Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!tut.cis.ohio-state.edu!cica!gatech!uflorida!ufqtp!sutherla
From: sutherla@qtp.ufl.edu (scott sutherland)
Newsgroups: comp.sys.amiga
Subject: Weighing Platters
Keywords: HD, lubricants
Message-ID: <625@orange6.qtp.ufl.edu>
Date: 2 Aug 89 18:38:12 GMT
Reply-To: sutherla@orange6 (scott sutherland)
Distribution: na
Organization: University of Florida Quantum Theory Project
Lines: 60


Charles Brown writes that, in order to determine the amount of lubricant
on a platter, do the following:

	1) Weigh the disk.
	2) Spread the lubricant.
	3) Weigh the disk again.
	4) Leave the rest as an exercise (with a smilie :)).


Okay, let's look at this:

WARNING, THIS IS AN EXCERSIZE IN RELATIVE ORDERS OF MAGNITUDES. HIT N IF 
YOU COULD CARE LESS. (PROBABLY 99% OF YOU) ;^)


	This is a back of the envelope (literally, I used my paycheck 
envelope) caculation to get a feel for what one is in for using the above
steps to figure out how much lubricant is on a disk.

Assumptions:

	Disk is 130 mm in diameter (~ 5.25").
	Layer is 5 molecules thick.
	Molecule is a fits in a cube of volume 1 cubic nm.
	Molecular weight is 1000 g/mole.


1) The area of the plate is pi*r*r, which is ~ 500 square cm  which is 
	500e14 square nm.
2) The volume of the layer is the area times the thickness, which is 
	500e14 times 5 molecules (or 5 nm) which is 25e16 cubic nm.
3) Since 1 molecular volume is 1 cubic nm, there are 25e16 molecules.
4) The number of moles of molecule is 25e16/6e23 (dividing by Avogadro's
	number), which is ~4e-7 moles. 
5) The mass of the layer is simply the M.W. (in g/mole) times the number
	of moles (from #4), giving 4e-4 g, or 0.4 mg.
6) Let the weight of the platter be ~ 3 oz, or ~ 400 g. (I don't have any
	idea if this is a reasonable guess for the platter weight.)


7) Then the weight of the platter without lubricant is 400.0000g and 
	the weight of the platter with lubricant is 400.0004g. Thus you
	are trying to measure one part in one million by using the 
	difference between two large numbers. This is a statistical 
	nightmare! (Any statisticians care to comment?). If the error
	in your scale is 0.0002g, you have a 50% error in your 
	measurement. NOT GOOD.


So, this ain't as easy as you might have first guessed. Please note that
I am in an analytical chemistry lab, so this type of calculation haunts 
me in my dreams. I couldn't resist posting this. It was a compulsion. ;^)


Scott Sutherland

"More lasers, more fun!"

"Orders of magnitude are my life!"