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On June 17 2015 19:22 Simberto wrote:Show nested quote +On June 17 2015 18:38 xM(Z wrote:that picture is obviously flawed. the phrase "same mass at same gravity levels" doesn't mean much. even if they have the same mass, they'll have different weights('cause they'll be in different gravitational fields). you'll need to prove that Xmass at 9.8 moves at the same speed as Xmass at 1.6. i think you'll fail, get different speeds/accelerations and your L1 will move/shift. it can not be balanced unless actively changing the loads per need/time. i'll move to this now: On June 17 2015 02:24 Acrofales wrote:On June 17 2015 02:04 xM(Z wrote:On June 17 2015 01:35 Simberto wrote:![[image loading]](http://i.imgur.com/3NE6JlR.jpg) I drew a picture. I forgot to link all of the train cars up, though. They should all be attached to each other with a long line with wheels at each end. meh, but then you have to use energy to move them ... (instead of using energy to move the water you also use energy to move the dust). hmm, i don't really get that idea. https://en.wikipedia.org/wiki/Space_elevator the space elevator won't work either since it requires energy to push something up the wire/ribbon; BUT, let's assume the energy required is negligible(very low). - we assume that magic will keep the water liquid on the moon and we have the means to transport and dump it in liquid form. - we assume that the elevator will withstand the moon's gravity when it passes by it(with the load attached to it)... - we assume we can compensate for the moon's elliptical orbit and just let go to some more wire... - the elevator is fixed on earth which means it will pass by the moon once a day (a little more than a day since the moon orbits in the same direction as earth rotates so that will add some more time/day) - the weight, at any one time, that can be moved up the ribbon depends on how much the ribbon can take and not break; (the current proposed material = carbon nanotubes) - hmm, how the fuck do we get the load back to earth?; nvm, after unload we just dump it in space. who cares!. we load the thing with water; we give it momentum; it goes up the wire; reaches geostationary altitude; when the moon get's closer to the load we push(consuming energy) the load to enter moon's gravitational pull(to skip/overcome any L1 shenanigans); we fail ... (we would need to constantly consume energy to land(WE NEED TOUCHDOWN) the same mass but of different weights in a different gravitational pull. even if we ignore all the mass/weight/speed/acceleration/drag/resistance bullshits, we would still have to consume energy to beat earth's rotation timer with respect to the moon. (there's a fixed amount of time in which the earth passes by the moon; if we don't unload in that time, we're useless). just, just ... repeat this until you believe it - + Show Spoiler + Your understanding of physics is quite lacking. The phrase "same mass at same gravity level" means exactly what you claim it does not. It means that there is the same mass, for example 1 ton of water moving up and 1 ton of moonrocks moving down, at the same height above earth. Which means that the gravitational acceleration at that point is the same for both of them. Meaning the total force on a looped tether is 0, since you have the downwards-moving moonrocks pulling generating energy by giving up gravitational potential, while the water consumes energy by gaining gravitational potential, and the total difference in energy is 0. You are still making the same error that i mentioned earlier, and that i thought my picture with giant red circles showing you which masses to compare made very clear is not necessary. You do not need to compare the mass that moves up at earth with the mass that moves up on the moon. (Of course, if you just sum up ALL of the force on the different train cars with at the same time, the result will still be 0, but you are not doing that either). You simply select two random train cars, see that the forces on them are not equal, and thus decide that the total force can not be 0, which is obviously nonsense. You also have this idea that moving things needs energy. That is incorrect. Accelerating things needs energy. Changing the potential of things needs energy. Moving things does not. For an example of this, take a look at earth. As you might know, earth is rotating around the sun, and quite quickly. It has been doing that for a few billion years (Or a few thousand if you are a creationist  ) If it were constantly using up energy doing that, it would have long fallen into the sun, and you wouldn't be here discussing this. Of course, this is not consistent with your everyday experience, because in your everyday experience, you are used to friction. Friction means that if you move something, you are constantly accelerating some other things (mostly air molecules) to get them out of the way of the thing you are moving. Which requires energy. This is generally not a problem in space, as there are no things that need to be moved out of the way. Regarding the Space Elevator, obviously it requires Energy. That does not mean that the thing doesn't work, it just means it requires you to put energy into it to move things up. Moving things up always requires energy, as you are changing their potential (Workaround see above, but there we still require the energy, we just get it from moving other stuff down). The whole point of a space elevator is that it will require a lot LESS energy to move stuff up compared to our current way of doing it, which is using grossly inefficient rockets, constantly fighting against a lot of friction, and moving up additional fuel to move more fuel further up.
Just made me wonder if there is any possibility to be gained through using the elevator concept mixed with a trebuchet?
Illustration:
+ Show Spoiler +
Also made me think about using Musk's Hyperloop tech to launch something vertical?
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@ Simberto
![[image loading]](http://i.imgur.com/MnbywoM.jpg)
sense?.
the only reason i got into this L1 business is because i thought the weights would slingshot around it but the way you drawn it ... it's just no way.
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Hm? The sense is there, if you just imagine the carts to be linked up in a circle. And of course have a few more of them, so you can approximate the whole business as a continuous mass distribution.
One big problem of understanding i see in your picture is that you somehow seem to assume that the water randomly changes into rock in the middle of space for no apparent reason. You obviously have water in all of the bottom carts moving towards the Moon, and rocks in the top carts moving towards earth. Since all carts are filled with equal mass, this is not really that relevant for the Force observations, but for the consistency of the whole idea it is obviously necessary.
As you seen in your picture, the two boxes to the left, the top one being filled with moon rocks and the lower one being filled with water, move in opposite directions. One of them is aided by gravity, the other is hindered. If you now attach them together by a rope that goes over a wheel on earth and another one on the moon, those two forces will work in opposing directions on that rope. And as Work (Energy)= Force * length (Usually you would need the integral of the Force over a way, but we are being lazy here and it is not really necessary to do this for the argument), and the total force is 0, the energy expended for this transfer is obviously also 0.
If you want to get fancy, you could even load up a few more moonrocks into each cart, and thus gain the energy required to move the carts against friction from the potential energy difference between Moon and Earth.
The point of the L1 station is to keep the whole thing in place, and for that, you would ideally want the mass distributed in such a way that the center of mass of the whole construction is in L1.
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- it's because i loaded 2 carts before unloading the other 2; it doesn't change anything though.
- what i saw in your drawing, were literally, tracks on a 2 way railroad. 2D
you now said it's a circle around L1(which was what i thought initially). it still doesn't work. you get screwed by the distance earth <- L1 -> moon (L1 is way closer to the moon). i won't go in details ('cause i'm bored of telling you that it won't work) but for your plan to work you need L1 at half way moon<->earth.
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I hate appeals to authority, but i am currently studying physics and maths to become a teacher of those, and believe me, either there is a gigantic misunderstanding between me and you regarding what i am talking about, or you are incorrect. Please consider this to be a possibility. I have considered the possibility of you being right, but it does not fit with basic Newtonian physics and the law of the conservation of Energy.
The idea is to have two railroad tracks of linked carts. The gravitational forces negate because the force pulling one down is linked via a cable to pull the other up.
It is irrelevant if there is a station at L1 or not for this. The form of the construct is irrelevant too, as long as the same mass is moving up and down at the same height at the same time. You can make it straight, circular, oval, a pretzel or whatever else you like.
The only point of the station is to make sure that the center of gravity of the whole construct is in L1 (Which actually means that the station would have to be closer to the moon) so it stays in place (This is actually very simplified, to figure out exactly where you need mass you would have to do some major calculations, it is not enough to have the center of gravity in L1, you need to make sure the total gravitational forces negate each other. But since we already assume a material of infinite strength for this and ignore several other major flaws of the plan, i didn't see the point of going into detail here)
Under the assumptions that we are working off, there are two things you need to make sure the whole thing works and does not require energy to move mass:
a) The same amount of mass moves up and down at all times and heights. All of these masses need to be linked in some way, through a rope or whatever.
b) The total gravitational forces on the whole construct must lead to it staying in between the Moon and the Earth.
(This does not exclude the possibility of the construct rotating away from either object, so you would need to fix that somehow, too)
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You need to attach wheels to the vacuum tower, as it needs to race around earth once every day to always face the moon. After that, no more complaints.
There are some problems with trajectories not overlapping with your sewsaw, but that is surely easily solvable, after all, that is just engineering, the science is figured out now.
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- How do you get new water into the vaccum tower at a high enough rate so that the whole thing doesn't stall and the vacuum does not drop?
- How do you shoot the first two masses?
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ok Mr.Authority:
The idea is to have two railroad tracks of linked carts. The gravitational forces negate because the force pulling one down is linked via a cable to pull the other up.
that doesn't make sense, especially the bolded parts. up/down doesn't really exist; but let's say it's motion against the gravity or with the gravity you'd still have to split it and differentiate between them gravities 9.8 - 0 - 1.6
meh, fuck that.
this,you need to make sure the total gravitational forces negate each other. can never happen because the moon has no atmosphere, so unless you spend a fuckton of energy to negate those forces you have nothing.
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Nothing you are saying right now makes even the slightest amount of sense. Maybe i am using the wrong words in english?
Of course the gravitational field is not constant in the whole volume of space. But at each single point in space (relative to the earth/moon system), it is. And in a point very close to that one, it is almost exactly the same. For simplicity, i will now look at a spot close to earth, but the exact same argument can be made for any other spot in space, as long as you define "Up" as "in the direction opposite of the gradient of gravitational potential field" and "down" as "in the direction of the gradient of the gravitational potential field".
If you have two carts right next to each other, the gravitational Force onto both of them is almost exactly the same on both of them. Now, if one of them is moving "down" (reducing it's gravitational potential Energy), and the other is moving "up" (Thus increasing its potential Energy in the gravitational potential), you will notice that since the gradient of the gravitational potential field at both spots is almost exactly the same, the potential energy increase in one of them is exactly identical to the potential energy decrease in the other one. If you now manage to somehow link them up in a way that lets both of these effects counteract each other while keeping the trajectories of the objects intact, (For example via a rope descending from one to a wheel, and then ascending back up to the other, leading to another fixed wheel at the moon, then returning to the first cart), the total change in potential Energy is 0.
Regarding the second part, you are once again incorrect. The atmosphere of the moon has absolutely nothing to do with any gravitational forces whatsoever. And since the gravitational force is pulling towards the moon at some spots, and towards earth at other spots, it is quite obvious that with the correct distribution of mass you can produce an object onto which the total pull of gravity is 0. As a though experiment, think of an object in the L1 spot. This object is in a (semi) stable position regarding the Earth-Moon system. Now extend a long stick towards the moon, and another long stick of appropriate mass (You will have to do calculations here) towards earth. If you calculated correctly, the force pulling the one stick towards the moon is equal to the force pulling the other stick towards earth. Thus, the total gravitational forces negate each other. You can extend this concept to construct arbitrarily complicated objects.
And the reason i use simple terms like "up" and "down" is that people understand them, and most people are not very comfortable with the concepts of vector fields, skalar fields, and their various different derivatives like gradient, rotation, divergence. From the understanding of physics you have so far displayed i highly doubt that you belong to those people to which those concepts mean anything, but of course i could be wrong here. Plus they are not in the slightest bit necessary to understand the problem at hand, for that you just need basic newtonian physic, knowledge about the conservation of Energy and a very rough understanding of gravity.
Also, there is no such thing as multiple gravities. Gravity always works the same, you just need to calculate the potential field. The gravitational acceleration on the surface of Earth is ~9.81 N/kg or m/s² towards the center of Earth, on the surface of the moon it is ~1.6 N/kg towards the center of the Moon, and in between there (and pretty much everywhere else) is a continuous force field (Force fields in science are not the thing you see on Star Trek). For calculations, it is usually a lot simpler to use the gravitational potential field, since that is skalar as opposed to a vector field, but still contains all of the same information about gravity that the force field (Which is the gradient of the potential) contains. Meaning you have a lot less stuff to handle when calculating stuff.
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And i thought this couldnt get any better :D
Best topic evah ?
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On June 17 2015 20:42 Simberto wrote:
Hm? The sense is there, if you just imagine the carts to be linked up in a circle. And of course have a few more of them, so you can approximate the whole business as a continuous mass distribution.
A link of fixed length between the wagons is just useful to make sure all wagons move at the same speed, so that you have time to empty/fill them on the surface. As long as you have the tracks in place and no friction, a cart starting around L1 will loop around on its own (for the same reason that the linked chain works).
You could have a cart stopped on both sides, that you empty/fill in before another one comes in at high speed and bumps into it to send it on its way.
Just needs a way to fully transfer the kinetic energy without heating, but that's just engineering, like the tensile strength of the rope in the linked case.
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On June 17 2015 23:46 Simberto wrote:
Nothing you are saying right now makes even the slightest amount of sense. Maybe i am using the wrong words in english?
Of course the gravitational field is not constant in the whole volume of space. But at each single point in space (relative to the earth/moon system), it is. And in a point very close to that one, it is almost exactly the same. For simplicity, i will now look at a spot close to earth, but the exact same argument can be made for any other spot in space, as long as you define "Up" as "in the direction opposite of the gradient of gravitational potential field" and "down" as "in the direction of the gradient of the gravitational potential field".
If you have two carts right next to each other, the gravitational Force onto both of them is almost exactly the same on both of them. Now, if one of them is moving "down" (reducing it's gravitational potential Energy), and the other is moving "up" (Thus increasing its potential Energy in the gravitational potential), you will notice that since the gradient of the gravitational potential field at both spots is almost exactly the same, the potential energy increase in one of them is exactly identical to the potential energy decrease in the other one. If you now manage to somehow link them up in a way that lets both of these effects counteract each other while keeping the trajectories of the objects intact, (For example via a rope descending from one to a wheel, and then ascending back up to the other, leading to another fixed wheel at the moon, then returning to the first cart), the total change in potential Energy is 0.
Regarding the second part, you are once again incorrect. The atmosphere of the moon has absolutely nothing to do with any gravitational forces whatsoever. And since the gravitational force is pulling towards the moon at some spots, and towards earth at other spots, it is quite obvious that with the correct distribution of mass you can produce an object onto which the total pull of gravity is 0. As a though experiment, think of an object in the L1 spot. This object is in a (semi) stable position regarding the Earth-Moon system. Now extend a long stick towards the moon, and another long stick of appropriate mass (You will have to do calculations here) towards earth. If you calculated correctly, the force pulling the one stick towards the moon is equal to the force pulling the other stick towards earth. Thus, the total gravitational forces negate each other. You can extend this concept to construct arbitrarily complicated objects.
And the reason i use simple terms like "up" and "down" is that people understand them, and most people are not very comfortable with the concepts of vector fields, skalar fields, and their various different derivatives like gradient, rotation, divergence. From the understanding of physics you have so far displayed i highly doubt that you belong to those people to which those concepts mean anything, but of course i could be wrong here. Plus they are not in the slightest bit necessary to understand the problem at hand, for that you just need basic newtonian physic, knowledge about the conservation of Energy and a very rough understanding of gravity.
Also, there is no such thing as multiple gravities. Gravity always works the same, you just need to calculate the potential field. The gravitational acceleration on the surface of Earth is ~9.81 N/kg or m/s² towards the center of Earth, on the surface of the moon it is ~1.6 N/kg towards the center of the Moon, and in between there (and pretty much everywhere else) is a continuous force field (Force fields in science are not the thing you see on Star Trek). For calculations, it is usually a lot simpler to use the gravitational potential field, since that is skalar as opposed to a vector field, but still contains all of the same information about gravity that the force field (Which is the gradient of the potential) contains. Meaning you have a lot less stuff to handle when calculating stuff.
i think we're just talking past each other.
all i'm saying is that the bolded part is incalculable; you can not negate the forces: earth->L1 and L1->moon, in order to have L1 as origin. (I'm assuming here the cards are in a (close to) a free fall, else energy ...)
the moon has no atmosphere = no terminal velocity = the moon-card keeps increasing its speed.
the earth-card once it hits terminal velocity has 0 acceleration and ~10m/s speed.
the moon-card increases its speed so in turn its traveled distance/time, while the earth-card keeps them constant/time.
how can you negate a constant with a variable(distance traveled and time) when you:
-can't move L1
-can't dump the water
-can't use energy
-both cards are hard linked to each other = the distance traveled by the one on the moon has to be equal with the distance traveled by the card on the earth.
or, i think we just have a totally different apparatus in mind here.
(with the gravities - previously, i was repeating the same thing over and over and figured a change in wording would make a difference to my boredom levels. it obviously didn't...)
+ Show Spoiler +i can only make sense of your post when you're talking in halves: earth to L1 and L1 to moon. i just can see them canceling each other without some serious interfering which would make the whole idea not worth it to begin with)
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Ok, firstly, i was talking without friction, as air friction pretty much stops almost anything from working (hence the vacuum tower in Cascades picture)
And i think i finally found the problem in our understanding. I do not mean that carts going down on the moon are cancelling out carts going up on earth. That does not work, as the gravitational potential of the surface of the moon is still a lot higher than on the surface of the earth, and thus a cart going down to the moon from L1 does not cancel out a cart going up from Earth to L1.
However, a cart going down from L1 to the Earth DOES cancel out a cart going up from Earth to L1.
And a cart going down from L1 to the Moon cancels out a cart going up from the Moon to L1.
The above is obviously only true if the carts have the same mass, and similarly clearly it does not matter if instead of sending up and down water on one side, and moon rocks up and down on the other side, you swap the carts in the middle so on Earth Water goes up and moonrocks go down, and on the Moon water goes down and Moonrocks go up.
Thus, if you set things up correctly (the energy transfer between the carts is the crux here, i suggested a rope with wheels on the ends, but other solutions are also workable. (and ignore friction or make the carts move slow enough that friction is not that relevant), both the Moon - L1 side of the apparatus and the Earth - L1 side are energy neutral, and thus the whole thing is also energy neutral.
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On June 18 2015 01:44 xM(Z wrote:Show nested quote +On June 17 2015 23:46 Simberto wrote:
Nothing you are saying right now makes even the slightest amount of sense. Maybe i am using the wrong words in english?
Of course the gravitational field is not constant in the whole volume of space. But at each single point in space (relative to the earth/moon system), it is. And in a point very close to that one, it is almost exactly the same. For simplicity, i will now look at a spot close to earth, but the exact same argument can be made for any other spot in space, as long as you define "Up" as "in the direction opposite of the gradient of gravitational potential field" and "down" as "in the direction of the gradient of the gravitational potential field".
If you have two carts right next to each other, the gravitational Force onto both of them is almost exactly the same on both of them. Now, if one of them is moving "down" (reducing it's gravitational potential Energy), and the other is moving "up" (Thus increasing its potential Energy in the gravitational potential), you will notice that since the gradient of the gravitational potential field at both spots is almost exactly the same, the potential energy increase in one of them is exactly identical to the potential energy decrease in the other one. If you now manage to somehow link them up in a way that lets both of these effects counteract each other while keeping the trajectories of the objects intact, (For example via a rope descending from one to a wheel, and then ascending back up to the other, leading to another fixed wheel at the moon, then returning to the first cart), the total change in potential Energy is 0.
Regarding the second part, you are once again incorrect. The atmosphere of the moon has absolutely nothing to do with any gravitational forces whatsoever. And since the gravitational force is pulling towards the moon at some spots, and towards earth at other spots, it is quite obvious that with the correct distribution of mass you can produce an object onto which the total pull of gravity is 0. As a though experiment, think of an object in the L1 spot. This object is in a (semi) stable position regarding the Earth-Moon system. Now extend a long stick towards the moon, and another long stick of appropriate mass (You will have to do calculations here) towards earth. If you calculated correctly, the force pulling the one stick towards the moon is equal to the force pulling the other stick towards earth. Thus, the total gravitational forces negate each other. You can extend this concept to construct arbitrarily complicated objects.
And the reason i use simple terms like "up" and "down" is that people understand them, and most people are not very comfortable with the concepts of vector fields, skalar fields, and their various different derivatives like gradient, rotation, divergence. From the understanding of physics you have so far displayed i highly doubt that you belong to those people to which those concepts mean anything, but of course i could be wrong here. Plus they are not in the slightest bit necessary to understand the problem at hand, for that you just need basic newtonian physic, knowledge about the conservation of Energy and a very rough understanding of gravity.
Also, there is no such thing as multiple gravities. Gravity always works the same, you just need to calculate the potential field. The gravitational acceleration on the surface of Earth is ~9.81 N/kg or m/s² towards the center of Earth, on the surface of the moon it is ~1.6 N/kg towards the center of the Moon, and in between there (and pretty much everywhere else) is a continuous force field (Force fields in science are not the thing you see on Star Trek). For calculations, it is usually a lot simpler to use the gravitational potential field, since that is skalar as opposed to a vector field, but still contains all of the same information about gravity that the force field (Which is the gradient of the potential) contains. Meaning you have a lot less stuff to handle when calculating stuff. i think we're just talking past each other. all i'm saying is that the bolded part is incalculable; you can not negate the forces: earth->L1 and L1->moon, in order to have L1 as origin. (I'm assuming here the cards are in a (close to) a free fall, else energy ...) the moon has no atmosphere = no terminal velocity = the moon-card keeps increasing its speed. the earth-card once it hits terminal velocity has 0 acceleration and ~10m/s speed. the moon-card increases its speed so in turn its traveled distance/time, while the earth-card keeps them constant/time. how can you negate a constant with a variable(distance traveled and time) when you: -can't move L1 -can't dump the water -can't use energy -both cards are hard linked to each other = the distance traveled by the one on the moon has to be equal with the distance traveled by the card on the earth. or, i think we just have a totally different apparatus in mind here. (with the gravities - previously, i was repeating the same thing over and over and figured a change in wording would make a difference to my boredom levels. it obviously didn't...) + Show Spoiler +i can only make sense of your post when you're talking in halves: earth to L1 and L1 to moon. i just can see them canceling each other without some serious interfering which would make the whole idea not worth it to begin with)
Why are we talking about terminal velocity? Sure, when taking the atmospheric friction into account, it becomes clear that we will need to add a bit more energy (perhaps by adding more rocks on the moon side than the mass of water we scoop up on the earth side), but the whole point of the carts going towards the earth is to serve as a counterweight to the ones going towards the moon, thereby making it one gigantic elevator (if you look in an elevator you see those blocks of lead up near the wall, hanging on a cable? That cable is the same one that connects to the box you stand in. They are calculated to weigh the same as the elevator (with an average load in it), so as to minimize the amount of energy required to keep the whole thing moving. You still need energy to accelerate at the start, and counteract friction, but you don't have to counteract gravity when moving the elevator up, because that's what the lead is for. That is the exact same function as the blocks of moonrock serve in the railway system.
Terminal velocity is irrelevant.
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On June 17 2015 15:53 OtherWorld wrote:Show nested quote +On June 17 2015 15:42 Thieving Magpie wrote:The flaw with Capitalism is that it assumes there is no correct or wrong cost to any object. Well to be fair, how would you put a correct or wrong cost to an object?
Well, absolute value is usually arbitrated (It is X because Y said so)
But what I was talking about was having stricter guidelines on how to price things to prevent overinflation and to provide protection for lower class communities. The original idea of capitalism was that companies would make things cheaper or more expensive depending on the needs of people in that market. This is no longer true.
Right now, local cost is connected to non-local demands creating a market where costs are dictated by the people outside of the community that has to face the brunt of that evolving market. The main side effect is that the lower class are getting kicked out of their communities despite having a stable market prior.
Example: Housing Market Spikes in the Silicon Valley due to immigration affecting local housing for the residents of the bay area.
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On June 18 2015 02:22 TMagpie wrote:Show nested quote +On June 17 2015 15:53 OtherWorld wrote:On June 17 2015 15:42 Thieving Magpie wrote:The flaw with Capitalism is that it assumes there is no correct or wrong cost to any object. Well to be fair, how would you put a correct or wrong cost to an object? Well, absolute value is usually arbitrated (It is X because Y said so) But what I was talking about was having stricter guidelines on how to price things to prevent overinflation and to provide protection for lower class communities. The original idea of capitalism was that companies would make things cheaper or more expensive depending on the needs of people in that market. This is no longer true. Right now, local cost is connected to non-local demands creating a market where costs are dictated by the people outside of the community that has to face the brunt of that evolving market. The main side effect is that the lower class are getting kicked out of their communities despite having a stable market prior. Example: Housing Market Spikes in the Silicon Valley due to immigration affecting local housing for the residents of the bay area.
Ah ok I get it
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On June 18 2015 01:59 Simberto wrote:
Ok, firstly, i was talking without friction, as air friction pretty much stops almost anything from working (hence the vacuum tower in Cascades picture)
And i think i finally found the problem in our understanding. I do not mean that carts going down on the moon are cancelling out carts going up on earth. That does not work, as the gravitational potential of the surface of the moon is still a lot higher than on the surface of the earth, and thus a cart going down to the moon from L1 does not cancel out a cart going up from Earth to L1.
However, a cart going down from L1 to the Earth DOES cancel out a cart going up from Earth to L1.
And a cart going down from L1 to the Moon cancels out a cart going up from the Moon to L1.
The above is obviously only true if the carts have the same mass, and similarly clearly it does not matter if instead of sending up and down water on one side, and moon rocks up and down on the other side, you swap the carts in the middle so on Earth Water goes up and moonrocks go down, and on the Moon water goes down and Moonrocks go up.
Thus, if you set things up correctly (the energy transfer between the carts is the crux here, i suggested a rope with wheels on the ends, but other solutions are also workable. (and ignore friction or make the carts move slow enough that friction is not that relevant), both the Moon - L1 side of the apparatus and the Earth - L1 side are energy neutral, and thus the whole thing is also energy neutral.
yes, i get it now; i missed the old switcheroo you have going there. i'll think about it
@Acrofales: there were 2 elevators there going in different directions not one going from the moon to the earth.
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Energy efficiency does seem to be a problem, if not a severe one. We only have about 50 times as much moon as we have oceans. But that raises another question; if we send half the ocean to the moon and dump half the moon into the ocean, do we win?
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EPT![[image loading]](http://i.imgur.com/j99e11Z.png)
![[image loading]](http://i.imgur.com/PLIGn8A.jpg)
