Pythagoras2
- Thread starter empip
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Is this it
http://www.math.umn.edu/~nykamp/m2374/readings/detareavol/index.html
Edit
I think this a better site.
http://www.shodor.org/interactivate/textbooks/chapter/18/659/
http://www.math.umn.edu/~nykamp/m2374/readings/detareavol/index.html
Edit
I think this a better site.
http://www.shodor.org/interactivate/textbooks/chapter/18/659/
Depends on the size of the sphere, I think.....
Surface area of a sphere = 4 pi r^2
Surface area of cylinder = pi r^2 h = pi r^2 2r (in this case)
So the sphere's surface area is 2/r times larger than that of the cylinder....
(EDIT: Toasty's answer reveals what I have just written to be complete nonsense - please ignore it)
Surface area of a sphere = 4 pi r^2
Surface area of cylinder = pi r^2 h = pi r^2 2r (in this case)
So the sphere's surface area is 2/r times larger than that of the cylinder....
(EDIT: Toasty's answer reveals what I have just written to be complete nonsense - please ignore it)
T
toasty
Interesting,
So the surface area of the sphere is 4.pi.r^2
The area of the cylinder's curved surface is height.2.pi.r
but height = 2r so the area is 4.pi.r^2
So they are the same!!
Nice puzzle, I never realised that.
-Dan
So the surface area of the sphere is 4.pi.r^2
The area of the cylinder's curved surface is height.2.pi.r
but height = 2r so the area is 4.pi.r^2
So they are the same!!
Nice puzzle, I never realised that.
-Dan
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Blimey DiyIF bit complex that !
Here is a calc'. http://www.1728.com/diam.htm
More to come Toasty..
Cylinder circumference = Pi x diameter
Sphere diam = 2 x radius = 2r = Cylinder diam = cylinder height (h)
Area inner face of cylinder Pi x d x h
In terms of the sphere Pi x 2r x 2r = 4.Pi.r² OR pi.d²
The area of the sphere and the cylinder are equal.
Take the 'contact' to cylinder, line (circle) around the sphere as the 'equator' ... any cut parallel to this marks a zone of the sphere, this area is equal to the corresponding part of the cylinder.
See the red radius line?
What if the Earth were a perfect sphere and the angle from the equator was about 23.5° (North or up) The Sine of 23.5° is approx 0.3987 So the area of the sphere within the zone would be 0.3987 of the total area.
I think that might be close to a tropical zone maybe Tropic of Cancer or Northern tropic...
Here is a calc'. http://www.1728.com/diam.htm
More to come Toasty..
Cylinder circumference = Pi x diameter
Sphere diam = 2 x radius = 2r = Cylinder diam = cylinder height (h)
Area inner face of cylinder Pi x d x h
In terms of the sphere Pi x 2r x 2r = 4.Pi.r² OR pi.d²
The area of the sphere and the cylinder are equal.
Take the 'contact' to cylinder, line (circle) around the sphere as the 'equator' ... any cut parallel to this marks a zone of the sphere, this area is equal to the corresponding part of the cylinder.
See the red radius line?
What if the Earth were a perfect sphere and the angle from the equator was about 23.5° (North or up) The Sine of 23.5° is approx 0.3987 So the area of the sphere within the zone would be 0.3987 of the total area.
I think that might be close to a tropical zone maybe Tropic of Cancer or Northern tropic...
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