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...
I have some new toys to share in this installment of 'Yes, the Earth really is Curved even though you can see one distant mountain peak from another', so this should be fun! This time we're going to look at a view of two distant mountains from Pic de Finestrelles (2826m) in the Pyrénées , taken by Marc Bret of Beyond Horizons (see also the Flickr album ). Pic Gaspard (3880m) in the Massif des Écrins range at a distance of 443 km. Grand Ferrand (2758m) at a distance of 392.48 km. Our view is right around 42.414466°N, 2.132839°E at about 2826 meters elevation, looking right along the coast. Pic Gaspard/Grand Ferrand from Pic de Finestrelles in the Pyrénées, image by Marc Bret The EXIF metadata shows this image was taken by a Panasonic DMC-FZ72 with a focal length of 215mm -- given the 5.62 crop factor of the 1/2.3" sensor in this camera, this gives you a 35mm equivalent focal length of 1200mm giving a full frame view of 1.644° wide and 1.215° high. Since 1521 x 101...
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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