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Wizard
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Wizard
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April 5th, 2020 at 4:43:53 PM permalink
I've spent entirely too much time on this problem and still have a long ways to go. I see others have posted since I was last here and are probably ahead of me. However, I want to post what I have accomplished, hopefully correctly, for my own benefit at least.



Suppose the selfish walker starts at pt A = (0,0) and walks right along the X axis. Thus, the unselfish walker starts at point E = (12,0).

The circle in this diagram is the 6-ft. radius of safety.

I submit that the unselfish walker should aim for a point to be soon on the tangent of this circle.

When the selfish walker reaches point B, the unselfish walker should reach the tangent of the circle. In other words, AB = EF

Here are some things we know:

  • Angle BFE is 90 degrees, because the unselfish walker walked straight to the tangent point.
  • BF = 6, because B is the center of the circle and F is along the circle's edge.
  • BE = 12 - EF, because AE = 12, AB + BE = AE, and AB = EF


From the Pythagorean forumula:

BF^2 + EF^2 = BE^2
6^2 + EF^2 = (12 - EF)^2
36 + EF^2 = 144 - 24EF + EF^2
24EF = 108
EF = 9/2 = 4.5

There are two ways to measure the area of triangle BEF. One is half the square of sides length BF and EF. That would be (1/2) * 6 * 4.5 = 13.5

The other is (1/2) * base * height.

We have established that EF = 9/2, and AF = EF, so AB = 9/2. Thus BE = 12 - AB = 15/2.

The formula for the area that way is (1/2) * CF * 15/2 = (15/4) * CF.

Equating the two formulas for the area of triangle BEF:

13.5 = (15/4) * CF
CF = 13.5 * (4/15) = 18/5 = 3.6.

Using more simple applications of Pythagorean, we find:

AB = 4.5
BC = 4.8
CE = 2.7

So, when the unselfish walker reaches the circle of danger of the selfish walker, he will be at point (9.3,3.6).

Part 2, will be much harder. I assume the unselfish walker will walk along the circle as it moves right along the x axis. Here is where we'll need some calculus.

I think this square wheel video will come in handy.

Last edited by: Wizard on Apr 6, 2020
It's not whether you win or lose; it's whether or not you had a good bet.
Gialmere
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April 5th, 2020 at 5:17:23 PM permalink
I actually came across this problem when I was a kid except both walkers were selfish. The solution was...

...to build civilization around them.



Never budge! That's my rule. Never budge in the least!
Not an inch to the west! Not an inch to the east!
Have you tried 22 tonight? I said 22.
Wizard
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Wizard
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April 5th, 2020 at 5:35:25 PM permalink
Quote: Gialmere

I actually came across this problem when I was a kid except both walkers were selfish. The solution was...

...to build civilization around them.



Never budge! That's my rule. Never budge in the least!
Not an inch to the west! Not an inch to the east!



I've seriously wondered what happens when this happens, no joking aside. It happens a lot that two men are headed straight towards each other and one will absolutely not veer out of the way, even a little. When this happens, I'll take the moral high ground and get out of the "selfish walker"'s way. However, I'd like to see what happens when two of them have a face off.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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April 6th, 2020 at 3:58:43 PM permalink
Quote: Wizard

So, when the unselfish walker reaches the circle of danger of the selfish walker, he will be at point (9.3,3.6).



After some calculus, which I'll explain later, I show that when the selfish walker reaches his destination (12,0), the unselfish walker will have been pushed back 4.5 units, and risen sqrt(15.75)-3.6 = 0.3686 units to y = sqrt(15.75) =~ apx. 3.9686. In other words, he will be at point (13.8, sqrt(15.75)). This is 14.3593 directly away from the destination. Considering he must go up further to get around the danger circle, it will take much longer.

He could have just gone 6 units up the Y axis to start with and have been in better shape.

So, I think my whole strategy is flawed. As I think about it, he is mainly walking ahead to just get pushed back.

Sorry to have wasted your time.
It's not whether you win or lose; it's whether or not you had a good bet.
petroglyph
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April 6th, 2020 at 4:30:46 PM permalink
Have you noticed when in a "clean" room, meaning no one has smoked in it recently, to the point that it doesn't smell like cigarette smoke. And then someone lights up, how far the smoke pollution from just one cigarette will carry amazes me. Especially if viewed in sunlight rays coming through a window.

Not sure I read these links correctly but, if a smoke particle is twice the size of a Covid 19 particle, would it be reasonable to assume that when in a room, if you can smell cigarette smoke [or see it] that were the smoker carrying the virus and you can smell the smoke, you have a good chance of inhaling the virus along with the secondhand smoke?

Anyway, that's how I see secondhand smoke, as carrying a plague.

https://smartairfilters.com/en/blog/coronavirus-pollution-masks-n95-surgical-mask/

https://www.ncbi.nlm.nih.gov/pubmed/2751166

I think these links indicate a Covid particle is about 1/2 the size of a smoke particle.

My reason for even posting this is, we all know how far smoke travels in a room, I consider that the virus travels at least half as far as smoke. I think that droplets of virus on the floor can re aerosol from being stomped on, so they get around further than 6 ft?

Ever come home from hours in a Vegas casino, or a few hours in a smoke filled bar, toss your clothes in a pile and shower, just to come out of the shower and smell the stinky pile? Think of the stink as virus, take off your outside clothes and put them in the washing machine.

First time I ever saw a laundry [washer and dryer] in a bathroom was at my girlfriends [wife] apartment. It made so much sense to me, to disrobe in the shower room, toss dirty clothes in the machine, instead of having a laundry basket and saving them up for laundry time. Step out of the shower and press, wash. Easy peesy.
ChesterDog
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April 7th, 2020 at 4:55:31 PM permalink
Quote: Wizard

...


Let's say the horizontal movement is straight ahead along the sidewalk, as if the other guy wasn't there.

I think you will start out walking in either direction 60 degrees away from straight ahead. After walking 6 feet, you'll be 6 feet away from the rude guy walking straight ahead.

Then it gets tricky. I think you'll take some kind of an altered cycloid path -- getting pushed back horizontally and up vertically. I think at some point you'll be 6 feet vertically from the straight walker. Then continue on this curved path a little until you can leave it along its tangent line and head straight to the point 12 feet directly ahead of where you started.

I think this is going to involve parametric equations and arc length. Here is where the air starts to get thin for me.

I tend to think I won't be the first to solve this one, knowing the brain power here. Whoever solves it first, please give a good solution.



Using your idea of walking in a direction 60 degrees from straight ahead for six feet, then keeping a constant 6 feet from the straight walker, and then leaving this curved path straight to the destination, I get a total distance of 17.4469 feet.

I used the following numerical method. I considered a moving 6-foot-radius circle centered on the straight walker, and a circle of radius 1/1000 foot about the curved-path walker. One of the intersections of the two circles gives the next approximate location of the curved-path walker. Finding the slope of the segment between two adjacent points and comparing it to the slope of the line from the path to the destination tells when to end the curved path.

The total path of this method gives 6 feet for the initial straight path + 2.7240 feet for the curved path + 8.7229 feet for the final tangent walk for a total of 17.4469 feet.

The above method assumes both people walk at the same speed as each other. (They can both speed up or slow down, but they must do it simultaneously.) However, if the 2nd walker is allowed to pause while the other continues walking, he can do the initial 60-degree walk but all the way to 6 feet from the center line and pause, letting the other guy pass. Then he can walk straight to the destination. This total walk would be only 17.3619 feet.

And if the 2nd guy is allowed to run, he can run to the point (6,6), let the 1st guy pass, and then walk to his destination for a total of 16.9706 feet.

I guess a calculus method can be derived from the numerical method, but I'm stuck.
ssho88
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April 7th, 2020 at 6:19:55 PM permalink
deleted
Last edited by: ssho88 on Apr 7, 2020
charliepatrick
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April 7th, 2020 at 8:33:48 PM permalink
If the Walker heads off at 60 degrees then when has gone 6 feet the Walker has to move out (i.e. away). Thus it is best if the walker heads off at more than 60 degrees. Using a spreadsheet, not the best way but to only way I could think of to get near any sensible solution, I created an arc and then looked for the point where is was tangential to the starting point, for joining it, and to the finishing point, for leaving it. There was an element of recursion as one didn't exactly know where the Person would be until you knew the distance the Walker needed to travel to get to the 6 foot line.

Wizard
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April 8th, 2020 at 1:01:03 PM permalink
I get an answer of 21.5344. Given Charlie comes in much less, I won't bother to explain mine. I'll just summarize.

1. Both walk 3.3981 feet. The unselfish walker will walk at such an angle he touches the danger circle after the 3.3981 seconds.
2. The unselfish walker will walk at an angle twice that of the tangent line to the direction of travel of the selfish walker. The reason for doubling the angle is to keep from entering the danger circle, as it is approaching you. The unselfish walker will do this until he reaches the top of that circle. Using the coordinate system I explained before, the unselfish walker will be at point (11.55007666,6) at this time.
3. The short walker will walk along the back side of the danger circle a very short time, until his direction would take him directly to point (0,0), at which point he will do that.

I have not fussed with the "very short time" part of the puzzle and had the unselfish walker walk straight back when he got to the top of the circle. Total time spent is 21.5344.

I hate to question Charlie, but the curved part of your graph doesn't seem to reflect that the selfish walker is moving towards you. I think the part from Q to S should be steeper and from S to U more flat.
It's not whether you win or lose; it's whether or not you had a good bet.
charliepatrick
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Thanks for this post from:
tringlomane
April 8th, 2020 at 3:17:52 PM permalink
Quote: Wizard

...I hate to question Charlie, but the curved part of your graph doesn't seem to reflect that the selfish walker is moving towards you. I think the part from Q to S should be steeper and from S to U more flat.

Please do as my assumptions may be false.

I started where you (the Walker) were on the six foot path opposite the selfish one (the Person) (DS) and then worked out how you could gradually move, getting nearer to the path but maintaining the six foot distance.

In the first case the Person continues Northbound (in my diagram upwards from D) while you head Southbound (down from S). As the Person moves a distance (say .001) North, the curve you take will have length of .001, ideally maintain six feet apart at all times (my estimate actually ensured you used a straight line at right angles to where the Person was, so it gradually got further away but then occasionally made a correction to back back closer), and move you towards the path.

If the curve thus drawn keeps you at six foot away (or as near as makes no difference) while the person walks North, then by reflected symmetry the same curve (reflected) will work as the person is still walking towards D and you are have yet to reach S.

The only difference is when you join/leave the curve as your distance from the start will be shorter than your distance to the finish. This is why you join the curve earlier, and leave it sooner, as the straight lines PQ is steeper (or makes less progress South) than the UV line.


A few interesting ideas might come from looking at a "tractix" ( https://en.wikipedia.org/wiki/Tractrix ), the problem I see is the distance covered by the puller is much greater than the end of the pole.

There's also a list of curves at https://www.matematica.pt/en/useful/list-curves.php and the concept of an involute. Cycloid looked interesting, take a point on a circle and roll the circle along a straight line, look at how the point moves. Then consider extending the diameter 2r. The problem is the cicrcle moves 2 Pi r whears the point moves 4r.

I think the trick is to try and find which curve it is!

Edit: Just found this - might this be an idea!
Last edited by: charliepatrick on Apr 8, 2020

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