Surface Area of a Cylinder

Hi guys/gals, I need some help with this.

I know how to find the surface area of a cylinder, but how do I compute it, while taking thickness of the material used into consideration?


I have an engineering project to do, and I have to come up with the dimensions of the most cost effective cylinder that will hold 250CC of fluid.

The thickness of the material is required to be .01cm.


I dont know how to figure it out.

 

Do not know if this helps but 1 CC is equal to 1 ML which is equal to 0.06102 Cu. Inches....so you would need a container that is 15.255 Cu. Inches. A standard Aluminum Pop can is 21.357 Cu. Inches, those minnie Pop Cans I believe are 250ML/CC Volume = πr²h were r = Radius & h = Length

 

No, doesn't help

Thanks though

I'm graphing my results, and there are a ton of other factors too, that I didn't mention. Its just this one is hanging me.

 

You're looking for a cylindrical container whose internal dimensions provide the least possible surface area while still containing a volume of 250cc. So that means there is one combination of height and diameter that requires less material than any other, if that is the definition of cost-effective. I suppose the thickness could be considered after the fact, adding it to the previously determined dimensions to come up with the outer surface area, if that's what the problem is asking you to find. Or are there complexities involved not evident (to me at least) in your original post?

 

I just need to know how to calculate the material area.

There are quite a few factors, price wise, but thats the only hang up.

In other words, I will be calculating many different possibilities, I just need the know the general form, to include thickness.

 

Surface Area of a cylinder = 2πr(r+h) then times the thickness of material chosen, but again I may be missing the point.LOL

 

Ill take a look at that, thanks

 

ummm well its kinda hard without having a pen and paper but ill try to offer some advice though you prolly know a lot more than me lol.

looking from the top of a cylinder; take the area of the top (a circle) using the outside as the diameter. subtract the inner area using the diameter to the edge of the inside of the cylinder (which compensates for the thickness of the material). im forgetting the formula at the moment but then just find the outside SA of the can or whatever.

maybe im missing the point here (i probably am) lol cause it seems pretty simple

here is a crappy pic of what im talkin about
http://i62.photobucket.com/albums/h98/viper_boy403/can.gif

then multiply that by 2 and use BC's formula

 

2(pi)(r)(L)

the circumference of the cross-sectional circle multiplied by the length of the cylinder.

 

I appreciate the help everyone, I got it handled, and finished

I appreciate it.

wolf...where is the thickness component I mentioned?

 

it's easy, no?
you have a few equations, say r is radius in cm, h is height in cm

(PI)(r-0.01)^2(h-0.02)=250

and graph that. now with that data calculate the SA(3 variable graphing will help, if you can) say that 2(PI)r^2+h(2(PI)r)=SA, and find the lowest possible SA.

was that what you did/what did you do? I'm interested now

 

I'll attach the spreadsheet in a few minutes