03 Jan 17
5 edits
So suppose you had a giant liquid mirror say in a flat desert, I think the numbers are you drop 6 feet or 182.88 cm or 1828.8 mm or 1,828,800 microns from center to edge of a liquid mirror 14 miles in diamber, and that drop from center to edge.
So it stands to reason, at least I think so, if you have a two foot diameter flat liquid reflector like mercury or 304.8 mm diameter it would show a drop at edges of about 50 microns. I would think therefore, if you put that mercury flat in a vacuum chamber and a vibration table to reduce ground vibrations, you should be able to measure that 49 odd micron change in height from edge to center.
Anyone have thoughts about that? isn't that about 49,000 nanometers which would be 10 wavelengths of 490 nm light so it should be easily measured.
Am I missing something?
So the numbers: 7 mile radius, 36960 feet, 443520 inches, 11265408 mm and a drop from center to edge of 6 feet, 72 inches, or 1828.8 mm or a ratio of 6160:1. So dropping that down to a 2 foot diameter, 1 foot radius, 304.8 mm or 304800 microns/6160 should show a drop from center to edge of 49.4805 microns which, using a laser height measurment tool at at one micron wavelength (IR band) would show 49 odd wavelengths of that light center to edge droop.
Sound reasonable?
Also, would it be possible to see, using very short wavelengths, the change in the rotation speed of Earth (we added one second to the year) so that would be a change of center to edge of one part in about 33 million (that many seconds in a year, roughly) Would that be possible to measure at any reflective wavelength?
03 Jan 17
Originally posted by sonhouseNot sure what you are on about here. Can you describe it a bit more?
Also, would it be possible to see, using very short wavelengths, the change in the rotation speed of Earth (we added one second to the year) so that would be a change of center to edge of one part in about 33 million (that many seconds in a year, roughly) Would that be possible to measure at any reflective wavelength?
03 Jan 17
1 edit
Originally posted by twhiteheadFor a while I thought if you were say on a spinning merry go round there would be a shift around the perifery, a force like that that would say bend a rope with a weight on it.
Not sure what you are on about here. Can you describe it a bit more?
But then it was finally pounded into my head that there would be no such force unless there was some kind of acceleration, say from 0 to 100 RPM or whatever, so during the acceleration there would be a bending of the rope around the perifery, the end of the rope with the weight would not now be in a straight line away from the radius but bending away from the direction of spinning motion.
Now thinking about Earth, a year is about 32 odd million seconds but it undergoes a tiny change in circular velocity since we add one second to the end of 2016. So that would represent an acceleration of one part in 32 odd million, a very tiny amount. That would seem to me to distort the spherical nature of the bulge by that amount. Measurable with lasers? The bulge would now be slightly off center by the amount of acceleration or decel.
So as to the main measurments, did you go over my numbers, it seems the curve ratio is 6160 to 1 and that should be linear so it seems you should get a bulge in the center of a liquid reflector like mercury and of course you would have to set up the experiment to be in a vacuum and vibration isolation table. But if the numbers are right and a 300 mm disk filled with mercury would give a bulge of about 50 microns in the center compared to the edge, it should be readily measurabe since even using IR at 1 micron wavelength, would record a change of about 50 wavelenths of light at that wavelength, 1 micron.
But the acceleration/decel of Earth I figure would be the one second to 32 odd million seconds in a year so the acel/decel number would be the radius/divide by 32 milion.
A 300 mm radius is 300,000 microns, or 300 million nanometers so 300 million divide by 30 million roughly would be about a shift of 10 nm of dead center of the bulge, at least that is what it looks like to me. Or about 100 angstroms. The problem I see there is figuring out exactly and I mean within a few angstroms of where the center of bulge is. Then measure a 100 ansgtrom shift in the possision of the bulge which it would not have if Earth was not changing it's spin rate.
03 Jan 17
Originally posted by sonhouseTechnically, deceleration or negative acceleration as the earth is slowing down.
So that would represent an acceleration of one part in 32 odd million, a very tiny amount.
That would seem to me to distort the spherical nature of the bulge by that amount.
No, I don't think so. At best it would result in the liquid being pushed ever so slightly to one side of the container.
So as to the main measurments, did you go over my numbers,
Numbers don't matter until you get the concepts right.
As for your original plan, your mention of the earth slowing reminded me of tidal forces. But they might not change the surface curvature significantly.
03 Jan 17
Originally posted by twhiteheadGoing back to the original test to prove curvature of Earth, do my numbers or my concept make sense? It seems if a large body of water bulges in the center then a small body would bulge with the same ratio of radius to bulge which I calculated to be 6160 so I assumed that ratio would hold up pretty much no matter the sizes, in that range.
Technically, deceleration or negative acceleration as the earth is slowing down.
[b]That would seem to me to distort the spherical nature of the bulge by that amount.
No, I don't think so. At best it would result in the liquid being pushed ever so slightly to one side of the container.
So as to the main measurments, did you go over my number ...[text shortened]... wing reminded me of tidal forces. But they might not change the surface curvature significantly.
So a small body would bulge the same ratio as a large one it seems to me at least.
Make sense so far? If so the bulge for a 300 mm radius would be about 50 microns which should be easy to measure.
Have you seen anything like this as a probe to the roundness of Earth as opposed to the flatlanders nonsense?
04 Jan 17
1 edit
Originally posted by sonhouseNo, not correct.
It seems if a large body of water bulges in the center then a small body would bulge with the same ratio of radius to bulge
I am too lazy to work out the correct formula, but we are talking about the relationship of h and c on this page:
http://www.ajdesigner.com/phpcircle/circle_segment_chord_t.php
They are not directly proportional.
Have you seen anything like this as a probe to the roundness of Earth as opposed to the flatlanders nonsense?
Except that what you are actually probing is gravity, not roundness, and as I said, local gravity anomalies might be significant.
Further, it is not the sort of experiment a person can carry out at home, thus of no use in a flat earth argument (as they discount all third party claims not in their favour).
04 Jan 17
1 edit
Originally posted by twhiteheadI think your point about surface tension is spot on, it probably would mask any such bulging effect. It is not an experiment you could do at home but if your company used optical measuring tools, it could be done. It seems though that there would be some size that could be built that would show a central bulge, for instance, if you built a swimming pool 14 miles across it would have that 6 foot bulge in the middle, so if you built one 1.4 miles across I would still expect to see a half foot bulge, right?
No, not correct.
I am too lazy to work out the correct formula, but we are talking about the relationship of h and c on this page:
http://www.ajdesigner.com/phpcircle/circle_segment_chord_t.php
They are not directly proportional.
[b]Have you seen anything like this as a probe to the roundness of Earth as opposed to the flatlanders nonsense?
Exc ...[text shortened]... f no use in a flat earth argument (as they discount all third party claims not in their favour).[/b]
Even if you were on a lake on a wind free day and it is 14 miles across you would get that 6 foot bulge so it should be possible to show with a series of lasers, say 6 of them all pointing out across the water and a pole in the center and poles at say the 3.5 mile mark and the pole say 3 feet out of the water, the top laser would hit the top of that pole as well as the pole clearly 3 feet out of the water but the other lasers, being spaced at 1 foot vertica intervals would say, at the 5 foot high level would run into water and thus would not hit the pole sticking out at the halfway point 6 feet above the water line. I guess you could just use floats with a plastic stick going up out of the water 6 feet high at the halfway point and 3 feet out of the water at the quarter distance and so forth, where you could see clearly the laser beams are still a foot apart vertically but the pole in center would not be able to read the other light beams. If at the quarter way mark the laser beams were shown to still be a foot apart that would eliminate diffraction from the atmosphere and such.
05 Jan 17
Originally posted by sonhouseAs I said above, they are not directly proportional, so you might need to actually work that out. A circle tends towards a straight line the smaller the section you take.
It seems though that there would be some size that could be built that would show a central bulge, for instance, if you built a swimming pool 14 miles across it would have that 6 foot bulge in the middle, so if you built one 1.4 miles across I would still expect to see a half foot bulge, right?
Even if you were on a lake on a wind free day and it is 14 miles across you would get that 6 foot bulge so it should be possible to show with a series of lasers, say 6 of them all pointing out across the water and a pole in the center and poles at say the 3.5 mile mark and the pole say 3 feet out of the water, the top laser would hit the top of that pole as well as the pole clearly 3 feet out of the water but the other lasers, being spaced at 1 foot vertica intervals would say, at the 5 foot high level would run into water and thus would not hit the pole sticking out at the halfway point 6 feet above the water line.
You are seriously over complicating things and have a problem with wind, waves etc. What you are actually measuring is the difference in gravity between two points. The easiest is to simply use two levels at a distance from each other and surveying equipment. I do not know whether or nor surveying equipment is sensitive enough to pick up earths curvature.
05 Jan 17
2 edits
Originally posted by twhiteheadI was under the assumption that if you have for instance, a perfectly spherical body covered with water, you wouldn't be seeing differences in gravity since the body would be homogenous, but just differences in height V distance apart, looking at a chord where if you could have a totally focused laser beam going through the water (seems to be impossible ATT) Assuming beam to be perpendicular to a radius line, that the beam would be buried 6 feet underwater at the 7 mile distance and if you started at a reference point and went 14 miles the beam would reappear. Isn't that just a geographic situation having nothing to do with gravity?
As I said above, they are not directly proportional, so you might need to actually work that out. A circle tends towards a straight line the smaller the section you take.
[b]Even if you were on a lake on a wind free day and it is 14 miles across you would get that 6 foot bulge so it should be possible to show with a series of lasers, say 6 of them all ...[text shortened]... do not know whether or nor surveying equipment is sensitive enough to pick up earths curvature.
05 Jan 17
Originally posted by sonhouseIt has everything to do with gravity. The surface of water is a reflection of the direction of gravity. Water levels have been used since ancient times for this very reason.
Isn't that just a geographic situation having nothing to do with gravity?
As I say, all you need is two water levels (or something more accurate perhaps) at a distance and a line of sight between them. Measure the difference in the levels and you are done. No need for lasers or large bodies of water at all. At best you need a telescope and some well calibrated equipment such as surveyors use.
You might even be able to do it with a pair of sufficiently accurate laser levels. Start with them both together pointing the same direction to verify they agree, then take them some distance apart and point them towards each other. If both point above each other then you are done.
05 Jan 17
Originally posted by twhiteheadHas it been shown you can do measurments accurately over that distance, that is to say where atmospheric refractions and interactions with the water surface would tend to deflect the laser beams and destroy the measurment?
It has everything to do with gravity. The surface of water is a reflection of the direction of gravity. Water levels have been used since ancient times for this very reason.
As I say, all you need is two water levels (or something more accurate perhaps) at a distance and a line of sight between them. Measure the difference in the levels and you are done. ...[text shortened]... tance apart and point them towards each other. If both point above each other then you are done.
05 Jan 17
Originally posted by sonhouseYou are not paying attention. No water surface is required.
Has it been shown you can do measurments accurately over that distance, that is to say where atmospheric refractions and interactions with the water surface would tend to deflect the laser beams and destroy the measurment?
And what distance? With an accurate enough laser level you could do it in a few hundred metres.
05 Jan 17
Sorry, I haven't read through all the posts, so if this has been covered then sorry. A problem with this is going to be controlling for non-gravitational effects (waves etc.), things like tides, and local gravitational variation (mountains etc.). I think that there are easier ways of working out curvature than this. Just travel between three (distant) fixed points (using accurate direction finding with radio for example), and measure the difference in angle from 180 degrees that one has to turn through at each station point. Clearly one only needs the radio waves to travel. This method only hangs on the precision one can determine the source of the radio waves to.
06 Jan 17
Originally posted by DeepThoughtThat's certainly a good idea, but I believe identifying the direction of non-line of sight radio waves is extremely difficult as they tend to reflect rather easily.
Just travel between three (distant) fixed points (using accurate direction finding with radio for example), and measure the difference in angle from 180 degrees that one has to turn through at each station point. Clearly one only needs the radio waves to travel. This method only hangs on the precision one can determine the source of the radio waves to.
Here in Southern Africa, the most popular satellite TV comes from a couple of satellites. One could measure the angle of satellite dishes. I do know for a fact that dishes in Cape Town point closer to the horizon than those in Zambia, but that would be the case even on a flat earth as the satellites are to the north.