https://www.youtube.com/watch?v=f_EiiCXAmOA&ab_channel=NickHall <font color="#00b0f0">Waves 5.3 - Tidal Periods and Amplitudes</font> []() [[../../Attachments/Tides 2 1 1 1 1.pdf]] [[Library/Database Tabs/Tides - Waves, lunar and Solar Tides, Tidal periods and Amplitudes, Tidal Forces, and Tides in Ocean Basins 1]] https://www.youtube.com/watch?v=eMX--80J_5o&ab_channel=EarthRocks%21 https://youtu.be/bPhhYhN0FAc?si=2ofHwcZCevZESvjF&t=446 [[[obsidian://open?vault=Obsidian%20Vault&file=Tides%20PDFs]]] []() Daily tidal cycle variations - 2 high tides and 2 low tides every day the amplitude varies throughout the month, 2 maxima per month indicative of two different frequencies with a similar but not overlapping periodicity ![[Attachments/Pasted image 20240124132445 1 1 1 1 1 1.png]] <font color="#00b0f0">The globular theory is solar and lunar gravitational waves</font> They need to explain a force relationship as if the sun and the moon were the same size and same mass, because this is the relationship between the frequencies we attribute to them they dance around the common center of mass which is very close to the center of the earth. not at the center. the moon is also doing a circuit as a result of this relationship ***fundamental assumption, everything is in equilibrium all bulges are balanced all the time, all bulges pointed towards the moon all the time. ![[Attachments/Pasted image 20240124132936 1 1 1 1 1 1.png]] ![[Attachments/Pasted image 20240124133449 1 1 1 1 1 1.png]] The moon is pulling the ocean to create a tidal bulge when it aligns with the centrifugal force which itself is also oscillating around the earth. The moon's tide generating force is twice as important as the sun ______________________________________________________________________That is actually also very important. Because this force here, even though it's lifting the ocean, it's got a huge job to do. It can't compete with the gravity of the Earth. It's tiny compared to the Earth's own gravity. So that wouldn't actually make it tight. But the fact that you have these converging horizontal components, that's what makes the tides possible. Because they can allow the ocean to flow in sideways and create this bulge. So the horizontal forces and the vertical forces are important in creating what we're going to see is a kind of tidal bulge. So that's the next thing. So in summary, what have we got here? We've got a difference between two forces. Radius of the Earth is 6,400 kilometers. The distance to the moon is 380,000 kilometers on average. And so these two terms are different enough to have an effect. You can combine those two terms to write the equation in full here and then simplify it. And it's approximately equal to this. ![[Attachments/Pasted image 20240124133928 1 1 1 1 1 1.png]] A spherical object inside an oblate spheroid A football rotating inside a fixed rugby ball WE still have an oscillating semi diurnal tide, 2 tides per day ***fundamental assumption, everything is in equilibrium all bulges are balanced all the time, all bulges pointed towards the moon all the time. <font color="#00b0f0">Tidal Period =</font> Earth rotates within the bulge of water and the bulge which produces two tides per day points to the moon, 24 hours 50 mins [[Library/Database Tabs/Tides - Waves, lunar and Solar Tides, Tidal periods and Amplitudes, Tidal Forces, and Tides in Ocean Basins 1]] ![[Attachments/Pasted image 20240124123845 1 1 1 1 1 1.png]] the period of the lunar semi-diurnal tide 12 hours 25 mins to return so a tidal wave is whipping around the earth at that speed but the bulge cannot keep up with the moon, so its not just an equilibrium tide it will catch up somehow, and do some non equilibrium thing that will allow it to catch up and point at the same position the same time the next day (real answer) but, the moon is not in the equatorial plane = 28.5 degree between moon and sun (23.5) 28.5 tilt is not perpendicular to the axis of the earth plane of the moons orbit is 5 deg off from the plane of the earth orbit around the sun [[]] ![[Attachments/Pasted image 20240124123436 1 1 1 1 1 1.png]] ***fundamental assumption, everything is in equilibrium all bulges are balanced all the time, all bulges pointed towards the moon all the time. this results in repeating oscillations of a diurnal and semidiurnal tides, where at the same latitude, highest high tides rotate with lowest low tides to form a concurrently rotating system resulting in extremes of at higher latitude, the big high tide would be no bigger than the low tide, so... one tide per day. (all because of the tilt in the plane of the earth's orbit) another complication, the moons orbit is not circular ![[Attachments/Pasted image 20240124124258 1 1 1 1 1 1.png]] sometimes the moon is closer (stronger tide), and sometimes the moon is further away, (weaker tide). when the moon is opposite the sun you get a full moon. Shape of this ellipsoid imposes an 18.6 year procession on the tide cycle 13% difference in the distance of the moon between each end of the ellipse if it goes faster when its further, https://agupubs.onlinelibrary.wiley.com/doi/10.1029/2021JC017570 # A Seasonal Harmonic Model for Internal Tide Amplitude Prediction # Tidal Patterns https://www.youtube.com/watch?v=HBQgm1r2RYc&ab_channel=EarthRocks%21 ![[Attachments/Pasted image 20240124124940 1 1 1 1 1 1.png]] <font color="#00b0f0">**lunar tidal period 12 hours 24 min**</font> <font color="#00b0f0">**solar semidiurnal tidal period is 12 hours**</font> (ignoring the equation of time) this creates a modulation of the bodes kepler effect - differing speeds of the earth in its elliptical orbit around the sun. when earth is closer to the sun, day is longer, when it is further away, less than 24 hours which alters the solar tidal frequency ![[Attachments/Pasted image 20240124125529 1 1 1 1 1 1.png]] twice a month stronger tides and twice a month weaker tides This just shows when the new moon intercedes with the sun you get a strong tide (Spring tide) a week later, the solar and lunar (growing half moon) tides are canceling each other out, but since the lunar tide is twice as strong as the solar tide there wont be zero tide, but as weaker tide. a week later a full moon reinforcing again for a stronger tide ![[Attachments/Pasted image 20240124130531 1 1 1 1 1 1.png]] adding another oscitation that has to be accounted for is the lunar orbital period this motion also adds a dimensional wobble to earths orbit The Phases of the moon also change on a period of 29.5 days . which dictates a change in maximum tidal amplitude. the time between a new moon and a full moon, the time between these maxima of tidal amplitude of 15 days ![[Attachments/Pasted image 20240125132725 1 1 1 1 1 1.png]] ![[../../Attachments/Animation of the tides around the globe 1 1.mp4]] ![[../../Attachments/9convert.com -Global Ocean Tides_1080p 1 1.mp4]] ![[../../Attachments/9convert.com - M2 Tidal Constituent_480p 1 1.mp4]] ![[../../Attachments/9convert.com - Animation of the global tide_1080pFHR 1 1.mp4]] ![[Tidal Nodes Overlay Wiith live current 1.mp4]] All right, let's see if we can quickly go through some of Robert Bennett's math here. So we have course up here trying to find the acceleration with respect from a to O, which is over here. And that would be a to O minus a o O so. And then this one minus this one and a with respect to beo. So that would be this one. To this one and we're going to get, you know, the gravitational influences at the end of substituting all of these values, which of course, are form the heliocentric model, but we are going to pretend that they are real. And we're going on to the next slide here. Essentially, we're going to look. Look at O and X being at the center of the Earth, of course, and in the center of the moon, O and X. And then the radius of Earth will be R e standing for the radius of Earth, the distance between Earth and the Moon just BD and the mass of the Earth and the Moon be M E and respectively. So mass of Earth, mass of moon. Will be using that frequently. Acceleration to the right will be positive and left will be negative. So the acceleration at point a which would be opposite facing the moon right? Here is it got to be given by accelebration of a. Equals the gravitational force of the mass of the Earth over the radius of the Earth squared plus the gravitational force of the mass of the Moon over the distance of between the Earth and the Moon plus the radius of the Earth squared. Approximately equal to the gravitational force of the mass of the Earth over the radius of the Earth squared plus gravit. Gravitational force of the mass of the moon over the distance to the Earth to the moon squared, all multiplied by one minus two of two times the radius of the Earth over the distance from the Earth to the Moon on. So that's. That's all. Acceleration of a. Then we're going to go back and calculate acceleration of B, which would be roughly the same, but gravitational force of the mass of the Earth over the radius of the Earth squared plus gravitational force of the mass of the moon over the distance from the Earth to the moon minus the radius of the Earth squared. OOPs. We reset the slide. No problem. It's approximately equal to the gravitational force of the mass of the Earth over the radius of the Earth squared. Or the negative rather approximate a negative that and then plus the gravitational mass of the moon over the distance squared, all multiplied by 1 plus two times the radius of the Earth over the distance from the moon to the earth and moving on. We have again acceleration from point B opposite to the moon to be over. Here given by acceleration of B equals negative gravitational force of the mass of the Earth over the radius of the Earth squared plus gravitational force of the mass of the moon over. The distance of the. From the Earth to the Moon minus the radius of the Earth all squared approximately equal to the negative of the gravitational force of the mass of the Earth. Over the radius of the Earth squared plus gravitational force of the mass of the moon. Over the distance between the Earth and the moon multiplied by one plus two times Radius of the Earth over the distance between. Radius over the distance between Earth and the Moon again. And then finally we have the acceleration at point a just given as you know G gravitational force of. Gravitational force times a mass of the Earth over the radius of the eth squared plus the gravitational force of the Moon over the distance from the Earth to the Moon plus the radius of the Earth squared. Approximately equal to the gravitational force of the mass of the Earth over the radius of the Earth squared plus the gravitational force of the mass of the Moon over the distance of mass over the distance Earth to the Moon squared times one minus two time the radius of the Earth over the distance from the Earth to the Moon. And this is again approximately equal to acceleration of O would be roughly equal to the gravitational force of the mass of the moon over the distance to the from the Earth to the moon squared. And moving along here, just to show exactly where we were driving these from, this would be a B. We already went over that, but that's what this is. This was a of a again on this side where we started. And of course, the celebration with respect to the center was always there. So just making sure we know where we're calculating. So we have the acceleration at point B with respect to O this would be from this point to the center of the Earth and we have acceleration of B vers with respect to O equals celeration of B minus acceleration with respect to equals negative or approximately negative. The gravitational force of the mass of the Earth over the radius squared. The rate radius of the Earth square plus two times the gravitational force of the mass of the moon times the radius of the Earth over the distance between the Earth and the Moon cube. And then we have the acceleration of this point a with respect. Again. Concluding this side. And then we already did this side. This is just acceleration of a with fish back to zero equals clibration of a minor accelbration of O approximately eating equaling gravitational force of the mass of the Earth over the radius of the Earth squared minus two times the gravitational mass of the moon times radius of the Earth over the distance between the Earth and the moon cubed. So we do want to combine these forces. Actually, let's go back here, check something. Right? So, yeah, we do want to combine these forces. Considering the order of magnitude to the radius of the Earth from the Earth and the Earth between the moon, we need to calculate the difference. So. So the combined effect of the moon and the sun when acting in the same straight line. And again, this is indicative of the constructive interference theory of the equilibrium theory. Assuming you know all the assumptions that we're assuming currently of the universal static response to the force of gravity. You know, one one tidal basin, no other bodies of water, no Earth rotation, et cetera. But we're going to assume that when the moon and the sun are close together, and the geocentric models to be, you know, when the moon is next to the sun, they're going to increase the gravitational force exerted by the celestial bodies on the sides, like so. Right, so they're going to be combining the effect of the mass. I'm sorry. Combining the effect of the moon and the sun. When acting in the same straight line and again accelebration of respect to with A O would be what we went over before which would be exactly the those formulas right? But me to go over them again. They're right there. And this we're just going to quantify the acceleration difference due to the gravitational attraction of the Earth, Moon and Sun. Meaning that cergy all lined up. And again, the geocentric model Earth obviously doesn't move, can't be lined up with celestial bodies. But let's just assume that the alignment of the Moon and the Sun also indicates the alignment of the so called Earth, right? So we need to find this quantification with considering the order of magnitude of the radius of the Earth over the distance between the moon and the Earth and the radius of the Earth over the distance between the Earth and the moon. So the difference in acceleration from the near side to the far side to be with the acceleration of egg with respect to B equals the acceleration of a with respect to O minus the acceleration of a with respect to B. Of zero equals about two times little. A over. The acceleration of O equals 2 all times the gravitational force of the mass of the Earth over the radius of the Earth squared minus four times the force of the mass of the Moon times the radius of the Earth over the distance between the Earth and the Moon. Cubed minus four times the gravitational force of the mass of the sun times the radius of the Earth over the distance between the Earth and the Moon. Cubed now. Getting there. Almost done. Hang with me here. Finally, we're going to replace the Moon with the Sun and repeat the calculations. So that when we do this, we're substituting a mass of the Moon could be m m of m with a mass of the Sun Ms m m s and the Earth Moon distance d d with now the Earth Sun distance big dd. Okay, so with the same variables, essentially, we're going to have them mean different things. Combine the gravitational effects. And again, when the moon and sun act along the same line. We already went over these formulas, but we're going to go ahead and calculate them again. And an acceleration with the difference across Earth. Is this given by this formula over here? Celebration of the acceleration of eg. With respect to B equals a of respect to a with O and then minus a of O to B equals two times little a and acceleration of O equals two times a. Gravitational force of the mass of the Earth over the radius of the Earth squared minus four times the gravitational mass of the Moon times the radius of the Earth over the distance between the Earth and the Moon. Cubed minus four times the gravitational force of the mass of the Sun times the radius of the Earth over the distance between the. The Earth and the moon cube. All right, moving right along here, we need to find the ratio of the lunar acceleration to the solar acceleration, which would be acceleration of M and solar acceleration of S. And again back to the same diagram we started with, based only on the static gravitational forces above. The ratio could be given as again a of m with over as would be what we went over before. Essentially with just the substituting of the mass of the moon with the same formula, but this over the mass of the sun and the radius of the Earth both using radius of the earth equals approximately the distance between the Earth to the sun and the distance between the Earth and the moon cube, right? So that ratio right there, stick with us, we're almost there. The math's almost done, guys. And again, this is the heliocentric position. The best, the man that we can get to make sure we're all on the same page. Giving them the benefit of the doubt. So that we have the ratio of title to lunar acceleration translated to the MTS unit. So we get, you know, all these substitutions. We can finally get our answer. And what we get when we fill in the substitutions is the acceleration of the Due to geostatics would be 2.35. Which would be 2.35 times the gravitational force of the Sun. So the moon exerts that much more tidal force than the Sun. Very contrary to what the math would suggest or what the other model would suggest. Certainly contrary to what the equilibrium theory would suggest. Okay. Okayol.