to be beating a dead horse, but I would now like to revisit the issue of the tacking duel. In particular, if we suppose that in the very first second of time, Ogre moves CW and Maiden moves at angle a, as determined above. Now suppose that in the next second, Ogre reverses direction ("tacks") and steps back to O's start point, moving CCW.

(1) What is M's move? Does she do that recalculation you mentioned? If so, it's a different calculation, as she is now no longer directly opposite Ogre, and (in a large enough pond) hasn't gotten to a point where she has a guaranteed straightline solution.

(2) If she does not recalculate, I predict the Ogre will continue moving CCW. WHEN will M do her recalculation, and what will it look like then?

• 0

### #43 Prime

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Posted 27 February 2013 - 12:27 AM

This is a good clear thorough analysis of the tangential escape--a better linear escape than the straight-ahead escape or my wait-until-Ogre-has-committed-22-degrees escape.

I'm sorry to be beating a dead horse, but I would now like to revisit the issue of the tacking duel. In particular, if we suppose that in the very first second of time, Ogre moves CW and Maiden moves at angle a, as determined above. Now suppose that in the next second, Ogre reverses direction ("tacks") and steps back to O's start point, moving CCW.

(1) What is M's move? Does she do that recalculation you mentioned? If so, it's a different calculation, as she is now no longer directly opposite Ogre, and (in a large enough pond) hasn't gotten to a point where she has a guaranteed straightline solution.

(2) If she does not recalculate, I predict the Ogre will continue moving CCW. WHEN will M do her recalculation, and what will it look like then?

I see the makings of a computer game Ogre vs. Maiden.

Since both Ogre and Rower can turn on a dime and have infinite acceleration, any history of their moves means nothing. The Ogre's strategy is to spot the nearest landing point for the boat and run there along the shorter arch.

However, the maiden must be given some small advantage in the speed ratio. Ogre knows that, and he is also aware that the young lady has read our blog and knows all about the optimal path. So the Ogre deviates from his optimal strategy, giving his opponent an additional advantage. But if she is so dumb as to continue on the path she started, she'll run straight into the Ogre's paws.

• 0

Past prime, actually.

### #44 CaptainEd

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Posted 27 February 2013 - 04:04 AM

If I read the picture properly, the Ogre can walk all around the lake in 8 steps. I agree that O would be foolish to tack. I was imagining a much larger lake. Sorry to disturb you guys.
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### #45 Prime

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Posted 27 February 2013 - 07:04 AM

If I read the picture properly, the Ogre can walk all around the lake in 8 steps. I agree that O would be foolish to tack. I was imagining a much larger lake. Sorry to disturb you guys.

I am not disturbed. Your question points to an essential unsolved part of the problem. How should rower recalculate her path if Ogre deviates from optimal.

I used the diagram to explain the idea. The picture is not to scale and very scetchy. I used large steps to make it more illustrative. The size of the lake does not matter, only the ratio of speeds does. The same concept applies if it takes the Ogre thousands of steps to run around the lake.

Anyone can still work on this problem to find more efficient and creative ways for recalculation of rowers path. As well as finding how long it takes to escape, and or how far away the Ogre would end up at the landing.

• 0

Past prime, actually.

### #46 bonanova

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Posted 27 February 2013 - 07:42 AM

I have finished a rigorous proof that the tangent to the inner circle is the

best path, straight or curved, that permits the girl and boat to reach shore

safely, with the lowest speed ratio compared with ogre, namely 0.2174,

or 1/4.6003.

It shows also that reversing direction is not part of either Ogre's or

the girl's best strategy.

Does it cover all the bases?

• 0

Vidi vici veni.

### #47 Prime

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Posted 02 March 2013 - 05:01 AM

I have finished a rigorous proof that the tangent to the inner circle is the

best path, straight or curved, that permits the girl and boat to reach shore

safely, with the lowest speed ratio compared with ogre, namely 0.2174,

or 1/4.6003.

It shows also that reversing direction is not part of either Ogre's or

the girl's best strategy.

Does it cover all the bases?

Looks like the maiden gets an unfair edge here. The calculated rate does not give the Ogre his chance, unless the lake is really small and Ogre's hands are really long.

• 0

Past prime, actually.

### #48 bonanova

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Posted 02 March 2013 - 10:14 AM

• 0

Vidi vici veni.

### #49 Prime

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Posted 03 March 2013 - 06:15 AM

So the Ogre and the Maiden problem is actually a solution/approximation method for tan(x) = x + π.

Spoiler for finer points and formula for f.

Edited by Prime, 03 March 2013 - 06:16 AM.

• 0

Past prime, actually.

### #50 bonanova

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Posted 03 March 2013 - 06:48 AM

So the Ogre and the Maiden problem is actually a solution/approximation method for tan(x) = x + π.
Spoiler for finer points and formula for f.

So on top of your last graph you would plot f = cos a and get both f and a at the intersection. Very cool.
Using the other equations you would intersect tan a with a + pi to get a, then take cos to get f. An extra step.
• 0

Vidi vici veni.

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