# Thread: strange request involving a CD

1. ## strange request involving a CD

I need to find the resistance of the aluminum layer in a CD for a project I'm working on. However, I don't have access to a multimeter right now, which poses a bit of a problem.

If anyone has a multimeter and some spare time, could you do this? Grab an AOL CD (or any CD you can afford to ruin) and carefully carve away enough of the plastic from the shiny side that the Al layer is exposed. You'll have to do this twice - once at each end of a diameter of the CD... so near the rim, please. Then just use the multimeter probes. I'm not even sure if this will work for measuring the R, but it's the best way I can think of right now. If anyone does this, please post the value R that you got. Thanks in advance.

edit: sufficiently careful scraping with a knife on the painted side will reveal the Al easier. Angle the knife so it doesn't dig in and scratch over it slowly.

2. Alternatively, does anyone know how to calculate the current induced in a loop WITHOUT knowing the resistance of the loop? I tried measuring the R in the Al layer today, and didn't have any luck. Here's the list of things that are known in my theoretical setup:

Area of CD (the CD is acting as the loop)
Power needed in the CD (655W)
delta T for the magnetic flux.

The CD is going to be over a solenoid. So I want to relate the current in the solenoid, number of turns and length of the solenoid to the power induced in the CD, which needs to be 655W. Then I could solve for the number of turns and amperage in the solenoid. Anyone with ideas?

3. i'm doing it now... just need some time... hold on a sec

//edit: \$\$\$\$ing leaf is a pain in the ass...

i'm getting around .65ohms, but i wouldn't take that for a constant...
I used an unburned memorex cd-r (700MB)

4. Originally posted by confuted
Alternatively, does anyone know how to calculate the current induced in a loop WITHOUT knowing the resistance of the loop?
You can calculate resistance from resistivity - here's a reference; the bottom of the page has a small inset table that includes the resistivity of aluminium (yes, that's how it's spelt on this side of the Atlantic. ) if you want to calculate it yourself but it seems that the same part of that site will do the calculation for you.

http://hyperphysics.phy-astr.gsu.edu...ric/resis.html

edit: For purposes of using resistivity calculation as noted above, the cross sectional area would be the area of your cd and the length would be the thickness of the Al layer.

5. But... but... is it still pronounced the same?

6. no - it is pronounced al-loo-min-ee-um

heh

7. I was afraid of that.

"Aluminium foil" just has a really odd sound to it.

8. Bleedin' septics wiv yer saaaaaaeeeeeeeeeedwalks and your tomaeeeedoes

It's called English, that means I'm right and you're wrong! (Or if you asked my old English teachers they'd probably say it's more likely we're both wrong!)

9. Since ENGLISH was first, and US butchered it later on, Aluminium should be considered 'technically' correct.

Colour and Faeries and Gaols oh my.

10. maybe the person that came up with the name should know?

11. After a fruitful search on google I have found this:

In 1807, Sir Humphry Davy (1778-1829) proposed the name alumium for this still undiscovered metal and later agreed to change it to aluminum. Shortly thereafter, the name aluminium was adopted by IUPAC to conform with the -ium ending of most elements.
It does go on to say that the American Chemical Society changed it to aluminum in 1925. But I care as much about that as I do about the ANSI standard. It's all about the International standards.

Therefore Aluminium is the correct spelling and pronunciation.

12. [celebration dance]
yay!
[/celebration dance]

13. the grammar police save the day!

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