Science a GoGo's Home Page Forums General Discussion General Science Discussion Forum Calculation of Surface Area

Hi

I was wondering how you go about calculating teh increase in surface area due to pitting ?

Please help

Drew

NOTE: just did a dimensional check and realized I had a problem. Below I've fixed it.

You should check my math as I'm doing this on the fly, but the basic idea is probably something like this:

Suppose pits are, on average, hemispheres (as good an assumption as any given no other knowledge).

d = density of pits (ave # per unit area)
r = ave radius of a pit

Surface area of sphere = 4 * pi * r^2
Surface area of hemisphere = 2 * pi * r^2

Area of circle = pi * r^2

For each pit, you are replacing the surface area of a circle with the surface area of a hemisphere (by our assumption).

The difference is:
2*pi*r^2 - pi*r^2 = pi*r^2

If the total (non-pitted) surface area is S,
then one estimate for the difference between them might be given by ...

pi * S * d * r^2

Make sure S, d, and r^2 are all in the same units.
(This is how I noticed my initial mistake.)

NOTE: you should still check the math.

Hi FallibleFiend,

Thanks for yur response.
No, it is not a homework assignemnt but rather a design issue.
I actually got to the 1/2* (4 pi*r^2) bit but was usure from there.
Many many thanks for your help
Drew

Hi Drew:

It really is amazing what can be done with applied mathematics. I confess I could not get off the ground with a potential answer because I would start to wonder if all the depressions werw the same diameter, the same depth, circular, dipped a little eliptically, included discrepencies of surface contour and had different shadings that might confuse my extremely sensitive equipment that was designed for the specific purpose to decipher this issue.

I am sure you will work it out. I couldn't. jw

With a little calculus (or algebra with limits) you can arrive at a similar measurement assuming elliptic pits.

I immediately took this as a fermi problem and gave a method for determining a zeroeth order estimate. If precision is desired, there are other possibilities, some of them pretty elaborate. It would probably be trivial to use a camera and a short computer program (probably less than 1000 lines of cod) to determine the density of pits on a substance, as well as the flat area at the top of the pits. To get at the depth of the pits and maybe estimate the surface area might require two cameras and a non-trivial program. I'm not sure exactly, but it might be 3 to 10K lines of code. (The geometry isn't that big a deal ... it's the pattern matching.)

An easier solution might be a laser rangefinder. I'm not sure of the specs on the things. I've seen them used on robots, but I'm not sure what kind of resolution you can get with them. I kinda suspect that if there exists one up to the task that it would be prohibitively expensive for most applications.

All that is way too much effort unless the problem solution yields dollars. For quick and easy, I'd settle for treating it like a fermi problem.

Fermi or no fermi it looks good to me.

When I was young and trying to teach myself a little workable math I focused on circles and squares. I remembered the Greeks taking the circumference of a circle bit by bit to work out a ratio value relationship to the circles diameter. I did it by putting a circle enclosed by a square. Say the square has sides of 3 inches. That was also the diameter of the circle.
I learned quite quickly that the perimeter of the square (12) equates to the cimcumference of the circle by .7854 of same. When you visualise this you can see a cube image may work quite similarly. 12 * .7854 = 9.4248 / 3= 3.1416.

This sort of thing is childs play for serious students of math but I was neither serious nor a student. jw

I should have said the simplistic method for the surface of a sphere is again a cube with an edge equal to the diameter so a side of 9 * 6 = 54 * .5236 = 28.2744. The result is the same as your 4*Pi*R^2 with R=3. Numbers are so strange. In this instance .7854 and .5236 occupy the oposite corners of the calculator pad which is meaningless? 3.1416^2 = 9.87 on the keypad at the top from right to left. Just curious.
jw