BrainDen.com - Brain Teasers
• 0

# An "Oldie" with a twist of lemon

## Question

An ogre with a top land speed of 4 mph, but incapable of swimming, stalks the shore of a circular lake. The fair maiden out in the boat can outrun the ogre on land, but her rowing speed is nothing to write home to mother about. How fast must she be able to row to effect an escape?

Make the assumption that the ogre is hungry [or otherwise desirous of the girl] and uses a best-strategy pursuit.

## 49 answers to this question

• 0

so you mean the Ogre looks South, and the Maiden takes off to the North, and the Ogre (a Zen master apparently) sees the inevitability, and continues the long way around, like Mark Twain's "Cooper Indian". If the Ogre does turn and go north, he can go four times further north than M can. Exactly when should M recalculate her new tangent? Will it really require as long a road for Ogre?

##### Share on other sites
• 0

so you mean the Ogre looks South, and the Maiden takes off to the North, and the Ogre (a Zen master apparently) sees the inevitability, and continues the long way around, like Mark Twain's "Cooper Indian". If the Ogre does turn and go north, he can go four times further north than M can. Exactly when should M recalculate her new tangent? Will it really require as long a road for Ogre?

It is a lot simpler than that for the zen master Ogre. The Ogre ignores where the maiden looks, where she goes, and where she intends to go. Ogre simply draws an imaginary straight line from the center of the lake, through the boat, to the shore; and heads for that spot using the shorter arch. Insofar as the optimum is concerned, it is the Ogre who determines the path -- not the maiden. The maiden reacts to Ogre's moves, not other way around.

If the maiden wanders back into her inner circle r, that could make the Ogre change the direction when the boat crosses straight line drawn from the Ogre to the center of the lake. Then the young lady would be going towards Ogre rather than away from him. (Not an optimal strategy.)

Presently, we are solving the sub-problem, where the maiden after leaving her inner circle goes in a straight line towards some point on the shore. We must find, that point. Bonanova suggested going off the inner circle on a tangent line. That gives a better ratio than going in a straight line to the nearest point on the shore. However, I don't believe it is the optimum.

The numbers I have given in the post #23 inside the spoiler are off. But then, as everyone knows, I am siding with Ogre.

##### Share on other sites
• 0

so you mean the Ogre looks South, and the Maiden takes off to the North, and the Ogre (a Zen master apparently) sees the inevitability, and continues the long way around, like Mark Twain's "Cooper Indian". If the Ogre does turn and go north, he can go four times further north than M can. Exactly when should M recalculate her new tangent? Will it really require as long a road for Ogre?

It is a lot simpler than that for the zen master Ogre. The Ogre ignores where the maiden looks, where she goes, and where she intends to go. Ogre simply draws an imaginary straight line from the center of the lake, through the boat, to the shore; and heads for that spot using the shorter arch. Insofar as the optimum is concerned, it is the Ogre who determines the path -- not the maiden. The maiden reacts to Ogre's moves, not other way around.

If the maiden wanders back into her inner circle r, that could make the Ogre change the direction when the boat crosses straight line drawn from the Ogre to the center of the lake. Then the young lady would be going towards Ogre rather than away from him. (Not an optimal strategy.)

Presently, we are solving the sub-problem, where the maiden after leaving her inner circle goes in a straight line towards some point on the shore. We must find, that point. Bonanova suggested going off the inner circle on a tangent line. That gives a better ratio than going in a straight line to the nearest point on the shore. However, I don't believe it is the optimum.

The numbers I have given in the post #23 inside the spoiler are off. But then, as everyone knows, I am siding with Ogre.

Blame it on the cold, Prime.

I am not convinced that tangential is optimal - it may not be enough!

It's easy to overlook that r is one of the unknowns.

##### Share on other sites
• 0

so you mean the Ogre looks South, and the Maiden takes off to the North, and the Ogre (a Zen master apparently) sees the inevitability, and continues the long way around, like Mark Twain's "Cooper Indian". If the Ogre does turn and go north, he can go four times further north than M can. Exactly when should M recalculate her new tangent? Will it really require as long a road for Ogre?

It is a lot simpler than that for the zen master Ogre. The Ogre ignores where the maiden looks, where she goes, and where she intends to go. Ogre simply draws an imaginary straight line from the center of the lake, through the boat, to the shore; and heads for that spot using the shorter arch. Insofar as the optimum is concerned, it is the Ogre who determines the path -- not the maiden. The maiden reacts to Ogre's moves, not other way around.

If the maiden wanders back into her inner circle r, that could make the Ogre change the direction when the boat crosses straight line drawn from the Ogre to the center of the lake. Then the young lady would be going towards Ogre rather than away from him. (Not an optimal strategy.)

Presently, we are solving the sub-problem, where the maiden after leaving her inner circle goes in a straight line towards some point on the shore. We must find, that point. Bonanova suggested going off the inner circle on a tangent line. That gives a better ratio than going in a straight line to the nearest point on the shore. However, I don't believe it is the optimum.

The numbers I have given in the post #23 inside the spoiler are off. But then, as everyone knows, I am siding with Ogre.

Blame it on the cold, Prime.

I am not convinced that tangential is optimal - it may not be enough!

It's easy to overlook that r is one of the unknowns.

I did overlook that r is one of the unknowns because of cold and my partiality to Ogre's cause.

Still I don't see that the problem is solved even for a straight line escape.

By the way, I did not see the answer by Phaze to which you referred. Is BD not displaying some of the posts to me?

Edited by Prime

##### Share on other sites
• 0

so you mean the Ogre looks South, and the Maiden takes off to the North, and the Ogre (a Zen master apparently) sees the inevitability, and continues the long way around, like Mark Twain's "Cooper Indian". If the Ogre does turn and go north, he can go four times further north than M can. Exactly when should M recalculate her new tangent? Will it really require as long a road for Ogre?

It is a lot simpler than that for the zen master Ogre. The Ogre ignores where the maiden looks, where she goes, and where she intends to go. Ogre simply draws an imaginary straight line from the center of the lake, through the boat, to the shore; and heads for that spot using the shorter arch. Insofar as the optimum is concerned, it is the Ogre who determines the path -- not the maiden. The maiden reacts to Ogre's moves, not other way around.

If the maiden wanders back into her inner circle r, that could make the Ogre change the direction when the boat crosses straight line drawn from the Ogre to the center of the lake. Then the young lady would be going towards Ogre rather than away from him. (Not an optimal strategy.)

Presently, we are solving the sub-problem, where the maiden after leaving her inner circle goes in a straight line towards some point on the shore. We must find, that point. Bonanova suggested going off the inner circle on a tangent line. That gives a better ratio than going in a straight line to the nearest point on the shore. However, I don't believe it is the optimum.

The numbers I have given in the post #23 inside the spoiler are off. But then, as everyone knows, I am siding with Ogre.

Blame it on the cold, Prime.

I am not convinced that tangential is optimal - it may not be enough!

It's easy to overlook that r is one of the unknowns.

I did overlook that r is one of the unknowns because of cold and my partiality to Ogre's cause.

Still I don't see that the problem is solved even for a straight line escape.

By the way, I did not see the answer by Phaze to which you referred. Is BD not displaying some of the posts to me?

Good catch Prime, and I have no health excuse. Apologies.

I've edited my post 25 to properly attribute you as the solver.

Edit:

We can't continue indefinitely increasing the angle however.

At some point it will become to the ogre's advantage to reverse his direction.

Edited by bonanova

##### Share on other sites
• 0

...

Edit:

We can't continue indefinitely increasing the angle however.

At some point it will become to the ogre's advantage to reverse his direction.

For a straight line escape, it seems clear that the optimal angle is between the zero (straight line to the nearest point on the shore) and the tangential line escape. Past the tangential escape, maiden re-enters her inner circle going towards the Ogre causing him to change the direction.

You have calculated the precise rate O:M rate (f) for the tangential escape. And it is better than the nearest point escape.

However, for the other angles in between those two, calculation of maiden's trip is a bit more messy.

I think this problem still has a suspense left and warrants further investigation.

Edited by Prime

##### Share on other sites
• 0

...

Edit:

We can't continue indefinitely increasing the angle however.

At some point it will become to the ogre's advantage to reverse his direction.

For a straight line escape, it seems clear that the optimal angle is between the zero (straight line to the nearest point on the shore) and the tangential line escape. Past the tangential escape, maiden re-enters her inner circle going towards the Ogre causing him to change the direction.

You have calculated the precise rate O:M rate (f) for the tangential escape. And it is better than the nearest point escape.

However, for the other angles in between those two, calculation of maiden's trip is a bit more messy.

I think this problem still has a suspense left and warrants further investigation.

If re entering the small circle is the criterion that gives advantage to ogre to change his direction then I think we know the optimal straight-line escape path.

Reason: I postulate the advantage increases initially with angle from radial, and then does one of two things:

1. It increases monotonically with angle from radial to tangential.

If so, we're done. A very quick and dirty analysis shows the advantage is still increasing with angle at tangential.

2. It reaches a maximum at some angle between radial and tangential.

If this is true, the advantage would be decreasing with angle at tangential. But analysis says it's increasing. It seems clear from the diagram that every degree of angle added to the escape path adds proportionately to the ogre's path. While the rower's path increases to a much lesser degree. In particular, the 45 degree case was analyzed rigorously and shows an advantage intermediate to those of radial and tangential. Only when the ogre can advantageously change direction would further increase in angle be ill advised.

It would be interesting to show that going beyond tangential incents ogre to reverse direction. Rower would then reverse as well, putting us in zig zag mode.

##### Share on other sites
• 0

Then the problem is solved tor a straght line escape and it is the tangential line off the inner circle's perimeter. The new problem becomes how to feed the Ogre. Perhaps, if we convinced the maiden that there is such a thing as zig-zag mode, the Ogre could get his dinner after all.

The function of the Ogre/Maiden rate looks like this (unless I messed something up again):

Where a is the angle on Bonanova's diagram. The function increases continuously within its domain from a = o (nearest point escape) to the tangential line escape. The tangential line is the limit, after which the function is no longer valid, because the Ogre changes his direction. The angle a corresponding to the point where the tangential escape goes is approximately 77.5 degrees. The rate of speeds Ogre:Maiden f ~ 4.6033

One thing that seems obvious to me is the Ogre's strategy. At each instant, he spots the point on the shore nearest to the boat and runs there along the shorter arch. The optimal path for the maiden is fixed. Whenever she starts zig-zagging wandering off that path, she must start over, or end up as a main course at Ogre's dinner table. The best (and necessary) headstart for the maiden is where she is on the perimeter of her circle r=R/f and the Ogre is on the opposite end of the straight line drawn from the boat through the center of the lake. If the maiden wanders back inside her inner circle r, she loses her headstart advantage and must work to get it back.

The Ogre can determine the spot on the shore nearest the boat by drawing a straight line from the center of the lake through the boat to the shore.

Although, the straight line escape, most likely is not the optimum. The optimum must be some curve dictated by Ogre's movement. To prove (or disprove) that we could let her row half way along the tangent line and then figure out new optimal path from that position.

Edited by Prime

##### Share on other sites
• 0

Bonanova's solution says that O, once starting to go CW, is compelled to continue the long way around, because if he switched to CCW, M would be further away, and would plot a new tangential line.

I'd like to see M's algorithm. I don't see that it's that clearcut.

First, I question whether the Ogre's best strategy is merely to go towards the spot on the bank closest to M's current position. In many games, a player has to plan against more than one threat, sometimes choosing to respond to the most urgent, if there is conflict.

If this Ogre has met the requirements for a PhD (Doctor of Phagology), he may have already taken Diff Eq and may already have an iPhone tuned to this very forum. In that case, O is aware of this separate tangential threat. And O may be insulted at the thought of having to go the long way around. O may be inclined to address the more urgent tangential threat

Let's look at it in large infinitesimals to show my concern: suppose O takes one (large )step of length 1+pi while M takes her dainty row of length f (f is slightly less than 1, we're eager to find out how much less--I claim 0.95, BN claims 0.89). At this point, is M on the diameter O is now on? If she's on the NE side of that diameter, then O should head back north, as the threatened arrival site is closer that way, and so is M's path of shortest distance. Even if M is on that diameter (ie, a line through the Ogre and the Center also passes through M), O should reverse direction, so as to handle both threats with the same choice. Only if M is on the SW side of that diameter is there a question. Even then, it's a much shorter trip to M's new threatened landing site, and O's heading back north can threaten to capture her there.

At this point, Bonanova holds up the spectre of M's recalculating. Does she recalculate once O has reached O's starting point? has gone a step further? If so, she is no longer on O's diameter. I think her recalculation needs to be specified, so that we can see if she really has acquired enough of an advantage to get out.

I suspect that her recalculation would not be instantaneous, but might wait until O has committed enough to his new course. In my numerical exploration, I evaluated M's strategy of heading straight West until O had committed enough so that M could change course to a new straight line. I found that, when f=.947, M's only successful time window was to wait until O had moved about 23 degrees, then M should change course toward the spot on the other end of O's current diameter. This point was not the closest spot to M, but it was the only one she could reach before O could reach it going either way.

In this case, the win was guaranteed, so further zig-zagging did nobody any good. But for any smaller f, O could reach her one way or the other, regardless of her wait time and angle.

UNLESS she could do some zig-zagging to threaten to cover both of his threats, and he zig-zagged in response.

I think a straight-line solution is not The Solution.

##### Share on other sites
• 0

Define:

1. azimuth - the angle with regard to the center of the lake.
Zero at ogre's starting position and increasing in a clockwise direction.

2. gap - the angle from ogre to maiden, measured in the ogre's direction of pursuit.
3. w - the ogre's azimuthal speed = s/R where s = 4R/hour where R = lake radius.

4. r - the radius of safety that depends on maiden's top speed:
an inner circle inside which maiden can achieve w, and outside of which she cannot.

Pre-chase maneuvering:

Within the circle of safety, maiden achieves the optimal starting position for the chase:

1. Ogre's azimuth is 0
2. Maiden's azimuth is 180 on the safety circle perimeter
3. Gap = 180 [maximum]
4. Distance from ogre: R+r [maximum]
5. Distance to shore: R-r [minimum]

Objectives:

1. Maiden: Survival = reach shore before the gap becomes 0.
2. Ogre: Capture = reduce gap to 0 before maiden reaches shore.
3. Us: Find the lowest maiden speed for which she can achieve simultaneous arrival on shore.
This means find the best escape path to permit escape at the lowest maiden speed.

Progess so far:

Maiden speed = Ogre speed / 4.6033 gets maiden to shore on tangential path, arriving simultaneously with ogre.

Now let's try to improve on 4.6033.

Here are my conjectures: comments are welcome

1. When the chase begins, Maiden leaves, and then never re-enters the circle of safety.
[a] if she re-enters, she can only re-set the race at the optimal starting position
.

2. Ogre will maintain the maximum speed of w, meaning that at any moment his only strategy is to choose CW or CCW pursuit.
[a] If he were to stop, maiden improves her position by heading directly away from him,

eventually landing on the opposite shore with gap = 180.
If he were to slow down, maiden's relative speed increases, and she moves more efficiently toward shore.

3. Once the chase begins, outside the circle of safety, gap decreases monotonically from 180 degrees
[a] maiden cannot match ogre's angular speed of w unless she re-enters the circle of safety. She won't. See above.
if gap were to increase, it means ogre is moving in the wrong direction, which he instantaneously reverses.

4. Ogre will never zig zag, meaning that ogre's only strategy is initial choice of direction. Nothing else. Then run full speed.
[a] At the outset of the chase, an infinitessimal motion by maiden dictates ogre's initial choice of direction [CW or CCW].
At any point thereafter, if ogre changes direction he will increase the gap.
[c] At any point thereafter, if maiden changes direction she will decrease the gap.
[d] Therefore, neither will ever change their angular direction.

5. Maiden's best initial direction is tangential
[a] That direction initially matches ogre's angular speed of w.
It begins her radial journey toward R keeping the decrease of gap at a minimum.
[c] Thereafter, it remains to be found the best way to add a radial component to her escape path.

The next step is to answer these questions:

1. Will spiraling away from ogre get her to shore with a greater gap than for a straight line?
2. If so, what is the optimal curvature of that spiral?

Approach for determining the advantage if any for a spiral path:

Straight-path analysis was simple enough to permit solving directly for maiden speed that gave simultaneous arrival. For spiral path analysis to be equally tractable, its curvature [non zero] will have to be specified.

Analysis for spiral.

Write rm = r + m.THETA where THETA increases CW from maiden's initial position.
Landing point will be the THETA for which rm = R.
The PATH LENGTH of the spiral can be calculated.

THETA and PATH LENGTH are substituted into the equations used in the straight-line analysis to determine the speed ratio f.

So we do this several times, for different values of m.
One value of m will give a smallest maiden speed - i.e. highest value of f.
That will be the optimum spiral escape path.

The problem I'm having at present is integrating the spiral equation to obtain path length.

Disturbing result!

I ran the straight line analysis to send the boat to the 3 pi / 2 landing point.

That is just slightly farther down the shore for ogre.

The speed ratio decreased from the tangential value of 4.6033 to 4.500...

I did not expect that.

Perhaps tangential straight line escape is optimal.

##### Share on other sites
• 0

Define:

1. azimuth - the angle with regard to the center of the lake.

Zero at ogre's starting position and increasing in a clockwise direction.

2. gap - the angle from ogre to maiden, measured in the ogre's direction of pursuit.

3. w - the ogre's azimuthal speed = s/R where s = 4R/hour where R = lake radius.

4. r - the radius of safety that depends on maiden's top speed:

an inner circle inside which maiden can achieve w, and outside of which she cannot.

Pre-chase maneuvering:

Within the circle of safety, maiden achieves the optimal starting position for the chase:

1. Ogre's azimuth is 0
2. Maiden's azimuth is 180 on the safety circle perimeter
3. Gap = 180 [maximum]
4. Distance from ogre: R+r [maximum]
5. Distance to shore: R-r [minimum]

Objectives:

1. Maiden: Survival = reach shore before the gap becomes 0.
2. Ogre: Capture = reduce gap to 0 before maiden reaches shore.
3. Us: Find the lowest maiden speed for which she can achieve simultaneous arrival on shore.

This means find the best escape path to permit escape at the lowest maiden speed.

Progess so far:

Maiden speed = Ogre speed / 4.6033 gets maiden to shore on tangential path, arriving simultaneously with ogre.

Now let's try to improve on 4.6033.

Here are my conjectures: comments are welcome

1. When the chase begins, Maiden leaves, and then never re-enters the circle of safety.

[a] if she re-enters, she can only re-set the race at the optimal starting position

.

2. Ogre will maintain the maximum speed of w, meaning that at any moment his only strategy is to choose CW or CCW pursuit.

[a] If he were to stop, maiden improves her position by heading directly away from him,

eventually landing on the opposite shore with gap = 180.

If he were to slow down, maiden's relative speed increases, and she moves more efficiently toward shore.

Once the chase begins, outside the circle of safety, gap decreases monotonically from 180 degrees

[a] maiden cannot match ogre's angular speed of w unless she re-enters the circle of safety. She won't. See above.

if gap were to increase, it means ogre is moving in the wrong direction, which he instantaneously reverses.

Ogre will never zig zag, meaning that ogre's only strategy is initial choice of direction. Nothing else. Then run full speed.

[a] At the outset of the chase, an infinitessimal motion by maiden dictates ogre's initial choice of direction [CW or CCW].

At any point thereafter, if ogre changes direction he will increase the gap.

[c] At any point thereafter, if maiden changes direction she will decrease the gap.

[d] Therefore, neither will ever change their angular direction.

Maiden's best initial direction is tangential

[a] That direction initially matches ogre's angular speed of w.

It begins her radial journey toward R keeping the decrease of gap at a minimum.

[c] Thereafter, it remains to be found the best way to add a radial component to her escape path.

The next step is to answer these questions:

1. Will spiraling away from ogre get her to shore with a greater gap than for a straight line?
2. If so, what is the optimal curvature of that spiral?

Approach for determining the advantage if any for a spiral path:

Straight-path analysis was simple enough to permit solving directly for maiden speed that gave simultaneous arrival. For spiral path analysis to be equally tractable, its curvature [non zero] will have to be specified.

Analysis for spiral.

Write rm = r + m.THETA where THETA increases CW from maiden's initial position.

Landing point will be the THETA for which rm = R.

The PATH LENGTH of the spiral can be calculated.

THETA and PATH LENGTH are substituted into the equations used in the straight-line analysis to determine the speed ratio f.

So we do this several times, for different values of m.

One value of m will give a smallest maiden speed - i.e. highest value of f.

That will be the optimum spiral escape path.

The problem I'm having at present is integrating the spiral equation to obtain path length.

Disturbing result!

I ran the straight line analysis to send the boat to the 3 pi / 2 landing point.

That is just slightly farther down the shore for ogre.

The speed ratio decreased from the tangential value of 4.6033 to 4.500...

I did not expect that.

Perhaps tangential straight line escape is optimal.

It seems, you have proven that the tangential straight line escape is the optimal path. I'll try to put put a simpler perspective on this whole Ogre-Maiden business using BMAD's “Dog Freedom” recent problem in this forum.

The “gap” and Ogre's angular speed simplify things quite a bit.

Imagine a solid wall rising around the Maiden's inner circle r when she leaves its perimeter at the escape point E. A thin rigid rod pivoted at the center of the lake extends from Ogre across the entire lake. As Ogre runs, he rotates the rod painting the angle α. (Somehow, the wall does not interfere with the rod.)

The distance traveled by the maiden is α*r along the perimeter of her inner circle, or any other path outside it. The intersection point between the rod and tangential path is at r*tan α dangerously close to the distance α*r traveled by the maiden. Looks like if she just swerved to the right she could catch that rod and get herself a new better head start. But then our work would be invalidated.

Say, some optimal curve existed leading to the landing point L1, but then we could draw a straight line to the same point as shown on the diagram, thus invalidating that curve as optimal. Straight line is shorter, and as long as it does not catch the moving rod, a better valid path.

Suppose we found what we are looking for – the overall angle α traveled by Ogre and the optimal r. Then we could make a leash of the length α*r, tie one end to the point E, the other – to the boat. Pulling at the maiden's end of the leash, intersect it with the shore. If the end of the leash extends past the shore, rotate to the right while keeping the tension until the boat rests exactly at the edge of water, and, so happens, at the optimal α.

No curve constructed of the same leash can beat that. If the leash wraps a bit around the wall instead of just touching it at E, then it's too long. It simply resets the points E and O.

The lower boundary for the length of the leash is α*r >= R – r (otherwise it cannot reach the shore.)

In the post #34 I gave the formula for the straight line escape routes between nearest point escape L0 and tangential escape Lt. That function has its minimum value at the boundary point Lt.

It appears, there is no spiral escape. But don't tell the maiden.

##### Share on other sites
• 0

Define:

• azimuth - the angle with regard to the center of the lake.

Zero at ogre's starting position and increasing in a clockwise direction.

• gap - the angle from ogre to maiden, measured in the ogre's direction of pursuit.

• w - the ogre's azimuthal speed = s/R where s = 4R/hour where R = lake radius.

• r - the radius of safety that depends on maiden's top speed:

an inner circle inside which maiden can achieve w, and outside of which she cannot.

Pre-chase maneuvering:

Within the circle of safety, maiden achieves the optimal starting position for the chase:

• Ogre's azimuth is 0
• Maiden's azimuth is 180 on the safety circle perimeter
• Gap = 180 [maximum]
• Distance from ogre: R+r [maximum]
• Distance to shore: R-r [minimum]
Objectives:
• Maiden: Survival = reach shore before the gap becomes 0.
• Ogre: Capture = reduce gap to 0 before maiden reaches shore.
• Us: Find the lowest maiden speed for which she can achieve simultaneous arrival on shore.

This means find the best escape path to permit escape at the lowest maiden speed.

Progess so far: Maiden speed = Ogre speed / 4.6033 gets maiden to shore on tangential path, arriving simultaneously with ogre.

Now let's try to improve on 4.6033.

Here are my conjectures: comments are welcome

• When the chase begins, Maiden leaves, and then never re-enters the circle of safety.

[a] if she re-enters, she can only re-set the race at the optimal starting position.

• Ogre will maintain the maximum speed of w, meaning that at any moment his only strategy is to choose CW or CCW pursuit.

[a] If he were to stop, maiden improves her position by heading directly away from him,

eventually landing on the opposite shore with gap = 180.

If he were to slow down, maiden's relative speed increases, and she moves more efficiently toward shore.

• Once the chase begins, outside the circle of safety, gap decreases monotonically from 180 degrees

[a] maiden cannot match ogre's angular speed of w unless she re-enters the circle of safety. She won't. See above.

if gap were to increase, it means ogre is moving in the wrong direction, which he instantaneously reverses.

• Ogre will never zig zag, meaning that ogre's only strategy is initial choice of direction. Nothing else. Then run full speed.

[a] At the outset of the chase, an infinitessimal motion by maiden dictates ogre's initial choice of direction [CW or CCW].

At any point thereafter, if ogre changes direction he will increase the gap.

[c] At any point thereafter, if maiden changes direction she will decrease the gap.

[d] Therefore, neither will ever change their angular direction.

• Maiden's best initial direction is tangential

[a] That direction initially matches ogre's angular speed of w.

It begins her radial journey toward R keeping the decrease of gap at a minimum.

[c] Thereafter, it remains to be found the best way to add a radial component to her escape path.

The next step is to answer these questions:
• Will spiraling away from ogre get her to shore with a greater gap than for a straight line?
• If so, what is the optimal curvature of that spiral?
Approach for determining the advantage if any for a spiral path:

Straight-path analysis was simple enough to permit solving directly for maiden speed that gave simultaneous arrival. For spiral path analysis to be equally tractable, its curvature [non zero] will have to be specified.

Analysis for spiral.

Write rm = r + m.THETA where THETA increases CW from maiden's initial position.

Landing point will be the THETA for which rm = R.

The PATH LENGTH of the spiral can be calculated.

THETA and PATH LENGTH are substituted into the equations used in the straight-line analysis to determine the speed ratio f.

So we do this several times, for different values of m.

One value of m will give a smallest maiden speed - i.e. highest value of f.

That will be the optimum spiral escape path. The problem I'm having at present is integrating the spiral equation to obtain path length. Disturbing result! I ran the straight line analysis to send the boat to the 3 pi / 2 landing point.That is just slightly farther down the shore for ogre. The speed ratio decreased from the tangential value of 4.6033 to 4.500... I did not expect that. Perhaps tangential straight line escape is optimal.

It seems, you have proven that the tangential straight line escape is the optimal path. I'll try to put put a simpler perspective on this whole Ogre-Maiden business using BMAD's Dog Freedom recent problem in this forum.

The gap and Ogre's angular speed simplify things quite a bit.

Imagine a solid wall rising around the Maiden's inner circle r when she leaves its perimeter at the escape point E. A thin rigid rod pivoted at the center of the lake extends from Ogre across the entire lake. As Ogre runs, he rotates the rod painting the angle α. (Somehow, the wall does not interfere with the rod.)

OgreMaiden2.gifThe distance traveled by the maiden is α*r along the perimeter of her inner circle, or any other path outside it. The intersection point between the rod and tangential path is at r*tan α dangerously close to the distance α*r traveled by the maiden. Looks like if she just swerved to the right she could catch that rod and get herself a new better head start. But then our work would be invalidated. Say, some optimal curve existed leading to the landing point L1, but then we could draw a straight line to the same point as shown on the diagram, thus invalidating that curve as optimal. Straight line is shorter, and as long as it does not catch the moving rod, a better valid path. Suppose we found what we are looking for the overall angle α traveled by Ogre and the optimal r. Then we could make a leash of the length α*r, tie one end to the point E, the other to the boat. Pulling at the maiden's end of the leash, intersect it with the shore. If the end of the leash extends past the shore, rotate to the right while keeping the tension until the boat rests exactly at the edge of water, and, so happens, at the optimal α.

No curve constructed of the same leash can beat that. If the leash wraps a bit around the wall instead of just touching it at E, then it's too long. It simply resets the points E and O. The lower boundary for the length of the leash is α*r >= R r (otherwise it cannot reach the shore.)In the post #34 I gave the formula for the straight line escape routes between nearest point escape L0 and tangential escape Lt. That function has its minimum value at the boundary point Lt.

It appears, there is no spiral escape. But don't tell the maiden.

I'm still surprised...

Differential analysis at the tangential landing point should be able to confirm or deny it. That is, if the landing point is moved a very small amount clockwise. There might then be a tractable expression for differential change in speed ratio. First-order perturbation theory it's called. Like using only the first term in a Taylor series expansion of the various quantities. The same idea as L'Hopital's rule. As a last gasp I might look at it, but I feel a bit burned out.

##### Share on other sites
• 0

For the tangential straight line escape the problem has been solved. The method of escape dictated construction of the equations finding the values for

r and Ogre/Maiden speed ratio. That path also determines the angle the Ogre has to travel. For the landing points past the tangential and bigger angular travels by Ogre, consider the following.
The shortest path to such point is around the small circle perimeter and then on a straight tangential line. Any other curved or segmented path is longer and must be rejected by clever maiden. But then all it comes down to is the same tangential escape only delayed.

This puzzle bears some logical inferences, which allow to dispense with extra complex math. Just what we all like in puzzles.

##### Share on other sites
• 0

I agree with that logic.

I've now also verified with differential analysis that for angles greater than a the rower's path increases more than the ogre's path.

Details in next post.

##### Share on other sites
• 0

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.

##### Share on other sites
• 0

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?

##### Share on other sites
• 0

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.

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.

##### Share on other sites
• 0

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.

##### Share on other sites
• 0

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.

##### Share on other sites
• 0

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?

##### Share on other sites
• 0

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.

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.

##### Share on other sites
• 0

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?

##### Share on other sites
• 0

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.

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

Edited by Prime

##### Share on other sites
• 0

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.
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.

## Create an account

Register a new account