An experiment to disprove FET

Take a ship 10m tall and assume a 'horizon' 30km away. This means, according to an accelerating earth model, that in the time light travels 30,000m horizontally the Earth has moved 10m vertically; ie in about 1/10,000th of a second. Meaning that the vertical velocity of the Earth at this point in time is about 100,000m/s.

After a day accelerating at 1g (call it 10m/s/s) the Earth will now be moving at approximately 100,000+(10x3,600x24)=964,000m/s. The horizon distance of the same ship will now be, well about a tenth as far away, obviously.

I think that kills that

No. The boat, the ocean, and the observer are all moving at the same speed. The argument is that while the light is "in flight" that the ocean accelerates blocking the light's travel. The RE Primer works the math for a similar case, that of the setting Sun, in detail. In the 3,000 mile distance in that case, the FE could not have moved more than one inch.

No. The boat, the ocean, and the observer are all moving at the same speed. The argument is that while the light is "in flight" that the ocean accelerates blocking the light's travel. The RE Primer works the math for a similar case, that of the setting Sun, in detail. In the 3,000 mile distance in that case, the FE could not have moved more than one inch.

I tried to figure out how far away the boat had to be to lose 1m of it's image.
Okay this is my math:
First calculate the final velocity of the observer after it's traveled 1m upwards.

V_f² = V_i² + 2ad

V_f² = 0m/s² + 2(9.81m/s²)(1m)

V_f² = 19.62m²/s²

V_f = 4.430m/s

Then the average velocity of the 1m trip.

V_av = 1/2(V_f + V_i)

V_av = 1/2(4.430m/s + 0m/s)

V_av = 2.215m/s

Then the time it took to make that trip.

t = d/V_av

t = 1m/2.215m/s

t = 0.4515s

So in that time it took to travel 1m with an acceleration of 9.81m/s² I caclulated how far a beam of light traveled.

d = (v)(t)

d = (3.00x10⁸m/s)(0.4515s)

d = 135450000m

So in the boat would have to be 135,400km away to lose 1m of the boat.

Er, I'm terrible with math, but that seems....uh...wrong.

Quote from: Gulliver on August 13, 2007, 04:58:57 PM

No. The boat, the ocean, and the observer are all moving at the same speed. The argument is that while the light is "in flight" that the ocean accelerates blocking the light's travel. The RE Primer works the math for a similar case, that of the setting Sun, in detail. In the 3,000 mile distance in that case, the FE could not have moved more than one inch.

I tried to figure out how far away the boat had to be to lose 1m of it's image.
Okay this is my math:
First calculate the final velocity of the observer after it's traveled 1m upwards.

V_f² = V_i² + 2ad

V_f² = 0m/s² + 2(9.81m/s²)(1m)

V_f² = 19.62m²/s²

V_f = 4.430m/s

Then the average velocity of the 1m trip.

V_av = 1/2(V_f + V_i)

V_av = 1/2(4.430m/s + 0m/s)

V_av = 2.215m/s

Then the time it took to make that trip.

t = d/V_av

t = 1m/2.215m/s

t = 0.4515s

So in that time it took to travel 1m with an acceleration of 9.81m/s² I caclulated how far a beam of light traveled.

d = (v)(t)

d = (3.00x10⁸m/s)(0.4515s)

d = 135450000m

So in the boat would have to be 135,400km away to lose 1m of the boat.

I encourage you to refer the RE Primer for a simpler approach, but I do very much applaud your result. Good job.

(s= 1/2at²)

So narc is wrong, as usual. The acceleration of the FE cannot cause the effects of either sunsets or disappearing hulls.

Waves oscillate. If a wave appears to block the sail at one moment it should not do so the next. If Tom is right you would see a ship... No ship... A ship... No ship...

Is this what is observed?

Didn't think so.

It's a compound effect. We're not talking about a single wave obscuring the hull of a boat, we're talking about an increased frequency of hull-obscuring waves such that at any given moment the hull is invisible.

Quote from: Brennan on August 13, 2007, 01:57:37 PM

Waves oscillate. If a wave appears to block the sail at one moment it should not do so the next. If Tom is right you would see a ship... No ship... A ship... No ship...

Is this what is observed?

Didn't think so.

It's a compound effect. We're not talking about a single wave obscuring the hull of a boat, we're talking about an increased frequency of hull-obscuring waves such that at any given moment the hull is invisible.

Why would the frequency increase? 

Er, I'm terrible with math, but that seems....uh...wrong.

* divito waits for Gulliver.

Short wait.

Short wait.

Wasn't worth it. Although, he hasn't arrived in my other thread. Maybe he is preparing for his Nobel Prize.

Or a purpose.

I tried to figure out how far away the boat had to be to lose 1m of it's image.
Okay this is my math:
*snip math*

The problem with this whole argument is that it only considers light rays emitted horizontally. Some light rays are also emitted up at an angle, so that they bend down as the earth accelerates and still hit the observer. Quite simply, at no distance can acceleration of the earth cause all the light emitted from a distant object to be blocked by the ground without an intervening object in between - the light curves down, so actually distant objects will appear higher rather than lower, as the light observed to be coming from them is emitted up at an angle and then curves down so that it reaches the observers eye coming down at an angle.

Os course, the acceleration is too small compared to the speed of light to make any difference anyways. But my point is that even if the effect were large enough to be significant, it would still be in the wrong direction to explain the horizon.

No. The boat, the ocean, and the observer are all moving at the same speed. The argument is that while the light is "in flight" that the ocean accelerates blocking the light's travel. The RE Primer works the math for a similar case, that of the setting Sun, in detail. In the 3,000 mile distance in that case, the FE could not have moved more than one inch.

So the Earth isn't accelerating then? Make up your minds please...

What causes what we think of as gravitational acceleration then?

Where did you get that the FE wasn't accelerating from that?

Apologies, being dense.

The objection seems irrelevant though. The horizon distance is dependant upon the velocity of the Earth. If the Earth accelerates the velocity will be higher and the horizon distance becomes correspondingly smaller.

I've shown that this would be seriously noticeable at 1g (a 30km horizon becomes a 3km horizon in one year), yet we do not observe it.

Why would velocity affect it? 

The distance at which the Earth blocks horizontal light will vary proportionally to the vertical velocity of the Earth.

If the Earth is stationery and flat there would be no horizon, light would travel straight across the sea without being impeded.

If the Earth moved vertically 9.81 m (ie in the first second of it's alleged acceleration) Socrates' example holds.

My example shows the velocity needed for a 30km horizon (which is approximately what we now have IIRC) and extrapolates the horizon 1 day later given 1g of acceleration.

This is clearly a load of bunk. First of all, velocity has no effect, since we can do the math in an inertial reference fram where the earth is stationary, and then the ocean isn't blocking things at all. Secondly, even acceleration won't cause the ocean rising up and hitting light rays to stop you from seeing distant objects because of an apparent close horizon, because when you consider light rays emitted at all angles, it becomes clear that the effect is in the wrong direction, and in fact distant object appear higher up than they would otherwise, not lower down as would they would have to in order to disappear beneath a horizon.

Either you first statement is 'complete bunk' or you are assuming that light is being accelerated by the UA as well. At which point the argument I was refuting is no longer in use and you have are having to use an alternative explanation, in which case I claim a win.

Secondly: Your changed paradigm suffers from the same problem I have described: as the velocity increases over time, the apparent angles you are talking about would all be steadily increasing with the effect that the world would appear to be rising up around us. This obviously does not happen.

2 wins.

Your argument is so incoherent that I don't even know where to begin.

First of all, you are aware of special relativity, correct? In other words, you know that the speed of light is the same in all reference frames? It seems not, since you seem to be thinking that the speed of light is the speed at which light moves in one specific priveleged reference frame, and is modified accordingly in other reference frames. This is a 19th century worldview, and went out of fashion in the early 1900s, so I can only hope that I am mistaken. If not, you should read Wikipedia's introduction to special relativity (or another introduction, since Wikipedia's seems to leave out the Michaelson-Morley experiment and others which strike me as important introductory material).

If you are aware of special relativity, you should understand that if we neglect the supposed acceleration of the flat Earth, and focus instead on the velocity, we can just as well treat the FE in the model as stationary, and so there will obviously be no horizon effect due to the speed at which the Earth is moving. This is why my first claim is not bunk.

My second point is a bit more subtle, so I'll wait until you have mastered this before trying to explain it.

Your argument is so incoherent that I don't even know where to begin.

First of all, you are aware of special relativity, correct? In other words, you know that the speed of light is the same in all reference frames? It seems not, since you seem to be thinking that the speed of light is the speed at which light moves in one specific priveleged reference frame, and is modified accordingly in other reference frames. This is a 19th century worldview, and went out of fashion in the early 1900s, so I can only hope that I am mistaken. If not, you should read Wikipedia's introduction to special relativity (or another introduction, since Wikipedia's seems to leave out the Michaelson-Morley experiment and others which strike me as important introductory material).

If you are aware of special relativity, you should understand that if we neglect the supposed acceleration of the flat Earth, and focus instead on the velocity, we can just as well treat the FE in the model as stationary, and so there will obviously be no horizon effect due to the speed at which the Earth is moving. This is why my first claim is not bunk.

K.O

oh and one question (I read the FAQ but couldn't find it), does a FE rotate?

Your argument is so incoherent that I don't even know where to begin

This is a stupid and irrelevant dodge. My calculations do not in any way require the velocity of light to vary in any sense.

Then explain how it is that an Earth moving upwards experiences this horizon effect you claim is proved by your calculations while a stationary Earth does not, despite Galilean invariance which tells us that the laws of physics are the same in all inertial frames, and that no experiment can tell a closed system in constant motion apart from a closed system which is stationary.

Then explain how it is that an Earth moving upwards experiences this horizon effect you claim is proved by your calculations while a stationary Earth does not, despite Galilean invariance which tells us that the laws of physics are the same in all inertial frames, and that no experiment can tell a closed system in constant motion apart from a closed system which is stationary.

yup, right on that one

its all about your frame of reference, and what calculations did you do anyway?

See that term 'constant motion' you used?

In Universal Acceleration motion is not constant. I'd have thought that was stunningly obvious myself.

Take a ship 10m tall and assume a 'horizon' 30km away. This means, according to an accelerating earth model, that in the time light travels 30,000m horizontally the Earth has moved 10m vertically; ie in about 1/10,000th of a second. Meaning that the vertical velocity of the Earth at this point in time is about 100,000m/s.

After a day accelerating at 1g (call it 10m/s/s) the Earth will now be moving at approximately 100,000+(10x3,600x24)=964,000m/s. The horizon distance of the same ship will now be, well about a tenth as far away, obviously.

Perhaps you chaps would like to make it clear which FE paradigm you are representing btw. You certainly do not appear to agree with the poster I refuted in the first place.

For clarification: By means of what mechanism do we perceive a horizon?

The acceleration is constant.  The velocity is not.  Your method of calculating the answer is incorrect.

See that term 'constant motion' you used?

In Universal Acceleration motion is not constant. I'd have thought that was stunningly obvious myself.

Quote

Take a ship 10m tall and assume a 'horizon' 30km away. This means, according to an accelerating earth model, that in the time light travels 30,000m horizontally the Earth has moved 10m vertically; ie in about 1/10,000th of a second. Meaning that the vertical velocity of the Earth at this point in time is about 100,000m/s.

After a day accelerating at 1g (call it 10m/s/s) the Earth will now be moving at approximately 100,000+(10x3,600x24)=964,000m/s. The horizon distance of the same ship will now be, well about a tenth as far away, obviously.

You said

My example shows the velocity needed for a 30km horizon (which is approximately what we now have IIRC)

In other words, you are claiming that even without any acceleration at work, the velocity of the Earth would cause a horizon effect. This is complete bullshit. Due to the principle of invariance, which I just explained, observers on a flat Earth moving at a constant velocity would perceive things exactly the same way as observers on a stationary flat Earth. In other words, you are completely missing the point by doing your math in a reference frame in which the Earth is moving, and are confusing yourself. The problem with doing the math in a moving reference frame is that you need to consider a) light emitted at an angle so that it is heading towards where the observer will be, not where the observer is now, which the observer will see despite the movement of the earth, and b) the relativistic beaming effect which explains why this light, although it has traveled further, appears just as bright to the observer. You are insisting on doing the math in a moving reference frame without taking these effects into consideration, which is why your argument is bunk. You either need to take them into account, or else simply think about what would happen in a reference frame in which the Earth is stationary, which is equivalent anyways. Now do you understand why your calculation of "the velocity needed for a 30km horizon (which is approximately what we now have IIRC)" is completely wrong and based on incorrect assumptions? Once we have settled this point we can start talking about accelerating reference frames.

This argument has gone beyond my head .

This argument has gone beyond my head .

If I were you, I'd chalk that up to the fact that brennan is talking nonsense. If you ignore everything he said, you should be less confused.

Engineer: The acceleration is too small to make a significant difference over the timescales in question. At 100,000m/s i'm not going to bother calculating an extra 10m/s over 1/10,000 th of a second am I?

Scientist: The effect requires acceleration, I felt that this was clear. Your massive response does not appear to dispute that. The fact that I mentioned a specific instantaneous velocity was merely for illustrative purposes since the effect upon the calculations of including the acceleration (considering how small it is compared to the actual velocity) is negligible, as I said above.

You also seem to accept that light will be arriving at a different angle, so we appear to be broadly in agreement.

I never mentioned the brightness of the light in question, not sure why you brought it up.

This argument has gone beyond my head .

Focus on this then:

The acceleration is constant.  The velocity is not.  Your method of calculating the answer is incorrect.