It feels silly to ask, but a 'like' to this video (to the video, NOT to this comment) would be appreciated. Of course, only if you do believe it is good. The KZread system simply works that way. It isn't geared towards spreading truth; it's geared towards spreading advertising space. That's why popular videos do better. I am trying to spread this explanation of the tides - against the prevailing current of the abundantly presented popular but incorrect explanation. Your 'likes' can help spread the message.

Hi Tribus, _(or whatever your name is?)_ Seeing as you’ve now removed your ‘work in progress’ video !! :) I thought it best to rewrite my previous comment and post it here instead… This is exceptional work and your animation is perfect! What software package do you use uncle? ;) I have absolutely no problem with this excellent piece of tidal analysis and wish you all the best in your future endeavours. It’s nice to see the Earth remain stationary for a change, without any weird ‘wobbly’ movement. I just wanted to clarify for the record one last time, that the ‘flipping’ central vector in your video is actually a ‘fictitious’ force and results in a zero movement ‘towards’ or ‘away’ from the Moon _(ignoring the slight orbital eccentricity and conservation of angular momentum that is!)._ This is what Newton, Einstein and me! :) have been banging on about for ages! And I thank you for your efforts in explaining this - well done! 👍 The only movement left for the Earth to make, is in a tangential direction away from the Moon and around its barycentre, just like all the other gravitationally bound objects in the universe. I hope I was able to clear things up! but, if you’re still confused… then all I can say is keep digging and keep reading, because it is worthwhile. Kind regards WD.

@TribusMontibus

2 жыл бұрын

Hello Wavy Davey, Always a pleasure to hear from you. Besides, it was a bit of a revelation in this case. Thank you for your persistence! I had to read your remark a few times before it dawned upon me that you were absolutely right (initially I thought you were 'full of, …. the brown stuff'). I was so focused on strictly using the vector addition as a comparative tool that I hadn't realized that, in fact, that means that one is graphically creating a new reference frame, indeed using a fictitious force (or acceleration, if you will). Funny, how we sometimes need to have a mirror held in front of us to realize what we're doing. Thanks for holding up the mirror. Kind regards, Garret

Very nice explanation. And you did it all without saying the word "centrifugal" - (oh sorry, I've said it now ;). Love the skateboarding guardsman. Looking forward to the next one.

@TribusMontibus

2 жыл бұрын

Hello Nick, Thanks, I'm happy you like it. Indeed, I didn't mention that word, but I will be mentioning it in the future, although not in this series. I believe that addressing the 'centrifugal controversy' here might divert too much attention from what I am trying to get across. An American professor, who is very much opposed to any centrifugal analysis, told me that 'this doesn't help students visualizing what's really happening'. I disagree with him, and from what I've seen in your videos; I think you do too. I heard Feynman say in a video once, that there isn't a 'one-size-fits-all' for teaching methods. I believe that many people will respond better to the explanation, when approached from the centrifugal-point-of-view. Thank you for noticing my guardsman. I've had so much fun with him. This little figure started his 'career' as a 1970's Britains toy soldier in my backyard. I took a picture of him on the garden table and imported it into a drafting program. Then I half closed my eyes to convert his face into just two skin tones. I used him for an animation in the OTMS-1 video, to step on a scale. But then, for OTMS-2, I wanted to improve on an explanation I've been using for some time; the skater in the truck's cargo hold (really just a variation of the famous elevator). So I found a cartoon of a skater boy on the internet, pictured how a uniform of the Coldstream Guards would look on him and slapped the guardsman's head on him. I'm rather fond of the result myself. As I'm a bit busy with work I can't spend as much time on my tidal-hobby as I'd like. So, OTMS-3 may be a while in the making. I think I will enjoy making this one even more than the previous ones as it is always my favorite part of the explanation, when the theoretical model meets the real world, doesn't behave as you might expect, yet - when you take a closer look - it all makes sense in the end. I want to take my time about it because I don't want to mess it up. In fact, I have to dive into quite a few of your videos for inspiration, as one of the most important elements of OTMS-3 will be right up your alley; waves. I may steal some of your words. The mainstay of understanding the tides on Earth, in my opinion, is the realization that, after a modest initial elevation, and a forced wave which can only exist in a limited band of latitudes on Earth, the phenomenon of the tides at our beaches is almost exclusively a matter of understanding how (very-) shallow water waves behave on a global scale. I've gotten long-winded again. I'm off to fetch a cup of coffee. Kind regards, Garret

@NickMJHall

2 жыл бұрын

@@TribusMontibus Yes definitely variety is the key with teaching. Different approaches work with different students, or even with the same student. If you can explain things without talking about inertial forces then that's very elegant, but not always straightforward. Sometimes you need a stepping stone to understanding. Feel free to take what you need from my lectures. I have new course notes, made by a colleague and very nicely put together to match the videos. Link in the banner on my channel. Cheers. Nick.

@TribusMontibus

2 жыл бұрын

@@NickMJHall Goodness me Nick; thanks for pointing that out. Last time I got my hands on a course like that it'd cost the tax payers a lot of money ☺. My compliments (to you and Serena Illig) as I've scanned through the course notes and they could pass for a stand-alone-paper. They will be very helpful to me, immersing myself into waves (sorry, couldn't resist the pun).

@wavydaveyparker

2 жыл бұрын

And you did it all without saying the word “barycentre” - (oh sorry, I’ve said it now ;) And just for the record. That’s a skateboarding ‘Scots’ guardsman! 💂‍♂️ And not a ‘Coldstream’ guardsman (who carry a ‘red’ plume in their ‘bearskin’)💂‍♂️ And it’s NOT called a ‘busby’- so, there! - Please make corrections immediately! :) Kind regards. (Of course, I was kidding about the corrections bit, as this is a very accurate description of the tides on Earth) :)

You get my thumbs up on both comment and video! Your contents are extremely interesting.

@TribusMontibus

2 жыл бұрын

Thank you Stefano. Your support is greatly appreciated. I've kept the biggest surprise for the next video (OTMS-3). That's when things will really start falling into place and the theoretical model of the equilibrium tides meets the real world. For most people I talk to, that part is the real eye-opener, and for me it's the favorite part of the explanation. So, I'm eager to make it a good video. I'm taking a little break now as I also have work duties and my family to attend to and plan to start making the next video in a few weeks time. Kind regards, Garret

I read ur comment on another channel and I was just searching for someone who also has such an idea of explaining this concept in the logical way not just saying the moon is explicitly pulling it up. Till now I was not aware of such tidal equations I thought maybe I was making somewhere mistake but in the end, people like u save the science community by speaking what they really know not a part of spreading misconceptions. I just want u to add the term shear stress that's what explains the later half of the video as fluids can't hold it and are thus displaced.

@TribusMontibus

2 жыл бұрын

Hello, Thank you for your kind comment. It is fascinating how few people (and KZread videos) explain the ocean tides this way. It is my personal interpretation and presentation, but the theory is, by no means, mine. It is simply the current scientific consensus on the ocean tides. Although very few people recognize it as such, because they have been so over-fed with the popular nonsensical explanation that they have come to believe that it is the correct explanation. This means that many will approach my videos, believing they have come across ‘some wild hare-brained speculation’. When it comes to the tides,you already need to know what you are looking for, in order to find it. I have very little hope that videos like mine will ever be popular on KZread, while nonsensical tides videos rake in the viewers by the thousands. On such channels viewers typically congratulate the makers for having provided such a clear explanation, unaware that they have just gotten a clear explanation of nonsense. But my KZread excursion has proven very useful to me, as I am getting a lot of nice reactions and good ideas as well. Your idea about shear stress is a good example. I hadn’t thought about approaching it from that viewpoint, but it’s certainly valid. It is very hard to implement it into an existing video though. The KZread algorithm is a fickle thingy. When you retract a video for a small change it takes forever to get it back up to speed, in the case of impopular videos like mine. But I will certainly be considering it in the future. Kind regards, Garret

This is simply the best video I’ve seen on the tides. The video of Neil degrasse Tyson makes me so made bc of how misconstrued it is. I can’t tell you how, as a marine science student, this has helped me thank you!

@TribusMontibus

Жыл бұрын

Thank you for your comment. This is one of those strange cases where many, otherwise very knowledgeable, people go around, simply repeating what they THINK is the correct explanation and are never confronted with its incorrectness. Astrophysicists like NdGT are very familiar with 'tidal forces' but clearly never bother to analyze how these forces apply to Earth's oceans. Apparently they are satisfied with the idea that their (over simplified) explanation is correct because they know that tidal forces exist and they know that seawater, at many beaches rises and falls periodically. The fact that the timing of these phenomena doesn't match is lost on them. A quick check would show them. But, in their defence, these misunderstandings are so widespread that nobody points this out to them. Once you know that the 'simplistic explanation' is incorrect you know what to look for and find out that knowledge of the correct explanation is widespread among oceanographers and has been known since Newton. Good reading on this are publications by Eugene Butikov of St. Petersburg, Erofeeva of Oregon State and the NOAA. They keep on telling people clearly. But most people are just not listening. Kind regards and the very best with your studies

I Liked the video. You put a lot of effort into making this show. How come large lakes don't have tides? Have a Magical Day.

@TribusMontibus

2 жыл бұрын

Hello, Some large lakes do have discernible tides. But that's not just a measure of lake size or water quantity. It has a lot to do with geographical position, lake dimensions and orientation. The same tractive components which work on the oceans, also work on lakes. This can set a lake in motion, ever so slightly. When this motion is roughly 'in phase' returning around the time the next 'pull' comes around, the tidal movement can build up over time to discernable highs and lows. Lake Michigan is a good example. It is often approximately 45 degrees away from the sublunar point (the most effective zone) for a considerable time. An existing rise is then reinforced and then makes its way north. The (2-inch) high tide at the lake's southern shore occurs some 6 hours apart from the high tide at the northern shore. When this wave-movement returns, some twelve hours later, it's in time to be reinforced. Lake Victoria, in Uganda, on the other hand, has moon and sun 'passing overhead' while the same tractive components never get the time to concentrate and build up a rise. Before that happens, the tractive components are pulling in the other direction and 'lake mechanics' are never in phase with this alternating pull. So, no discernible tides at all.

@BuzzSargent

2 жыл бұрын

@@TribusMontibus Thank you. That is interesting.

Thank you TMO for the insightful video. I am many years navigation officer and thought I had a basic understanding of tides and oceanography, when in reality the theoretical framework I held was insufficient to fully describe tides. I still don't understand how the net acceleration for water on the opposide side of earth relative to the moon is away from the moon, it really goes against all intuition so there is something for me to think about.

@TribusMontibus

2 жыл бұрын

Hello, Thank you for your kind comment. I am happy to hear the video struck a chord. Yes, you are correct; it’s all a bit counter-intuitive. It isn’t all that hard to find analytical treatments in professional literature. An equation on paper may prove or disprove a point but it may not aid understanding, if not properly explained. I guess, for me, the best intuitive explanation for the far side tendency (because neither the far side, nor the near side ‘bulges’ really exist) is that a mass there still experiences an absolute gravitational acceleration towards the moon (when talking about the Earth-Moon system). But it is key to realize that the lunar gravitational accelerations are overpowered 10,000,000 over by Earth’s own gravity. This means that the total mass of everything on Earth roughly moves along a net path through space which is parallel to the path of Earth’s center of mass (which is being curved more towards the moon because, as a whole, it is closer to it than the far side mass). This means that our far side mass, even though it neatly moves along with the earth, like all other masses, does have a net tendency which appears to be a form of minute anti-gravity to an observer attached to the earth’s surface. And it is the integration of all these ‘tendencies’ along their correct directions which yields the force pattern which starts the ocean tides moving. What surprises most people, in my experience, is that it isn’t the more obvious vertical tidal components which end up doing the actual moving of the water, but rather the horizontal ‘tractive components’. Kind regards, Garret

I’m confused! Hope you get help with that?

@TribusMontibus

2 жыл бұрын

Hello Heavy Light, Haha, did you know that your avatar looks exactly like the 'confused teacher' on the Tribus Montibus Oceanography website? Anyway, what are you confused about? Kind regards

@heavylight5994

2 жыл бұрын

@@TribusMontibus Hello Garret, I’d seen website it is good! :) Now I understand it :) … Newton; Euler; Laplace; Descartes; Einstein; Lorentz Contraction :) … Feynman; Wavy and Tribus. :) … I do what I like and answer to no ONE !! :) The Earth accelerates ‘towards’ Moon, there’d still be tides :) BENDS it path! :) You’re welcome :)

@heavylight5994

2 жыл бұрын

@@TribusMontibus Haha! You are WRONG!! Tribus Montibus Oceanography. By subtracting the moon-earth gravitational force from the moon-ocean forces, you INVENT a non-inertial reference frame that gives FALSE signals regarding what is and what is not a real force. Centrifugal force is not REAL and it called a fictitious force. [insert wiki link here] I’m making a video about that now! More coming! :) You’re welcome :)

I have a question to the part at 4:40 in your video: You explain that locations which are the same distance away from the moon will have the same amount of force applied but the force points somewhere else. Until now everything is fine. But then it is explained that the netforce of the other locations will point downwards due to vectoraddition of the forces. This is where I have a problem: If we add the vectors of the force, it should be orientated downwards to the left, due to the vectors having the same length (the amount of force) and one being angled to the bottom right. The resulting forcevector isn't pointing downwards. The location where the netforce is pointing downwards should be closer to the moon than the center of earth. Then the angled forcevector would be a hypothenuse of a right triangle, therefore having a larger amount of force than the vector of the centre of earth which would be the bottomside of the triangle, guaranteeing that the netforce would be perpendicular to the bottomside (forcevector of the centre of earth). Am I missing something? Clarification would be appreciated. :)

@TribusMontibus

11 ай бұрын

Hello, No, you aren't missing anything. You are totally correct, if you are referring to 'the lateral point' (the point 90 degrees from the sublunar point). Here the net acceleration for a mass, due to the gravitational influence of the moon points slightly away from the moon due to the angle you refer to. The actual 'neutral point' where the resultant acceleration points straight downward is slightly forward of the lateral point. Mind you, what I am really adding (subtracting) are vectors of acceleration, not really forces. A mass in space may have an absolute acceleration due to a certain gravitational attraction from a mass (in this case; the moon). But when you are trying to figure out the net influence of that attraction in an environment, you need to correct for the acceleration of that environment (in this case; the earth underneath) due to the same gravitational attraction. That's what the tides are all about. That's also where many people fail a real understanding, because they remain fixed on absolute attractions and forces rather than the comparison. Another caution; don't lose sight of the comparative dimensions. The average Earth-Moon distance is measured in hundreds of thousands of kilometers while Earth's radius is measured in only thousands. The angles we are talking about are incredibly small in magnitude. I'm on vacation in the French Alps (riding my bike up hills which are a bit too steep for me) right now so I don't have my laptop with me. But I could look up in my spreadsheet where the neutral point is and just how small the angles are. Typically you have to read it two or three times before realizing how small such numbers are. Many draw the tidal force envelope as symmetrical. But as you correctly deduce, that isn't correct. It isn't exactly, longitudinally, symmetrical. But many an 'explainer' is already so happy that he/she can get people to understand why 'the far side' also experiences tidal forces which raise the water that he/she takes this tiny, negligible asymmetry for granted. Kind regards

@ManifoldMold

11 ай бұрын

Thanks for the fast answer! But I actually didn't mean the point that is 90° from the sublunar point. I mean the point that is on the surface of the earth and is the same distance away from the moon as you show it in your video. Should be around 80-85° (?), marked by the dashed line in 4:40 . Also wanted to say that your video is superb! :D Just one more question: Why vectors of acceleration and not forces? Sure you can explain the tides with comparisons of the accelerations. But the reference point (the centre of earth) doesn't fall into the moon. The centripetal/centrifugal force is counterbalancing the gravitational attraction of the centre of earth to the moon. Therefore we can speak of forces (?) Or is it because F=m*a ? Because we can't apply a particular force to a single point on a planet, due to not knowing what the mass of a single point is? But even if that's the case, one could just "measure" the forces of a grain of sand in different locations giving a broad estimate of the envelope now made up of forcevectors.

@TribusMontibus

11 ай бұрын

Hello again, To be honest, the geometry of the vectors, within the range you are talking about, gets complicated. It is no longer a matter of trigonometry with fixed values of the vectors which are involved. They change with every next iteration as a varying position-to-moon distance requires a recalculation of the 'local' acceleration vector. This becomes an exercise in differential equations which would have made my head spin during my student days. Either way, when I wrote my spreadsheet, that wasn't the problem I was trying to solve. So the spreadsheet isn't especially suited to finding that exact answer. But I can tell you that you are off by a significant amount, in your estimate. The spreadsheet does allow me to fumble around with the numbers a bit. This shows the 'neutral point' to be in the vicinity of 89.1299 degrees away from the sublunar point. My answer to your second question is a lot easier. There are very good, practical reasons for comparing accelerations rather than forces. The aim is to find out what effect the moon's gravitational attraction has on any given mass in any given position on or within the earth. The earth moves roughly like a huge blob through space. The deformations of the planet and variations of various masses, on and within, are insignificant within our context. Therefore you can easily calculate the acceleration vector, for the entire earth, due to the moon's gravitational attraction. Mind you, this certainly doesn't mean that the earth is getting closer to the moon. But the moon is constantly pulling the earth away from the elliptical orbit which the earth would have described around the sun if the moon hadn't been there. In the end it's really the Earth-Moon barycenter which describes the orbit around the sun. This means that you end up with an acceleration vector which is the same for any position on or within the earth. It would, at this stage, be rather meaningless to use forces anyway because the magnitude of the gravitational force on the entire earth isn't useful for any local comparison. But it's most important to realize that, for tidal purposes, you need to compare the gravitational influence on a local mass to the gravitational influence on the entire environment (the earth). By comparing both (the vector addition) you are left with a net acceleration which can be added up for local masses (using F=m*a) to get meaningful results. Kind regards

@ManifoldMold

11 ай бұрын

@@TribusMontibus thanks again! :) Yeah the 85° were just eyeballed from the schematics from your video, were the distances of moon and earth were changed to visualize the stark contrast of the vectors. And with the fact that I should bear in mind that the scales of the radius and the distance from earth to moon are magnitudes away from each other, as you state in your previous comment, I should have realised that it should be really close to 90° . Loved also the explanation for the use of vectors of acceleration instead of forcevectors. Thanks for clarifying!

@ManifoldMold

11 ай бұрын

@@TribusMontibus hello again! After some thought I tried to calculate the tangential force that pushes the water and came up with my own formula. I checked my results with the values of your video and they matched (hurray!) . I came back to ask you if you know a much more simple formula for these forces. I calculated them over many steps with long terms and tried to find a simple version on the Internet, but couldn't find any. Do you know any? Even if it is really oversimplified, I would love to see it.

Hi, Max here. Having watched your video, I'm afraid I'm not able to follow your explanation. Some specific questions (or challenges) for now: 0:37 I think this is a misconception. When talking about the tides, i.e. the bulges of the water ball outside of the Earth, everything is still considered as a balanced condition in general at any given moment. But your analogy here is misleading because it seems to suggest that to cause a rise of water level, i.e. a bulge at one side, one must apply a force counteracting the gravity and *greater* than the gravity. This is incorrect because in this case, the net force (>0 along the direction opposite of the gravity) will cause the object (water in the case of tide) to fly away with an acceleration. 1:39 I don't understand what you mean by saying "the tiny forces in the ocean". 2:40 I'm quite confused by your use of "acceleration" here. I think you mean the gravity force caused by the Moon (like that caused by the Earth)? 4:31 Can you please clarify what the green arrow at the centre of the Earth denotes? Why should it be inverted before vector addition? Thank you.

@TribusMontibus

2 жыл бұрын

Hello Max, Thanks for your reply. You make some good points. I hope I can explain myself better in this message. 0:37 "To lift any object you will, at least, have to apply a force which overcomes its weight." No, that, most certainly isn't a misconception. I'd challenge you to come up with any example where a mass is moving in an upward direction, against earth's gravity, without the sum of all the forces on it matching its weight. That is why I made the cutaway drawing of the 'block of water' in the harbor in Nova Scotia. To emphasize that it is being held up by pressure from below (which provides the at-least-equal-force at that time). But you misunderstand me when you think I am saying that an imbalance in forces and accelerations exists at any time. I made this statement especially to counter the misconception that rising tides are about 'the moon's gravitational attraction pulling stuff upwards', which many people mistakenly believe. The statement is meant to get people to realize that the water is moving up because of water pressure, not because of a vertical force applied by the moon. But, yes, while that water is rising, there will momentarily be a net positive upwards force. But it is tiny, as the acceleration of moving water upwards by meters in hours is very small. That net force will be provided by water pressure from below. 1:39 Yes, I might have made that clearer. I mean the tiny forces on individual water particles caused by the moon's gravitational attraction. But as I use the phrase 'gravitational attraction by the moon' so often in this video I chose to just say "tiny forces" this time. 2:40 I see how this can be confusing. But, no, I do mean 'acceleration' and not force. Both the earth and a drop of water react to the gravitational attraction by the moon. But you will agree that the attractive force on the earth will be much greater than the force on a drop of water. The absolute forces are therefore quite useless for any comparison. But it is important, in tidal analysis, to realize that the tides are about relative attractions and accelerations (my OTMS-1 video goes deeper into that) and not about absolute forces. By the way, I am talking about accelerations which the attractive force by the moon would WANT to cause. The rest of the video tries to make it clear that water in that position isn't free to accelerate at all and I try to explain how these forces result in pressures. 4:31 The green arrow here is the acceleration vector for the earth. I invert it because this is a comparison of accelerations. If the earth underneath your feet experiences exactly the same acceleration which you experience, you experience a zero acceleration with respect to that surface. You could visualize this by subtracting the earth's acceleration vector from yours. Inverting it and putting it at the tip of your vector is one method of visualizing that. I hope this clarifies the video. Kind regards, Garret

@maxwang2537

2 жыл бұрын

@@TribusMontibus Thanks for taking the time to clarify - much appreciated. On my first point. I did and do see where you are coming from, and understand people can easily get it wrong with things like 'lifting an object'. I would not go into this too far because I see what you mean except - I think it's a bit too far for you to say what you referred to is a misconception (although I do agree the video in your mind when saying this is nonsense). And to respond to your request to give an example of applying a force that is less than the weight of an object to lift it - This CAN happen when we think this way. First remember we are talking about only static or quasi-static problems, not dynamic problems involving accelerations, i.e. generally everything is in a balanced condition without non-zero net force. Everything on the top of the Earth is all sitting on top of something with a certain degree of elasticity not strictly completely rigid. This is particularly the case for liquid such as the ocean. But to simplify the problem and isolate the focal point, let's consider an object sitting on a spring. By applying ANY non-zero force on it upwards will move it to a higher position, how far it's 'lifted' depending on the rigidity of the spring and the applied force. This is effectively *LIFTING* it, as is the case when the gravitational force of the Moon exerted on the ocean *LIFTS* the ocean to some extent and forms a bulge. So in this context, lifting does not mean lifting and moving away to another place, but rather lifting it to a new, higher, position of balance. The second - Thanks for your clarification. Now I see what you mean. The third - if you do mean acceleration, that's ok but can you please clarify what this a is? is it the centripetal a needed to maintain the orbiting of the Earth around the Sun (within the ecliptic plane), or the centripetal a associated with the rotation of Earth? I would not associate the gravitational force exerted by Earth on the water with any acceleration because the water is relatively still at any moment along the line connecting the drop of water in consideration and the centre of the Earth. You said it's "accelerations which the attractive force by the moon would WANT to cause". This is really difficult for me to follow, particularly "WANT". I do agree the gravitational force exerted on a drop of water by the Earth with be much much greater than that generated by the Moon - to a degree that, when putting both together, I believe the latter is almost negligible. The Fourth - the questions raised above remain. Again I don't understand "I invert it because this is a comparison of accelerations", because after inverting you are doing vector addition to get the resultant combined gravitational force on a drop of water (to illustrate or demonstrate the direction in which the drops of water will go). I'm unable to follow your explanation here.

@TribusMontibus

2 жыл бұрын

Hello Max, Thanks for your reply. I don't mean to take up much of your time with lengthy explanations but I am very interested to hear when an idea of mine isn't coming across the way I intended it to. Let's start with your example; the weight on the spring. Fair game, because I did ask for ANY example. A spring, as you realize, will do something which more rigid objects will not do to the same extent. As you apply a small upward force the spring will extend and still provide an upward force from below. Simple example. A 1 kg mass rests on a spring with a downward force of approx. 10 N. In fact the spring had to compress until it 'produced' that reactive upward force of 10 N before this system came to rest. When you now apply an upward force of, say, 3 N to the mass, it will indeed be lifted to a higher level. That's because directly at the moment of that application an upward force of 13 N works on the mass (exceeding its weight). The mass will continue to move as long as the sum total of your applied force + the force which the spring provides from below exceed its weight. Of course the dynamics of this experiment are slightly more complex, but still, you probably see my point. The mass will now come to rest at a new equilibrium where you provide the 3 N upward, while the spring, at its new extension, provides 7 N. Funny enough, I used a spring myself in the animation at 0:37 where the 1 kg mass is lifted upward from the scale. If you pay close attention you'll see that, as the spring extends, the weight on the scale decreases, until the force in the spring exceeds the weight of the mass. Only then does it start rising. Yes, I will repeat again and again that it is the greatest misconception held by people (and a very persistent one too) that ocean waters go up 'because they are being pulled upwards by the moon'. They most certainly aren't! The acceleration I mean is simply the stand-alone-acceleration which the gravitational attraction of the moon would cause, if the mass was free to respond. That is actually why I say "want". Because that isn't going to happen, for the water drops and the sand grains. The point of this video is to explain just that. Mass on Earth gets attracted (ever so slightly) but the response isn't to fly off into space (forces too small) or to start frolicking around (movement restricted). The response is that pressures build up in the materials which are capable of doing so. Be careful not to think in labels too much. A centripetal acceleration is only 'centripetal' because it happens to occur in a situation where this acceleration exactly accounts for the acceleration which is necessary to accommodate a curved orbit. Physically speaking it is still just an acceleration, like any other accelerations. I'm afraid that, if you insist to put a label on this acceleration, you may get a pre-conceived notion of what it is supposed to do. I'm trying to keep viewers of this video open minded about what's going to happen. On the subject of inverting vectors. Please be careful. Because if you think that vector addition means that I am combining forces to get a resultant gravitational force, then we're in danger of grave misunderstanding. That is what you do when two (vector-) forces work on the same object. That is not the case here. Btw in that case you would have to put the second vector onto the tip of the first, without inverting. This is vector subtraction. It's a visual method to subtract the acceleration of the earth underneath from the acceleration of an object. That way you can judge what the acceleration of that object across the earth's surface will do. If you are looking at the acceleration, purely through the force exerted on the object, you're left with how that force would attempt to accelerate that object through inertial space. Not very useful when analyzing ocean tides on Earth. This is why I did the animation of the skater in the truck. If you look carefully you will see that the acceleration of the skater, with respect to the road is constant. His downhill speed is constantly increasing. But the truck, although initially accelerating slower, picks up and surpasses the skater's acceleration. You could point the skater's constant acceleration vector forward and constantly subtract the truck's acceleration vector. Initially you'll subtract a shorter truck-vector. The skater's net acceleration along the truck's floor is forward. Eventually the truck accelerates so much that you'll have to subtract a greater vector. The skater's net acceleration along the cargo floor is aft. I hope this explains it better Kind regards, Garret p.s. By the way, have you watched my video: kzread.info/dash/bejne/qoqu0a6NiNKXmrg.html because it specifically goes into these issues?

@maxwang2537

2 жыл бұрын

@@TribusMontibus Hi Garret, again thanks for your comprehensive response - and Merry Xmas! Unfortunately however, I don't agree with you AT ALL and believe you are incorrect - you tripped yourself off without realising. I didn't intend to get into the simpler point of 'lifting' but now I think I will focus on this one without going into the more complex topics. This is because if we cannot agree on simpler and more fundamental things there is no point in talking into more complex topics. The problems of your argument I can see, around this point, include: 1) the scenario of lifting an object is now different from what is illustrated in your own video. There the object is lifted OFF the scale, which indeed would normally require an extra lifting force, upwards, no less than the gravity (acting downwards). You are now changing the problem in the discussion. 2) when saying applying an extra force in an attempt to lift an object, I presumed the focus was only the additional, or extra, external force beyond the ubiquitous force of gravity and corresponding reaction force supporting the object from falling (and this presumption is not unreasonable). It's a strange argument, to me, to say the applied force if >= the weight of an object by adding this reactional force and the additional lifting force. Ok, if let me accept this way of thinking, then in the other argument of saying the water is 'lifted' by the gravitational force that you refuted - and using the same method - the TOTAL force applied by a drop of water will also definitely be >= its own weight (the forces including the pressure of the surrounding water applied to this drop to keep it from falling, plus the additional force by the Moon, however tiny it might be). In this case, because the sum of force is >= its weight then the lifting of it can indeed happen, according to yourself. But this defeats your refute of this 'theory' of lifting. 3) From a logical point of view - and using the proper terms in the area of logic - you are "committing" a "fallacy" of "Equivocation" in your argument. That is changing the underlying sense, or semantics, of a term in the argument, which is the 'force' in this case. When refuting what you considered as a misconception, the meaning of the force was the additional external force applied on the object (the gravitational force by the Moon), not including the reacting force which was already existed before applying this additional force. But then being challenged, this term is changed to include this pre-existing reaction force and take the 'force' as the sum of these two forces. (Sorry for using these 'academic' terms which seems pretentious. I couldn't help doing this because this is the proper language I just recently learned from reading a book - the book I mentioned in our conversation earlier. This defect of argument was known to me and I knew the term in another language Chinese, but I was excited to learn it in a second language of English. By the way, this is probably I would say the best book I've ever read in my lifetime - A Concise Introduction of Logic by Hurtley. I have not finished it yet. And the counterpart of this term in Chinese is "偷换概念 (tou huan gai nian)", which literally means "changing the term in stealth". This book is freely downloadable online, at least its earlier editions. I've downloaded its 7th and 11 editions and am now reading the 7th edition. The latest edition is probably 13 or 14. Highly recommended.) On the point of acceleration, your explanation is totally confusing to me. I don't know, when there is only a force (though indeed a force can have the potential of causing acceleration), why you need to consider the potential of it causing an acceleration. This only confuses (by introducing unnecessary more complex terms and unnecessarily complicating the problem) not only you but others.

@maxwang2537

2 жыл бұрын

I added one post yesterday but was (believed to have been) automatically deleted by YT probably because it included URLs. Reposting without including the links. I came across a very useful, and believed to be authoritative, reference, Physical oceanography [Hill, M. N. (Maurice Neville), 1919-] on the website of Internet Archive. Pdf version is available for free download. Tides are covered in its Chapter 23. The big problem, though, is that this is a very serious book and the theory is probably beyond the comprehension of the majority. But I will give it a try. After a glimpse, this theory seems to align with a discussion on Physics Stackoverflow. Again because I'm not allowed to include a link here, I would rather copy and paste the best answer here - [QUOTE] The tides are a result of the response of the Earth's oceans to the tidal forces exerted on the water by the Moon and the Sun. The responses are vastly complicated by the Earth's rotation about its axis, by the physical geography of the Earth, and by the nature of the orbits of these bodies. Of key interest with regard to this question are the inclination of the Moon's orbital plane and the Earth's equatorial plane with respect to the ecliptic. Consider the forcing at a point on the Earth when the Moon is straight overhead (at the zenith point). 24 hours and 50 minutes later (a lunar day), the Moon will once again be straight overhead, more or less. What about the midway point, 12 hours and 25 minutes after that first instance in time? If the Moon was in an equatorial orbit, the tidal force at this point would be almost identical to the force when the Moon was straight overhead. This is the source of the semidiurnal tide, with a period of 12 hours and 25 minutes. This is called the "principal lunar semidiurnal" component of the tides, or M2 for short, where "M" denotes the Moon and "2" denotes two high tides per day. However, the Moon's orbit is not equatorial. It's orbit instead is close to the ecliptic, about 5 degrees. Thanks to the Earth's axial tilt of about 23 degrees, the tidal force on the point of interest at 12 hours and 25 minutes after the first instance is less than it was at the first instance. This induces a 24 hour and 50 minute frequency response in the tides as well as the 12 hour and 25 minute frequency response. (Actually, this is an oversimplification. There are a number of different frequency responses with a period of about 24 hours that are a result of the Moon's gravitational pull, the Moon's orbit, and the Earth's rotation.) The Sun raises tides as does the Moon. The largest component is the principal 12 hour semidiurnal solar tide, or S2 for short. All told, there are hundreds of different frequency components in the Earth's tides. The response of the Earth's oceans to these different components varies from place to place, and is different for each component, primarily because of geography. The response to any one component results in a set of "amphidromic systems" for that component. An amphidromic system comprises a central point called an amphidromic point at which the response to that frequency is zero and tides that rotate about that point, with amplitude increasing with distance from the amphidromic point. In addition to these amphidromic points where the tidal response to a component is zero, there are areas where the tidal response is broadly suppressed. The Gulf of Mexico is somewhat isolated from the Atlantic and is shaped in a way that keeps the M2 and S2 tides very small. Places where the M2 and S2 tides are suppressed but the diurnal tides are not suppressed see one tide per day. In addition to those places in China mentioned in the question, tides in the Gulf of Mexico are diurnal. These places the see a diurnal tide tend to be in or near the tropics because the forcing function for these diurnal tides are at their greatest in those areas where the Moon and Sun can be straight overhead. For the Sun, these are the Tropic of Cancer and Tropic of Capricorn. For the Moon, the latitude varies between 18 and 28 degrees over the course of 18.6 years. [UNQUOTE] I'm not officially endorsing this explanation but only noticed the terms and concepts used here are largely in line with the reference I mentioned. Of course, this is very complicated. And after some pondering, I've reached my own simple explanation (specifically for the two bulges of the water ball, i.e. the oceans, rather than only one bulge) which I now believe is correct -- These are due to the existence of both the gravitational attraction of the Moon and the 'centrifugal force' caused by orbiting of the Earth (relative to the Moon). The effect of this of pulling an elastic ball (represented by the oceans covering the surface of the Earth) is intuitively an elliptical shape with two bulges [This is obviously really a very, probably overly, simplification, of the phenomenon but it should work if merely for explaining the two bulges. The whole topic of tides is much much more complex and involves at least the interactions between the Sun, Earth, and the Moon. And other important things include 1) tides are a result of multiple harmonics of the periodic rise and fall of the ocean due to these interactions. 2) the complexities are due to, at least, the inclination of the Earth's rotational axis relative to its ecliptic plane, within which it orbits around the Sun (this inclination is the reason for the seasons), as well as the inclining angle between this eclectic plane the orbiting plane of the Moon around the Earth.] I have two main points to clarify to preempt any hasty and premature laughing and attack. 1) 'centrifugal' force - some people are allergic to this term because they think this is not a real force. This is true, there is no force such as a centrifugal force. This is instead only a treatment - treating a dynamic problem where acceleration is involved as a static problem with the object in balance - in the problem of an object moving along a curve requiring a centripetal force to maintain its centripetal acceleration. By this treatment, from the perspective (or in the frame) of the object, it can be considered as still under the force of a centripetal force and the same, but opposite, centrifugal force. This treatment does not suggest the centrifugal force is a real force. 2) The Earth orbiting around the Moon. This seems weird and breaches what we were taught (It's the Moon that is orbiting around the Earth). The fact is there is no such thing that is absolute as A orbiting around B or the other way round. These are all relative. The Moon is orbiting around the Earth in the eyes of the Earth and vice versa.