A little math help, please ...

I wonder if someone would be so kind as to double-check my math.

You have a disk, 100mm in diameter and 10mm thick. The disk has a 10 mil coating. What is the volume of the coating?

A mil is 0.0254mm, so 10 mil is .254mm. Taking this away from the original volume we have a disk 99.492mm in diameter and 9.492mm thick.

The surface area of that disk would be 2*(PI*radius2)+(PI*diameter*thickness)=18515.627mm2

Now I multiply that by the 10 mil (.254mm) and I get a volume of 4702.9692mm3

But, if I take the volume of the original 100mm disk (78539.816mm3) and subtract the volume of the disk without the coating (73780.568mm3) I get 4759.2479mm3.

Which way is correct? Are they both wrong? The figure seems kind of high to me either way.

reply to post by VictorVonDoom


If your gonna coat the disk take the larger, so if your math is wrong it will still be enough!
Hope it helped.
Got you a star, for yjr hassle.

do your own homework.

It's been many a decade since I've done homework. I was just trying to figure out a way to determine how much coating would be on a disk given the size of the disk and the thickness of the coating. It seems to me that either approach should work, but they give different answers and in both cases the volume seems kind of high to me.

reply to post by VictorVonDoom


The best way to approach this is:

to take the volume of the disk (your first measurement given) = Vd

add measurement of coating to original disk measurement and recalculate volume = Vt

Vt - Vd = Vc (volume of coating)

And I think he is doing his own homework, just needs input.

You are better off working with volume than surface area. After all, the answer you want is a volume.

Providing that the final dimensions (after the coating) of the disk are 100mm x 10mm, this is my working:

Area of top (or bottom) of disk 100mm x 10mm
= pi * (R^2)
= pi * 50 * 50
= 7853.9816 mm2

volume of disk
= area * depth
= 7853.9816 * 10
= 78539.816 mm3


Area of top (or bottom) of disk 99.492mm x 9.492mm
= pi * (R^2)
= pi * 49.746 * 49.746
= 7774.3878 mm2

volume of disk
= area * depth
= 7774.3878 * 9.492
= 73794.4896 mm3



Volume of 10mil coating
= V1 - V2
= 78539.816 - 73794.4896
= 4745.3264 mm3


.

Oh and that answer may seem large, but when converted it is only 4.74 millilitres.

V = (pi * r^2)*h

r1 = 50mm
h = 10mm
so V1 = (pi * 50^2) * 10


V2 = (pi * (50+2(mil))^2)*(h+2(mil))
= (pi * 50.508^2)*10.508

i can't be bothered calculating it, but the volume of the coating is just V2 - V1

errrr i read wrong, i thought you were adding the coating, so replace all those +2mil with -2mil and you should have what the above poster has, i.e. the correct answer

also: can i suggest plugging this into excel? that way you can google the appropriate formulae, and then plug in the variables, and you'll have all the answers you need.

reply to post by VictorVonDoom


1. Calculate the volume of the inner disk.
2. Calculate the volume of the whole thing.
3. Subtract the first value from the second.

reply to post by VictorVonDoom


The correct answer for the coating volume (assuming the disk is uniformly coated on all surfaces) is 4,745.327 mm^3 as calculated below.

Best regards,
Z

[atsimg]http://files.abovetopsecret.com/images/member/1937299e2cf8.gif[/atsimg]

How can you guys speak in riddles such as this and understand one another,

yet on this very same site

there are sooooo many people who can't work out the simplest of things?

This mathcake is a lie.

reply to post by VictorVonDoom

I`m only using pi as 3.142 as I dont have it on my calculator,just an office type.

And I know its the same thing but I looked at it as a cylinder not a disc.

Volume of a cylinder
formula =pi x r2 x hieght

volume of the inner cylinder was 73804.04mm

volume of the total cylinder (including the coating) was 78550mm

Minus the inner from the total =4745.98mm volume of the coating

Had fun though cheers.

Edit to add............

Duh I`ve had a computer for 17 odd years and just remembered its got a calculator on it

So same process using the pi key I got 4745.3272mm

("pi key" lol any English about?)

reply to post by badw0lf

We are party to esoteric knowledge that is dangerous in the wrong hands.

Such as, for example, eighth-grade arithmetic.

I appreciate all the replies, thank you. After getting a little sleep, I realized the problem with the surface area approach. Since a picture is worth 1000 words:




Anyway, it was probably close enough considering the volume I was getting. I was using a spreadsheet for the formulas. I love working with spreadsheets. What I was acutally working on was a way to calculate how much gold was on a gold-plated coin given the size of the coin and the thickness of the plating. Given a coin 39mm in diameter and 2.41mm thick plated with 10 mils of gold, my spreadsheet shows that the coin would have 12.8605 grams of gold. That seems like quite a lot. With the spot price of a gram of gold is $49.45, the coin would have $635.95 worth of gold on it, but that can't be right.