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# Thread: Mathematical flattening of a sphere

1. ## Mathematical flattening of a sphere

If i were to take a sphere and flatten it perfectly, is there a forumla to calculate the resultant area and circumference that is produced by the disc that remains.

in other words if i took a perfectly spherical ball of putty and flattened it perfectly from the top with a plate, the only thing that would remain would be a putty disc....i need to calculate the resultant circumference of the putty disc given only the dimensions of the original sphere.

there has got to be someone on this forum that can help....its driving me up the wall, can't find the answer anywhere on the net

2. That would be kinda difficult, because if you press harder the putty would just widen out more... would it not? From a realistic standpoint of course.

3. I guess it would depend on the chemical properties of the sphere itself. A sphere made from putty will more than likely produce a different disc to a sphere made from steel, due to elasticity, tension, cohesion etc.

4. why dont you just calculate the surface area of the sphere 4 x Pi x r^2

5. Originally Posted by The_Pumpkin_King
why dont you just calculate the surface area of the sphere 4 x Pi x r^2
The surface area of the sphere would be for the entire sphere. He wants to calculate the area of the sphere as viewed from above it.

6. Easy solution: Calculate total surface area. Surface of a circle = Pi(r)^2, right?
So total surface area of sphere = Pi(r)^2 and you can work out r, thus getting the size of the circle. 2Pi(r) = circumference. Problem solved.
That's the simple solution.
More complex...well, they'd need to specify the thickness of the resultant disc. If they don't, then the disc could have a thickness approaching zero, which means the radius would approach infinity, as the volume of the sphere would need to be the same as that of the circle. Unless the sphere is made up of a specific compound. Then you could use the height of the smallest element of said compound and use that as a height.

I dunno...*shrug*

7. Originally Posted by Smooth Criminal
The surface area of the sphere would be for the entire sphere. He wants to calculate the area of the sphere as viewed from above it.
Umm...Surface area of sphere as viewed from above? That's exactly the same as viewing a circle from above, isn't it? So just use the sphere's radius to do the math as a circle?

8. Surface area != area.

Area = 2D plane.
Surface area = 3D plane.
Sphere = 3D object.
Originally Posted by Messugga
Umm...Surface area of sphere as viewed from above? That's exactly the same as viewing a circle from above, isn't it? So just use the sphere's radius to do the math as a circle?
Yeah but it's not exactly that - he wants the area of the resultant object if the sphere is flattened.

9. then he needs the volume and density of the sphere

10. Originally Posted by Smooth Criminal
Surface area != area.

Area = 2D plane.
Surface area = 3D plane.
Sphere = 3D object.

Yeah but it's not exactly that - he wants the area of the resultant object if the sphere is flattened.
Meh, then I misunderstood his question. I think a rephrase from the OP's side and some sleep from my side, may aid in the solution of this little problem

Edit - Looking at it again, he says "flatten it perfectly". So he wants to turn a 3D object into a 2D object. However, since he mentions a real substance - the putty - which most certainly has a minimum thickness greater than 0mm, the resulting disc will have a height and thus be 3D. This disc's volume just has to be the same as the original sphere's while not having a thickness less than the minimum thickness of the material it is constructed from. The area can then be calculated as viewed from above, like one would do with a circle.

Edit - In my first post I suggested using surface area. This is incorrect. Volume would be the important factor to keep constant here...

11. A bubble or sphere? If it is a sphere... Then the qeustion is how* flat.* The thing is if you work in 2d like a bubble's flat round surface, it already has a area and depth, you can lay it open flat. Now you ask for putty sphere then you must say how flat do you want to go and also how flat can you go. So it is simple, restate your qeustion or add 'how flat do you want to go' to your math? you can do this by filling your sphere with millions of tiny spheres and then adding the empty space that goes with thm

12. Firstly, in order to be mathematically correct, are you actually working with a sphere? Is it a perfectly round ball? If not, you're working with an ellipsoid.

Secondly, do you simply want to take the cross-section of the sphere and get the resultant circumference of the shape that remains, or do you want to physically squash the sphere and flatten it out? If it is the latter, then there is no 'easy' way of solving your problem. That is to say, not without some integration and very precise measurements. There is unfortunately no single, simple formula you can use.

13. Provided the volume stays constant then as suggested above, 4/3 Pi r^3 (volume of a sphere with radius r) = Pi R^2 h (volume of a disk with radius R and thickness h). The circumference of the disk would have a length of 2 Pi R.

If you are transforming the sphere into an ellipsoid or an oblate spheroid then you could probably use some of these formulae ...
http://home.att.net/~numericana/answ....htm#ovalsolid

14. Wouldn't the volume and density remain the same no matter if it is square, round or flat? You aren't adding anything or taking anything away, just changing it's shape. And if you take a round ball of prestik and hammer it as flat as you can it doesn't change all that much. It might stretch yes but we can't assume this sphere in the OP stretches.

I would say the area/volume would remain the same. If flattened perfectly round the 3D circumference would also stay the same (more flat but thinner sides cancelling it out for lack of better explanation).

15. i've subsequently realised what i asked for can't be done, converting 3d to 2d.....

i initially calculated the volume of the sphere and then used it in the volume of a cylinder and reduced the height component of the formula close to zero....all that would happen as stated above is that it would move the resultant value to infinity....it all depends on the height i wanted the resultant disc at then it can be done.

i was just trying to see if there is a way to confirm something i read relating to the ratio of the circumference of a sphere to the circumference of the resultant disc produced when the sphere is flattened perfectly.

apparently what i read....was a [email protected]

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