Length of coiled metal...

[smcadman]

I need help in finding a formula to calculate the length of coiled metal. It is .031 painted steel (used for roofing). Anyway, the coil is 16" I.D., 24" O.D.
Here is one way I tried this, but I would like a more accurate formula:
16+24=40
40/2=20 (median)
20*pi=62.83 in.
4"/.031=129 "revolutions" (if you will)
129*62.83=8,105.07"
8,105.07"/12"=675.42 ft.

Any better suggestions?

Flores

 

[MechEng13]

Flores,

The way I figure the length of coiled material is by using a wheel supplied by a Feed System supplier. It takes the ID/OD and gives you a lbs/in width. You take that number and using the strip width will give you the coil weight. Once you set lbs/in width to find total weight you read off the length that matches up to your thickness...

in your case it would be
~70 lbs/in width
assuming 4" wide
~280 lbs total weight
~650 ft

of course this isn't exact



Alan M. Etzkorn
Manufacturing Engineer
Hoffco/Comet Industries Inc.

http://www.hoffcocomet.com

 

[NickE]

Mecheng13-- yeah in teh steel industry thats what we do.

Flores -- try asking your supplier if you can have a circle calculator, many suppliers give them out as advertising.

nick


Nick
I love materials science!

 

[metman]

Flores,
I also used to do it your way for rough estimate but after reading Alan's comments, the following would seem an accurate calculation:

V = Volume
W = Width
L = Length
t = thickness
D = O.D.
d = I.D.
A = Area

V = AW = pi/4(D^2-d^2)W
V = LtW

LtW = pi/4(D^2-d^2)W
Lt = pi/4(D^2-d^2)

L = pi/4t(D^2-d^2)= 3.14/4*.031(24^2-16^2) = 8107.335882

8107.335882/12 = 675.6 ft

Huh! Looks like the not much dfference either way.

Since you have painted material this eliminates confusion of dual densities. Volume is volume.

Hope this helps.

Jesus is THE life,
Leonard

 

[smcadman]

Thanks for the replys everyone. I was hoping someone would supply a formula.

Flores

 

[MHWood]

If you want to supply your emal address to me, I will send you a copy of the spreadsheet I use to calculate coil length.

The calculation requires:
Coil I.D. & O.D.
Coil width
Material Gauge, and
type of material (for density)
The spreadsheet basically calculates the PIW and from that, the weight of the coil. The last cell of the spreadsheet uses this weight to calculate the coil length. For your example, I get a length of 675.6 ft.
Mike

 

[MechEng13]

MHWood,
I would be interested in that spreadsheet. My email address is in my profile.



Alan M. Etzkorn
Manufacturing Engineer
Hoffco/Comet Industries Inc.

http://www.hoffcocomet.com

 

[smcadman]

MHWood,
I am interested in your spreadsheet.
Send it to: sitecomments@sbcglobal"dot"net
(replace the "dot" with a ., do not want any email harvesters/spiders sending me spam)

Thanks in advance,
Flores

 

[etch]

please send also to Eric.clement@btinternet.com

 

[cdvk]

This may help some of you:

http://www.foilcenter.com/utilities/index.cfm

(online calculator)

 

[aluminumphil]

Here is a simple formula:

L=t/(pi(R^2-r^2))

where:
L = strip length
t = strip thickness (including paint)
R = OD/2
r = ID/2

This formula assumes that the product is tightly wound and the shape is good. It will give you a quick approximation.

Phil

 

[Christian62]

Hmm! .. aluminumphil!

There must be a mistake in the formula or my calculator is broken?

The above coil would give:
R = 12"
r = 8"
t = 0.031"

L = 0.031 / pi * (12²-8²)
L = 1.2 e-4 ???

May I suggest this:

L = pi * (R²-r²)/t

here you will get
L = 8107 " or 676 ft !

Chris

 

[metman]

Chris,
You have it right which is the same formula I derived on june 4th except I used diameter instead of radius. However my mathamatical notation was incorrect: I showed it as:

L = pi/4t(D^2-d^2)

It should read L = (pi/4t)(D^2-d^2) or
L = pi*(D^2-d^2)/4t

Actually this is more accurate than the method using density or wieght if the materail is coated like with paint in this case since this is a volumetric consideration rather than mass.

Jesus is THE life,
Leonard

 

[Maui]

Smcadman, there is an easier way. I assume that you can measure the total weight of the coil. If we define the following terms,

L = total length of coiled steel wire or strip in feet
G = total weight of coil in lbs
A = cross sectional area of the wire or strip in inches^2
rho = density of the steel in lbs/inches^3

The weight per foot of the steel wire or strip is equal to

weight per foot = A*rho*(12 in/ft)

As an example, if you have steel wire that is 0.031" in diameter, then the weight per foot is

weight per foot = pi*[(0.031"/2)^2]*(.283lbs/in^3)*(12in/ft)
= 0.00256 lbs/ft

If you have steel strip, then the calculation for the area will be different, namely the width times the thickness. Everything else in the above equation will be the same. After you calculate the weight per foot, divide this into the total weight of the coil to find the actual length.

L = G/(weight per foot)

This should give you an accurate answer. Note that I have ignored the weight of the paint on the steel. You can account for it if you like by altering the calculations appropriately.


Maui

 

[JimMetalsCeramics]

I ran it on a spreadsheet. The length is between 674 and 680 ft. long.

In Excel, put in 16" then each successive line add 0.062". The next collumn over run the first collumn multiplied by pi. Then add the right hand collumn. Assuming a very tight coil without burrs between layers, this is what you'd calculate.