The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Debate => Topic started by: Levi Dettwyler on April 05, 2013, 10:38:28 PM
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I've just realized a huge problem with the so-called "universal accelerator" (other than the obvious ones, like diverging to an infinite amount of energy). The asteroid that hit Russia not too long ago came in at a pretty low speed relative to what the earth would have to be traveling at now if under constant 9.8 ms^-2 acceleration, so obviously the universal accelerator would also have to have been acting on the asteroid, and it just happened to be moving toward us when the accelerator turned on or something. However, this isn't the problem.
The problem is how the universal accelerator chooses which objects to accelerate. It obviously accelerates the earth, but not the things on the surface, namely humans and rocks. Those all fall back to the earth. Even dirt dug out from the ground will fall back.
So maybe it's just pushing on the bottom of things? If that's the case, why doesn't it push on me when I jump? It pushes on the sun and moon, and they're not too far above the surface of the earth either.
Why was it accelerating the asteroid for a long time beforehand but randomly decided to stop to allow it to fall to earth? It can't just be pushing on the earth, or else we would have slammed into that asteroid at such a ridiculously high speed it would have torn the earth completely apart, since we gain 9.8 meters/second every second. So the universal accelerator decided to keep it accelerated along with us before stopping when it was just close enough for it to fall at a relatively low speed to earth's surface and break a few windows?
The universal accelerator makes no sense at all. Once I actually thought about it, I realized that it's the most contrived thing I've ever read on this forum. The things that it accelerates and doesn't accelerate (and when!) seem so random, if not just to specifically account for all the problems that arise when you make the earth flat and gravity not important. You have two explanations in front of you: A and B. Explanation A accounts for everything that you observe, and does so very simply. Explanation B on the other hand does not account for what you observe, and becomes immensely and arbitrarily complicated when you try (and fail) to make it work. Sane people would choose A.
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I do not understand either how air resistance works out to exactly 9.8 m/s^2 either. I would think you should not see a parabolic path described by an object going up and then down. At least not every time, since air resistance would be somewhat variable.
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Imagine a boat on a lake. The wind blows the sails. Now imagine a sailor stood behind the sails. Does the wind also blow him? No, because he is shielded from the wind the same way as we are shielded by the earth from UA. Before you decide on a semantic debate about sails and wind, a tactic in sailing races is to 'steal' the wind of other competitors, ie get in the way of their wind.
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Imagine a boat on a lake. The wind blows the sails. Now imagine a sailor stood behind the sails. Does the wind also blow him? No, because he is shielded from the wind the same way as we are shielded by the earth from UA. Before you decide on a semantic debate about sails and wind, a tactic in sailing races is to 'steal' the wind of other competitors, ie get in the way of their wind.
OK, so a phenomenon of the universal accelerator is that it doesn't push on things if something else is in its way. Does it push on the sun and moon then when they're directly above the earth? How high are they... 3,000 miles, compared to the diameter of the earth, which is... what, 25,000 miles (the approximate circumference of a round earth, so lain flat, it should be the diameter, or close to it)? If it's truly like wind, then the effect on the sun and moon, if any effect at all, should be very small compared to the effect on the earth, since the earth is getting in the way so much. Therefore, we should have crashed into them a long time ago.
Or does it make a "special exception" for the sun and moon?
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At 3000 miles there is sufficient distance for the UA to lick around the earth and push celestial bodies. They are not close enough to be protected from the UA stream.
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At 3000 miles there is sufficient distance for the UA to lick around the earth and push celestial bodies. They are not close enough to be protected from the UA stream.
Really? 3000 miles versus 25,000 miles that the earth extends to? Perhaps I can diagram this... http://grab.by/lpfC (http://grab.by/lpfC)
The sun is shown on the left, 3 units from the surface of the cylinder. The cylinder has a diameter of 25 units.
It sure doesn't look like there's sufficient distance. Maybe a LITTLE BIT of the "UA stream" will lick around, but only a very small amount, so it should be accelerating the sun a lot less than the earth, correct?
Unless, of course, you tell me that the earth is only 10 miles wide or something. Then I might believe it. But at only 3000 miles away, given a reasonable diameter, the earth is simply too wide to not be significantly blocking it.
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If you were to look at this in terms of fluid dynamics the angle at which the UA could turn would be based on its momentum and the Reynolds number. As you do not know and have not asked the Reynolds number of UA, I fail to see how you can scoff and say 3000 miles is not enough.
Similar shapes can give wildly differing wakes depending on that Reynolds number.
(http://www.see.ed.ac.uk/~johnc/teaching/fluidmechanics4/2003-04/fluids14/image41.gif)
(http://www.classicairshows.com/Education/Aerodynamics/AeroDynamicsImages/AerodynamicsOne2.gif)
Point M will move depending on the viscosity of UA as shown by the formula
(https://upload.wikimedia.org/math/2/3/0/23032e25451927df6094806dcd3ba4dd.png)
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Does FE have a solution then that gives 3000 miles for the point M yet maintains a reasonable diameter for the earth?
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A Reynolds number between 5 and 10 would allow the UA to make the turn around earth and still push the sun and moon.
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A Reynolds number between 5 and 10 would allow the UA to make the turn around earth and still push the sun and moon.
"Still push," as in, push the sun as it pushes on the earth, giving an equivalent acceleration, i.e. behaving as if the earth isn't there at all?
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Yes, if you like.
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A Reynolds number between 5 and 10 would allow the UA to make the turn around earth and still push the sun and moon.
The Reynolds number must be close to 0!
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Does this also take into account the fact that the fluid must also be accelerating? This isn't a fluid with a constant velocity. The velocity of the fluid is always changing in order to be able to always be applying 9.8ms^-2 of acceleration on anything caught in it.
The formula you posted simply doesn't seem adequate for the actual nature of the problem. It has a variable for a relative velocity, v, but everything in the system is accelerating. What are you plugging in for v?
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A Reynolds number between 5 and 10 would allow the UA to make the turn around earth and still push the sun and moon.
The Reynolds number must be close to 0!
No, if it was 0 then surface objects like you would also be accelerated. This would make you 'weightless', which you are not.
Does this also take into account the fact that the fluid must also be accelerating? This isn't a fluid with a constant velocity. The velocity of the fluid is always changing in order to be able to always be applying 9.8ms^-2 of acceleration on anything caught in it.
What difference does it make?
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The formula you posted simply doesn't seem adequate for the actual nature of the problem. It has a variable for a relative velocity, v, but everything in the system is accelerating. What are you plugging in for v?
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v would depend on t. at t=0 it is 0. At t=1 it is 9.81 m/s. You said it yourself 'relative velocity'. So with relativity you may assume the earth to be stationary and the datum.
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A Reynolds number between 5 and 10 would allow the UA to make the turn around earth and still push the sun and moon.
The Reynolds number must be close to 0!
No, if it was 0 then surface objects like you would also be accelerated. This would make you 'weightless', which you are not.
What happens if the Reynolds number is negative?
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A Reynolds number between 5 and 10 would allow the UA to make the turn around earth and still push the sun and moon.
The Reynolds number must be close to 0!
No, if it was 0 then surface objects like you would also be accelerated. This would make you 'weightless', which you are not.
What happens if the Reynolds number is negative?
The universe will turn inside out and it will rain frogs.
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What you just described is impossible. Remember, v = earth's velocity - UA's velocity. This is what makes it relative (to the fluid).
At t=0, according to you, the relative velocity is 0. Therefore, if the earth's velocity is 0, then the UA's velocity is also 0 (0-0=0). This makes sense.
However, at t=1, you say the relative velocity is 9.81 m/s. This means that the velocity of the earth minus the velocity of the UA is 9.81. Therefore, if the earth's velocity is 9.81, which is consistent with the earth's known accleration (9.81 ms^-2), the UA's velocity is still 0, which implies that the UA isn't accelerating, which doesn't make sense.
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You are implying that at t=1 then v is still 0. Well good luck punching that into the equation. The UA accelerates around the earth and on to push the sun and moon. This is the v you need.
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You are implying that at t=1 then v is still 0. Well good luck punching that into the equation. The UA accelerates around the earth and on to push the sun and moon. This is the v you need.
No I'm not. v is RELATIVE VELOCITY. I'm using the numbers you gave me. You said that at t=1, v=9.81. This is relative velocity, which means if the earth's absolute velocity (relative to the medium rather than the UA) is 9.81, which is what it should be after 1 second, since that's what its measured acceleration is, then the velocity of the UA would still have to be 0, since 9.81-0=9.81.
Perhaps you mixed up your variables? Let's be explicit. v is relative velocity, i.e. absolute velocity of earth - absolute velocity of the UA, which is the "v" in your equation. Absolute is just relative to the medium that this is all taking place in.
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I'll try this again.
t=0, v=0 (as you said), v₁ (earth's velocity) = 0 (we're assuming this), v₂ = 0 (UA's velocity).
t=1. We know the acceleration of the earth is 9.81m/sē, so at t=1 the velocity of the earth MUST be 9.81 m/s. At t=1, you said v (the relative velocity) is also 9.81, which means the UA's velocity (v₂) hasn't changed at all. v₁-v₂=v, so 9.81-0=9.81. The numbers you gave don't make sense. Perhaps you meant the earth's velocity is 9.81 at t=1 rather than the relative velocity?
So I ask again, what value for v (the relative velocity, i.e. v₁-v₂) are you using in your fluid dynamics equation, and are you taking into account the fact that it necessarily changes, since the system is under acceleration, and it takes nonzero time for the UA to impart its momentum energy onto the earth due to inertia?
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(I'm done making edits to the above comment now)
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A Reynolds number between 5 and 10 would allow the UA to make the turn around earth and still push the sun and moon.
The Reynolds number must be close to 0!
No, if it was 0 then surface objects like you would also be accelerated. This would make you 'weightless', which you are not.
What happens if the Reynolds number is negative?
The universe will turn inside out and it will rain frogs.
Does that result from the formula you provided?
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Does that result from the formula you provided?
Technically, it could. If the formula implies something known to be false (let's say A), then you can construct the following implication:
A ⇒ the universe will turn inside out and it will rain frogs.
By the definition of implication, the above statement is correct. A false statement can imply essentially anything you want.
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Anyone else care to elaborate on this? Posting an equation from wikipedia and failing to account for the values that you used when doing your calculations, does not an explanation make.
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Actually, now that I think about it more, the earth still should have crashed into the sun and moon a long time ago. Why? Because in order to keep accelerating the earth, the UA stream itself must be accelerating. We can think about it in terms of two objects sitting in a dry riverbed. If you open the floodgates (acceleration), the water will reach object A first, taking a little longer to reach object B, since they're spread apart. Thus, it should carry A closer to B before it begins carrying them both at the same speed.
But if you continue to accelerate the flow of the water, it's always going to push object A before object B. Thus, the earth should be getting pushed faster before the sun and moon. Ergo, the sun and moon should have crashed into us a long time ago. It's even worse than that though: since the closer they get, the further away from the point "M" that Thork pointed out they get, so the less of an effect the UA stream will have on it. As a result they should not just crash into us, but do so at high speed! According to the flat earth hypothesis in its current state, we are all dead.
Makes sense!
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Imagine a boat on a lake. The wind blows the sails. Now imagine a sailor stood behind the sails. Does the wind also blow him? No, because he is shielded from the wind the same way as we are shielded by the earth from UA. Before you decide on a semantic debate about sails and wind, a tactic in sailing races is to 'steal' the wind of other competitors, ie get in the way of their wind.
I really cannot accept that you are as stupid as your own comments seem to suggest. Get on a boat and make your own observations, or even better, accept that you are a lot more intelligent than this and look for another analogy.
Wind is everywhere. On both sides of the sail, over the deck, around the hull. Your oversimplification of the subject is so overwhelmingly vast that you just cannot do it with a straight face.
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A Reynolds number between 5 and 10 would allow the UA to make the turn around earth and still push the sun and moon.
Don't we require a laminar flow to push sun and moon at a constant rate?
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According to the flat earth hypothesis in its current state, we are all dead.
No, the dead live at the bottom.
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Oh come on, don't tell me I'm going to have to start a new thread just to discuss the UA killing us. I even updated the title!