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Crossing The River: Newsletter Challenge (05/06/08)

Posted May 04, 2008 5:01 PM

Welcome to May edition of Monthly Challenge Question from Specs & Techs by GlobalSpec:

You use your boat to cross a 500 m wide river from Point A to Point B which is 750 m upstream. The river current is 4 km/h, and your boat speed in still water is 10 km/h. How long this trip will take?

And the Answer Is...(June 3, 2008: 3:45 PM EST)

The above figure shows the movement of the boat across the river. The boat, moving at a constant speed of u, starts by making an angle a respect the positive vertical axis. Let's find the two components of the boat speed.

where v is the speed of the river current. Now, let's calculate the time that it takes the boat to reach point B. Let this time be T. Let X and Y be the horizontal and vertical components respectively of the distance from A to B. Then, we have

To get the time, solve these two equations by T. Get

(1)

By substituting the values of u and v , and rearrange the above equation, we get

(2)

This equation can be solved numerically or graphically as is shown in the next figure

As you can see the solution is

a = 1.2065 rad

Substitute this value into Eq.(1) to get

T = 0.14 h = 8.4 minutes

Notes:

(1) In still water the time the boat will travel the same distance of 0.9014 km in 5.41 minutes.

(2) The angle b is given by

b = tan-1 (0.75/0.5) = 0.9828 rad

The difference between the two angles a and b is, then

a - b = 0.2237 rad = 9.7226 deg

This is the adjustment that the boat must make in order to reach exactly point B in the other side of the river, given the speed of the river.

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Blink
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#95
In reply to #93
 Find in discussion

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/09/2008 11:53 AM

Wow!

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SPIJman
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#128
In reply to #95

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/21/2008 3:21 PM

Damn, we're good! Now if only NASA was looking for a few good men...do they test?

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#96

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/09/2008 1:26 PM

This challenge is a really simple one which can be solved graphically or even freehand if you have some square-ruled paper and a pencil at hand. No need really for a calculator (or even straightedge and compass**).

The benefit of drawing or sketching the solution is that the whole picture with proper relative magnitudes and directions can be clearly visualised. The likelihood of making calculation mistakes or conceptual mistakes is reduced. The effect of changing a parameter is also readily apparent -- some additional line-work could even help to estimate certain maximum or minimum limits.

Of course you must know the basic maths (and principles of mechanics -- velocity triangles in this case). All that may suffer is 'accuracy', which in many practical cases is utterly meaningless. Do you really need to know boat travelling and docking time to the nearest second even?

So here goes (the description of the procedure being far far more cumbersome than actually doing it***):

The river banks are 500 m apart (mark off say 20 or 40 squares to use up a decent width of paper). Point B is 750 m upstream, hence 30 or 60 squares up. A freehand line AB tells us the direction the boat has to take. The resultant boat speed ED must 'obviously' be along this line as well.

We need another arbitrary scale for the velocity triangle. Somewhere along AB mark the stream velocity of 4 km/h as CD (say 10 or 20 units long). Boat speed of 10 km/h in still water is along EC (25 or 50 units), with E located by estimation or drawing an arc with C as centre, even a rough one ****.

The resultant velocity ED is now properly directed along AB. We don't need its value in this problem though. The 'across' component EF is the crossing speed, and we know its magnitude by counting squares. The river width of 0.5 km divided by the value of EF in km/h times 60 is the crossing time in minutes. If you can't estimate this reasonably without a calculator or slide rule, go ahead. A protractor can be used if you need to know any angles.

We can clearly see that as the boat speed decreases, EC gets steeper and EF gets shorter, so the crossing takes longer. I couldn't be bothered with posting an 'answer' as too many have already flogged it to umpteen decimal places. I was only interested in presenting this kind of approach to a problem. I know of course that it is not suitable for presenting on CR4. What I could emphasise is that the scales can be conveniently chosen, and changed if the points go off the paper. Naturally it helps to have a scale and compass. If you don't have square ruled paper the figures can always be constructed on plain paper.

=TeeSquare=

** Should that be compasses, or even a pair of compasses?

*** Computer graphics is a huge hassle for me. I've spent over an hour on that figure in MSWord97 Draw, and saved as jpg in Paint, but the dotted line EF just vanished! I didn't dare attempt drawing a grid, but just marked an arbitrary scale.

**** Use fingers and pencil as a compass. Set the radius from a fingernail placed on the centre, and spin the paper. Works well for me!
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#97
In reply to #96

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/09/2008 4:14 PM

Great Answer!

I agree that the graphic approach is a great aid to understanding and avoiding conceptual mistakes. Coincidentally, on my screen, your drawing scales to a convenient 1/4 inch per km/h, and on that basis, your result looks both correct and close enough to have a very good idea of how long the trip will take.

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#98
In reply to #96

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/10/2008 9:38 AM

*** I came up with this after a half hour tinkering with some freeware I found. It churned out a resultant speed of 6.4225 km/hr etc. That, plus the (also derived) distance, works out OK. My layout and scaling is a bit rubbish, but it seems to work as a tool. I need a bit of practice. Wish I'd found this when the trisection was still hot, it's ideal for such tinkering.

If it meets with any approval, I may even name the software ! Back later.....

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#101
In reply to #98

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/13/2008 2:26 AM

Not free, but, have you seen Cinderella:-

http://cinderella.de/tiki-index.php

€50 $70 ish

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#102
In reply to #101

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/13/2008 5:07 AM

Hey, that looks really cool ! Thanks Randall, nice one.

PS - the thing I used is called GeoGebra. It seems perfectly suited to Euclidean construction. You can draw stuff and it will tell you the angles/lengths, or vice versa.

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#99

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/11/2008 10:50 AM

I find now that maressiellada (in post 92) had anticipated most of what I wished to convey, and expressed it rather neatly too in his(/her) very first post. In fact I merely did a cursory skim through before firing my blunderbuss! Anyway, thanks for acknowledging my viewpoint.

I wish more people would realise the benefits of just walking down the road (essentially with pencil and paper) instead of juggling with the controls of a car (calculator, as in post 1, 21, etc.) or a battle tank (computer, as in post 4, 42, etc.) just to get from A to B if the distance is say 901 m. Thor Heyerdahl had demonstrated that even those who were deemed savages by the 'civilised' world were capable of accurate navigation over substantially greater distances several centuries ago, without any riverbanks or shorelines to guide them. But there I go ranting again ....

The 53-63-77-86 exchanges were an informative spin-off from this otherwise near-trivial thread. Looks like even the silliest of questions will be able to generate interesting off-shoots. Or maybe that's the secret!

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#100

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/12/2008 2:24 PM

Trust me...it's so much easier with a handheld flight computer and you're always right. Let's ya keep yer mind on rowing the boat, err...plane.

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#103

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/13/2008 10:06 AM

7 min 29,1 sec
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Time_traveler
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#104

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/13/2008 11:26 AM

I assumed the point on the far shore was initially 0.75 km upstream and was moving upstream at 4 km/h - realtivity. Then I solved the quadratic equation from the distance traingle and found both a positive and a negative root. So, assuming you have a time machine like mine in your boat, you can go backwards in time a little over 4 minutes and be on the other side of the river, which is shorter than the 8+ minute answer. The current runs backwards in negative time and helps you get up river - probably have to point the boat backwards too. This is absurd, but it kind of makes you think doesn't it?

The real question is why did I cross the river? Is there a brew pub over there?

Oh, and aside from significant digit accuracy, the current needs a direction to be a true vector. Upstream means where the water typically is sourced from, but the current could be going in either direction, and not necessarily parallel to the shore.
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Anonymous Poster
#105

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/13/2008 1:57 PM

ans-3 min.
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Anonymous Poster
#106

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/13/2008 3:38 PM

This question reminded me of a story from Cryptonomicon, by Neal Stephenson:

He went back to Iowa State, considered changing his major to mathematics, but didn't. It was the consensus of all whom he consulted that mathematics, like pipe-organ restoration, was a fine thing, but that one needed some way to put bread on the table. He remained in engineering and did more and more poorly at it until the middle of his senior year, when the university suggested that he enter a useful line of work, such as roofing. He walked straight out of college into the waiting arms of the Navy.

They gave him an intelligence test. The first question on the math part had to do with boats on a river: Port Smith is 100 miles upstream of Port Jones. The river flows at 5 miles per hour. The boat goes through water at 10 miles per hour. How long does it take to go from Port Smith to Port Jones? How long to come back?

Lawrence immediately saw that it was a trick question. You would have to be some kind of idiot to make the facile assumption that the current would add or subtract 5 miles per hour to or from the speed of the boat. Clearly, 5 miles per hour was nothing more than the average speed. The current would be faster in the middle of the river and slower at the banks. More complicated variations could be expected at bends in the river. Basically it was a question of hydrodynamics, which could be tackled using certain well-known systems of differential equations. Lawrence dove into the problem, rapidly (or so he thought) covering both sides of ten sheets of paper with calculations. Along the way, he realized that one of his assumptions, in combination with the simplified Navier-Stokes equations, had led him into an exploration of a particularly interesting family of partial differential equations. Before he knew it, he had proved a new theorem. If that didn't prove his intelligence, what would?

Then the time bell rang and the papers were collected. Lawrence managed to hang onto his scratch paper. He took it back to his dorm, typed it up, and mailed it to one of the more approachable math professors at Princeton, who promptly arranged for it to be published in a Parisian mathematics journal.

Lawrence received two free, freshly printed copies of the journal a few months later, in San Diego, California, during mail call on board a large ship called the U.S.S. Nevada. The ship had a band, and the Navy had given Lawrence the job of playing the glockenspiel in it, because their testing procedures had proven that he was not intelligent enough to do anything else.

The sack of mail carrying Lawrence's contribution to the mathematical literature arrived just in the nick of time. Lawrence's ship, and quite a few of her sisters, had until then been based in California. But at just this moment, all of them were transferred to some place called Pearl Harbor, Hawaii, in order to show the Nips who was boss.
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Anonymous Poster
#114
In reply to #106

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/14/2008 10:54 AM
This ones deserves to be On Topic and Voted on. Guest #

106
Re: Crossing The River: Newsletter Challenge (05/06/08) 05/13/2008 12:38 PM

This question reminded me of a story from Cryptonomicon, by Neal Stephenson:

He went back to Iowa State, considered changing his major to mathematics, but didn't. It was the consensus of all whom he consulted that mathematics, like pipe-organ restoration, was a fine thing, but that one needed some way to put bread on the table. He remained in engineering and did more and more poorly at it until the middle of his senior year, when the university suggested that he enter a useful line of work, such as roofing. He walked straight out of college into the waiting arms of the Navy.

They gave him an intelligence test. The first question on the math part had to do with boats on a river: Port Smith is 100 miles upstream of Port Jones. The river flows at 5 miles per hour. The boat goes through water at 10 miles per hour. How long does it take to go from Port Smith to Port Jones? How long to come back?

Lawrence immediately saw that it was a trick question. You would have to be some kind of idiot to make the facile assumption that the current would add or subtract 5 miles per hour to or from the speed of the boat. Clearly, 5 miles per hour was nothing more than the average speed. The current would be faster in the middle of the river and slower at the banks. More complicated variations could be expected at bends in the river. Basically it was a question of hydrodynamics, which could be tackled using certain well-known systems of differential equations. Lawrence dove into the problem, rapidly (or so he thought) covering both sides of ten sheets of paper with calculations. Along the way, he realized that one of his assumptions, in combination with the simplified Navier-Stokes equations, had led him into an exploration of a particularly interesting family of partial differential equations. Before he knew it, he had proved a new theorem. If that didn't prove his intelligence, what would?

Then the time bell rang and the papers were collected. Lawrence managed to hang onto his scratch paper. He took it back to his dorm, typed it up, and mailed it to one of the more approachable math professors at Princeton, who promptly arranged for it to be published in a Parisian mathematics journal.

Lawrence received two free, freshly printed copies of the journal a few months later, in San Diego, California, during mail call on board a large ship called the U.S.S. Nevada. The ship had a band, and the Navy had given Lawrence the job of playing the glockenspiel in it, because their testing procedures had proven that he was not intelligent enough to do anything else.

The sack of mail carrying Lawrence's contribution to the mathematical literature arrived just in the nick of time. Lawrence's ship, and quite a few of her sisters, had until then been based in California. But at just this moment, all of them were transferred to some place called Pearl Harbor, Hawaii, in order to show the Nips who was boss.

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Anonymous Poster
#115
In reply to #106

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/14/2008 10:59 AM
Guest

This one deserves to be On Topic and voted on without handicap.

Different poster

 #

106
Re: Crossing The River: Newsletter Challenge (05/06/08) 05/13/2008 12:38 PM

This question reminded me of a story from Cryptonomicon, by Neal Stephenson:

He went back to Iowa State, considered changing his major to mathematics, but didn't. It was the consensus of all whom he consulted that mathematics, like pipe-organ restoration, was a fine thing, but that one needed some way to put bread on the table. He remained in engineering and did more and more poorly at it until the middle of his senior year, when the university suggested that he enter a useful line of work, such as roofing. He walked straight out of college into the waiting arms of the Navy.

They gave him an intelligence test. The first question on the math part had to do with boats on a river: Port Smith is 100 miles upstream of Port Jones. The river flows at 5 miles per hour. The boat goes through water at 10 miles per hour. How long does it take to go from Port Smith to Port Jones? How long to come back?

Lawrence immediately saw that it was a trick question. You would have to be some kind of idiot to make the facile assumption that the current would add or subtract 5 miles per hour to or from the speed of the boat. Clearly, 5 miles per hour was nothing more than the average speed. The current would be faster in the middle of the river and slower at the banks. More complicated variations could be expected at bends in the river. Basically it was a question of hydrodynamics, which could be tackled using certain well-known systems of differential equations. Lawrence dove into the problem, rapidly (or so he thought) covering both sides of ten sheets of paper with calculations. Along the way, he realized that one of his assumptions, in combination with the simplified Navier-Stokes equations, had led him into an exploration of a particularly interesting family of partial differential equations. Before he knew it, he had proved a new theorem. If that didn't prove his intelligence, what would?

Then the time bell rang and the papers were collected. Lawrence managed to hang onto his scratch paper. He took it back to his dorm, typed it up, and mailed it to one of the more approachable math professors at Princeton, who promptly arranged for it to be published in a Parisian mathematics journal.

Lawrence received two free, freshly printed copies of the journal a few months later, in San Diego, California, during mail call on board a large ship called the U.S.S. Nevada. The ship had a band, and the Navy had given Lawrence the job of playing the glockenspiel in it, because their testing procedures had proven that he was not intelligent enough to do anything else.

The sack of mail carrying Lawrence's contribution to the mathematical literature arrived just in the nick of time. Lawrence's ship, and quite a few of her sisters, had until then been based in California. But at just this moment, all of them were transferred to some place called Pearl Harbor, Hawaii, in order to show the Nips who was boss.

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Anonymous Poster
#107

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/13/2008 3:52 PM

What is the size of the boat?

What if the boat is 500m long? Do you count the nose or the toes?
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Anonymous Poster
#108

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/13/2008 8:39 PM

ty = 0.75 km / (vy - 4 km/hr) ..................(given) (1)

tx = 0.5 km / vx ............................................(given) (2)

tx = ty ................................................(shortest time - assumed) (3)

vboat = vx2 + vy2 = (10 km/hr)2 ..............(vector addition) (4)

vy = (3/2)vx + 4 ...................................(units omitted) ((1), (2), and (3), substitution) (5)

vx2 + ((3/2)vx + 4)2 - 100 = 0 ................((4) and (5), substitution) (6)

(13/4)vx2 + 12vx - 84 = 0 ......................((6), simplification) (7)

vx = 3.5625833326913679156759911705825 km/hr ...............((7), quadratic formula) (8)

tx = .5 km / 3.5625833326913679156759911705825 km/hr.......((2) and (8), substitution)

= 0.14034759423361277217825578157377 hr

= 8.4208556540167663306953468944261 minutes

= 8 minutes 25.251339241005979841720813665564 seconds

Discounting unreasonble accuracy, the direct route from Point A to Point B takes a little under 8 and a half minutes.
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Anonymous Poster
#122
In reply to #108

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/20/2008 1:38 PM

vboat = vx2 + vy2 = (10 km/hr)2 ..............(vector addition) (4)

should read

v2boat = vx2 + vy2 = (10 km/hr)2 ..............(vector addition) (4)

Please excuse the typo.
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makhdum_786
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#109

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/14/2008 1:33 AM

it will take 9 minutes and 69sec approx
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#111

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/14/2008 4:21 AM

My river was frozen, I just walked straight across.

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#112

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/14/2008 8:01 AM

The trip will take as long as the beer lasts, if it's a six-pack, only 20-30 min, for a case of beer or larger extrapolate accordingly.
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Anonymous Poster
#113

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/14/2008 8:42 AM

901.4 m with a speed of 6km/h = 9.013 minutes.

The distance had been identified being the hypotenuse AC= 901.4 m, the speed of the boat is at 6km/h (speed of the boat in still water 10km/h impeded by river's current at 4km/h).
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bobf
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#117

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/16/2008 8:31 AM

As an avid long distance sailor with the equivalent of more than 6 circumnavigations under his belt and also a former private pilot, I looked for a simple solution to this challenge. No more quadratic equations for me, since it has been almost 50 years since University days, and I saw no need to disturb those learnings assigned to the mothball stage of my memory. Therefore, I drew up a quick Maneuvering Board, plotted the information directly on it, and came up with a very quick answer.

To do this, draw a series of concentric circles, one inch apart, for a total of 10. The last one will have a radius of 10in, but you can use any scale you want. Ten inches was desirable, as you will see in the following. Draw a North/South line thru the center, and also an East/West line. From 0 degrees (the top of your vertical line) mark a series of angle marks, say every 2 degrees, towards the right, or your E/W line, on the outer circumference.

Now, plot your basic 0.5, .75 triangle as per the problem. You can solve by simple math, the length of the hypotenuse, which as others have found, is 0.901km. When you plot the triangle, start at the center and go .5 to the right (East), and .75 up (North).

Draw the hypotenuse all the way out to the outer circle and determine the angle. I plotted it to be 56 degrees. This line now can be labeled as CMG, or Course Made Good, since that is the line you will be traveling on to reach you goal, after the influence of the South-going current of 4 km is accounted for.

Because your circles are 1 inch apart, you can use dividers to pick up 4km right off the board. Take those dividers, which are now "4km apart" and draw a line parallel to the N/S line so that the "north" end of the 4KM line touches the outer circle, and the "south" end of the line touches your CMG line. If you now draw a line from the center of the circles to the point where this 4km line touches the outer circle, it will be 10 units long, representing the boat speed. (Now you know why I made the outer circle @ 10 inches!) This line will also represent the CTS, or Course to Steer, in order to counteract the influence of the 4km current.

Now, go to the CMG line, and measure from the center of the circles to where the vertical 4km line touches it. This distance represents the SMG, or Speed Made Good towards your destination that lies 0.75 upstream of your departure point. It measures on my board 6.4 (Autocad shows it to be 6.43).

So, now you have all you need to solve the problem, in addition to knowing how to steer the boat to make it all work. You know the distance to the goal (0.901), and now you know how fast the boat is going towards the target (6.4), and you solve for how long it will take to get there.

The answer is: 8.4 minutes, or 8 minutes 24+ seconds.

If someone could show me how to add in the diagram of the Maneuvering Board, I will be glad to include it and show the graphical way of solving the problem.

Incidentally, it took me exactly 3 minutes to draw the board and solve the problem. No complicated math involved, for this quick and easy solution. Note that the boat speed given in the problem is taken as SOG, or Speed over Ground. Thus, the 6.4 is also SOG.
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#118
In reply to #117

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/16/2008 4:25 PM

As one sailor/pilot to another: Welcome Aboard! I hope you stick around, making great contributions like this one.

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#119
In reply to #117

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/19/2008 2:04 AM

Sparkstation did a good post on inserting graphics:-

http://cr4.globalspec.com/thread/15743

Basically you just save the required picture as a .jpg or other acceptable format then when your in the CR4 editor click on the little green camera then browse to the picture you want.

Like this: initial scale for Course Made Good is 100m per inch; second scale for Course to Steer is 1Km/hr per inch.

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SPIJman
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#130
In reply to #117

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/22/2008 8:32 AM

Excellent contribution. Got my vote.

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#120

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/20/2008 12:14 PM

It will take 9 minutes for the trip
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#124

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/20/2008 5:00 PM

This problem could be solved by vectorial addition of 3 velocities :

1) Water velocity of 4kM/hour

2) Still water speed of boat @ 10kM/hour, and

3) Resultant velocity V of boat along line AB which makes an angle of tan-1(500/750)

=33.69 degrees to the water-flow.

Using the Cosine law for a triangle,

(10)^2 = (4)^2 + (V)^2 - 2*4*V*Cos(180-33.69)

100 = 16 + V^2 + 6.6564*V

V^2 + 6.6564*V - 84 = 0

This is a quadratic equation which can be solved as follows:

V = (-6.6564 + ((6.6564)^2 + 4*84))^0.5)/2

= 6.4225kM/hour

Distance AB = (0.5^2+ 0.75^2)^0.5 = 0.9014kM

Therefore, time to cross river from A to B = 0.9014/6.4225 = 0.1403 hour =8.4208 mins = 8 mins, 25.2 seconds

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blisterbill
Member

Join Date: May 2008
Posts: 9
#125

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/20/2008 11:36 PM

Being that Trigonometry and calculus is simply a memory to me from long long ago in a land far away, I felt the need to simplify.

Shore to shore the speed is 10km/h. Upstream reduces your speed by 40% because of the current. I just added 40% to the upstream length and solved for a constant speed of 10km/h left to right and downstream to upstream.

Pythagoras and a calculator tell me it will take 7.24 minutes.

It's just a hunch but it feels right.
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blisterbill
Member

Join Date: May 2008
Posts: 9
#126

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/21/2008 12:24 AM

OK

Did my math wrong before but if you account for the reduced upstream speed by increasing the right bank distance proportionally, then you can solve for a constant speed.

Boat speed (10) divided into upstream speed (6) equals 1.6666....

a= 500m

b=750m multiply by 1.666... and you get 1250

adj=500 new opposite=1250 so hyp=1346

divide hyp 1.346 km by speed 10 km and you get .1346 hours or eight minutes and 4 seconds.
Reply
Anonymous Poster
#129

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/21/2008 9:41 PM

My answer:

The boat final speed should be deduct upstream that 4 km/h, so actually it's speed is 10-4=6km/h, and the upstream distance is 0.75km, so 0.75/6km=0.125hrs will spend on the trip...
Reply
Anonymous Poster
#132

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/27/2008 2:00 PM

13 min
Reply
Anonymous Poster
#133

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/27/2008 8:49 PM

8.4 minutes.

the first order approximation is 7.5 minutes.

Most of the velocity points upstream, approx 10km/h. Net upstream speed is approx 6km/h transverse speed is therefore approx 6 * (500/750) = 4 km/h, giving 7.5 minutes for the 500m width.

Exact answer is slower, because upstream component is < 10km/h.

solving quadratic gives 8 minutes 25 seconds
Reply
Anonymous Poster
#134

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/27/2008 11:51 PM

9.0139 MIN.
Reply
Vasilios
Participant

Join Date: Dec 2006
Posts: 4
#135

Re: Crossing The River: Newsletter Challenge (05/06/08)

05/29/2008 1:30 PM

River Crossing Problem

This problem is very common to sailors in strong tidal waters e.g. around UK.

It is a simple Physics problem of relative velocities and can be solved by simple vector arithmetic in one of the two frames of reference i.e. either

(a) the ground (fixed frame of reference) or

(b) ) the water (moving frame of reference - with respect to the ground).

It is perhaps easier to understand the solution in the (a) frame.

The first thing to do is to add the velocities (vectorially) for the river (vr) and the boat (vb), in order to find the boat velocity relative to the ground (vg).

The Fixed frame of reference (space dimensions in metres x100) can be seen in Fig.1 below,

AC corresponds to the width of the river (5 m) and point B is its final destination.

Fig 1

The angle (theta) can be computed from its tangent : tan(theta) = BC / AC = 1.5

Therefore theta = 56.3°

The velocity diagram can be seen in Fig. 2, below.

Fig. 2

AD = vr DE = vb and AE = vg

The ground velocity vector, vg, must be at a direction (theta) = 56.3° so that the boat arrives directly to point B travelling in a straight line (the shortest route) relative to the ground (fixed frame of reference). At the risk of confusing the readers, I have combined the space diagram of Fig. 1 with the velocity diagram of Fig. 2 by extrapolating AE to become AB. If one ignores the x100 scaling factor, then one can see that vg is coincident with the space vector AB; this fulfils the above requirement. All we have to do now is to solve the triangle DAE, in order to determine the unknown vg. In this triangle we know 2 sides and one angle viz. AD, DE and b. Unfortunately, this is not the included angle (which guarantees solution always) but this triangle does have one solution because angle (beta) is obtuse.

(beta) = 90° + (theta) = 146.3°

We can use the sine theorem to work out sin(gamma) and then calculate(alpha), since (alpha) + (beta) + (gamma) = 180°

The angle solutions are: (alpha) = 20.88° and (gamma) = 12.82°

vg = vb* sin(alpha) / sin(beta) =6.423 km/h

Therefore, the time taken for the boat to reach point B is:

t = AB / vg = 0.90138 / 6.423 = 0.14034 hours = 8 min 25 sec

Where AB has been computed by using Pythagoras theorem on the triangle ABC

AB = square root (AC2 + BC2) = 0.90138

For those that feel masochistic, the transit time can also be computed in the moving frame of reference (the river water).

Thus, t = DF / vb , where DF represent the "water track" (see Fig. 2)

DF can be computed from triangle DFG as:

DF = DG / sin(alpha) = 14.0287; this has to be scaled by x100 to convert it to meters.

Thus, DF = 1402.87 m = 1.40287 km and therefore t = 1.40287 / 10 = 0.140287 hours

This is identical to the previous answer, within the rounding errors of the angles etc. and gives, again, the crossing time as about 8 minutes and 25 seconds.

Didn't I tell you it was very simple ?
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