Flat Earth Self-Debunk: Danyang–Kunshan Grand Bridge meme shows Hangzhou Bay Bridge


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Deconstructed - Auguste Piccard: “It seemed a flat disk with upturned edge”

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Deconstructed - Auguste Piccard: “It seemed a flat disk with upturned edge” Is it possible he means it looked like an upturned disk?  As in, you see a bit of horizon curvature? Stratobowl image from 1935 The human eye has a wide field of view so you see a bit more of the Horizon Circle than most images and thus more curvature, all else being equal. One reason for this is that this is NOT the "curvature of the Earth" -- this is the curvature of the Horizon Sagitta viewed nearly on edge .  This is a mistake I constantly see people making.  So it's not a circle of 3959 miles diameter that curved downward, but an OVAL that you are in the middle of with, in this case, a 300 mile radius and viewed on edge at a 4.8° angle -- and this oval curves 360° around YOU. As to why it looks flat, it's because the terrain is far away and the curvature is fairly slight and you're looking down on it.  We just cannot see the geometry under these conditions.  You can't even see the...
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Resources on Coriolis forces use in external ballistics

Over and over and over and over and over again the experts tell you that at VERY LONG ranges the Coriolis and Eötvös effects matter and are used in the calculations used to aim ballistic artillery and rifle fire... to deny the VAST body of evidence on the word of some anonymous & unverified source is utterly ridiculous and just shows profound epistemic bias. Application to Long-Range Artillery - Shelling Paris in WW1 Military: THE FIELD ARTILLERY JOURNAL (1918) Fire Control Fundamentals (287389 O-54-4) Long Range Shooting: External Ballistics – The Coriolis Effect Earth's Curvature and Battleship Gunnery (Blog) US Navy: OP 770 -- RANGE TABLES FOR 16"/50 CALIBER GUN 4DOF™ Ballistic Calculator
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How much should the moon appear to shift between two positions on Earth?

Let's say we have two people viewing the moon who are 10300km apart along the same line of latitude at the same time. How much of a shift against the more distant background stars should the moon appear to shift? This approximation assumes that the sublunar point is roughly between the two observers. Well, first we need to find their actual linear distance (rather than the distance over the curvature of the Earth, which is what you get from Google Earth). Let's define our variables: \[ R = 6371393 m \;\; \text{Earth Average Radius} \] \[ d = 10300000 m \;\; \text{Earth distance along curvature} \] We can find the angle in radians from the arc length using: \[ \theta = d/R \] and we can find the length of a chord using the half-sine rule: \[ \text{crd} \, \theta = 2 \sin \frac{\theta}{2} \] Plugging that in we find that the straight-line distance is slightly shorter than around the curvature: \[ 6371393 \times 2 \sin(\frac{1}{2} \frac{10300000}{6371393}) \approx 9214000m \] Nex...
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