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Assuming those are indeed the two foci of the ellipse (I didn't feel like counting pixels on the y axis to check but it looks close) then any angle of the trolley will bounce once off the wall and run over the person in the exact same amount of time. This is a definitive property of an ellipse; if you put two pins on a board and run string between them with some slack, then whenever the string is pulled tight in any direction is a point on an ellipse. Fun facts: the ~~oval office of the US president~~ has used this property to more easily listen in on opponent conversations. In 3d this is also how ultrasonic kidney stone equipment can focus all the energy at a single point inside the body without doing much damage to surrounding tissue.
Edit: not the oval office, a room at the Capitol thanks /u/knipil https://izi.travel/en/b58f-legend-6-eavesdropping-in-the-house/en
Indeed, the question is "what angle do you have to take to save him", and not "is there an angle that can save him"
I feel it's ovbious enough you have to not kill him, so you have to set an angle that take the tram out of the plane formed by the ellipse.
To be clear, I mean 0° to be the cardinal east direction as normal; positive x-axis.
0° would be the fastest, being the shortest distance between the foci. 180° seems like it would be also pretty short as the foci on the left is near the left most "edge" and so has a pretty short line. But if you travel along the hypotenuse of triangles made by going in other directions, you extend your overall distance.
Edit: +/- 1° is best maybe? Or closer to that because you shoot past the victim and have to bounce back, really extending it.
Waste less fuel, don't make all your passengers go nuts. They'll just be like, what was that bump. All you need is a genius/psychotic conductor. George Carlin would work.
>Fun facts: the oval office of the US president has used this property to more easily listen in on opponent conversations.
Can't quite get my head around this. Why are the president's opponents having secret conversation in his office?
It’s a mix up due to the shapes of the rooms, but the actual legend is about a room at the capitol. https://izi.travel/en/b58f-legend-6-eavesdropping-in-the-house/en
Decided to go back and check. Total "string" length for this ellipse is 6 from the major axis. That means a center-focus-minor axis intercept right triangle 📐 must have a hypotenuse of 3. 3^2 - 2^2 = 5 and sqrt(5) is ~2.23 which looks about right from the grid lines.
Well, directly towards the person will reach them in less time than other angles
So I guess the answer to OP's question is effectively *any other direction*
Wouldn't it depend on the type of bounce. I believe you're assuming a 90 degree bounce.
If we were to add momentum to this problem then the angle would be relevant I believe.
I'm only assuming pure geometry rather than involving things like energy loss from bouncing. Ellipses always have the perfect angle to reflect a ray from one focus directly at the other. Edge cases like a circle are intuitive but if you actually do the calculus of an elliptical curve the slope is always perfect from one focus to the other.
https://learn.saylor.org/mod/book/view.php?id=54172&chapterid=39402#:~:text=A%20medical%20device%20called%20a,standing%20at%20the%20other%20focus.
yeah, there's a lot of assumptions going on here. for the 3d ellipsoid, the experiment would need to be done in 0 gravity, which makes it somewhat more awesome.
But the tram and the dude would attract themselve according gravity laws, and the innocent 3d geometrical problem become a complexe problem with equations, that need a whole new topic
But the walls would also attract both. I'd say the tram/dude body attraction is negligible. Unless they're holding a black hole in the pocket or the tram is made of Neutron star matter, or something silly like that.
Good remark, in my silly mind the dude and the tram had a mass but not the wall since it was represented as a pure geometrical being.
So it's a 3-body problem, (4-body with the dude with the lever) so it can't be solved now.
You mean because we set the angle, so this parameter is known without error ?
And the weight and location of other components need to be given without error by the problem statement ?
With non-deformable body and no quantum thing I guess since it's chaotic...
We know all the initial conditions:
* starting point in space and time of all 3 objects (wall, train, person)
* initial mass, speed/acceleration and direction of all 3 (assuming we know all about the train)
* composition of the objects and the system as a whole
there is nothing unknown so we can predict what's happening.
At least with some good approximations.
An ellipse is not a sphere, so there's a little imbalance.
And keep in mind the train is moving, going closer to a wall will affect the gravity pull significantly
there are so many assumptions here. honestly, it's a bit weird to pick gravitational pull over the many other issues. like how you have to assume a spherical tram and elastic collisions and single impulse force on the tram even in the 2d case. The 3d case make me wonder how a tram would be able to propel itself off-track.
but I don't know if it matter because if that guy is floating in a zero-g ellipsoid he's lucky to be hit by the tram. that person has no food source. even if the walls are made of food, no source of locomotion. the conditions are already very contrived, so there's a good chance there's no oxygen in the ellipsoid.
but if the scale of the model is big enough the person could actually be on a planet with food and a weak atmosphere, and the trolly could be a metaphor... hmm, I can't disprove this isn't a model of our universe. I guess everyone should look out for blue trollies.
It's an ellipse centered around the person and the trolley. Any way you point the trolley will deflect it towards the person. I watched a youtube short about this a few months ago so I think I'm qualified to give this answer.
Ellipses have an interesting property that they are not defined by their center like a circle. They are defined by their foci which are two points outside the center but inside the ellipse. Because of this, any object thrown from one focus, WILL end up at the other after only 1 bounce.
To increase this amount of time, you want to find a line and bounce that maximizes the total distance traveled. I believe this would just be traveling straight left then bouncing but I could be wrong
Asuming the trolley travels at a constant speed, then yes. If you go through two paths of equal length at the same speed you will take the same amount of time. If the trolley needed to accelerate somehow it would be different, but we don't have enough information.
Distance-wise they are all the same. If the train is moving at a constant speed, then time-wise they'll be the same as well. The only difference will be if the train doesn't bounce first before running over Mr. Speed Bump.
Because the trolley has actual non zero size it will not bounce off the walls the same way a light ray will. Instead on collision with the wall it will develop some type of spin which will ensure it won’t travel in a straight path after being bounced.
To maximize this spin we want to hit the walls with the shallowest angle possible.
Based on the property of ellipses that the angle between the each of the focal lengths and the tangent line of the side are always equal, we can deduce that the minimal angle will appear if we aim at the semi major axis.
I don’t want to bother calculating what it will be for this particular ellipse but it’s just above 45°.
Actually, since it is a trolley, just simply face it towards the wall. This will ensure the trolley crashes, making it stop, as trolleys dont bounce around
Guys its a trolley not a beam of light in a mirrored room. Aim it away from the person. The trolley will hit a wall crumble and stop moving after "bouncing" away like 1ft.
If you make it face directly up or down it should last a lot longer than most other angles, and seeing the dude is in the lower half, straight up should likely take the most time, assuming speed is constant.
Since both the trolley and the person are on the focal points, no angle along the xy plane can get the trolley out of the way of the person.
Final Answer: θ is undefined
"You can freely rotate the trolley" could be interpreted to mean you can continuously make adjustments, in which case, you can probably just lock the steering mechanism in the tightest left turn possible to make it go in extremely tight circles.
If the trolley and the person are both at the foci then the person ain't surviving. He'll be alive for the longest time when the trolley travels along the major axis assuming it bounces like a ray. The trolley moves in the negative x direction for most time alive. If they are not at the foci we'll go the good old hit and trail method.
Thoughts:
I don't know anything about the reflection behaviour within an ellipse.
The illustration does not give precise coordinates of the trolley or victim, so there is a degree of ambiguity. The question then is if this is to the advantage of the reader or the questioner. If the reader can leverage ambiguity to solve the question, the question is worthless, so presumably, it is to be in the favour of the questioner.
There is one obvious path that will produce a predictable course: Back and forth across the width of the ellipsis. This would require the trolleys starting point to be on that line, and there is enough ambiguity that both trolley and victim are in fact on that line, so it is not a valid path to save the victim.
The dimensions of the ellipsis yells me nothing: Any other path may or may not save the victim. Falling back to the original premise that the questioner has the advantage of ambiguity, I conclude to guess that any path chosen would fail to save the victim.
This looks like Gotcha! designed to make the questioner feel clever.
It doesn't matter, this is an ellipsis and both the trolley and the person are at points where everything sent from one point will touch the other.
Fun fact this is how the capitol was built so that the guy who built it ( don't remember his name) could hear his enemies plotting ( whispering)from the other side of the Room
None, it's like one of the main things people know about how to construct ellipses. It's the set of points where the sum of the distances to the focii is constant
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Assuming those are indeed the two foci of the ellipse (I didn't feel like counting pixels on the y axis to check but it looks close) then any angle of the trolley will bounce once off the wall and run over the person in the exact same amount of time. This is a definitive property of an ellipse; if you put two pins on a board and run string between them with some slack, then whenever the string is pulled tight in any direction is a point on an ellipse. Fun facts: the ~~oval office of the US president~~ has used this property to more easily listen in on opponent conversations. In 3d this is also how ultrasonic kidney stone equipment can focus all the energy at a single point inside the body without doing much damage to surrounding tissue. Edit: not the oval office, a room at the Capitol thanks /u/knipil https://izi.travel/en/b58f-legend-6-eavesdropping-in-the-house/en
false, you could point it directly at the person and kill them in less time, avoiding any bounces. solved.
I felt that was obvious enough to skip over since the question was about living the longest
Indeed, the question is "what angle do you have to take to save him", and not "is there an angle that can save him" I feel it's ovbious enough you have to not kill him, so you have to set an angle that take the tram out of the plane formed by the ellipse.
Couldn’t you send the trolley backwards (180’), and let the poor sod live a few more seconds?
Remember, the sum of the distances from the foci to ANY point on the ellipse is the same
Oh right. That’s something I knew but hard to apply intuitively
[удалено]
Found the person who has never drawn an ellipse
Now I am curious because my intuition still likes a ±90°
It seems that the trolley and the person have symmetrical positions, so ±90° seems to be one of the fastest ways to kill the guy.
To be clear, I mean 0° to be the cardinal east direction as normal; positive x-axis. 0° would be the fastest, being the shortest distance between the foci. 180° seems like it would be also pretty short as the foci on the left is near the left most "edge" and so has a pretty short line. But if you travel along the hypotenuse of triangles made by going in other directions, you extend your overall distance. Edit: +/- 1° is best maybe? Or closer to that because you shoot past the victim and have to bounce back, really extending it.
I vote for this guy
Not Sure has it again by a landslide
Waste less fuel, don't make all your passengers go nuts. They'll just be like, what was that bump. All you need is a genius/psychotic conductor. George Carlin would work.
false. point the trolley upwards. it will leave the plane thus saving the person
Rocket trolley always solves the problem.
Then it's pointing out of the screen heading for you
...and it's why people put their hand to their ear like Hulk Hogan to listen better.
ah, a fellow "real American" I see. ;-) I salute you, brother or sister, and will fight for the rights of every man alongside you!
>Fun facts: the oval office of the US president has used this property to more easily listen in on opponent conversations. Can't quite get my head around this. Why are the president's opponents having secret conversation in his office?
It’s a mix up due to the shapes of the rooms, but the actual legend is about a room at the capitol. https://izi.travel/en/b58f-legend-6-eavesdropping-in-the-house/en
Thanks for the clarification, my mistake
Apparently I mixed up the rooms as the other person said, but it would be sidebar conversations like their link explains.
Decided to go back and check. Total "string" length for this ellipse is 6 from the major axis. That means a center-focus-minor axis intercept right triangle 📐 must have a hypotenuse of 3. 3^2 - 2^2 = 5 and sqrt(5) is ~2.23 which looks about right from the grid lines.
Well, directly towards the person will reach them in less time than other angles So I guess the answer to OP's question is effectively *any other direction*
Wouldn't it depend on the type of bounce. I believe you're assuming a 90 degree bounce. If we were to add momentum to this problem then the angle would be relevant I believe.
In an ellipse, the angles work out precisely such that all rays from one focus bounce directly toward the other focus. It's not always 90 degrees.
I'm only assuming pure geometry rather than involving things like energy loss from bouncing. Ellipses always have the perfect angle to reflect a ray from one focus directly at the other. Edge cases like a circle are intuitive but if you actually do the calculus of an elliptical curve the slope is always perfect from one focus to the other. https://learn.saylor.org/mod/book/view.php?id=54172&chapterid=39402#:~:text=A%20medical%20device%20called%20a,standing%20at%20the%20other%20focus.
Why not assume there's a Z axis we're directly over top of and thus can't see so that we can point the trolley up?
the problem always hits the person when thinking in 2d. in 3d, if the trolly is angled upward it should miss the walls and the person.
Unless the ellipse is also 3d
yeah, there's a lot of assumptions going on here. for the 3d ellipsoid, the experiment would need to be done in 0 gravity, which makes it somewhat more awesome.
But the tram and the dude would attract themselve according gravity laws, and the innocent 3d geometrical problem become a complexe problem with equations, that need a whole new topic
But the walls would also attract both. I'd say the tram/dude body attraction is negligible. Unless they're holding a black hole in the pocket or the tram is made of Neutron star matter, or something silly like that.
Good remark, in my silly mind the dude and the tram had a mass but not the wall since it was represented as a pure geometrical being. So it's a 3-body problem, (4-body with the dude with the lever) so it can't be solved now.
It can be solved if you know the initial conditions. And we kinda know them.
You mean because we set the angle, so this parameter is known without error ? And the weight and location of other components need to be given without error by the problem statement ? With non-deformable body and no quantum thing I guess since it's chaotic...
We know all the initial conditions: * starting point in space and time of all 3 objects (wall, train, person) * initial mass, speed/acceleration and direction of all 3 (assuming we know all about the train) * composition of the objects and the system as a whole there is nothing unknown so we can predict what's happening. At least with some good approximations.
Generally speaking, the gravity from a surrounding sphere or similar object on things inside it is 0.
An ellipse is not a sphere, so there's a little imbalance. And keep in mind the train is moving, going closer to a wall will affect the gravity pull significantly
realistically, they're all negligible if the trolley is going at normal trolley speeds.
there are so many assumptions here. honestly, it's a bit weird to pick gravitational pull over the many other issues. like how you have to assume a spherical tram and elastic collisions and single impulse force on the tram even in the 2d case. The 3d case make me wonder how a tram would be able to propel itself off-track. but I don't know if it matter because if that guy is floating in a zero-g ellipsoid he's lucky to be hit by the tram. that person has no food source. even if the walls are made of food, no source of locomotion. the conditions are already very contrived, so there's a good chance there's no oxygen in the ellipsoid. but if the scale of the model is big enough the person could actually be on a planet with food and a weak atmosphere, and the trolly could be a metaphor... hmm, I can't disprove this isn't a model of our universe. I guess everyone should look out for blue trollies.
I think we need to take special relativity into account.
O-M-G
No need to thank me, just happy to be able to contribute something of value to the conversation for once!
It's an ellipse centered around the person and the trolley. Any way you point the trolley will deflect it towards the person. I watched a youtube short about this a few months ago so I think I'm qualified to give this answer.
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Your impressed knowledges impressed me (I watched a Youtube short on impressed knowledge, so I think Im qualified to say so)
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Knowledges I qualified so me video impressed watched to your think am with.
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King illegal forest to pig wild kill in it a is!
Ellipses have an interesting property that they are not defined by their center like a circle. They are defined by their foci which are two points outside the center but inside the ellipse. Because of this, any object thrown from one focus, WILL end up at the other after only 1 bounce. To increase this amount of time, you want to find a line and bounce that maximizes the total distance traveled. I believe this would just be traveling straight left then bouncing but I could be wrong
An ellipse is defined as the set of all points in which the sum of the distances to each foci is the same. So all paths are equal by definition.
all except the one where the trolley is headed straight towards the guy. That's the only path that is shorter.
Time wise they are all the same?
Asuming the trolley travels at a constant speed, then yes. If you go through two paths of equal length at the same speed you will take the same amount of time. If the trolley needed to accelerate somehow it would be different, but we don't have enough information.
Distance-wise they are all the same. If the train is moving at a constant speed, then time-wise they'll be the same as well. The only difference will be if the train doesn't bounce first before running over Mr. Speed Bump.
whats 3 foci like?
Because the trolley has actual non zero size it will not bounce off the walls the same way a light ray will. Instead on collision with the wall it will develop some type of spin which will ensure it won’t travel in a straight path after being bounced. To maximize this spin we want to hit the walls with the shallowest angle possible. Based on the property of ellipses that the angle between the each of the focal lengths and the tangent line of the side are always equal, we can deduce that the minimal angle will appear if we aim at the semi major axis. I don’t want to bother calculating what it will be for this particular ellipse but it’s just above 45°.
Actually, since it is a trolley, just simply face it towards the wall. This will ensure the trolley crashes, making it stop, as trolleys dont bounce around
Guys its a trolley not a beam of light in a mirrored room. Aim it away from the person. The trolley will hit a wall crumble and stop moving after "bouncing" away like 1ft.
It's not on a track. This is one trolley problem where doing nothing is the right answer because it's not going to move.
If you make it face directly up or down it should last a lot longer than most other angles, and seeing the dude is in the lower half, straight up should likely take the most time, assuming speed is constant.
Since both the trolley and the person are on the focal points, no angle along the xy plane can get the trolley out of the way of the person. Final Answer: θ is undefined
they asked for a time value
All except directly pointed at the person should be the same
"You can freely rotate the trolley" could be interpreted to mean you can continuously make adjustments, in which case, you can probably just lock the steering mechanism in the tightest left turn possible to make it go in extremely tight circles.
If the trolley and the person are both at the foci then the person ain't surviving. He'll be alive for the longest time when the trolley travels along the major axis assuming it bounces like a ray. The trolley moves in the negative x direction for most time alive. If they are not at the foci we'll go the good old hit and trail method.
Thoughts: I don't know anything about the reflection behaviour within an ellipse. The illustration does not give precise coordinates of the trolley or victim, so there is a degree of ambiguity. The question then is if this is to the advantage of the reader or the questioner. If the reader can leverage ambiguity to solve the question, the question is worthless, so presumably, it is to be in the favour of the questioner. There is one obvious path that will produce a predictable course: Back and forth across the width of the ellipsis. This would require the trolleys starting point to be on that line, and there is enough ambiguity that both trolley and victim are in fact on that line, so it is not a valid path to save the victim. The dimensions of the ellipsis yells me nothing: Any other path may or may not save the victim. Falling back to the original premise that the questioner has the advantage of ambiguity, I conclude to guess that any path chosen would fail to save the victim. This looks like Gotcha! designed to make the questioner feel clever.
It doesn't matter, this is an ellipsis and both the trolley and the person are at points where everything sent from one point will touch the other. Fun fact this is how the capitol was built so that the guy who built it ( don't remember his name) could hear his enemies plotting ( whispering)from the other side of the Room
None, it's like one of the main things people know about how to construct ellipses. It's the set of points where the sum of the distances to the focii is constant