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

I agree with your answer. I keep thinking there must be some trick to this as it is too easy.

1. It certainly asks for shortest time, otherwise you may go head directly towards opposite bank and travel,reach some point downstream. Have your lunch there and then have some rest. Then travel upstream. There are N no. of ways other than shortest time.

For actual travel you eed to keep pointing to point B, without any calculations. This path is not little longer, as you say, but is the shortest. The boat will take its appropriate shortest time as calculated by you in your first post.

If you choose to cross the rever straight, you will have to head upstream at angle 23.58 deg. With this your speed along the stream will be zero and across the stream 9.1652 kms/hr and you wil reach opposite bank withing 0.5/9.1652 = 3 minutes 16.4 sec. Then afterwards, you can travel along the bank as per my other (joking) post withing 7.5 minutes.

Thus total time taken will be 7 minute 30 seconds + 3 minutes 16.4 seconds = 10 minutes 46.4 seconds (with no fishing or swimming.... oh! you don't swim)

How did you calculate time for blue & black path?

For the black line, the boat is headed straight across. It takes 0.5 km / 10 km/hr or 0.05 hours (3 minutes) to get to the other side. During this time the boat is taken 0.05 hours x 4 km/hr or 0.2 km down stream. The distance straight upstream to point B is 0.2 km + 0.75 km. The time to go up stream is 0.95 km / (10 -4) km/hr = 0.15833 hr or 9.5 minutes. Total time is 12.5 minutes.

For the blue line I set up an Excel spread sheet that bootstrapped forward in time. At the beginning of each time step the X and Y positions were known. Started at X = 0, Y = 0. A pointing angle toward Point B was calculated and global velocities in X and Y were calculated based on the pointing angle. dX/dt = cosΘ x 10 km/hr and dY/dt = sinΘ x 10 km/hr - 4 km/hr, where Θ = angle upstream of straight across. Once X = 0.5 km, Y = 0.75 km, and the time was whatever it was. I kept using smaller and smaller time steps until the time didn't change significantly. I'm sure someone could do the math for an exact answer but the most elegant solution is the one that lets you move on to the next problem.

Thanks,

Jim

Very perceptive.

As you no doubt know, pilots make these calculations all the time, and VFR pilots can be fairly obsessive about following the straight line path from point-to-point, because if you drift off course, you can quickly find yourself lost -- which can make you feel profoundly stupid. If you have a straight line course drawn on the chart, at the instant when you stop recognizing stuff, at least you have a very good clue as to where you might be.

There is another approach () pilots use... on approach. Here, typically you have not precisely calculated a wind correction angle because you are no longer worried about being lost -- you don't set up an approach unless you are darn sure you're very near the airport. Assuming a crosswind, if you point the nose of the plane at the runway, you follow the blue line curved course you've drawn, meaning that when you arrive at the runway threshold, you have to look out the side window to see the runway. So, instead, you pick a reasonable correction angle guess, and fly it. You observe drift, and then angle sharply back to the extended runway course, and adjust the wind correction angle. You do this several times, and by the time you arrive at the runway you are both in the right place and headed in the almost right direction (the runway heading plus or minus the correction angle), which is eminently more comfortable than trying to land the plane really sideways.

Few people will do this in a boat, because observers will point and laugh and assume you are so drunk you can't even hold the wheel straight.

I think you should go out there with one of those leather hats and flight jackets with a scarf and goggles, tell them you are instrument certified and that this Delta flight 510 leaves for Beijing in 30 minutes. Tell them they should straighten up and fly right if they want to achieve the stature you have achieved and that there is no more room in 1st class.

Oh, and don't worry about getting lost because the place that you rented the boat from is probably down river and they will find you sooner or later and return you to your car. They probably do this all the time. Have fun, drink responsively, and don't throw your cans in the river.

Given the available information, your (voted good) answer presents the best available lower bound on how long it might take - but perhaps a partial reality check is in order?

This is an unusual river - same water velocity all the way across. If you accept this for now, with a real boat you would have expect to head along the bank (upstream) initially, and dock in the same direction (plus there must be some acceleration and deceleration time). So you would try aim for a point slightly downstream of the docking point; presumably you would make a guess at the optimum direction of travel, and then use a marker some distance behind it (to check whether you have the correct alignment) and adjust direction accordingly.
When the water velocity varies across the river, the fastest time is achieved by pointing further upstream in the slower-moving water.

Finally, my boat is a cable-driven ferry; therefore the time will be just a little greater than 5-minutes 24.5 seconds (the additional time would be to account for additional length of cable that is needed allow sufficient bowing to prevent it snapping).

I also came up with the same answer which doesn't necessarily provide the slightest degree of verification of correctness.

I don't think this is the correct answer. A straight line path is not always the quickest route.

My instinct tells me that the shortest route is curved e.g. start out going upriver and then gradually change till you are going more across the river. Possibly two solutions with the go across river and then go more upstream as the second solution. maybe there is a symmetrical solution in the maths.

I may be wrong, but I may be right. I unfortunately don't have the time to do the integral maths which is to give the velocity of the boat as measured perpendicular to the bank.

Other approach is to make the river stop and to make the banks move. What does the traced path look like now?

you can give answers like 505.25133912 etc. , but the answer cannot be that precise, since significance is one digit the answer should be : 5*10^2 s.

Funny to see your solving method. I did 100=x^2+(1.5x+4)^2 then 500/3.6x

I agree (more or less with your time), I got 8.27 minutes using vectors to get the resultant velocity and a/Sin A = b/Sin B and so-on to solve the angle of the heading needed to actually get to point B. In note someone else thinks this it too easy, so maybe there's a catch? Oh well, time for coffee, Cheers

This is the first time I have tried one of these challenges. Help if I am doing this wrong. For the answer to the challenge I have 14.4 minutes. I derived that by calculating that the distance from point to point was 901.388 meters, and at an angle of 56.310 degrees from horizontal (across the river). I TRIED calculating the resultant speed of the boat by subtracting the 4000 meters per second speed of the water from the 10000 meters per second speed of the boat (netting 6000 meters per second) and then taking a percentage (don't know if I can do this) of the actual degree traveling (56.310) divided by 90, giving me an actual speed of 3754 meters per hour. 901.388 meters divided by that was .2401 hours = 14.4 minutes. Yes/No ?

Your numbers assume the current exactly opposes the motion. Actually, only part of the rivers velocity opposes the motion, as much of the motion is across the river. If you go through the "voted good" answers, you will find various ways of accounting for the difference in directions. Although personally I would go for Randall's as being the first simple implementation, different people have different ways of thinking, so one of the others (such as resolution of velocities into components along and across the river and solving the equations) may initially be more to your taste.

KISS principle, PYTHAGOREAN theoerm:

750m against the current = 7.5 min.

500m (straight) accross the river = 3.27 min.

901.4m diagonal course to destination = 8 minutes, 10 seconds.

First issue is to find the straight-line distance from A to B. The Pythagorean Theorem gives us 901.39 meters, which we can simplify to .9 km.

Now the boat will not move at 10 km/h along this line because of the current. Vector addition comes very close to 7 km/h for the effective velocity.

(I admit I let AutoCAD do this calculation after I arranged the elements.)

.9 km / 7 km/h = .129 h = 7.7 min = 7 minutes 42 seconds. We will have to continually steer to the right 18.4° or we will fall short.

I think there is a problem with your answer:

the width of the river (AC) is 500m, and the destination point (AB) is 750m upstream (triangle ABC). the vector angle (BAC) of the destination is 56.3 deg. The resultant distance (hypotenuse) is 901.4m. The angle ABC is 33.7deg (180-90-56.3). A second triangle (ABD) is described by the speed of the boat (AD), the opposing current speed (BD), and the distance of the hypotenuse AB (by pythagoreus, 901m). The angle ABD is supplementary to ABC (180-33.7=146.3). We dont know the actual distances of AD, or BD, nor the angles DAB and ADB, but we do know the distances are a function of the speed; 10t, and 4t. Using the law of sines, sin146.3/10t=sinDAB/4t. This gives angle DAB as 12.8, and therefore angle ADB as 20.9. Again using the law of sines to compare distance; sin20.9/.901m=sin146.3/x, or 1.401km at 10km/hr or 8min, 24.5 sec.

Glad to see someone agreed with my answer!

sb

Time to toss a wrinkle into the problem! The velocity of the water across the width of the river is not uniform. The water will be running slower beside the shore and there are standard equations to provide the relationship. I am not a mechanical type but I do remember this from my college days taking a Chem E class. So even if you assume the river has a square bottom, it will still be quite pervalent that the water near the shore will run slower and if you use a a typical tapered shore of say 45 degrees, this agrument will become even more significant. Thus it could be to an advantage to travel most of the distance upstream beside the shore.

It will take 7.5 minutes only.

You never said that point B is on opposite bank. Thus A and B are 750 m apart on same bank. This will be travelled at 6 kms/hr. speed.

It reads Crossing The River:!!!!!?

Slight oversight here..

You use your boat to cross a 500 m wide river

So, Point B must be on the other side from Point A

ARGHHhhhh couldn't turn Italics offf :P

I'd have to say I like this answer. You are quite right! The question did not say point B was on the opposite bank from point A.

29.08 min

Shishir Kumar

My answer is 8.42103717176754 minutes. Heck, call it 8 minutes and 25.2622 seconds.

Gosh, with all those long involved answers, I feel left out by just giving my (correct) answer in a few simple words. So I feel moved to tell how I did it. I just laid it all out in PowerCADD (an excellent 2D drafting/design program for the Mac that I have used for the past 12 years)(unlike AutoCad which was originally discovered in King Tut's tomb). A big triangle with a 500m base and 750m altitude yielding a hypotenuse of about 901.387m. Then I added a 4km/h vector up from the top of the first triangle, the B point. Then added a second hypotenuse from the A point to the top of the 4km/h vector. This second hypotenuse represents the time/distance vector supplied by the boat moving at 10km/h. It is also the proper heading that the boat operator would take if he or she had sat down and planned it all out before starting the trip. The length of the second hypotenuse represents actual km/h speed, and dividing that into the length of the first hypotenuse yields the answer 8.421037176754 minutes.

A vector solution seems easier

To go across with no current will take 3 minutes (10km/hour and 500m)

To go upstream will take 7.5 minute (Actual speed is 10km-4km=6km/hour and 750m)

To go diagonally across the time is Square Root of (3 squared x 7.5 squared) = 8.078

8 minutes and 5 seconds

It may be possible only on Planet of Apes. On earth you can not stop the current while crossing the river and start it again while going upstream.

The current flows down the river. Hence I deducted off the current when going up the river

There is no current going across the river. Hence no correction needed

A vector solution seems easier

To go across with no current will take 3 minutes (10km/hour and 500m)

[at the same time the current of 4 will cause you to arrive at the other bank 200 downstream]

To go upstream will take 7.5 minute (Actual speed is 10km-4km=6km/hour and 750m)

[adding the 200 to the 750 = 950; therefore 9.5 minutes]

To go diagonally across the time is Square Root of (3 squared x 7.5 squared) = 8.078

[3^2 + 9.5^2 = 9.9624294225856375699831295947436^2]

8 minutes and 5 seconds

[= 9 minutes 57.75 seconds] [or 12.5 minutes if your compass only has the four cardinal points ei. straight up then straight across]

i believe this is the good answer except!

i don't have my nautical books nearby, but off the top of my head i believe that any boat of less the 25 feet will start hydroplaning at 6 knots (roughly 10km/hr) which will reduce the effect of the current (though not the 10km/hr which is a given in the question)

in other words, if Hercules were skipping a stone he would aim straight at point B, whereas a full draft boat would aim at 200m upstream (and hold that TRUE bearing). So size of boat must be known to find the actual time between 8m5s and 9m57s.

Theoretically you may be right. But practically, the boat crossing the stream will be dragged much more downstream, as stream flow will push the boat on all its length.

You need to use CFD for the real solution. Or you travel actually and find out what happens, (in various ways suggested by many people)

Being an avid boater and fisherman driving a boat at an angle to the current rather than straight on would cause the boat to run slower, more of the boats hull surface would be facing the current. To maintain 10 km/h you would have to add more hp. Is there a way to calculate the additional force applied to the side of the hull?

Yes, it is amazing, having lived along the Ohio river for a number of years, you do occasionally hear about an unfortunate, un-informed boating who is speeding up along side a barge and decides to cut across in front of it before placing a safe distance between them. The boater finds themselves in a life and dead race to save their skins and sometimes will lose the race. The moral is you can not stop a 200 ton barge on a dime. But the draft of the boat does place a lot of drag on the current.

You really didn't need to take your pad into the river to figure the answer.

Nice, completely-presented answer, albeit a bit fuzzy. You get my GA vote.

An elegant solution

When you solve for "t" you get a quadratic equation which has two roots. The positive one is 8.421 minutes and is the correct answer.

The second root gives a negative value of -4.135 minutes which (ignoring the negative sign) is exactly the time it takes to make the return journey, where the current assists.

I got the same two values using trig.

I suspect that many people solved the equations with a "computer solver package" and don't realize that there is a negative root, which has a meaningful interpretation in this case.

Travelling distance of boat = 901.38m

River speed = 4 kmph

River speed in the direction of diagon = 4 X sin (56.31) = 3.33 kmph

Speed of boat in still water = 10 kmph

So, total speed of boat = 10 + 3.33 = 13.33 kmph

Time for sailing through a total distance of 901.38 m at 13.33 kmph

= (901.38/1000) / 13.33 = 0.0676 hr = 4 min

OH my dear! You started boating from B to A.

Original question is for A to B.

A little correction

total speed of boat = 10 - 3.33 = 6.67 kmph

Time for sailing through a total distance of 901.38 m at 6.67 kmph

= (901.38/1000) / 6.67 = 0.0676 hr = 8 min 6 sec.

You got the correct answer. Guess we followed the same logic

A good strategy is to first estimate what a reasonable answer would be. Suppose you simply checked the time required to travel from A to B with no current. This would have to be .0901 hours, which equals 5.4 minutes. Given that the speed of the boat over the ground is slowed substantially by the adverse current, then you'd expect that answer to be more not less than 5.4 minutes.

lets see now, allowing for the curvature of the Earth....

You will need calculator of many many decimal points to calculate the effect of curvature of earth on the distance and difference of timings may be of mili or microseconds over this small distance of less than a kilometer

And the tide...which could make the river wider or the current faster, and...what day and time was this crossing going to take place?

The time required to cross the river is really a function of how many beers you have packed in the cooler.

This is absolutely correct

After 2 beers I will navigate a course which gets me there in the shortest time of 8min 5 seconds (I will point my boat at a spot further upstream than my destination)

After 6 beers I will just point my boat at the destination point and this takes 8 min 25 seconds

After 12 beers I will have forgotten why I was going there, and will turn around and go home

Whose home?

After all that beer, does it matter what direction home is, the turning around part may be a bit colourful, and the fish are then feeding on beer flavoured Burley

After 6 beers I will just point my boat at the destination point and this takes 8 min 25 seconds

Actually, to get straight to the destination across the ground, you must aim the boat upstream by about 13 degrees from the course over ground. It is under this condition that the 8.4 minute solution is correct. If, instead, you point the bow of the boat at the destination, and hold that heading, you'll miss your target. If you continually adjust heading to keep the boat pointed toward the destination, you follow a curve, as has been described and calculated by others.

There is no heading that can get you to point B (assuming it is on the opposite shore -- even though the question fails to state that clearly) in less that 8.4 minutes. If you are sceptical, there are e6b calculators available online which do these calculations. The one I referenced in another post needs to have the input figures multiplied by ten to provide adequate resolution on the outputs (otherwise, 6.4 kph shows up as 6).

After 6 beers I will just point my boat at the destination point and this takes 8 min 25 seconds

CORRECTION

I get 8 min 34 seconds

I think your calculations will be wayyyy off.

6 beers in 8m34sec

are you sure your going to find your distination?

AB = (500² + 750²)^1/2 = 901.38 m

With reference to the figure, we can see that the shortest path is AB; in order to follow this direction, the boat will have to compensate the river current OD, and so the vector of velocity of the boat will be oriented as line OC.

The component along OD of the vector OC will compensate the river current. In formula:

|OC|*cos β-|OD| = |OR|*cos α => cos β = (|OR|*cos α + |OD|) / |OC|

|OC|*sen β = |OR|*sen α => sen β = (|OR|*sen α ) / |OC|

But it is also : sen² β + cos² β = 1 and then :

(|OR|*sen α )² / |OC|² + (|OR|*cos α + |OD|)² / |OC|² = 1 =>

|OR|² + 2*|OD|*cos α * |OR| + (|OD|² - |OC|²) = 0

This is a 2^nd degree equation with unknown |OR|, |OD| = 4 km/h, |OC| = 10 km/h and

α = atan (500/750) = 33.69 deg.

We have 2 solutions taking in account that the river current could be favourable:

1) |OC| = 6.422 km/h

2) |OC| = 13.078 km/h

In our case the good solution is 1) and the required time for trip is :

T = AB/|OC| = 0.90138 / 6.422 = 0.14 h = 8 min 25 sec.

what is OC? OD? OR? etc.?

0.14 HOUR OR 8'25"

The time is 10.4 minutes. Posted by jfb.

The time will be 8.428 minutes. The boat has to be angled 20.85 degrees upstream. Resultant velocity is 6.417 km/h

Regards

E Murphy

Would have been more elegant with a 3, 4, 5 triangle.

THE CORRECT ANSWER IS: 9 MIN 7 SEC.

Must adjust for additional force placed on hull driving diagonally to current!!!!!

Has any one here actually driven a boat on a river with a current. First you can't steer the boat diagonal to current with out adjusting for the currents push on the side of the boat. If no adjustment is made the current would push the front of the boat down stream. You must turn into the current to keep the boat on a straight course to reach the other side. You must calculate the additional energy used to keep the boat on course. If it was as simple as steering directly into the current then you would subtract current by speed of boat in still water (which is how most of you have calculated). More of the boats hull is in contact with the current driving at a diagonal than straight into the current. Sorry if my math background does not allow me to calculate this addition to the problem, but experience on the river tells me that this is a factor that must be calculated to get a real time.

Thanks all Big Ed. If I'm wrong about this please explain. I'm willing to listen and learn.

In addition to my previous novel:

Neither aircraft (in coordinated flight) nor boats can be made to "crab" relative to the fluid in which they move (other than when a boat is anchored, or otherwise tied to the ground). (Both often crab relative to the ground however.) A plane can be made to crab (relative to the air) by applying elevator in one direction and rudder in the opposite (called slipping or uncoordinated flight). This is only done (commonly) for two reasons. One is to increase drag to steepen a flight path for a landing on something like a grass strip. (It's never done in airliners for several reasons, but if it were, the passengers would be uncomfortable, because it feels like the plane is leaning to one side.) The other common usage (in both small planes and large) is to align the fuselage of the plane with the runway after you have arrived there by flying a wind correction angle along the approach path. In a small plane you might arrive at the runway, traveling along its extended centerline, but heading to the left by 10 degrees. If you simply land like that, the tires screech or you run the risk of going off the runway. So you drop the left wing, and apply right rudder, which aligns the fuselage with the runway and you touch down, banked, on the left wheel. When you learn to do this it makes you squirm. Hold the "slip" too long and the plane remains parallel to the runway, but drifts sideways.

So, other than for a couple seconds here and there, wind never "pushes" on the side of an airplane in flight -- the flow is always perfectly straight over the nose. The same is true for a boat in a river. The entire mass of water is moving along, and the boat has it's own motion relative to that mass, with the water always coming straight into the nose.

Thanks ken

Big Ed

It's a long time since I've been sailing, but I have a distinct recollection of considerable slippage downwind when sailing in shallow water (because I've needed to raise the centreplate).
Even for motorised craft on a shallow river, wouldn't the effective velocity of the water depend on your angle of attack?

I don't know whether you would care to describe either of these effects as crabbing?

Sailing adds considerable complexity. There is a good vector drawing in Marchaj's classic book "The Aerohydrodynamics of Sailing." I've been unable to find mine (which is sad, because I had to search several rare books stores to get it). But I think it was on page 750something of the last edition that he provides the vector diagram of the various components at work. Maybe 12-15 arrows?

Only sailboats routinely have a continuous side force on them. That is what enables them to sail on anything other than a perfectly straight course downwind. On my Windrocket

when sailing at its optimum efficiency (about 3 times wind speed, and just below a beam reach) the side force was about 4 times the forward (propelling) force. (When you subtract the rearward vector of all the various drags, you come away thinking that it's a miracle that the things sails at all.)

Even with the board down, a sailboat must sail with leeway (and the board cannot generate a lift force to resist that leeway unless there is an angle of attack other than 0 degrees, assuming a symmetrical board profile. Nut cases like me and others have made asymmetrical boards, and boards w/ adjustable angle of attack... but sane designers are content with the entire boat being dragged through the water partly sideways, making about 2 to 4 degrees of leeway [and much more at very low boat speeds, and especially when accelerating from low speed]) So yes, I'd describe a typical sailboat as crabbing all the time (other then when headed straight downwind with a spinnaker perfectly squared -- which, on my boat, is it's slowest point of sail other than, of course, when it is "pinching" into the wind)

Other boats (unless using bow thrusters and the like) cannot do anything but move en masse with the current (in somewhat the same way that people near the equator cannot help but move along with the Earth's surface at 1000 mph. There is no side force on our feet, because we are not being accelerated by earth contact; there is no side force on the boat, because it is not being accelerated by the current.) A typical motor boat would require fore and aft rudders both set to parallel angles to crab. A kayak or canoe can be made to crab by "parrying" but in the ordinary straight line movement of such boats, there is no relative side flow component at all. (When turning, a boat's hull generates lift to the side -- otherwise, the boat would only yaw, but keep moving in the same direction through the fluid).

So, in ordinary straight line travel, the typical motor boat or paddle boat has a zero degree angle of attack with its fluid: all flow is straight over the nose, just as it is for an airplane in flight, no matter the perceived crosswind from the ground perspective.

Even for motorised craft on a shallow river, wouldn't the effective velocity of the water depend on your angle of attack?

If you mean angle of attack with the water, it is zero, no matter what the current (ignoring turbulence). If you mean angle of attack with the wind, then yes, a motor boat crabs (generates leeway) just as a sailboat does, if there is any wind relative to the boat. (I put a link elsewhere in this thread re sailing in zero wind -- from ground perspective -- in a large river with current: if you can't see the shores, you'd never know you weren't sailing with a nice breeze in the middle of the ocean.)

Of course, one simplifying assumption (among many) that we've been making with this challenge question is that either the air is massless, or it is moving at the same speed and direction as the current.

"Even for motorised craft on a shallow river, wouldn't the effective velocity of the water depend on your angle of attack?
If you mean angle of attack with the water, it is zero, no matter what the current (ignoring turbulence)."

Presumably the water moves slower near the river-bed. So a shallow river will display considerable vertical sheer, whether turbulent or not - though I imagine 4-km/hour would be turbulent unless the river was very shallow indeed (can't be bothered to do the sums, even if I knew what Reynolds number to use). So the lower parts of the craft will experience different water velocities to the upper - and the downriver component of the velocity will depend on the relative weighting versus depth - which will presumably change with the direction in which the craft is pointing.

We may be drifting far from the intended true course of the question. But such is the ebb and flow of the discussions we encounter in the river of life.

For the purposes of the question, I suspect we can assume the water velocity to be uniform at every point across the river and through its depth. However, if there is a significant velocity gradient from waterline to the keel, then, depending upon the depth at which the current velocity was measured, the boat could appear to be crabbing (i.e., having a nonzero angle of attack) at the surface or at some point below. I suppose an obsessive experimenter might measure the current at the depth of the boat's center of lateral resistance. If that were the case, then on the surface, the angle of attack would be slightly upstream and at the keel, slightly downstream. The boat would heel slightly.

I'm beginning to get sea sick.

Here's another way to skin this cat. Assuming your boat is round and the current will effect it consistantly regardless of the angle of attack you take. Also assuming you aren't going to take the time to figure out the speed of the current, or formulate the vector quantities that you must set out at in order to offset the current, and land exactly at point B. If you were to just jump in the boat, start the motor, aim at point B, and correct the angle, say every 10 seconds-what would happen? At the end of the first 10 seconds, your boat has moved relative to the water 27.78m at 56.31 degrees from the perpendicular width of the river. The distance you have covered widthwise is COS(56.31)*27.78 or 15.41m. The distance you have covered upstream is SIN(56.31)*27.78 or 23.11m, but the water has moved 11.11m relative to the shore. So your net upstream is 12.00m. So after 10seconds, you have now another 484.59m widthwise to cover, and 738m upstream to cover. If you take ArcTAN(738/484.59), your new angle to point B is 56.71deg. If you repeat this every 10 seconds, you will end up at 509 seconds, 10.48m downstream of point B, and 20cm away from shore. At 515 seconds you will arrive at the dock. (If you set up the equasion to correct every second, the time is 514.5seconds). So my answer for this method is 8 minutes, 34.5 seconds.

It seems with all these avriations of timings, we are reaching no where.

I'm afraid we already reached the correct answer way back in post #1. If we assume the reasonable simplifications of constant current from shore to shore, instantaneous acceleration to the quoted boat speed, ignore the affect of apparent wind on the superstructure of the boat, etc, then the situation is as drawn in post #19. Several others have come up with the same answer: .14 hr.

Pilots do this sort of calculation all the time, and they used to use an e6b calculator for the purpose.

Nope all wrong!

The question didn't state minimum time it would take, just how long will it take.

My answer is 5 hours.

Guys, if your missus let you out in the boat for a day would you either:

1. Get to the other side as quick as you can to prove a point OR

2. Invite a mate over, buy a 24-pack (slab/box/carton) of beer, grab your rod, reel and tackle and go for a bit of a fish and get a bit drunk.

Mmmmmmm.

Hard answer.

Mate, if it was me, I'd be trying to get to my gf on the other side of the river as fast as I can.

Is the boat big enough to carry a fox a goose and a bag of grain too? Or just two at a time?

Del

What with the diversion to biofuels, I couldn't afford the bag of grain or the goose. Had to make do with a goat and a cabbage. (Oh, and there was a problem getting diesel for the boat, so had to row - just about managed 3-km/hr - fortunately the water was flowing slower near the banks).

P.S. I'm jiggered

Correction: I miscalculated solution of quadratic equation in my previous entry.

Proper ansewer is: 8 min 24 sec.

The easy way is to solve the problem grafically: I draw the triangle formed by the wideness of the river (side A), the distance upstream (side B) and the distance to travel (side C). Measuring or calculating by Pitagoras' side C is 901m. in point A I draw a vector proportional to current velocity (4); from arrow of this vector I open a divider for a velocity of 10 (in the same scale) and intersecate the path across the river (side C). The distance from starting point to the point so found is the advancing velocity: 6.3km/h; the time taken from the trip will be 8m35s.

around 8.4 seconds

Miles per hour, not miles per minute?