I want to welcome a new member!

[Joe Matt]

You never thought I would do this, but, I want to welcome Dante Sparda. So send him a PM and ask him questions about the legality of sponge, how many 1/4 inch holes you need to cut off 1 lb of you bot, and other FIRST stuff.

[FotoPlasma]

(1/4in)/2 = 1/8in
pi(1/8in)^2 = 0.049087385212341in^2
0.049087385212341in^2 * .125in = 0.0061359231515426in^3
1/0.0061359231515426in^3 = 162.9746617261 quarter inch holes per in^3
(I love my TI-89)
solve(((162.9746617261holes)/(0.098lb/in^3))=((x)/(1)),x)
According to my calculations, 1663.0067523071 holes per lb of aluminum...

If any of that looks wrong, please alert me to it immediately...

I hope that's right... it's kinda late...

[Stephanie]

heh, i coulda sworn that there were already a few posts on here during the build season about the number of holes per lb of aluminum...
good work, jim. i'm too lazy to actually *check* your calculations, but 'll trust you on it ;)

[Harrison]

I am moderately frightend by that math...

*shudders at math* lol

[chellyzee93]

So many numbers!!! ahhhhhhhhh~!
*runs around screaming* AHHHHHHHHHHHHHHHHH!
*finds a book*...hehehehe... no more numbers

[Madison]

Originally posted by FotoPlasma
(1/4in)/2 = 1/8in
pi(1/8in)^2 = 0.049087385212341in^2
0.049087385212341in^2 * .125in = 0.0061359231515426in^3
1/0.0061359231515426in^3 = 162.9746617261 quarter inch holes per in^3
(I love my TI-89)
solve(((162.9746617261holes)/(0.098lb/in^3))=((x)/(1)),x)
According to my calculations, 1663.0067523071 holes per lb of aluminum...

If any of that looks wrong, please alert me to it immediately...

I hope that's right... it's kinda late...

Two things, I think.

First, wouldn't it have been easier to find the weight of one 1/4" dia. x 1/8" thick piece of aluminum, and divide 1 lb. by that number? It seems like there's less math involved, and it omits having to make assumptions about the number of holes per in^3.

Therein, I think, lies the second problem. First, the number of holes should be measure in square inches, rather than cubic inches, because we're assuming that each hole goes entirely through the thickness of the aluminum. Second, while the combined area of 162... holes may approximate 1 in^2, the reality is that it may be a better approximation to treat each hole as a slightly larger square, thus insuring that you can properly punch the specified number of holes, and also providing you with a chance at having a piece of aluminum left when you're done.

If that doesn't make sense, I can try to clarify further :)

EDIT: Oh oh oh. I see what you did. It was just backward from what I did, kinda, but it's the same in the end.

Duh. Still, for what it's worth, assuming a 1/16" sliver of clearance on each side of the holes, you can fit 7 hole per square inch. 1600+ holes are required to lose one pound. That'd mean you'd have to continuously and uninterruptedly hole up 19.41 square feet of aluminum to lose 1 pound :)

[FotoPlasma]

M:

The only assumption I made was that you were drilling in 1/8" thick aluminum. It matters not to me what the 1663.0067523071 holes be in, only the fact that it takes that many to equal a pound.

You probably put more thought into the question than I did, but I was only concerned with the amount of holes, not where they are/were/would be...

Anyway... that was a completely useless question to answer in the first place, but fun, nonetheless...

/me chuckles...

[Scorpion515]

dude, that's way way way way to many numbers! ahhhhhhhhhhh!!!!!!!!!!!! lol :D

[Wetzel]

In 2000, we drilled over 2200 holes in the FRAME of the robot with a drill affectionaly known as Halfie...

Erik counted them once. Why, I don't know.
Dave, correct me if I am wrong.


Wetzel
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E:\Anabolic Frolic\Happy 2b Hardcore Chapter Six\Anabolic Frolic - 14 - Come Together.mp3

[Mike Schroeder]

I cant count past 11 cause then i forget what i am doing

[Dante Sparda]

Actually, with the 1/4 scale blx or whatever, you would just put the equation of 1232.0099388212/883.2993807736 sqrrt34239.233 • 233882.394873 and that would give you an answer of about 2388992378.30988483 rounded off or so, and that would be about enough to fit the requirements.

[FotoPlasma]

Originally posted by Dante Sparda
Actually, with the 1/4 scale blx or whatever, you would just put the equation of 1232.0099388212/883.2993807736 sqrrt34239.233 • 233882.394873 and that would give you an answer of about 2388992378.30988483 rounded off or so, and that would be about enough to fit the requirements.

I don't get it...
Are you joking, or something?

[Dante Sparda]

Uh.....No.
;)