swansont Posted June 26, 2017 Posted June 26, 2017 Surely figure A is total nonsense on any level, unless I'm missing something. Primary School logic: if the moon orbits the Earth in 28 days, then it moves 360/28 = 13 degrees per day. At an eclipse, the angle between the Earth-Sun and Earth-Moon lines is zero. So one day later that angle is about 13 degrees. How the hell can it be 90 degrees after 4 hours? Different angle. You're quoting the advance in the position in the sky from day to day, along the orbital path. I'm talking about the position with respect to the ecliptic. IOW, look at the position an hour earlier in the day. It will not have changed very much. But no matter, since the geometry presented is nonsense, and you are still correct that the angle is impossible.
Bjarne Posted June 26, 2017 Author Posted June 26, 2017 (edited) Your figures are bogus. The moon is never more than ~5º above or below the ecliptic. Yo have it at almost 90º in "A". The moon is also not 4000 km nor 8000 km away from the earth under any circumstances. These diagrams are meaningless. You misunderstood the illustration, - it shows that the Earth have moved - "to the right" relative to the Moon, Because the orbit speed of the earth (around the Sun) at this time is faster than the speed of the Moon. Therefore the resulting force vector (pulling the Earth upwards) has also changed.. This is the main reason to that the anomaly duration is total about 24 hours. - The upwards or downwards motion of the moon, play a secondary role, in that question. ( I have changed my mind few days ago, and also wrote it) This is the primary answer to why the Allais Effect only is true by (some) solar and lunar eclipse. https://www.youtube.com/watch?v=W47Wa7onrIQ What is true, however, is that the moon can be at whatever angle it has at an eclipse, during *any* new moon or *any* full moon, since it crosses the ecliptic twice during each orbit, and the angle will have changed only a small amount each day before and after those days. If the effect is due to the vertical component of gravity from the location of the moon, it means that you have not shown that an eclipse need be involved. plus, your math is of course wrong. sin45 = .707. sin22.5 = .3827. The ratio is 1.84. Close, but not a factor of 2. If you can't apply simple trigonometry here, you are always going to be wrong, even if your idea had any merit. Perhaps you should spend some time learning the basics. You misunderstood that too, - "C" is angle C, - (see the image below) to find the resulting upwards acceleration, you shall not use Sin*angle - but Vector Addition Calculation. - My calculation seems to be correct How are you defining upwards? Is that north, or directly overhead for someone on earth? North relative to ecliptic If the moon is 4000 km above the ecliptic, the angle is the same in all cases. In this case, you aren't even measuring with the same coordinate system. Very sloppy. More trigonometry fail. The new discovery in this thread is that the upwards (or downwards) motion of the moon (discussed last week) play a secondary role. The primary point is now; - the fast orbit speed of the earth (relative to the Moon) right before solar eclipse that within 12 hours brings the Earth to a position where the resulting upwards acceleration is pointing so much vertical as possible. 4 hours before and 4 hours after will the resulting upwards force have lost 50% and about 8 hours later 75% of the upwards pull. 4 hours after max eclipse the resulting force / aacc begin to point more and more horizonal.. - Not because of the motion of the moon, but because the motion of the Earth. THIS is why only (some) solar and (some) lunar eclipse is able to expose DFA. No other constellation (except lunar eclipse) will come near that magnitude of upwards acceleration. 1 week after the eclipse, - now the resulting upwards force is alomust 0, - the force is now pointing more than 99% horizontal Edited June 26, 2017 by Bjarne
swansont Posted June 26, 2017 Posted June 26, 2017 You misunderstood the illustration, - it shows that the Earth have moved - "to the right" relative to the Moon, Because the orbit speed of the earth (around the Sun) at this time is faster than the speed of the Moon. Therefore the resulting force vector (pulling the Earth upwards) has also changed.. This is the main reason to that the anomaly duration is total about 24 hours. - The upwards or downwards motion of the moon, play a secondary role, in that question. ( I have changed my mind few days ago, and also wrote it)[/size] The moon will never be more than ~5º away from the ecliptic, and you show it being almost 90º in A, and still quite a large angle in B. The anomaly duration is nowhere near 24 hours in the data. The force vector is a function of the relative position of the moon above or below the ecliptic. It has this position twice every orbit. You misunderstood that too, - "C" is angle C, - (see the image below) to find the resulting upwards acceleration, you shall not use Sin*angle - but Vector Addition Calculation. - My calculation seems to be correct Vector addition uses trigonometry. North relative to ecliptic It's one or the other, but it can't be both. The earth is tilted, so these two are different by ~23.5º The new discovery in this thread is that the upwards (or downwards) motion of the moon (discussed last week) play a secondary role. The primary point is now; - the fast orbit speed of the earth (relative to the Moon) right before solar eclipse that within 12 hours brings the Earth to a position where the resulting upwards acceleration is pointing so much vertical as possible. 4 hours before and 4 hours after will the resulting upwards force have lost 50% and about 8 hours later 75% of the upwards pull. 4 hours after max eclipse the resulting force / aacc begin to point more and more horizonal.. - Not because of the motion of the moon, but because the motion of the Earth. THIS is why only (some) solar and (some) lunar eclipse is able to expose DFA. No other constellation (except lunar eclipse) will come near that magnitude of upwards acceleration. 1 week after the eclipse, - now the resulting upwards force is alomust 0, - the force is now pointing more than 99% horizontal You keep saying as vertical as possible but also say that the effect goes away if the force is too big. So yet another contradiction from you. Vector addition does not support your claim that this vertical acceleration happens only in an eclipse, unless you are proposing some new theory of gravity. If the moon is 1º above the eclipse, it will exert the same vertical component of force regardless of its position relative to the sun. And that happens twice every orbit (near the new moon and near the full moon) You are predicting the effect will occur over at least 8 hours. The data do not agree with this prediction.
Bjarne Posted June 26, 2017 Author Posted June 26, 2017 (edited) It's one or the other, but it can't be both. The earth is tilted, so these two are different by ~23.5º Upwards = Ecliptic northpol Edited June 26, 2017 by Bjarne
swansont Posted June 26, 2017 Posted June 26, 2017 What unholy creation is this? You do realize that this is a two-dimensional depiction, and the sun, moon and earth will never have this orientation, right?The only way this makes sense is if the sun is 150,000,000 km deep into the screen, and we are looking at a projection of it. (at 4000km the sun would be (mostly) blocked by the earth, whose radius is ~6400 km) The 45º angle makes no sense. Why are you picking a point 4000 km from the earth to define it? Who cares abut the sun at all, since it exerts no force in the vertical direction — ever? It is always on the ecliptic. By definition. Your model, as you have presented it, only depends on the position of the moon.
Bjarne Posted June 26, 2017 Author Posted June 26, 2017 The moon will never be more than ~5º away from the ecliptic, and you show it being almost 90º in A, and still quite a large angle in B. The anomaly duration is nowhere near 24 hours in the data. The force vector is a function of the relative position of the moon above or below the ecliptic. It has this position twice every orbit. What unholy creation is this? You do realize that this is a two-dimensional depiction, and the sun, moon and earth will never have this orientation, right?The only way this makes sense is if the sun is 150,000,000 km deep into the screen, and we are looking at a projection of it. (at 4000km the sun would be (mostly) blocked by the earth, whose radius is ~6400 km) The 45º angle makes no sense. Why are you picking a point 4000 km from the earth to define it? Who cares abut the sun at all, since it exerts no force in the vertical direction — ever? It is always on the ecliptic. By definition. Your model, as you have presented it, only depends on the position of the moon. You keep saying as vertical as possible but also say that the effect goes away if the force is too big. So yet another contradiction from you. Vector addition does not support your claim that this vertical acceleration happens only in an eclipse, unless you are proposing some new theory of gravity. If the moon is 1º above the eclipse, it will exert the same vertical component of force regardless of its position relative to the sun. And that happens twice every orbit (near the new moon and near the full moon) You are predicting the effect will occur over at least 8 hours. The data do not agree with this prediction. I am just trying to explain you (and illustrate) that when the moon, sun and earth is aligned , the resulting upwards acceleration / force is so strong as it can be, - 4 hours later 50% of the vertical pull is gone. And that strong upwards pull in the Earth will only get back by solar and lunar eclipse.
Mordred Posted June 26, 2017 Posted June 26, 2017 (edited) Well as I didn't see any denial that this is about dark flow and not Allais effect. I will have to assume I am correct in my assessment. Particularly as you have not shown you can generate the specific anomoly on the pendulum. Swansont is already doing a good job pointing out all the math errors in your trig vector additions. Also pointing out what the orbits actually are. lmao Edited June 26, 2017 by Mordred
DrKrettin Posted June 26, 2017 Posted June 26, 2017 I am just trying to explain you (and illustrate) that when the moon, sun and earth is aligned , the resulting upwards acceleration / force is so strong as it can be, - 4 hours later 50% of the vertical pull is gone. And that strong upwards pull in the Earth will only get back by solar and lunar eclipse. But this is nonsense, as I said earlier. Four hours later the moon will have moved about two degrees, not 90 degrees.
swansont Posted June 26, 2017 Posted June 26, 2017 I am just trying to explain you (and illustrate) that when the moon, sun and earth is aligned , the resulting upwards acceleration / force is so strong as it can be, - 4 hours later 50% of the vertical pull is gone. And that strong upwards pull in the Earth will only get back by solar and lunar eclipse. During an eclipse, the moon is very near the ecliptic. The upward component of the force is the smallest it will be during the moon's orbit. It will be much larger 2 weeks later, when it's 5º above the ecliptic. I've pointed this out, and you said it's too strong then. So I don't understand why you keep saying these conflicting things.
Bjarne Posted June 26, 2017 Author Posted June 26, 2017 (edited) You keep saying as vertical as possible but also say that the effect goes away if the force is too big. So yet another contradiction from you. Yes I say so vertical upwards pull of the Earth as possible. However in a way where the moon NOT is above the test body, because the test body then not is free to follow DFA. It’s not a contradiction but 2 necessary requirements Vector addition does not support your claim that this vertical acceleration happens only in an eclipse, unless you are proposing some new theory of gravity. If the moon is 1º above the eclipse, it will exert the same vertical component of force regardless of its position relative to the sun. And that happens twice every orbit (near the new moon and near the full moon) It have still not a lot to do with the angle to the Moon, but about how an outsider will see the angle of the force triangle (acting on Earth) relative to ecliptic This triangle will be 90º by solar eclipse, and therefore the upwards acceleration of the Earth will at that point be 100% of the possible, (based on a certain moon/earth constalation) 4 hours later the force triangle have tilted 45º , whereby the resulting upwards acceleration also have decreased 50% .. The following 4 hours the same will happen , now the angle is only 22,5º - 75% of the upwards acceleration is lost. The exact same, but opposite happens before solar eclipse. Its exactly such pattern Allais research reveals. Edited June 26, 2017 by Bjarne
swansont Posted June 26, 2017 Posted June 26, 2017 Yes I say so vertical upwards pull of the Earth as possible. However in a way where the moon NOT is [/size]above the test body, because the test body then not is free to follow DFA.[/size] It’s not a contradiction but 2 necessary requirements [/size] If you don't want the moon to be "above" the test body, then you do not want the maximum force from the moon. It is, in fact, a contradiction to say so. This is the sort of thing that could be, and should be, described by an equation. But it confirms yet again that the sun plays no part in this, and all that's important is the position of the moon. It have still not a lot to do with the angle to the Moon, but about how an outsider will see the angle of the force triangle (acting on Earth) relative to ecliptic [/size] This triangle will be 90º by solar eclipse, (realtive to ecliptic) and therefore the upwards acceleration of the Earth will at that point be 100% of the possible, (based on a certain moon/earth constalation) The sun is less than ~1º above the ecliptic during an eclipse and never gets north or south of ~5 degrees with respect to it. The upward force of the moon is smallest at this time (if it's on the ecliptic, as the sun is, this component is zero) So you're very confused here, or aren't explaining things very well at all. Or both. 4 hours later the force triangle have tilted 45º , whereby the resulting upwards acceleration also have decreased 50% .. The following 4 hours the same will happen , now the angle is only 22,5º - 75% of the upwards acceleration is lost. The exact same, but opposite happens before solar eclipse. Its exactly such pattern Allais research reveals. Let me ask you this: where is the solar eclipse in August is going to be seen? The sun only gets to 90º on the tropics at the solstices. It's about halfway between the solstice and the equinox, so the earth's tilt is directed at about 45º to the sun, so the sun is overhead at noon for people at around 12º-13º or so of latitude. The eclipse is going to hit Oregon at around 44-45º of latitude (i.e. >30 degrees further north). The sun will peak at an angle of below 60º above the horizon. The moon, of course, will in the same angular position as the sun. It, too will never get more than 60º above the horizon. (The eclipse hits after 10AM local time, and it's daylight saving, so the sun will start lower and go higher for almost 3 hours) So yeah, I have no idea what your diagrams are supposed to show, but AFAICT it's fantasy and not reality, or it's unrelated to showing the vertical force (in which case why show it?)
Bjarne Posted June 27, 2017 Author Posted June 27, 2017 (edited) If you don't want the moon to be "above" the test body, then you do not want the maximum force from the moon. Correct, I do not want the maximum upwards acceleration of the Moon, - only certain constellation can trigger Allais Effect. One of the requirement is that the moon cannot be too high. I will later show you how much weaker the affect will be for each 1000 km the moon is too high above the ecliptic. What I want is the maximum upwards acceleration based on a constellation where the Moon is between 4 and 5000 km above the Earth. This will be perfect for Pendulum measurement of the anomaly. If the moon is higher than this the anomaly will be weaker, and if very much higher than 6500 km not possible to detec. However the best constellation for combined gravimeter measurement experiment is much more flexible If meassuremnet tak place near the artic area, the moon can be between 4000 and 6000 km above the ecliptic . Even when 7000 or 7500 km above, the effect can still be measured,, - but now the upwards pulls in the test body means that the exposed DFA is weaker If the Moon is much more than 8000 km above the ecliptic, the anomaly will be very weak, and sooner or later not possible to detect / distinguish . But it confirms yet again that the sun plays no part in this, and all that's important is the position of the moon. Yes this is most likely correct, but still we need to calculate the resulting force The sun is less than ~1º above the ecliptic during an eclipse and never gets north or south of ~5 degrees with respect to it. The upward force of the moon is smallest at this time (if it's on the ecliptic, as the sun is, this component is zero) So you're very confused here, or aren't explaining things very well at all. Or both. DFA is as I see it angular relative to the orbit of the Earth. This is based on all the analyses I did. I have considered this already and this is my conclusion.Off course you can may ask why ? - but let’s discuss that later. Its to early. Let me ask you this: where is the solar eclipse in August is going to be seen? The sun only gets to 90º on the tropics at the solstices. It's about halfway between the solstice and the equinox, so the earth's tilt is directed at about 45º to the sun, so the sun is overhead at noon for people at around 12º-13º or so of latitude. The eclipse is going to hit Oregon at around 44-45º of latitude (i.e. >30 degrees further north). The sun will peak at an angle of below 60º above the horizon. The moon, of course, will in the same angular position as the sun. It, too will never get more than 60º above the horizon. (The eclipse hits after 10AM local time, and it's daylight saving, so the sun will start lower and go higher for almost 3 hours) So yeah, I have no idea what your diagrams are supposed to show, but AFAICT it's fantasy and not reality, or it's unrelated to showing the vertical force (in which case why show it?) Let’s just simplify this All what matter is the "DFA interaction axis" (DFAIA) Even at the north pole the DFAIA is just perfect but only for the mentioned gravimeter experiment, however the worse possible setup for a pendulum test. Pendulum, measurement are perfect near the 50º +/- 10º latitude. So let us not make this unnecessary more complicated and technical than it really is, - in order to let so many as possible understand what we are talking about. We can off course discuss from now on and the next 100 years whether latitude 40º is better than 60º or opposite or whether 50º is perfect in August etc......... . But the fact is we have only poor data so far. To solve this once and for all, - put 2 gravimeter near artic the 20 of August and we are all much wiser. Edited June 27, 2017 by Bjarne
swansont Posted June 27, 2017 Posted June 27, 2017 Let’s just simplify this All what matter is the "DFA interaction axis" (DFAIA) Even at the north pole the DFAIA is just perfect but only for the mentioned gravimeter experiment, however the worse possible setup for a pendulum test. Pendulum, measurement are perfect near the 50º +/- 10º latitude. So let us not make this unnecessary more complicated and technical than it really is, - in order to let so many as possible understand what we are talking about. We can off course discuss from now on and the next 100 years whether latitude 40º is better than 60º or opposite or whether 50º is perfect in August etc......... . But the fact is we have only poor data so far. To solve this once and for all, - put 2 gravimeter near artic the 20 of August and we are all much wiser. How about not going off any tangents? DFA isn't even part of the discussion yet. I was commenting on your claim that the moon will be at 90º to the earth, and that's not physically possible. Is this digression an admission that you have no rebuttal?
Bjarne Posted June 27, 2017 Author Posted June 27, 2017 (edited) How about not going off any tangents? DFA isn't even part of the discussion yet. I was commenting on your claim that the moon will be at 90º to the earth, and that's not physically possible. Is this digression an admission that you have no rebuttal? Still you got it wrong Perspective; From the dark side of the Earth. The Sun is (invisible) behind the Earth. The Moon is 4000 km above the ecliptic (the angle to the Moon is less than 1°). The Moon is NOT at the angle 22,5° - 45° and 90° The Moon is between the Sun and the Earth The moon is always at the front side of the Earth, seen from this perspective. The red arrow illustrate how the orientation / angle of the force vector-triangle (between earth, -> moon and -> sun) is changing - relative to ecliptic. starting from mainly horizontal (22,5°) relative to ecliptic, 4 hours after reaching 45° (50% vertical / 50% horizontal) 8 hours later, reaching the maximum vertical orientation, and therefore maximum vertical pull in the Earth - (Eclipse). The same, but same but opposite order happens 8 hours after eclipse. No other constellation (where the moon also is 4000 km above the Earth) can compete with the magnitude / the upwards pull of the Earth that this configuration excerts on Earth (except lunar eclipse) Edited June 27, 2017 by Bjarne
swansont Posted June 27, 2017 Posted June 27, 2017 Still you got it wrong Perspective; From the dark side of the Earth. The Sun is (invisible) behind the Earth. The Moon is 4000 km above the ecliptic (the angle to the Moon is less than 1°). The Moon is NOT at the angle 22,5° - 45° and 90° The Moon is between the Sun and the Earth The moon is always at the front side of the Earth, seen from this perspective. Your drawing has LABELED THE MOON, and shows these angles. The moon cannot simultaneously be between the earth and sun (which you say is behind the earth) and also be in front of the earth. The red arrow illustrate how the orientation / angle of the force vector-triangle (between earth, -> moon and -> sun) is changing - relative to ecliptic. starting from mainly horizontal (22,5°) relative to ecliptic, 4 hours after reaching 45° (50% vertical / 50% horizontal) 8 hours later, reaching the maximum vertical orientation, and therefore maximum vertical pull in the Earth - (Eclipse). The same, but same but opposite order happens 8 hours after eclipse. No other constellation (where the moon also is 4000 km above the Earth) [/size] can compete with the magnitude / the upwards pull of the Earth that this configuration excerts on Earth (except lunar eclipse)[/size] If the moon is 1º above the ecliptic, the resultant force vector will never be larger than 1º above the ecliptic. It will be much, much smaller.
Bjarne Posted June 27, 2017 Author Posted June 27, 2017 (edited) Your drawing has LABELED THE MOON, and shows these angles. I have labeled the Moon , - Yes I have NOT shown the angle to the moon. I have shown the angle to the vector-forces-triangle acting of Earth If the moon is 1º above the ecliptic, the resultant force vector will never be larger than 1º above the ecliptic. It will be much, much smaller. I am not saying the resulting force it is larger than like 1º - but have calculated and shown much less in fact 0.0034º actually The moon cannot simultaneously be between the earth and sun (which you say is behind the earth) and also be in front of the earth. The Sun is behind the Earth, you cannot see it The Moon is (always) between the Earth and the Sun, - on the Sun side of the Earth to be excact, at this perspective This shows how the resulting force will point (on a central circle on the Sun) Edited June 27, 2017 by Bjarne
swansont Posted June 27, 2017 Posted June 27, 2017 The Sun is behind the Earth, you cannot see it The Moon is (always) between the Earth and the Sun, - on the Sun side of the Earth to be excact, at this perspective So why can I see the moon in this diagram? The force will point toward the moon and sun, roughly, at the time of eclipse. So the sun is behind the earth, and yet you are showing a top view in your graph, is that it? That's not how coordinate systems work. I have labeled the Moon , - Yes[/size] I have NOT shown the angle to the moon.[/size] I have shown the angle to the vector-forces-triangle acting of Earth[/size] The force acts toward the sun and moon, as a vector sum. The force from the moon act's in the direction of the moon. If the moon is 1º above the ecliptic, then so is the force. If what you are showing is the direction of the component of the force that's in the plane of the ecliptic, then this angle is the rotational angle, but that force can NEVER exert an upward or downward force, since it's perpendicular. Since your effect allegedly depends on the vertical component, this has zero effect. Why are you bringing it up?
Bjarne Posted June 27, 2017 Author Posted June 27, 2017 (edited) So why can I see the moon in this diagram? Let us just say the Moon is 6500 km above the ecliptic, now you can see the moon from night side perspective of the Earth The force will point toward the moon and sun, , roughly, at the time of eclipse. Right and this is what I calculated and we discussed last week. So the sun is behind the earth, and yet you are showing a top view in your graph, is that it? That's not how coordinate systems work. No I am not showing a top view, but a view from an observer seeing the Earth and Moon,(but not the Sun) - from the night side of the Earth The force acts toward the sun and moon, as a vector sum. The force from the moon act's in the direction of the moon. If the moon is 1º above the ecliptic, then so is the force. If what you are showing is the direction of the component of the force that's in the plane of the ecliptic, then this angle is the rotational angle, but that force can NEVER exert an upward or downward force, since it's perpendicular. Since your effect allegedly depends on the vertical component, this has zero effect. Why are you bringing it up? I am not showing the direction of the force, but only the direction of the resulting force - and what time the resulting force is pointing mainly vertical or horizontal. In addition to the illustration above, - take a look at the one below also, - this shows the same point. By eclipse the resulting the force is pointing upwards (not 90º upwards) (but 0,0034º upwards) as already discussed The labeled degree (90º)is only to illustrator the direction / tilt of the 2 (moon and Sun) vector triangle responsible for of the resulting force, nothing else This image shows where the resulting force will point (on a central circle on the Sun) 8 hours before and after an eclipse. As you can see, - by eclipse, the resulting force is pointing upwards. Upwards means 90 degree relative to the ecliptic. It does not mean that the moon is 90ºdegree above the earth Also not that the force angle is 90º, - but only that now is the 0,0034º resulting force pointing upwards, nothing else than this, Which also mean that the upwards accelerating of the earth by eclipse is 50% faster (upwards) as 4 hours earlier, and 75% faster (upwards) as 8 hours earlier. Edited June 27, 2017 by Bjarne
swansont Posted June 27, 2017 Posted June 27, 2017 I am not showing the direction of the force, but only the direction of the resulting force - and what time the resulting force is pointing mainly vertical or horizontal. In addition to the illustration above, - take a look at the one below also, - this shows the same point. By eclipse the resulting the force is pointing upwards (not 90º upwards) (but 0,0034º upwards) as already discussed The labeled degree (90º)is only to illustrator the direction / tilt of the triangle responsible for of the resulting force, nothing else This image shows where the resulting force will point (on a central circle on the Sun) 8 hours before and after an eclipse. As you can see, - by eclipse, the resulting force is pointing upwards. Upwards means 90 degree relative to the ecliptic. It does not mean that the moon is 90ºdegree above the earth Also not that the force angle is 90º, - but only that now is the 0,0034º resulting force pointing upwards, nothing else than this, Which also mean that the upwards accelerating of the earth by eclipse is 50% faster (upwards) as 4 hours earlier, and 75% faster (upwards) as 8 hours earlier. The triangle responsible for the upward force consists of a vector pointed toward the moon, and the vertical and horizontal projections of that vector. The vertical component is the upward force. The angle is found using the distance the moon is above the ecliptic divided by the distance to the moon. Fvertical = Fmoon*(dvertical/dmoon) so that's 4000 km/384,000 km, which is about 0.01 IOW, the upward component is about 1% of the total force. This will be true whenever the moon is 4000km above the ecliptic. There is no part of this geometry that includes a distance of 8000km. The moon moves a total of 5 degrees above the ecliptic in about a week, meaning it takes almost a day to move that much. There is no part of this triangle that includes a 45, 22.5 or 90 degree angle.
Bjarne Posted June 27, 2017 Author Posted June 27, 2017 (edited) The triangle responsible for the upward force consists of a vector pointed toward the moon, and the vertical and horizontal projections of that vector. The vertical component is the upward force. The angle is found using the distance the moon is above the ecliptic divided by the distance to the moon. Fvertical = Fmoon*(dvertical/dmoon) so that's 4000 km/384,000 km, which is about 0.01 IOW, the upward component is about 1% of the total force. This will be true whenever the moon is 4000km above the ecliptic. There is no part of this geometry that includes a distance of 8000km. The moon moves a total of 5 degrees above the ecliptic in about a week, meaning it takes almost a day to move that much. There is no part of this triangle that includes a 45, 22.5 or 90 degree angle. You still haven’t understood the point, - the 8000 km is where the resulting force (based on the force of the moon + the force of the sun), is pointing, that force is pointing....... 8000 km above the equator of the Sun by eclipse 4000 km above the equator of the Sun, - 4 hours before eclipse 2000 km above the equator of the Sun, - 8 hours before the eclipse 4000 km above the equator of the Sun, - 4 hours after the eclipse 2000 km above the equator of the Sun, - 8 hours after the eclipse Maybe you had better understand the point by looking at the image below. - It’s the exact same point. This tells you that you only get maximum upwards acceleration of the Earth, - by solar (and lunar) eclipse And it tells you a lot oabout how fast the anomaly inclines and declines. Excatly that is what the "Shoot circle" below, - situated on the central sun, - also illustrate Edited June 27, 2017 by Bjarne
swansont Posted June 27, 2017 Posted June 27, 2017 You still haven’t understood the point, - the 8000 km is where the resulting force (based on the force of the moon + the force of the sun), is pointing, that force is pointing....... 8000 km above the equator of the Sun by eclipse 4000 km bove the equator of the Sun, - 4 hours before eclipse 2000 km bove the equator of the Sun, 8 hours before. 4000 km bove the equator of the Sun, - 4 after the eclipse 2000 km above the equator of the Sun, - 4 after the eclipse Maybe you had better understand the point by looking at the image below. - It’s the exact same point. This tells you that you only get maximum upwards acceleration by solar (and lunar) eclipse And it tells you a lot of how fast the anomaly inclines and declines. The force is directed at the moon. The force can't point further north than the moon is. Since the moon isn't 8000 km above the ecliptic, the force is not directed there.
DrP Posted June 27, 2017 Posted June 27, 2017 If the effect is real.. (I have only done minor reading around it on the net and most of what I know is from wiki and reading/following this thread... wiki said that it was speculative and there have been mixed results in trying to replicate it)... I liked the interference of gravitational waves from the 2 sources speculation as an explanation of the effect (touched upon on the wiki page). I am not buying this 'dark flow acceleration' explanation - it seems unrelated to me. The interference pattern in the waves could explain why sometimes you can measure it and other times you cant... depending upon where the measurement system is set up (on a peak or a trough of the gravitational wave). As we now 'know' that gravity waves are a real thing (or at least have been detected), would it be better to assume that this effect has something to do with interference of these waves? Rather than some 'dark flow'... what does that even mean? - I must have missed something somewhere... I'll go back and read this through again if I get the time. Sorry if I am missing the point - I don't want to disrupt your conversation... I'll sit out and keep following the thread. I am none the wiser as to what this Dark Flow is supposed to be though, I will re-read and look it up some more sometime.
Bjarne Posted June 27, 2017 Author Posted June 27, 2017 (edited) The force is directed at the moon. No the illustration shows the resulting force pointing to the Sun The force can't point further north than the moon is. Its not The 8000 km íllustrated altitude in on the Sun Since the moon isn't 8000 km above the ecliptic, the force is not directed there. No its not the moon that is 8000km above, please read the post agian You misunderstood agian The red line shows the resulting force pointing to the Sun This cirkel, is on the Sun, it have not much with the moon to do If the effect is real.. (I have only done minor reading around it on the net and most of what I know is from wiki and reading/following this thread... wiki said that it was speculative and there have been mixed results in trying to replicate it)... I liked the interference of gravitational waves from the 2 sources speculation as an explanation of the effect (touched upon on the wiki page). I am not buying this 'dark flow acceleration' explanation - it seems unrelated to me. The interference pattern in the waves could explain why sometimes you can measure it and other times you cant... depending upon where the measurement system is set up (on a peak or a trough of the gravitational wave). As we now 'know' that gravity waves are a real thing (or at least have been detected), would it be better to assume that this effect has something to do with interference of these waves? Rather than some 'dark flow'... what does that even mean? - I must have missed something somewhere... I'll go back and read this through again if I get the time. Sorry if I am missing the point - I don't want to disrupt your conversation... I'll sit out and keep following the thread. I am none the wiser as to what this Dark Flow is supposed to be though, I will re-read and look it up some more sometime. In the end of the day it is not what most people believe or believe they know, but the scientific methods that counts. The good thing here is that this theory have a unique and powerful prediction, - the mentioned relative and absolute gravity test near artic. So this theory is very dangerous and I am sure that many don’t like it. But science should not (always) be about what most people like to hear, sometimes someone have to dare to think new thoughts. Edited June 27, 2017 by Bjarne
Mordred Posted June 27, 2017 Posted June 27, 2017 (edited) If the effect is real.. (I have only done minor reading around it on the net and most of what I know is from wiki and reading/following this thread... wiki said that it was speculative and there have been mixed results in trying to replicate it)... I liked the interference of gravitational waves from the 2 sources speculation as an explanation of the effect (touched upon on the wiki page). I am not buying this 'dark flow acceleration' explanation - it seems unrelated to me. The interference pattern in the waves could explain why sometimes you can measure it and other times you cant... depending upon where the measurement system is set up (on a peak or a trough of the gravitational wave). As we now 'know' that gravity waves are a real thing (or at least have been detected), would it be better to assume that this effect has something to do with interference of these waves? Rather than some 'dark flow'... what does that even mean? - I must have missed something somewhere... I'll go back and read this through again if I get the time. Sorry if I am missing the point - I don't want to disrupt your conversation... I'll sit out and keep following the thread. I am none the wiser as to what this Dark Flow is supposed to be though, I will re-read and look it up some more sometime. Well much like you my research on tbe Allais effect shows that it is questionable. As far as dark flow acceleration, well there is a far better methodology to test for its presence. Any form of acceleration upon a multiparticle system affects thermodynamic temperature. A DFA would be detectable via anistropy in the Planck dataset. Lol the original dataset. Caused a huge wave of "see evidence of DFA. However this turned out to be a calibration error in accounting for all the movements our solar undergoes. That being said its ok to speculate provided one learns how to properly model and proper science as they develop their models. The worse mistake is assuming a solution without checking if the solution is a viable via math. Hence why I stressed so much on detailing the particulars on the Measured Allais effects. You need measured results as a confirmation. Without that its nothing more than an assumption. That's precisely why I Asked which dataset was being used for comparison. Edited June 27, 2017 by Mordred
swansont Posted June 27, 2017 Posted June 27, 2017 No the illustration shows the resulting force pointing to the Sun The sun is on the ecliptic, so it doesn't exert an upward force. It can be ignored. You only have to look at the influence of the moon to find the vertical component of the force. According to what you've claimed, this is what matters. As we now 'know' that gravity waves are a real thing (or at least have been detected), would it be better to assume that this effect has something to do with interference of these waves? Rather than some 'dark flow'... what does that even mean? - I must have missed something somewhere... I'll go back and read this through again if I get the time. Gravitational waves are way too small of an effect to be noticed here.
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