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(65^{2}+ 40^{2}) = 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(65^{2}+ 40^{2}) = 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(65^{2}+ 40^{2}) = 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.