## Fact are facts, Watson !

Posted by rajesh on October 17, 2011 – 8:22 pm

“As per Inspector Lestrade, *the accused house is on the bank of a river which is quarter mile wide and has current of 2 mph*“. said Dr. Watson

“Don’t forget Watson, Just opposite his house, on the other bank, is victims house. The accused can swim across to his house.” Sherlock Holmes added.

“But its impossible for him to reach there in less than ten minutes”. exclaimed Dr. Watson.

“The fact are, the accused can swims at 2.5 mph in still water and his walking speed also is 2.5 mph.”

“Can he do it any faster by a combination of swimming and walking?

October 17th, 2011 at 11:11 pm

If the accused swam directly across at 90 degrees to the near bank he would reach the far shore in 6 minutes but in doing so he would be swept 0.2 miles downstream and would therefore need a further 4.8 minutes to reach the victim’s house on foot, totalling 10.8 minutes. (Why couldn’t he run?).

If, however, the accused were to apply Pythagorean geometry and swim at an angle upstream as well as across then to land directly at the victim’s house he would have to swim a distance proportional to the hypotenuse of his and the current’s speeds, i.e. sqrt(2.5^2 + 2^2) = 3.2. So, instead of swimming 0.25 miles he would need to cover 0.25 x 3.2 / 2.5 = 0.32 miles, requiring 7.68 minutes.

October 18th, 2011 at 4:49 pm

If he swims upstream to a point 900 ft from the house it will take him 7.26 minutes (approx) then to walk the 900 feet it will take about 1.71 minutes, making the total time less than 9 minutes.

October 18th, 2011 at 4:54 pm

But then Watson says “Swimming speed the same as walking? … I think that wrong… he could not swim that far that fast!”

October 18th, 2011 at 5:20 pm

Hi Wiz. I take it you were intending that he should swim at an angle such that he ends up actually going straight across. In that case his speed across the river is sqrt(2.5^2 – 2^2) = 1.5 mph. So he’d take 0.25/1.5*60 = 10 minutes – leaving no time to do steal the portable walrus polishing kit. You were attempting to give the hoped for sucker’s answer (tee-hee).

Ragknot has given the correct answer (with a small rounding error), but hasn’t given us a clue about how he did it (or how come working in feet came about).

I calculate that he could do it in approx 8.97997772825745932540099695755971832421444753866894467112898912156808

4375136653667144343748466510972 minutes.

October 18th, 2011 at 7:57 pm

Solve it?

Change variables to feet and minutes.

River width = 1320 ft

Swim speed = 220 ft/min

Walk speed = 220 ft/min

Water current = 176 ft/min

,,

Say he swims straight across the river. It would take 6 min,

but the current would take him 1056 downstream, that would take

4.8 minutes to walk upstream toward the house. Total time – 10.8 min,

..

Ok, he will need to swim at an angle upstream, it will take longer, but

at least he won’t have to walk as far.

,,

The guy will have to figure the angle to continuing swimming. But

we don’t have to, we just need the distance upstream to compute

the time for that angle. Let’s say the distance upstream from the house is X

Then the computation would be…

..

Swim min = sqrt(1320^2+X^2) / 220

WashDown ft = swim time * 176

Walk min =(Washdown – X)/220

..

Let guess X = 500 ft

Swim min = 6.416018178

Washed ft = 1129.219199

Walk min = 2.860087269

Tot min = 9.276105447

That would less than ten minutes. You can then solve for a minimum time, if you want, but from a quarter of a mile away, and across the river, the X won’t matter much it you figure to the nearest foot.

October 18th, 2011 at 8:50 pm

I’ll post a mathematical solution tomorrow, unless someone beats me to it.

October 18th, 2011 at 9:09 pm

I reviewed to find a minimum time at 881.9621 feet, which gave me 8.979977728 minutes.

October 19th, 2011 at 2:39 am

Hi Chris and Ragknot,

I initially read your answers to mean that the swimmer lands 900 feet UPSTREAM from his destination then walks the 900 feet downstream. That doesn’t make any sense. If on Chris’s figures it takes 10 minutes to land straight across then it must take longer to land further upstream. So I assume that you really mean that he lands 900 feet DOWNSTREAM and then walks the 900 feet upstream.

October 19th, 2011 at 3:02 am

If you want to picture this with movement, let’s figure his movement per second. The starting place is x=0,y=0 and the house is at (1320,0) in feet. He will swim x=+3.04858 feet and y=+2.037037 each second, but the water will push him at y=-2.933333 feet.

..

In 432.9632 seconds his position will be at (1320,-388.0633)

…. (which is 388 feet from the house,105.835 seconds walking, for a total time of 538.7987 seconds)

If the water speed was zero, he would be at (1320,881.96208)

… which gives his “swimming angle direction” but of course the water would be pushing him 1270.025 feet in the -y direction.

October 19th, 2011 at 3:14 am

Hi Ragknot, So the swimmer “aims” a little bit upstraem of his destination but the current carries him a little bit downstream, and 882 feet is the optimum distance downstream.

That’s all I was asking. Thanks.

October 19th, 2011 at 3:15 am

No, Wiz… the 900 feet was for the direction (or angle) of his swim. I picture the house at 1320 feet due east of his starting point. He would begin swimming toward a point 900 feet north of the house (and hold that same angle), but the water would be pushing him south and he would end up at 388 feet south of the house.

I should have said 882 foot north of the house… that would then end at 388 south.

October 19th, 2011 at 3:18 am

LOL, Wiz, I wrote that while you were also writing.

October 19th, 2011 at 9:56 am

copy/paste

Let the suspect swim at angle A (with respect to the water, not the ground). A = 0 => he’s swimming straight across and A = 90° => swimming exactly upstream. (If A = 0, an observer on the bank would see the swimmer moving downstream at 2 mph and crosswise at 2.5 mph).

His speed across is 2.5 cosA. So it takes time Ts = 0.25/(2.5 cosA) = 0.1/cosA to cross the river. He will be swimming slightly against the current with a speed 2.5 sinA (with respect to the water), and so will have a total ground speed (downstream) of

2 – 2.5 sinA. So he will have moved downstream (with respect to the ground) a distance, d = (2 – 2.5 sinA) 0.1 /cosA. He then has to walk to the house. That takes time Tw = d/2.5 = 0.08 / cosA – 0.1 tanA

The total time, T = Ts + Tw = 0.18 / cosA – 0.1 tanA

At the minimum T, dT/dA = 0 i.e. dT/dA = 0 = 0.18 sinA / cos²A – 0.1/cos²A

=> sinA = 1/1.8 => T = 8.98 mins (after converting hours to minutes and some sordid calculating ).

The Win 7 calculator gives 8.9799777282574593254009969575597

Wolframalpha agrees.

NB d/dA (1/cosA) = sinA/cos²A and d/dA (tanA) = sec²A = 1/cos²A (= 1 + tan²A)

October 19th, 2011 at 3:07 pm

Nice work Chris. I have to admit that my own calculus is but a distant memory. I would have had to do some serious revision to come up with something like this. But I can see where my assumption that swimming straight to the house was too lazy.

October 19th, 2011 at 3:37 pm

Hi Wiz. Thanks for the compliment. But, it’s pretty basic stuff really.

Also, I only really took the Michael because you went for the quickie solution that I went for when I first saw the problem.

Now check my last post (to you) in Girls, girls, girls

October 19th, 2011 at 4:40 pm

It is not necessary to use calculus to find out the approximate answer. The solution posted by Chris writes the equation for three different steps and then works to the derivative to find the minimum total time. I broken the solution into different parts, swim distance, downstream wash distance and the walk distance. Each part included the “angle”, as the distance upstream from the house. I found that each distance from about 10 degrees to almost 55 degrees would be less than 10 minutes.

..

And the angle of about 33.75 degrees or 882 feet upstream from the house would be slightly less than 9 minutes. The distance to that point gives the swimming time, and the swimming time gives how far the water would push him downstream, which was 388 downstream from the house. That gave the walking time.

..

If he could have painted a line on the water surface from the point of beginning to 388 feet downstream of the house, he would have his path marked, but to stay on that he would have to direct his swim to the 33.75 degrees upstream to stay on the path while the current moved him more downstream.

..

That painted path would be a straight line that combined the current push to his swimming push at 35.75 degrees toward upstream. It is simple to figure these separately and combine them.

To figure the problem, you should understand the current velocity depends on the depth and distance from the bank, making the usual middle 1/3 to be much faster than the outer 2/3 of the total width. Sherlock Holmes emailed me that he knew that already.

Thanks Rajesh for the great ToM, and Thanks Chris for your neat solution, and showing how that complex solution works.

October 19th, 2011 at 5:05 pm

Hi Ragknot. You’ve have to admit that the maths way is more stimulating than trial and error.

But you did good in realising the possibilities and getting the result.

October 19th, 2011 at 5:22 pm

Hi Chris, I didn’t realise that Girls x 3 was still going after all this time (9 months). I’ll have to catch up on all the 132 posts and see if I can come up with a suitable reply.

October 19th, 2011 at 6:18 pm

Hi Wiz. Just check the last three. Alhough there’s possibly quite a lot of good stuff throughout.

October 30th, 2011 at 11:39 am

he can walk upstream on the bank then get in the water going the direction of the victims house and swim at the same time