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Dear forum users,
Thank you for your patience with the Photo galleries. We've got your galleries online!
We have finished writing a custom script to migrate the PhotoPlog to vBulletin5’s albums.
Unfortunately V-Bulletin killed the "Photoplogs" in their software upgrade which was unforeseen and we're the first development group to have written a script for getting the galleries back... that said, it took some time to reverse engineer the code and get the albums to move over seamlessly!
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To create this migration path we used vBulletin5’s default album structure. Unfortunately, it won’t work like the “PhotoPlog” but is an album/gallery component on the forum now.
Let's start out simple and add complexity from there.
So, let's assume a perfect hemisphere (half a sphere). Surface area of a sphere is 4*PI*r^2 ~= 12.6*r^2. r = D/2 = 42in/2 = 21in. Thus, the surface area of a 42in sphere = 12.6*21^2 = 12.6*441 = 5556.6 in^2...but we want a hemisphere, so dividing by two yields 2778.3 in^2. That excludes the surface area of the floor mind you. That seems very far off of metalmaster's calculation, so I'm not sure what's going on there. Perhaps a mistake in my math!
Including the surface area of the floor (PI*r^2) would yield 2778.3 + 1385.4 = 4163.7 in^2, which is very close to metalmaster's estimate, so I expect that is approximately how he did it. People should always include their math so other people know where it came from. Just my opinion.
Alternatively, instead of using a sphere, we can calculate the surface area of an oblate ellipsoid (a vertically shortened sphere, like a discus, lentil, or GO game-piece) of the prescribed diameter and height. I don't like this kind of terminology because it suggests that circles aren't ellipses and spheres aren't ellipsoids, which is not correct; they are examples of such, but anyway...Let's calculate the surface area of an oblate ellipsoid 42in across, and, um, how high is your dome? I'll assume 18in (in between the Pompei low and high vault heights).........um, okay, I looked up the surface are of an oblate ellipsoid and decided I don't feel like calculating it. It's bonkers. Screw this, I'm moving to the next idea...
Alternatively, we can consider more precise descriptions of the dome's shape. We can account for a vertical soldier course by "building" our surface area out of two pieces. The lower piece would be the surface area of a very short cylinder, 42in diameter and soldier course height. The second piece would be the surface area of a spherical cap, which is a sphere with a planar slice through it. A hemisphere is, by definition, a spherical cap where the planar slice goes through the center of the sphere, but that plane could "slide" up the sphere resulting in a shallower cap, if you see my meaning. The question is, where should that plane go in our example. Consider that we know the height of the dome, and given the height of the soldier course, we can calculate the height of the spherical cap (= dome height minus soldier height). The diameter of the spherical cap remains our original full diameter of 42in since the soldiers don't "tilt in". I won't elaborate further here, but in short, one can calculate the surface area of a spherical cap given the measurements described so far, namely, the height and base-diameter of the cap. The spherical cap surface area equation is easy to find online. This is how I calculated the surface area of my own dome (on the exterior mind you) when estimating how to distribute my InsWool HP insulation with a relatively even thickness. I'm not going to run the numbers on your dome here and now, sorry, but you have the necessary equations so you can do it if you want.
Finally, the most precise description (assuming your dome is approximately an oblate spheroid atop a short cylinder) would be a combination of the two approaches mentioned above: a two-piece calculation consisting of a cylindrical soldier region and a oblate ellipsoid cap. I haven't even bothered to look up the surface area of an oblate ellipsoid cap. Have fun with that one!
In short, I am uncertain why my original hemispherical estimate and metalmaster's calculation were in such disagreement. The hemispherical calculation strikes me as a very good estimate since it is simultaneously too high and too low, which I would expect to cancel out to some extent. It is too high because an oven is shorter than its horizontal radius, thus shorter than a true hemisphere. It is too low because an oven rises vertically on the soldier course before tilting in.
I would love to know what metalmaster's equation was.
its really simple, i punched my numbers into the wrong spot on my sphere calculator.
you are correct sir.
my mumber is is for the total surface area of a sphere with a 42" dia which includes
the flat base area.