Social Distancing in an Elevator
By the way, I don't know the answer to this one - at least, not yet...
"Apparently," the standard size of a hotel elevator is 80 inches wide and 65 inches from front to back.
Assume the door opening is 60 inches wide, centered on the long side.
Three people get into an empty elevator, one at a time.
What is the largest possible "clear space radius" that any of the three can maintain from when they get on until when they get off?
Assume that the order in which the three get on and off maximizes the answer.
Quote: MoscaThey form an equilateral triangle, with the first guy going to one of the far corners.
I have a feeling that, if the first person goes into a far corner, the second one goes against the other side wall but not quite all the way to the back, and the third stays as close to the front as possible but not quite centered, then the shortest distance between two of them would be larger.
In NYC, many places have what we call "Jewish" elevators. On the Sabbath, some Jews are not allowed tp use phones or push a button and the elevators are programmed to stop o every floor.
Quote: billryanWho pushes the button ?
In NYC, many places have what we call "Jewish" elevators. On the Sabbath, some Jews are not allowed tp use phones or push a button and the elevators are programmed to stop o every floor.
It's probably like the elevators at Thed - er, The D - where you select a floor before you get in the elevator.
Let A be in the back left corner, B be in the back right corner, and C be in front, halfway between the two sides.
A and B are 80 inches apart; A and C are sqrt(652 + 402) = 76.322 inches apart, as are B and C.
If, say, C moves forward, then the distance from B to C is less than 76.322, unless C moves to the left, but this would make the distance from A to C less than 76.322.
Therefore, any solution requires that A and B be in the back somewhere.
Suppose B moves to the center of the back. The maximum distance from B to C that would be less than the distance from A to C is 76.322, when B is in the front right corner, but the distance from A to B is only 40.
Moving B to the left shortens the distance from A to B; moving B to the right shortens the distance from B to C.
76.322 inches appears to be the optimal result, with two in the back corners and the third in the center of the front.
Quote: billryanWho pushes the button ?
In NYC, many places have what we call "Jewish" elevators. On the Sabbath, some Jews are not allowed tp use phones or push a button and the elevators are programmed to stop o every floor.
Doesn't getting into the elevator at all cause it to consume more electricity? I thought the spirit of the Sabbath rules were shun modern conveniences and not consume power.
But if they are wearing masks. It would be more important to not bump into someone.
Quote: ThatDonGuyIt's probably like the elevators at Thed - er, The D - where you select a floor before you get in the elevator.
Let A be in the back left corner, B be in the back right corner, and C be in front, halfway between the two sides.
A and B are 80 inches apart; A and C are sqrt(652 + 402) = 76.322 inches apart, as are B and C.
If, say, C moves forward, then the distance from B to C is less than 76.322, unless C moves to the left, but this would make the distance from A to C less than 76.322.
Therefore, any solution requires that A and B be in the back somewhere.
Suppose B moves to the center of the back. The maximum distance from B to C that would be less than the distance from A to C is 76.322, when B is in the front right corner, but the distance from A to B is only 40.
Moving B to the left shortens the distance from A to B; moving B to the right shortens the distance from B to C.
76.322 inches appears to be the optimal result, with two in the back corners and the third in the center of the front.
Its a viewpoint as to whether the button constitutes the use of fire on the Sabbath or something like that.Quote: WizardDoesn't getting into the elevator at all cause it to consume more electricity? I thought the spirit of the Sabbath rules were shun modern conveniences and not consume power.
Quote: WizardDoesn't getting into the elevator at all cause it to consume more electricity? I thought the spirit of the Sabbath rules were shun modern conveniences and not consume power.
I believe some of the Ultra-Orthodox subscribe to that, but most observant Jews that I know don't. A few years back, some influential Rabbi ruled against their use but then a bunch went the other way. I'm sure the brightest minds are contemplating this and will come to a consensus in the next few decades. I think the decision on call waiting is almost done.
Quote: WizardDoesn't getting into the elevator at all cause it to consume more electricity? I thought the spirit of the Sabbath rules were shun modern conveniences and not consume power.
I have a very religious brother. We've talked about this in the past.
There are two types of elevators. A) Those that run slower as the load increases and B) Those that use more energy to maintain speed with higher loads.
Note that because of the weight of the elevator itself as well as the counterweight, the change is relatively small. Regardless, the rabbis debated it and decided the first type of elevator is acceptable for use as a Shabbat elevator.
Regarding the "shun modern conveniences" point: No. It's only about directly affecting / changing things. Therefore, while you can't turn lights on, the light from fixtures that are left on, or even those that are on programmed timers is acceptable. Similarly, we can benefit from heat & AC even though our presence, and opening/closing doors and windows may cause the thermostat to kick in sooner, because that is indirect and consequential.
On a side note, this year the rabbis gave serious consideration to allowing families to have Passover Seder via FaceTime or Zoom, etc. In the end they decided against it because A) there would be a temptation to handle the device for whatever reason, and B) people would change their procedures, speak louder, whatever, because they were in a broadcast conversation.
Quote: MoscaQuote: ThatDonGuyIt's probably like the elevators at Thed - er, The D - where you select a floor before you get in the elevator.
Let A be in the back left corner, B be in the back right corner, and C be in front, halfway between the two sides.
A and B are 80 inches apart; A and C are sqrt(652 + 402) = 76.322 inches apart, as are B and C.
If, say, C moves forward, then the distance from B to C is less than 76.322, unless C moves to the left, but this would make the distance from A to C less than 76.322.
Therefore, any solution requires that A and B be in the back somewhere.
Suppose B moves to the center of the back. The maximum distance from B to C that would be less than the distance from A to C is 76.322, when B is in the front right corner, but the distance from A to B is only 40.
Moving B to the left shortens the distance from A to B; moving B to the right shortens the distance from B to C.
76.322 inches appears to be the optimal result, with two in the back corners and the third in the center of the front.Wouldn't it still have to be an equilateral triangle? A in the far left corner. B in the far right corner. C in the front. Then they adjust so that they are equidistant from each other, along the walls to maximize the triangle. B would move forward a little, along the wall, increasing his distance from A, C would move a little to his left. Also, they enter and leave in reverse order. A goes to a corner first, leaves last. Then B, then C. But again, the answer has to be an equilateral triangle.
If they are in an equilateral triangle, then they are all 75.06 inches apart. By spreading A and B out to the corners, they are 80 inches from each other, and 76.322 inches from C.
Quote: ThatDonGuyQuote: MoscaQuote: ThatDonGuyIt's probably like the elevators at Thed - er, The D - where you select a floor before you get in the elevator.
Let A be in the back left corner, B be in the back right corner, and C be in front, halfway between the two sides.
A and B are 80 inches apart; A and C are sqrt(652 + 402) = 76.322 inches apart, as are B and C.
If, say, C moves forward, then the distance from B to C is less than 76.322, unless C moves to the left, but this would make the distance from A to C less than 76.322.
Therefore, any solution requires that A and B be in the back somewhere.
Suppose B moves to the center of the back. The maximum distance from B to C that would be less than the distance from A to C is 76.322, when B is in the front right corner, but the distance from A to B is only 40.
Moving B to the left shortens the distance from A to B; moving B to the right shortens the distance from B to C.
76.322 inches appears to be the optimal result, with two in the back corners and the third in the center of the front.Wouldn't it still have to be an equilateral triangle? A in the far left corner. B in the far right corner. C in the front. Then they adjust so that they are equidistant from each other, along the walls to maximize the triangle. B would move forward a little, along the wall, increasing his distance from A, C would move a little to his left. Also, they enter and leave in reverse order. A goes to a corner first, leaves last. Then B, then C. But again, the answer has to be an equilateral triangle.
If they are in an equilateral triangle, then they are all 75.06 inches apart. By spreading A and B out to the corners, they are 80 inches from each other, and 76.322 inches from C.
I think the answer is more interesting if the elevator is a square rather than the rectangle dimension you set up. In other words, if the shortest distance is between the two people in the corner as opposed to the two legs of the iscosoles triangle, I think (but haven’t worked out) that then only one person stands in the corner).
Quote: WizardDoesn't getting into the elevator at all cause it to consume more electricity? I thought the spirit of the Sabbath rules were shun modern conveniences and not consume power.
I think it depends on what sect of Judaism is involved. I once worked a Bar Mitzvah where they delayed the start to sundown because of the women said "a few folks." I knew she meant orthadox, but she was being half evasive for who knows why. But point being yes, some shun much on the Sabbath. Others not so much.
The square root of 113.04 is 10.63, so an elevator of somewhere around 11 feet x 11 feet would be needed.
But you don't need a 6 foot circle. You only need to be six feet away from everyone else. So the space needed would be less than that, but I can't see in my head how that works out to the minimum size of an elevator car. I assume there's a variety of answers, depending on the "rectanglearity" of the box. I think we need some calculus. My daughter teaches math. She's at hme now with nothing to do (but of course her union contract mandates that she get paid anyway...) I'll ask her for the answer.
Here's an interesting thought problem for those of you who want to enforce any sort of mandatory rule, and look to elevators as a wonderful place to put your regulatory foot on the neck of other people. No need to let a good emergency go to waste.
Four people get in an elevator on the first floor.
It stops on the third floor, the door opens, and someone decides to enter the elevator.
He says, "you guys can do what you want. If one of you want to get out, that's fine, but I'm getting in and I don't care if there are a hundred people in here with me."
How is this rule and it's, of course, MANDATORY FINE, get enforced? The fifth guy wants to go "up" too, and he presses a button for the ninth floor.
Quote: racquetIf the minimum distance in "social distancing" of 6 feet, a circle of six feet in diameter is 28 square feet. (3^^2 * 3.14). Assuming that there are four people in the elevator, and ignoring the fact that the elevator is a rectangle (or square), the four people would need a total of 113 square feet. (3^^2 * 3.14 * 4 = 113.04).
The square root of 113.04 is 10.63, so an elevator of somewhere around 11 feet x 11 feet would be needed.
But you don't need a 6 foot circle. You only need to be six feet away from everyone else. So the space needed would be less than that, but I can't see in my head how that works out to the minimum size of an elevator car. I assume there's a variety of answers, depending on the "rectanglearity" of the box. I think we need some calculus. My daughter teaches math. She's at hme now with nothing to do (but of course her union contract mandates that she get paid anyway...) I'll ask her for the answer.
Here's an interesting thought problem for those of you who want to enforce any sort of mandatory rule, and look to elevators as a wonderful place to put your regulatory foot on the neck of other people. No need to let a good emergency go to waste.
Four people get in an elevator on the first floor.
It stops on the third floor, the door opens, and someone decides to enter the elevator.
He says, "you guys can do what you want. If one of you want to get out, that's fine, but I'm getting in and I don't care if there are a hundred people in here with me."
How is this rule and it's, of course, MANDATORY FINE, get enforced? The fifth guy wants to go "up" too, and he presses a button for the ninth floor.
Pretty simple. Hotel Security is monitoring the elevator, notes what room the guy goes to and sends up a bunch of jackboots to backroom him and then toss him out on the street naked. Post a video of the incident next to the elevator for all to see. The four people who didn't prevent the guy from entering must wear a big sign around their necks stating they don't observe social distancing.
Quote: ThatDonGuyQuote: MoscaQuote: ThatDonGuyIt's probably like the elevators at Thed - er, The D - where you select a floor before you get in the elevator.
Let A be in the back left corner, B be in the back right corner, and C be in front, halfway between the two sides.
A and B are 80 inches apart; A and C are sqrt(652 + 402) = 76.322 inches apart, as are B and C.
If, say, C moves forward, then the distance from B to C is less than 76.322, unless C moves to the left, but this would make the distance from A to C less than 76.322.
Therefore, any solution requires that A and B be in the back somewhere.
Suppose B moves to the center of the back. The maximum distance from B to C that would be less than the distance from A to C is 76.322, when B is in the front right corner, but the distance from A to B is only 40.
Moving B to the left shortens the distance from A to B; moving B to the right shortens the distance from B to C.
76.322 inches appears to be the optimal result, with two in the back corners and the third in the center of the front.Wouldn't it still have to be an equilateral triangle? A in the far left corner. B in the far right corner. C in the front. Then they adjust so that they are equidistant from each other, along the walls to maximize the triangle. B would move forward a little, along the wall, increasing his distance from A, C would move a little to his left. Also, they enter and leave in reverse order. A goes to a corner first, leaves last. Then B, then C. But again, the answer has to be an equilateral triangle.
If they are in an equilateral triangle, then they are all 75.06 inches apart. By spreading A and B out to the corners, they are 80 inches from each other, and 76.322 inches from C.
Okay, I get it now. Thanks.
Quote: billryanPretty simple. Hotel Security is monitoring the elevator, notes what room the guy goes to and sends up a bunch of jackboots to backroom him and then toss him out on the street naked. Post a video of the incident next to the elevator for all to see. The four people who didn't prevent the guy from entering must wear a big sign around their necks stating they don't observe social distancing.
I'm fine with all of that. It's when the jackboots are wearing police badges (or when they're police WITHOUT badges) that I have a problem.
If the casino wants to limit elevator traffic in its hotel to four people, good for them. And if they want to advertise this, noting that the casino next door does not protect its clients like they do, all well and good.
Silly rules enacted by all these newly empowered mall cops (hereinafter referred to as "governor") will be unenforceable, and will doom legitimate, justifiable rules to the same degree of ridicule.
Passengers on airplanes have to endure whatever rules TSA and DHS come up with, and most of them are appropriate and reasonable. If you don't like the rules, they control your access to airplanes. But now governors are falling in love with their daily press conferences telling us "here's what I've decided you can and cannot do today." They're going to want to control how many people are in elevators. Where will it end?
White shores, and beyond, a far green country under a swift sunrise.
Quote: racquetIf the minimum distance in "social distancing" of 6 feet, a circle of six feet in diameter is 28 square feet. (3^^2 * 3.14). Assuming that there are four people in the elevator, and ignoring the fact that the elevator is a rectangle (or square), the four people would need a total of 113 square feet. (3^^2 * 3.14 * 4 = 113.04).
The square root of 113.04 is 10.63, so an elevator of somewhere around 11 feet x 11 feet would be needed.
But you don't need a 6 foot circle. You only need to be six feet away from everyone else. So the space needed would be less than that, but I can't see in my head how that works out to the minimum size of an elevator car. I assume there's a variety of answers, depending on the "rectanglearity" of the box. I think we need some calculus. My daughter teaches math. She's at hme now with nothing to do (but of course her union contract mandates that she get paid anyway...) I'll ask her for the answer.
Here's an interesting thought problem for those of you who want to enforce any sort of mandatory rule, and look to elevators as a wonderful place to put your regulatory foot on the neck of other people. No need to let a good emergency go to waste.
Four people get in an elevator on the first floor.
It stops on the third floor, the door opens, and someone decides to enter the elevator.
He says, "you guys can do what you want. If one of you want to get out, that's fine, but I'm getting in and I don't care if there are a hundred people in here with me."
How is this rule and it's, of course, MANDATORY FINE, get enforced? The fifth guy wants to go "up" too, and he presses a button for the ninth floor.
A 6’ by 6’ square elevator works with everyone in the corner. Even better a 24’ by 1’ elevator (let’s assume 1’ as the minimum depth of an elevator someone can squeeze into). Taking up a svelte 24 square feet.
The better solution is huge banks of single passenger "Our Father" lifts. The carnage would be impressive.Quote: unJonA 6’ by 6’ square elevator works with everyone in the corner. Even better a 24’ by 1’ elevator (let’s assume 1’ as the minimum depth of an elevator someone can squeeze into). Taking up a svelte 24 square feet.
back wall
passenger #1
six feet
passenger #2
six feet
passenger #3
six feet
passenger #4
front wall
So assuming a one foot width, you'd only need 18 feet back-to-front, but how "deep" is a person? That depth would need to be added.
Assuming the same 1 foot requirment for depth (front to back) as in width (left to right):
back wall
passenger #1 (takes up 1 foot)
six feet
passenger #2 (takes up 1 foot)
six feet
passenger #3 (takes up 1 foot)
six feet
passenger #4 (takes up 1 foot)
front wall
27 feet front to back.,
Assuming 1 foot width, 27 square feet.
Do you only occupy a one foot square when standing? Does a well-endowed woman, or an obese man?
back wall
passenger #1 (takes up one foot)
1 foot
Quote: rxwineTechnical question - If the airspace of an elevator is not big enough when someone farts how can it be big enough for social distancing?
The fart smell arises from methane gas. A gas is more likely to travel farther in the atmosphere than a water droplet. I don't really know anything about the gas velocity of a fart but it does seem to stay in the air for a considerable time.
Edit: Farts are more complex than I realized. Hydrogen sulfide, methanethiol and dimethyl sulfide all contribute to the odor of farts. I believe those are all gases.
Apparently, in the future, chemical analysis of farts may be useful for diagnosing disease.
or one more TV commercial tells me
condescendingly to 'Stay Safe', I'm
going to punch somebody. Enough
already, I'm sick of it.
I find it annoying when casinos ban customers from using stairs (saying they are only for emergency) and require elevator use in towers... Waiting in line for elevators even under normal circumstances is absurd....
Quote: GandlerI would take the stairs.
I find it annoying when casinos ban customers from using stairs (saying they are only for emergency) and require elevator use in towers... Waiting in line for elevators even under normal circumstances is absurd....
But if your room is on the fourteenth floor? I'd attempt it going DOWN the staircase, but going up? A lot of us vulnerable folks would be gathering on the landings catching our breath.
The whole idea of elevator occupancy restrictions points out the potential for silly rules promulgated by the Nanny State. Such a rule would be unenforceable and be seen as silly. Reasonable rules will be obeyed, because common sense is what makes them reasonable. But foolish attempts to mandate things like social distancing in public places (will we see this extend to urinals?) will reduce all the rules, the goods ones as well, to jokes.
What's the average width of a rest room stall? How far apart are the sinks? How many towel dispensers or air dryers?
If the rule-makers go overboard, protecting us from ourselves, whether we want to be protected or not, they will defeat their intended purpose. But of course their purpose is not to protect us, but to extend and expand the ways they think they can tell us what to do.
Quote: utahslim1Perhaps an updated CONE OF SILENCE for baccarat.
With all of the worries, how about those chips? I always thought they were the most dangerous from a disease point of view, such that I always washed my hands before using the men's room, not after. I had a good idea where Mr. Binky had been, but not those chips.
