49 replies to this topic

[bonanova]

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.

[CaptainEd]

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?

[Prime]

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.

 

Spoiler for Maiden adjustments

Suppose, we gave the maiden a small advantage to ensure she has time to jump out of the boat before the Ogre grabs her, say, instead of our calculated 4.6033 factor, the beast can move at his best only 4.5 times faster than the rower.

Let's add another dimension to the game – limit rower's time. After so many game steps, Ogre's wife, a good swimmer, emerges from her underwater cave and grabs the young woman, whereupon the Ogre couple can have a romantic dinner.

 

Let's work on CaptainEd's suggestion. The wily Ogre deviates from the optimal path and on his second step goes back to the initial position. The maiden has gained some advantage, but she must recalculate her path. Not finding any clear instructions on how to do that, she may be lost.

 

We need not bother with Ogre's traveled distance – just the angle he has to travel to the nearest boat landing point.

The Ogre made his first step from O₀ to O₁ . The rower goes from M₀ to M₁. At this point the Ogre's remaining angle clockwise (O₁ to L₁ ) is less than π (half circle.) But he goes back to O₀ to confuse the young lady. At this juncture, the maiden could draw a line from the Ogre to his opposite point on the shore, then another line to L₂ at the same angular difference as it was after the first step, and step onto that line using her top speed. (See the diagram.)

Thus after two steps, the Ogre only has reduced the angular gap the same amount as after the first step. Whereas the maiden is closer to the shore than she was after her first step.

 

In such way, the rower clearly grabs the advantage afforded by Ogre. Whether the method is optimal for the rower depends on the objective: less steps, or increased distance from Ogre at the landing.

That smallest angular gap may not be maintained on consecutive steps along with advancement to the shore. What we could do is to calculate angular gap for each step in the optimal path and then maintain it in the zig-zag mode. Then the maiden should be able to land while the Ogre would remain pretty much at the opposite side of the lake.

 

 

[CaptainEd]

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.

[Prime]

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.

[bonanova]

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.

 

 ogre1.doc   837KB   31 downloads

 

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?

[Prime]

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.

 

ogre1.doc

 

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.

 

Spoiler for comments

That is some serious analysis. I liked the (b – a) angle, but somehow I don't see that it's been successfully used as a proof of the concept.

Page 3: “Clearly the blue landing point is advantageous to boat.”
Is the diagram an accurate scale model? If so, the above statement represents a numeric test, not an analytical solution.

Page 4: The angle between brown and black arcs is indeed b – a. However, if you build a triangle with vertexes at the gray and blue points and the intersection of brown arc with blue segment, the corresponding angle seems different, and the triangle is not a right-angled triangle. I can't even decide whether sin(b-a) is greater or smaller than the actual Δp/ΔO ratio.

Page 6: I guess, cos(a) = y/R = f is a typo. It should be cos(a) = r/R = 1/f.
Plugging in the value a = 77.445˚ ~ 1.35167 radians, yields f ~ 4.6003.

Now let's do an alternative calculation for f. Using diagram on page 5:
y = R*sin(a), that is the distance traveled by the boat.
The Ogre has traveled (π + a)*R. Thus Ogre:Maiden ratio = (π + a)/sin(a). Plugging in the same value for a, we get f ~ 4.6033, which is drastically different from 4.6003 – life vs. death for the maiden.
(Incidentally, the same result I got from evaluation of my function given inside the spoiler in post #34.)

I don't really know how to solve tan(a) = π + a, but for the purposes of our problem a ~ 77.4534˚ fits the bill better.

 

There is another typo, I guess, sin(a-b) = f should be sin(b-a) = 1/f. (f being Ogre/Maiden ratio.) Is that a true formula for the f instead of the one I've given in post #34? Do those formulas amount to the same thing? Note, b and a are dependent variables. a = function(b), or vica versa.

As for the “spiral escape paths,” I could not really follow the argument in point (3) and did not see the need for the argument in (2). I believe, the argument in (1) is sufficient. Apparently, I failed to explain clearly my point in the post #39 and the function in the post #34.

[bonanova]

Spoiler for Thanks for reading and for the comments

That is some serious analysis. I liked the (b – a) angle, but somehow I don't see that it's been successfully used as a proof of the concept.

Fair enough, if you don't see it. I'm not sure what it left undone.

Page 3: “Clearly the blue landing point is advantageous to boat.”
Is the diagram an accurate scale model? If so, the above statement represents a numeric test, not an analytical solution.

Accurate scale:
We both understand that the radius of the inner circle depends on f.
So if you mean exactly the right inner radius, then no. It's approximate.
If you mean are the circles and lines accurately drawn, then yes. They are.

Analytical solution:
My intent is a proof that pi/2 is the optimal bearing.
That is an adequate treatment of the straight-line escape path problem.
Bearing is the only variable.

 

In a nutshell,

What I showed was that for a straight line escape at any bearing 0 < b < pi/2 it is advantageous to increase b.
That gets us to the maximum allowable b = pi/2 without re-entering the inner circle, and re-starting the race.

Page 4: The angle between brown and black arcs is indeed b – a. However, if you build a triangle with vertexes at the gray and blue points and the intersection of brown arc with blue segment, the corresponding angle seems different, and the triangle is not a right-angled triangle. I can't even decide whether sin(b-a) is greater or smaller than the actual Δp/ΔO ratio.

Of course we are talking about derivatives, not deltas, and tangents, not arcs.
To draw a triangle with infinitesimal sides is not possible.
The blue line is drawn from the center of the brown arc.
A radius is always perpendicular to its circle.

Page 6: I guess, cos(a) = y/R = f is a typo. It should be cos(a) = r/R = 1/f.
Plugging in the value a = 77.445˚ ~ 1.35167 radians, yields f ~ 4.6003.

You are correct, that is a typo.
Also, when I did the analysis I used one definition of f and when I wrote it up I changed to the other.

So an early version of the document was inconsistent on that point.
The version of the document currently attached to the post I believe is consistent.
Throughout, I referred to the boat relative to the Ogre.
So f as the speed ratio is defined to be < unity.

These equations are the ones that use f, and I think they are consistent:

f = sb/so < 1
re = fR
f = 1/(1+pi)
f delta tb/delta to = delta p/delta O .
sin (a-b) = f
y = fR (pi + a) = r (pi + a)
cos (a) = f.
f = 0.24174 = 1/4.6003 - with corrected values.

Now let's do an alternative calculation for f. Using diagram on page 5:
y = R*sin(a), that is the distance traveled by the boat.
The Ogre has traveled (π + a)*R. Thus Ogre:Maiden ratio = (π + a)/sin(a). Plugging in the same value for a, we get f ~ 4.6033, which is drastically different from 4.6003 – life vs. death for the maiden.
(Incidentally, the same result I got from evaluation of my function given inside the spoiler in post #34.)

I don't really know how to solve tan(a) = π + a, but for the purposes of our problem a ~ 77.4534˚ fits the bill better.

You solve it iteratively - Newton's method works.

Actually I think this was another typo - 77.445 instead of 77.455.
But seriously a better solution is 77.453397562356507 deg
That gives the speed ratio as 0.2172336282 = 1/4.603338849

There is another typo, I guess, sin(a-b) = f should be sin(b-a) = 1/f. (f being Ogre/Maiden ratio.) Is that a true formula for the f instead of the one I've given in post #34? Do those formulas amount to the same thing? Note, b and a are dependent variables. a = function(b), or vica versa.

I define f as maiden/ogre speed < 1.

If by true formula you mean accurate, or valid, then yes it is..

As for the “spiral escape paths,” I could not really follow the argument in point (3) and did not see the need for the argument in (2). I believe, the argument in (1) is sufficient. Apparently, I failed to explain clearly my point in the post #39 and the function in the post #34.

I guess it's sufficient to say a straight line to the landing point is optimal.
Statements 2 and 3 were intended simply to put some of the equations into words.

In effect, statement 3 says you want to intersect the shoreline closer to perpendicular than to parallel.

A spiral path does just the opposite.

OK?

 

[Prime]

Spoiler for Thanks for reading and for the comments

That is some serious analysis. I liked the (b – a) angle, but somehow I don't see that it's been successfully used as a proof of the concept.

Fair enough, if you don't see it. I'm not sure what it left undone.

Page 3: “Clearly the blue landing point is advantageous to boat.”
Is the diagram an accurate scale model? If so, the above statement represents a numeric test, not an analytical solution.

Accurate scale:
We both understand that the radius of the inner circle depends on f.
So if you mean exactly the right inner radius, then no. It's approximate.
If you mean are the circles and lines accurately drawn, then yes. They are.

Analytical solution:
My intent is a proof that pi/2 is the optimal bearing.
That is an adequate treatment of the straight-line escape path problem.
Bearing is the only variable.

 

In a nutshell,

What I showed was that for a straight line escape at any bearing 0 < b < pi/2 it is advantageous to increase b.
That gets us to the maximum allowable b = pi/2 without re-entering the inner circle, and re-starting the race.

Page 4: The angle between brown and black arcs is indeed b – a. However, if you build a triangle with vertexes at the gray and blue points and the intersection of brown arc with blue segment, the corresponding angle seems different, and the triangle is not a right-angled triangle. I can't even decide whether sin(b-a) is greater or smaller than the actual Δp/ΔO ratio.

Of course we are talking about derivatives, not deltas, and tangents, not arcs.
To draw a triangle with infinitesimal sides is not possible.
The blue line is drawn from the center of the brown arc.
A radius is always perpendicular to its circle.

Page 6: I guess, cos(a) = y/R = f is a typo. It should be cos(a) = r/R = 1/f.
Plugging in the value a = 77.445˚ ~ 1.35167 radians, yields f ~ 4.6003.

You are correct, that is a typo.
Also, when I did the analysis I used one definition of f and when I wrote it up I changed to the other.

So an early version of the document was inconsistent on that point.
The version of the document currently attached to the post I believe is consistent.
Throughout, I referred to the boat relative to the Ogre.
So f as the speed ratio is defined to be < unity.

These equations are the ones that use f, and I think they are consistent:

f = sb/so < 1
re = fR
f = 1/(1+pi)
f delta tb/delta to = delta p/delta O .
sin (a-b) = f
y = fR (pi + a) = r (pi + a)
cos (a) = f.
f = 0.24174 = 1/4.6003 - with corrected values.

Now let's do an alternative calculation for f. Using diagram on page 5:
y = R*sin(a), that is the distance traveled by the boat.
The Ogre has traveled (π + a)*R. Thus Ogre:Maiden ratio = (π + a)/sin(a). Plugging in the same value for a, we get f ~ 4.6033, which is drastically different from 4.6003 – life vs. death for the maiden.
(Incidentally, the same result I got from evaluation of my function given inside the spoiler in post #34.)

I don't really know how to solve tan(a) = π + a, but for the purposes of our problem a ~ 77.4534˚ fits the bill better.

You solve it iteratively - Newton's method works.

Actually I think this was another typo - 77.445 instead of 77.455.
But seriously a better solution is 77.453397562356507 deg
That gives the speed ratio as 0.2172336282 = 1/4.603338849

There is another typo, I guess, sin(a-b) = f should be sin(b-a) = 1/f. (f being Ogre/Maiden ratio.) Is that a true formula for the f instead of the one I've given in post #34? Do those formulas amount to the same thing? Note, b and a are dependent variables. a = function(b), or vica versa.

I define f as maiden/ogre speed < 1.

If by true formula you mean accurate, or valid, then yes it is..

As for the “spiral escape paths,” I could not really follow the argument in point (3) and did not see the need for the argument in (2). I believe, the argument in (1) is sufficient. Apparently, I failed to explain clearly my point in the post #39 and the function in the post #34.

I guess it's sufficient to say a straight line to the landing point is optimal.
Statements 2 and 3 were intended simply to put some of the equations into words.

In effect, statement 3 says you want to intersect the shoreline closer to perpendicular than to parallel.

A spiral path does just the opposite.

OK?

 

 

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.

I worked off the initial version of your .doc file (opened it when you made the post, but got around to it couple days later.)
Still, the typo remains: sin(a-b), where you meant sin(b-a) (page 6.) Sorry for nitpicking.

It seems my actual formula for Ogre/Maiden speed rate is being summarily rejected here. I don't see why.

Here is an addendum to put some finer points on your sin(b-a) analysis and to reiterate the actual formula for f:

 

 Ogre1.pdf   96.72KB   37 downloads

 

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

[bonanova]

Spoiler for Thanks for reading and for the comments

That is some serious analysis. I liked the (b – a) angle, but somehow I don't see that it's been successfully used as a proof of the concept.

Fair enough, if you don't see it. I'm not sure what it left undone.

Page 3: “Clearly the blue landing point is advantageous to boat.”
Is the diagram an accurate scale model? If so, the above statement represents a numeric test, not an analytical solution.

Accurate scale:
We both understand that the radius of the inner circle depends on f.
So if you mean exactly the right inner radius, then no. It's approximate.
If you mean are the circles and lines accurately drawn, then yes. They are.

Analytical solution:
My intent is a proof that pi/2 is the optimal bearing.
That is an adequate treatment of the straight-line escape path problem.
Bearing is the only variable. In a nutshell, What I showed was that for a straight line escape at any bearing 0 < b < pi/2 it is advantageous to increase b.
That gets us to the maximum allowable b = pi/2 without re-entering the inner circle, and re-starting the race.

Page 4: The angle between brown and black arcs is indeed b – a. However, if you build a triangle with vertexes at the gray and blue points and the intersection of brown arc with blue segment, the corresponding angle seems different, and the triangle is not a right-angled triangle. I can't even decide whether sin(b-a) is greater or smaller than the actual Δp/ΔO ratio.

Of course we are talking about derivatives, not deltas, and tangents, not arcs.
To draw a triangle with infinitesimal sides is not possible.
The blue line is drawn from the center of the brown arc.
A radius is always perpendicular to its circle.

Page 6: I guess, cos(a) = y/R = f is a typo. It should be cos(a) = r/R = 1/f.
Plugging in the value a = 77.445˚ ~ 1.35167 radians, yields f ~ 4.6003.

You are correct, that is a typo.
Also, when I did the analysis I used one definition of f and when I wrote it up I changed to the other.So an early version of the document was inconsistent on that point.
The version of the document currently attached to the post I believe is consistent.
Throughout, I referred to the boat relative to the Ogre.
So f as the speed ratio is defined to be < unity.

These equations are the ones that use f, and I think they are consistent:

f = sb/so < 1
re = fR
f = 1/(1+pi)
f delta tb/delta to = delta p/delta O .
sin (a-b) = f
y = fR (pi + a) = r (pi + a)
cos (a) = f.
f = 0.24174 = 1/4.6003 - with corrected values.

Now let's do an alternative calculation for f. Using diagram on page 5:
y = R*sin(a), that is the distance traveled by the boat.
The Ogre has traveled (π + a)*R. Thus Ogre:Maiden ratio = (π + a)/sin(a). Plugging in the same value for a, we get f ~ 4.6033, which is drastically different from 4.6003 – life vs. death for the maiden.
(Incidentally, the same result I got from evaluation of my function given inside the spoiler in post #34.)

I don't really know how to solve tan(a) = π + a, but for the purposes of our problem a ~ 77.4534˚ fits the bill better.

You solve it iteratively - Newton's method works.
Actually I think this was another typo - 77.445 instead of 77.455.
But seriously a better solution is 77.453397562356507 deg
That gives the speed ratio as 0.2172336282 = 1/4.603338849

There is another typo, I guess, sin(a-b) = f should be sin(b-a) = 1/f. (f being Ogre/Maiden ratio.) Is that a true formula for the f instead of the one I've given in post #34? Do those formulas amount to the same thing? Note, b and a are dependent variables. a = function(b), or vica versa.

I define f as maiden/ogre speed < 1.If by true formula you mean accurate, or valid, then yes it is..

As for the “spiral escape paths,” I could not really follow the argument in point (3) and did not see the need for the argument in (2). I believe, the argument in (1) is sufficient. Apparently, I failed to explain clearly my point in the post #39 and the function in the post #34.

I guess it's sufficient to say a straight line to the landing point is optimal.
Statements 2 and 3 were intended simply to put some of the equations into words.In effect, statement 3 says you want to intersect the shoreline closer to perpendicular than to parallel.A spiral path does just the opposite.

OK?

 

 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.

I worked off the initial version of your .doc file (opened it when you made the post, but got around to it couple days later.)
Still, the typo remains: sin(a-b), where you meant sin(b-a) (page 6.) Sorry for nitpicking.It seems my actual formula for Ogre/Maiden speed rate is being summarily rejected here. I don't see why.

Here is an addendum to put some finer points on your sin(b-a) analysis and to reiterate the actual formula for f: Ogre1.pdf 

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.