Uh, so you're saying each point follows a circle around the barycenter, and that circle has constant radius? I don't disagree with that. But sure sounded to me like you said that every point (as in all the different possible points on Earth's surface) have the same radius about the barycenter. That's what I was calling 100% false and I hope you'll understand the misunderstanding of "every point". If i've interpreted all this correctly then i don't see what i said wrong in post #82 that you pointed out in post #89, when u said i was 90% there but missed something. Can't completely agree. There are 2 centrifugal forces (Cf) and i don't wanna confuse them, so here they are again: one from rotation and one from revolution about the barycenter. The rotational one doesn't play a role, since the same centrifugal force from it occurs everywhere on the equator. The revolutional one, however, could play a role (but i have to calculate it to be sure, it could be negligible), because different points on Earth's surface have different centrifugal force from revolution. The anti-Moon side of Earth will have the greater Cf since it's farther from the barycenter; the near-Moon side of Earth is closer to the barycenter so it will have a smaller Cf. Actually we never agreed to that. You're prolly thinking of something agreed between you and billvon. I suppose i could go find a video that doesn't show any Earth rotation while still orbiting the barycenter, but honestly it will be the same Cf from revolution. Just pause it at any point in the diagram and i hope it clearly shows to you that different surface points on Earth have different radii to the barycenter. Since the angular speed of the revolution is the same, they must have different centrifugal forces. So again, I can't see what you're pointing out in post #89 about what i said wrong is post #82. One thing I just realized from all this, however, is that the 2 different Cfs can interfere with each other so to hell with it ill just calculate them both right now. The centripetel acceleration at Earth's equator, from rotation, is constant (thank god). I calculate it at 6378000*(2*pi/86164)^2 = 0.033915057 m/s/s. I used siderial rotation period of Earth of 86,164 s and an equatorial radius of 6,378 km. Centripetel acceleration, from revolution, at sub-Moon point on Earth's surface = (6,378,000 - 4,600,000)*(2*pi/2,360,585)^2 = 1.259656718 e-5 m/s/s Centripetel acceleration, from revolution, at anti-Moon point on Earth's surface = (6,378,000 + 4,600,000)*(2*pi/2,360,585)^2 = 7.777565493 e-5 m/s/s The Ca's from revolution differ from each other by a factor of over 6, but they're still much much smaller than Ca from rotation. Summing or subtracting any of the last 2 values from the 1st one is not gonna make a huge difference. But now let's get the acceleration from the Moon's gravity. I copy them from post #82 so go there if u wanna see how i derived these: sub-Moon point on Earth: 3.431 e-5 m/s/s anti-Moon point on Earth: 3.211 e-5 m/s/s Low and behold we're in the same order of magnitude (10^-5). I guess I hafta thank you for that misunderstanding or i might never have calculated this. I hereby state, with reasonable certainty based on calculations, that when you google image "tide chart" and find many amplitudes that repeat a second, slightly higher peak, that higher peak represents the anti-moon bulge, and it's higher cuz of the centripetel acceleration due to Earth's revolution around the barycenter, which is over 6 times stronger on the anti-Moon point than the sub-Moon point. However, this is not the same thing as causing the anti moon bulge. It's just causing it to be higher than the near-Moon bulge. One more thing. Russ Watters one of your links led me here: http://www.lhup.edu/~dsimanek/scenario/tides101.htm On that page it says, "The moon's gravitational force on earth acts on all parts of the earth with nearly the same force, differing by only a fraction of a percent on the sides of earth nearest and farthest from the moon." But that's false. I just calculated it. Difference between them is 3.431 e-5 m/s/s and 3.211 e-5 m/s/s. Divide one by the other and you get 1.0685. This is almost 7%, and I'm afraid that's not "a fraction of a percent" as claimed in the link.