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  1. You're correct, I certainly haven't looked at and read everything. I took a quick skim through the last 20 pages and saw many things that I have seen before. Discussions are not much fun if both parties don't have a similar understanding of the underlying material. From what I have seen, I would need to explain undergraduate level physics and optics before beginning most discussions. Again I don't say this to be arrogant and do not think the people here are stupid by any means. I do think there is a lot of ignorance here though (note that ignorance simply refers to lack of knowledge). Incidentally, my comment did not apply to your questions, which seem quite reasonable. To answer your boat question, if several observers were equidistant from an island and all were the same height above the water then yes, the same portion of the base of the island would be obstructed for each viewer. I most certainly have not shown you that you are incorrect, as I have stated, this would take far more time than I am willing to commit. I just stopped in to take a look around. I think it's great that people are taking the time to ask questions and go out and do experiments themselves. All the best.
  2. "If in the video you posted the curvature is visible in front of you, it would be visible on the horizon from left to right" No. The distance to the horizon remains the same as you scan your eyes back and forth. This means that the angle between true horizontal and your line of sight remains the same. This angle would need to change as you panned the horizon in order for you to see a curve. "How do you explain the horizon at eye level when at very high altitudes shown in numerous high altitude ballon videos?" Even with the highest balloon flights the horizon drops less than 1.5 degrees below 'true horizontal'. Why would you expect to see this? Even if the horizon dropped several degrees, why would you expect the person's or camera's line of sight to be horizontal instead of being drawn to the horizon? I've seen several high altitude videos and the camera is always moving up and down, how do you know when it is perfectly horizontal? "Gyroscopes and airplanes are also a fun subject, and how they would work on a ball earth." I am unaware of a gyroscope used for navigation that is not self-righting. Gyroscopes adjust themselves frequently. Manuals are readily available that outline how this is accomplished (warning - the math is rather advanced). "Thank you again for joining us!" I am sorry to disappoint you, as I have just announced I'll be leaving. However, I have left my email if you'd like to run a proposal past me for my suggestions. There will always be issues with results not matching predicted curvatures due to refraction, however, a well-designed experiment over a suitable distance is unlikely not to show some curvature. Personally, I would focus more on astronomy and the motions of the Earth, stars and planets for evidence of the shape of the Earth. You can always go out and track satellites too (it's easy enough to do), but I don't anticipate that being a popular option. Cheers. Edit:Here's a decent paper on navigational gyroscopes, it includes some info on drift correction: https://www.cl.cam.ac.uk/techreports/UCAM-CL-TR-696.pdf. My understanding is that all navigational gyroscopes self correct a minimum of every 15 minutes (this is for small planes, it's more frequent for jets). Of course if you are looking for a true uniform inertial reference system, look no further than a Foucault's Pendulum. ;) I think the main issue here is scale. Any curve, if sufficiently large, will resemble a flat line if only a small portion of this curve is visible (this is basic calculus). The Earth is pretty darn big. You might find these links interesting: http://thulescientific.com/Lynch Curvature 2008.pdf http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/
  3. Water seeks a level for the same reason that all large celestial bodies are spheres - gravity. No doubt you are going to say that gravity does not exist, or you are going to say something ill-informed regarding density, I'll save you the trouble.
  4. I'm sorry guys, but I was expecting a little more I guess. I've gone through the thread, looked at some of the videos, and all I see is fundamental misunderstandings of very basic concepts. I really see no point in explaining high school geometry and physics and I think this is all I would be doing. I don't mean this to be arrogant, my apologies if this appears the case, but there is just too wide a chasm for us to have a discussion on these topics without an awful lot of ground work. Rothbard: A three foot wave near the horizon (where your sight line intersects a point three feet above the surface) will have a tiny angular (apparent) size - this is insufficient to account for any appreciable blockage unless the camera is right at water level. The effects of this are diminished as the height of the observer is increased. For example, if a camera is placed on the shore of a lake covered with three feet waves and the height of the camera is ten feet above water level (i.e. seven feet above the wave tops) the distance to where your line of sight first intersects the top of a wave will be 3.2 miles. At this distance the wave will have an angular size of 37 arc seconds. Refraction over water is very common. If you would like to limit this I'd suggest attempting your measurement when the air is dry, the pressure is high and the difference between the ground/water and air temperature is as small as possible (there should be no heat flux between the ground and the air). The video you posted earlier where the fellow is showing his table for the curvature is absolutely fine. Notice, however, that he clearly states this table and the online calculator are used for different tasks. In his table he is calculating the curvature from a line that is tangential to the curve at the point where the observer is standing. This is not useful when calculating how much of an object will be hidden by the horizon. In this case we must draw a line of sight between the observer and the object we are observing (the observer must be looking towards the object and not into the sky). If this line intersects the surface, part or all of the object will be obscured. If the object being viewed is in front of the horizon and nothing else is in the way, it will be visible in it's entirety. Perspective and the vanishing point do not account for the preferential obstruction of the base of objects. Anyway, sorry but I'm not going to stick around. If you'd like to run an experimental design by me or ask me something about a specific issue, you can email me at calm_blue_ocean-at-yahoo.com. Cheers.
  5. The Bedford experiment has been redone numerous times. People do indeed object to paying money, but if you don't like what Wallace has to say, several others have repeated it. Whether you like it or not, refraction is an observable and demonstrable phenomenon. The best video evidence for curvature comes from the effects of altering the height of the observer. The fact that raising the height of the observer allows one to see further cannot be explained by a flat Earth. Here's a decent example of this: At the Centre of the Earth you would, of course, be crushed and incinerated. But if somehow you could maintain a chamber, deal with the heat, and stop your fall, yes you would float. How is this illogical? This is a conclusion made from well-support physical laws. Let me ask you a question. What maintains the shape of your flat Earth while at the same time causing things to accelerate downwards at 9.81m/s/s? The calculator predicts the curvature quite nicely and the example posted by Mrthriveandsurvive is an excellent example of the curvature. At surface level, refraction can be significant. I fail to see evidence of the Earth being flat in the Rothbard videos I have seen. Do you have one in particular you think I should see?
  6. At the Centre of the Earth, the gravitational potential will be zero. Of course if this magic hole contained a vacuum and you didn't touch the sides you could make it to the other side without expending any energy at all. The shortcomings of the Bedford experiment are well known and the experiment has been repeated several times by Wallace and others. It seems to me that Rothbard has shown that the measured curvature fails to fit his predictions. I fail to see where he has shown there is no curvature.
  7. I can assure you there is nothing wrong with it Rothbard, this is high school geometry.
  8. Look at the math and the diagram. You are calculating the distance between the line of sight and the ground. The line of sight is a STRAIGHT line that runs tangential to the curve. The curve "drops" away from the point at which the line intersects the curve. The distance between the line of sight and the curve (the ground) DECREASES towards the horizon and INCREASES beyond it. We calculate the distance to the horizon only to provide us with our starting point. It is the distance from our horizon to the object that we are viewing that is used to calculate the "drop".
  9. The drop BEYOND the horizon determines what is being obstructed - we start at zero from the horizon. The trig is pretty straight forward and it is supplied in the link I posted. According to your logic, the bottom of Mt. Rainier would be obstructed if it was between the observer and the horizon, this simply is not the case. We are trying to calculate how much the ground drops below the line of sight - it only does this beyond the horizon. As for the Suez canal, gravity pulls towards the centre of the Earth. The surface of the canal is curved.
  10. The Earth curves between any two points. If the rise in this curve is sufficient, it will obstruct the line of sight between the two points. What we are trying to calculate is how much is obscured by the curve.
  11. Let me restate the Mt Rainer issue here for everyone. Please take a close look at the diagram shown here first: https://github.com/dizzib/earthcalc The horizon is the point at which your line of sight intersects the curvature of the Earth. Anything between you and your horizon will be entirely visible, anything beyond your horizon will be partially or entirely obscured. The higher up you are, the more distant your horizon will be. Mrthriveandsurvives’ math is very wrong. There are three calculations that need to be performed: 1) We need to calculate the distance from the observer on Mt. Brunswick to the horizon. When we do this, we find that the horizon is 93.8 miles away. Anything within this range will be unobstructed – i.e. if Mt Rainer had been only 93 miles away it would be entirely visible- we would be able to see the base of the mountain. 2) Mt Rainier is located a further 100 miles beyond the horizon (193.8 – 93.8 =100). It will be obstructed by the horizon by the drop calculated for this distance and this distance ONLY. This drop is 6,668 feet. 3) Mount Rainer is 14,416 feet tall. The bottom 6,668 feet will be obstructed. Therefore the top 7,748 feet will be visible. (14,416 - 6,668 = 7,748). The Earth curves between the two mountains and it is the rise in the middle that obstructs the view. Calculating the total drop between the two mountains is meaningless.
  12. I don't understand why you would expect to be able to "witness a hump". Gravity pulls towards the centre of the Earth - i.e. you are not driving 'uphill'. Any point on a flat sphere is identical to any other point on a flat sphere. As you move on a smooth sphere, the distance and angle to the horizon will always remain the same. The horizon will always be flat as you pan along it because the angle to the horizon will always be the same no matter what direction you look in. In order to see a 'hump' (i.e. a curve) the angle to the horizon would need to change as panned along it. I am originally from Toronto actually; I'm quite familiar with the photos and have witnessed this myself. The amount that is obscured by the horizon does change based on atmospheric conditions over the lake. This is a result of refraction and it varies with humidity. The fact that it is different from day to day and even hour to hour is indicative of this. The main point is that the base of the buildings is ALWAYS obscured from across the lake. This would not be the case on a flat Earth. What is even better evidence though is that increasing the height of the observer decreases the occlusion by the horizon. (Look for photos taken from lake level at Niagara on the Lake and photos taken from slightly further away from the top of the Niagara Escarpment). Perspective and the vanishing point cannot account for this. Despite what many misguided YouTubers claim, perspective results in a decrease of apparent (angular) size - it does not preferentially obscure the BASE of objects. As for seeing no curvature at 6-7 miles away, you need to know the height of the observer. The camera does not need to be very far above water level in order to account for this. I've seen many analyses of the Toronto skyline that fail to include the height of the observer and/or the height of the buildings above lake level. There are large treed islands in front of Toronto, these are almost never visible. You need to be able to see these and the breakwaters on the shore in order to claim that there is no obstruction. (Incidentally, Lake Ontario is often completely still. What look like waves on the horizon are the result of localized refraction (if these were waves they would be huge). I think the issue here is that flat Earthers tend to focus on the fact that the obstruction is not as predictable as they think it should be, while disregarding the fact that there should not be any obstruction at all if the Earth was flat. Ironically, think Mrthriveandsurvive's video was a very straight forward example of the curvature as refraction is unlikely to play a significant role at this scale and altitude. (Incidentally, there are more complex calculators that take refraction into effect - a value of 15% is often used). I have to run, but I will leave you with one question before I go. Why do we see different stars in the Northern and Southern Hemispheres and why do they appear to rotate in opposite directions around different pole stars?
  13. I'm afraid I'll need to check out the video later, but I'll answer your other question. You cannot see the curvature if you are sighting along it, you can only see the effects (i.e. what is hidden behind the curve). The horizon is simply the point at which your line of sight intersects the curve, and the curve is equal in all directions from any point. Let's say that you are standing on a large smooth sphere. In every direction you look, your line of sight will intersect the sphere at the same distance and the angle between true level and your line of sight to the horizon will be the same (on a sphere as large as the Earth, even at the top of a mountain, this angle is a minuscule fraction of a degree). If you were to turn around 365 degrees and follow the horizon, your eyes will always be at the same angle - i.e. you will trace out a straight line (the edge of a large flat circle). This is a bit hard to describe without a diagram, hopefully this makes sense. In order to see the curvature of the Earth you need to high enough that you can see it in profile. This does not occur until you are at about 70,000 feet (and even then it is not very noticeable).
  14. In answer to your questions: 1) "Is there any aspect of the "official" models of the globe and the universe that you doubt? If yes, please describe." Nothing major. Theories are always being tweaked as new data is uncovered, but the major tenets are solid. 2) "Do you believe that man landed on the moon as described by NASA? If not, what portions of the "official" story do you doubt?" Yes, I do believe this and I am quite familiar with the history and details of the program. 3) Do you believe that NASA has deliberately lied with respect to any mission? If yes, please describe." I think there have been very few miscommunications and these have been largely a result of poor media coverage and interpretation. No, I am unfamiliar with any outright lies on the part of NASA" Please state whether you believe the "official" story with respect to the following (simple, yes or no): 9-11? - yes (and I am well versed in this - particularly the physics involved). I think the US government jumped on this opportunity to fight an unjust and illegal war, but I don't think they brought down the buildings. I am quite certain that the buildings were not brought down by controlled demolitions. Sandy Hook? - yes (although I am only familiar with the major details). Orlando shooting? - yes (although I am only familiar with the major details). Boston Bombing? - yes (although I am only familiar with the major details). A couple things you should know about me. I'm a biology prof. I question things for a living and apply logic and Occam's Razor quite stringently. I'm not American, nor am I fan of American policies or politics - but I base my assessment of American events on evidence. Your conclusions need to be logical, you need to understand the material, and you need to be wary of non-sequiturs. Also, I may not be able to post frequently (and may not be able to post again today), but I'll do my best. I don't mean this to be rude, but you'll need to do a lot better than the post you began this thread with - the math and logic in this video is atrocious.
  15. Here's my response to "Mrthriveandsurvive" on his video (I've cut and pasted it below). I honestly can't say this bodes well for our "debate" Rothbard: "Your math is wrong here, I'm afraid. On your diagram at 9:00 you correctly show a drop of 5,865 feet over a distance of 93.8 miles to where the line of sight intersects the horizon. However, you then show that the surface drops an additional 19,174 feet over the next 100 miles beyond the horizon. It should be obvious that the drops (before and after) will be very similar as the distances are very similar (and the line of sight is tangential to the horizon). Your second value is about three times too large. Also, the total drop is irrelevant, it is the drop beyond the horizon that determines what will be obscured by the horizon. Google "Earth Curve Calculator" and you'll find a simple calculator that will combine both steps for you (YouTube won't let me post the link). There is a link to the equations used at the top of the webpage. From this you should find that the bottom 6,668 feet of mount Rainer are obstructed by the horizon. This means that the top 7,748 feet of the mountain will be visible (14,416 - 6,668 = 7,748). This fits very nicely with what we see in the photo. If we wanted to do this as accurately as possible, we should be using the heights of the mountains above the clouds to do our calculations because it is the cloud tops that are forming the curved surface that obstructs our view (assuming continuous cloud cover at a uniform height between the two mountains). I have no idea why you think you should be able to see the curvature of the clouds when sighting along them from this height. Cheers." Here's the direct link to the calculator (hopefully it posts here). https://dizzib.github.io/earth/curve-calc/

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