I need a mathmeticians help

[Yankee Grower]

Say I have a particle that's 15 microns in diameter and let's assume it's perfectly spherical. Then you break that particle (volume) up into particles that are 1 micron that are also perfectly spherical. What is the approximate increase in surface area going from 1 15 micron particle/sphere to multiple (amount ?) 1 micron particle/spheres that will fit into the original volume of 1 15 micron sphere? Seems like you will lose some volume with a bunch of 1 micron spheres stuffed inside a 15 micron sphere but think you get the picture. I'm looking for close but exact, supported by an explanation, would be PRIMO! If perfectly round 1 micron spheres how much volume is lost in the space between these spheres if stuffed into that original 15 micron sphere?

Yes this actually has to do with growing and a project I'm working on . I wish this could be a contest of some sort. I guess it's the challenge and honestly I'm at a bit of a loss as how to approach this one...lol. For sure a real mind bender though!

To recap I'm interested in knowing what the increase in surface area is going from 1 15 micron sphere to multiple (amount ?) 1 micron spheres that fit within that 15 micron space. My head hurts just thinking about it!

 

[gdtrfb]

42.

 

[GrouchyOldDude]

http://library.thinkquest.org/20991/geo/solids.html#spheres
Area Formula: A = 4(PI)r2

1 micron = 3.141 x 15 = 47.123
15 micron = 706.858

 

[TwoOhSix!]

Volume of 15 micron sphere = 4/3 pi r^3 = (4/3)(3.1416)(7.5^3) = 1767.15 cubic microns
Volume of 1 micron sphere = (4/3)(3.1416)(.5^3) = 0.5236 cubic microns

It depends on how the spheres are packed (http://en.wikipedia.org/wiki/Sphere_packing), but if you go with the most efficient way (hexagonal packing) it will give you 90.69% density.

90.69% of 1767.15 cubic microns = 1602.63 cubic microns
so you lose 164.52 cubic microns (the difference of the two volumes)

(1602.63 cubic microns) / (0.5236 cubic microns) = about 3061 1 micron spheres packed inside the 15 micron sphere.

To calculate the surface area increase:
Surface area of 15 micron sphere = 4 pi r^2 = 4(3.1416)(7.5^2)= 706.86 square microns
Surface area of 1 micron sphere = 4(3.1416)(0.5^2) = 3.14 square microns
3061(3.14 square microns) = 9611.54 square microns

If you count the full surface area of each 1 micron sphere inside the 15 micron sphere, you've increased your surface area by a factor of 13.6 [9611.54/706.86].

 

[Rip Bong Winkle]

You seriously can't google how to calculate surface area and do the math yourself? Are you retarded?

I'm amazed people play monkey and do his bidding for him.

 

[the_red_bull]

Well I would want to start by saying you can't take a 15... anything spherical object and break it down to 15 1...anything spheres. Sphere's don't fracture into other spheres, unless you break them down to smaller parts and reform them.

 

[Yankee Grower]

You seriously can't google how to calculate surface area and do the math yourself? Are you retarded?

I'm amazed people play monkey and do his bidding for him.

Retarded I'm not and far from it...just failed higher math in high school MISERABLY...LOL Calculating the surface area of a sphere is simple and agree. But take that 15 micron sphere and fill with X number of 1 micron spheres and figuring out the increase in surface area is a challenge. I joined a math focused forum to help figure this one out. Wish this could be a contest!

Nope...no way can figure this one out myself and thx to all helping out. For sure a mind bender!

Anyone that can help with this one is FAR from being a monkey and my hat is off to them!!!

 

[TwoOhSix!]

The concern here is sphere packing and how it would be done. Not simple surface area calculations. I assumed they would pack the most efficient way possible, but that may not be the case, there may be no real way of knowing, only guessing. What is the actual application of this scenario?

 

[netwerx]

42.

I'm with this answer.

 

[netwerx]

You seriously can't google how to calculate surface area and do the math yourself? Are you retarded?

I'm amazed people play monkey and do his bidding for him.

I don't see the reason in calling someone retarded because they need help in a subject.

Geez. I think you better have another hit.

 

[Shanti]

 

[Yankee Grower]

LOL!

After some research best I can figure, and maybe someone can check my math, is 1 15 micron sphere has a volume of 1767 microns. The surface area is about 706.858 microns. A 1 micron sphere has a volume of .52359877 microns and a surface area of 3.14159265 microns (guess that's pi). So it seems you can fit 3374.72 1 micron spheres inside 1 15 micron sphere. Then you take 3374.72 times the surface area of 1 1 micron sphere and you get about 10601.986 which you then divide by the surface area of 1 15 micron sphere and you get 14.9987. So it seems the answer I think I'm looking for is that 1 15 micron sphere filled with the same volume of 1 micron spheres has a surface area that is about 15x that of 1 15 micron sphere.

If 2 peeps are saying 42 seems like I'm way off? Like I said I suck at math and had to try and use an online calculator.

What is the actual application of this scenario?

For a nutrient type product that is sloooow release where the smaller the particle size the better and think I found better and just trying to compare and get a better understanding. It's not even about it breaking down/releasing but more about surface area but eventually it will break down.

Well I would want to start by saying you can't take a 15... anything spherical object and break it down to 15 1...anything spheres. Sphere's don't fracture into other spheres, unless you break them down to smaller parts and reform them.

Why not? If you have a ball that's 15" in diameter how many 1" balls can you fit in the same space? Then what's the difference in surface area of all those 1" balls compared to the original 15" ball? Basically what I'm trying to figure out.

My math doesn't seem to make sense cause can't see how you can fit 3374 1" balls inside 1 15" ball. I'll look at it again. I know that volume calculation, 3374 fitting into 1, isn't right because it assumes all of the original volume is available but it's not because of all the space lost between the spheres...is that true?

Thx for all the help so far!

 

[Sheriff Bart]

42 is a joke, read (or watch the BBC version of) the Hitchhikers Guide to the Galaxy.

otherwise, you are forgettin about something called UNITS
the units for distance in this case is the micron
the unit for volume, is cubed. the unit for area is the square
thats because for an area you take 1 micron x 1 micron and you get 1 sq micron
volume, 1micron x 1 micron x 1 micron = 1cubic micron
at least for a simple square and cube
the area formula earlier provided is incorrect because the 2 at the end is not superscripted meaning it is to be squared, which is should be.

 

[Strainhunter]

You seriously can't google how to calculate surface area and do the math yourself? Are you retarded?

I'm amazed people play monkey and do his bidding for him.

Dude why do you keep spreading your atrocities throughout this board?

Can you PM me a viable answer?

No wonder you have your rep disabled!

If you got nothing constructive to contribute...just keep quiet.

.

 

[Sheriff Bart]

damn im stoned you just need to read TwoOhSixs post he got it right

 

[bentom187]

well im completley hopeless when it comes to math i took algebra one 3 times in high school, so take this with a ton of salt. so one of your requirments -(Then you break that particle (volume) up into particles that are 1 micron that are also perfectly spherical.15/1=15 ,What is the approximate increase in surface area going from 1 15 micron particle/sphere to multiple (amount ?) 1 micron,) -in my eye i dont see a increase in surface erea from the original cause you say you have already divided it into 15 different sections with all that was there originally,so i think the surface erea is the wrong question if your asking for the volume.i think their is missing info here. basicly the equasion im seeing is 1/15 = 15 i think the missing question is how much volume does the 15-1 micron spheres displace, in a 15 micron big sphere. if it doesnt strike you as relivent then disreguard.

 

[MarquisBlack]

I bet you read alot of Gordon Wood, eh?

 

[Yankee Grower]

Volume of 15 micron sphere = 4/3 pi r^3 = (4/3)(3.1416)(7.5^3) = 1767.15 cubic microns
Volume of 1 micron sphere = (4/3)(3.1416)(.5^3) = 0.5236 cubic microns

It depends on how the spheres are packed (http://en.wikipedia.org/wiki/Sphere_packing), but if you go with the most efficient way (hexagonal packing) it will give you 90.69% density.

90.69% of 1767.15 cubic microns = 1602.63 cubic microns
so you lose 164.52 cubic microns (the difference of the two volumes)

(1602.63 cubic microns) / (0.5236 cubic microns) = about 3061 1 micron spheres packed inside the 15 micron sphere.

To calculate the surface area increase:
Surface area of 15 micron sphere = 4 pi r^2 = 4(3.1416)(7.5^2)= 706.86 square microns
Surface area of 1 micron sphere = 4(3.1416)(0.5^2) = 3.14 square microns
3061(3.14 square microns) = 9611.54 square microns

If you count the full surface area of each 1 micron sphere inside the 15 micron sphere, you've increased your surface area by a factor of 13.6 [9611.54/706.86].

Yeah that's it...thanx! I tried myself and came close but your explanation of the sphere packing. Kind of blew right by your post last night. After playing wit da numbers it all makes sense.

Thanx everyone for helping!!!