[Guest]

I have a sphere that was 10" in diamater and I wanted to drill a hole to the center. Being round it was hard to keep the 1/2" drill bit from slipping, so I decided to cut off a the top leaving a 3" diamater flat side to center my drill bit. I drilled all the way to the center and decided to do the same thing 90 degrees from the top. I cut off another 3" diamater flat spot and proceeded to drill to the center making a perfect 90 degree angle between both drill holes.

What is the volume left in my no longer sphere?

For fun:

If I repeated this to obtain a 6 flat sided (like a dice with round parts) what would the volume be then?

What is the volume an end cap if I drilled a hole through the center of it?

[Guest]

Volume of sorta sphere 507.43

Volume of sorta sphere with 6 flat edges and drilled holes 474.98

Volume of cap with hole in middle 6.53

[Guest]

Ah much better representation, and more data to process. It is now solvable; but matthew i feel needs to reiterate his answer, I don't know what the hell he's saying xD.

[Guest]

[i don't have permission to edit my own post?!)

(Pi is used in terms of 3.14)

My work: V = 4/3[Pi]r^3 -yields-> 523.3{repeten} = Volume of the sphere.

Surface area of the shaving = 1.5^2(Pi) = 7.065

As I am simply using a normal comp calculator; I don't think i can find the Volume of the shaven area (which would be doubled because of the 2 sides). I can't really figure out much more information, except that the most likely way to find the answer would be to find the volume of the drill beneath the shaving and exclude the area that the second drill will go through, and double that (and hold it for later). And then find the volume of the middle by changing the idea of the shape into a cylinder with a 45 degree angle cut through from the bottom (god if i had prodesktop or something like that i could make this xD), and calculate the volume of that (theres an equation for that somewhere), double it and add it to the doubled drill cylinder volume, and then add that to the doubled shaving volume. Ultimately the final equation would look like this (with variables):

2a + 2b + 2c = d --> V(of the final holed sphere) = v-d

Well I'm just a Senior in Highschool so i wouldn't put 100% trust into my own methods. Hope this work helps, but as for the my lack of devices that could solve for this I'd like to apologize Dx.

[Guest]

                Diameter   Volume	Radius

         Sphere	10	   523.60	5

     Drill Hole	0.5	   0.91	    	0.25

        End Cap	3	   0.82	    	1.5

   Slice Height	0.23

The slice height is calculated using the Pythagorean Theorem; nothing fancy there.

I am assuming there is an almost spherical hole in the very center of the big sphere caused by the intersection of the drilled cylinders.

The drill hole volume is calculated as a cylinder from the sliced off face to the center "hole" (which I calculate as approximately .07). The center "hole" is not exactly a sphere but it's darn close to one so I calculate it that way.

The volume of the end cap is calculated as: PI R² x + ²/₃ PI R³ - ¹/₃ PI x³, where x is based off the slice height. x is -R + the slice height or about -4.77.

So the answer (according to my work here) is the sphere volume minus the center "hole" minus 6 drill holes minus 6 end caps: 513.14.

I don't know if that's right. Feel free to point out any mistakes I've made. I'm confident with most of it but not all of it.

[plainglazed]

If we locate the two slices of sphere and the drilled holes 180 degrees instead of 90 degrees apart, we can break the problem down into three simple volumes and not have to deal with the volumes of the slices themselves: a cylinder of diameter 3" and height = h; a three dimensional "ring" shape of height = h, and a hole of diameter 1/2" and height = h. The height h = 2 √(5²-1.5²). The volume of the ring = 4/3πh³ (

), the volume of the cylinder is π1.5²h, and the volume of the hole is π.25²h.

Edited October 17, 2009 by plainglazed