Weird question I think but I Googled...

and I could not find the answer.

How many feet/meters in a pound/kg of 1.75mm and 3mm filament of the various kinds?

There is an online calculator somewhere I have forgotten but it is easy to calculate from the density of the plastic, which can be Googled.

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[www.hydraraptor.blogspot.com]

for 3mm its approx 90 meters per kg but depends a little on the filament.

you can calculate it fairly easy tho
calculate the volume of the filament per meter and multiply it by density of PLA or ABS that will give you a weight per meter

Lost me.

Density for PLA is 1.23 - 1.25 g/cm³ and the volume is what?

[www.quabbin.com]
[www.timbercon.com]

The problem is that none give you the dimensions of the spools. :/

Edited 1 time(s). Last edit at 07/21/2012 07:06AM by Dark Alchemist.

You can use the equation:

Volume = mass / density

Also understanding that

Volume = Area x Length

The equation becomes:

Length = mass / ( density x pi x ( diameter / 2 )^2 )

This only gives you a very rough theoretical guide, the actual products will vary. Also note that density is in kg/m^3, Volume is in m^3, diameter is in metres, and mass is kg

Edited 1 time(s). Last edit at 07/21/2012 07:13AM by MacAleJam.

Yeah, cool beans cause all of the formulas I found the missing part of the equation is how many wraps per layer and how many layers. Ahem, isn't that easily figured via math? [www.ehow.com] that link made my head swim.

Hmmmm, could you check this formula? 1kg 3D Printer PLA Filament 3mm. PLA is 1240 kg/m3 diameter of the filament is .003m mass is 1kg pi is 3.1415926535897 and I came up with 0.02928701

Edited 1 time(s). Last edit at 07/21/2012 07:44AM by Dark Alchemist.

i get 114meters

but you have to take in account that every spool is different. Al manufactures say they get 3.0 +-0.1 mm so a spool diameter maybe 3.1 or 2.9 then you get a length between 122.1 m and 106.8 m

Edited 1 time(s). Last edit at 07/21/2012 01:04PM by Weedz.

With:
mass = 1kg
Diameter = 0.003 m
Density = 1240 kg / m^3

The formula would become Length = 1 kg / ( 1240kgm^-3 * 3.14159265 * ( 0.003m / 2 )^2 ) = 114.0895649404 metres, as Weedz has pointed out.

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For Coils around Spool
The Figure below shows a cross section of a spool with some filament wrapped around it (represented by blue circles).

www.flickr.com

Reel Cross Section by MacAleJam, on Flickr

For demonstration purposes assume that:

• H_SPOOL = Spool wrapping height = 0.05 metres
• W_CORE = Diameter of the core (what the filament wraps on to) = 0.01 metres
• W_OUTER = Size of the flanges = 0.05 metres
• D_FIL = Diameter of filament = 0.003 metres

Finding The Number of Coils
We know the length of tube we have to wrap the filament around, H_SPOOL, and we know the diameter of the filament, D_FIL, so it is possible to work out the number of times the filament can lay side by side along the length of tube using:

N_LOOPS = H_SPOOL / D_FIL

For this example that would be N_LOOPS = 0.05 / 0.003 = 16 times (roughly). As H_SPOOL should not change as you add more layers the number of coils (or loops dependant on what you would rather call a revolution of filament) should stay the same on each layer.

Finding the Length of the First Layer
Now that we know the number of coils it is possible to find the length of filament used to create the first layer of coils. To do this the distance travelled by the filament for one coil must be found -- and it can as one coil of filament roughly follows the circumference of the core. Hence the length of one coil can be found using:

L_LOOP = pi * W_CORE

Multiplying this length by the total number of loops gives the overall length of the first layer ( L_LAYER1 = L_LOOP * N_LOOPS ). For the current example this would be L_LAYER1 = 16 * pi * 0.01 = 0.5 metres.

Finding the Length of Subsequent Layers
If you look at the image, specifically where the second layer starts, you can see that each layer is added over the previous layer. This essentially means that each completed layer adds to W_CORE by twice the diameter of the filament, 2 * D_FIL. This can be seen on the image cross section by the filament of the first layer on either side of the tube. Hence to find the length of filament required to complete a certain layer, N_LAYER you can use:

L_LAYERN = ( 2 * N_LAYER + W_CORE ) * pi * N_LOOPS

Continuing the example, the length of filament for the 2nd layer of the spool would be L_LAYERN = ( 2 * 2 + 0.01 ) * pi * 16 = 0.8042477193 metres.

Calculating the Maximum Number of Layers
This is essentially the same as calculating the number of coils, except this time W_OUTER is used as the ratioing edge:

N_LAYERMAX = W_OUTER / D_FIL

Hence the maximum number of layers for the example are N_LAYERMAX = 0.05 / 0.003 = 16 approximately. From here it would be possible to work out the total amount of filament the spool will hold or to create an algorithm to determine how many layers a certain length of filament would produce.

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Notes

• This information is a rewrite of This Link changing imperial to metric and, hopefully, being a little more descriptive.
• These equations are generalised, for example the one working out the length of filament per coil does not take into account the extra small amount required to create the coil shape rather than a circle.
• When I have a spare moment I shall try and create a spool calculator if desired.

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Hope this helps,
Mac

What has the spool got to do with how long a certain weight of plastic is?

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[www.hydraraptor.blogspot.com]

I see where I screwed up it was because I forgot to negative cube the 1240.

MacAleJam Wrote:
-------------------------------------------------------
> With:
> mass = 1kg
> Diameter = 0.003 m
> Density = 1240 kg / m^3
>
> The formula would become Length = 1 kg / (
> 1240kgm^-3 * 3.14159265 * ( 0.003m / 2 )^2 ) =
> 114.0895649404 metres, as Weedz has pointed out.
>
> For Coils around Spool
> The Figure below shows a cross section of a spool
> with some filament wrapped around it (represented
> by blue circles).
> [farm9.staticflickr.com]
> 37b395.jpg
> Reel Cross Section by MacAleJam, on Flickr
>
> For demonstration purposes assume that:
>
> [*] HSPOOL = Spool wrapping height = 0.05 metres
> [*] WCORE = Diameter of the core (what the
> filament wraps on to) = 0.01 metres
> [*] WOUTER = Size of the flanges = 0.05 metres
> [*] DFIL = Diameter of filament = 0.003 metres
>
>
> Finding The Number of Coils
> We know the length of tube we have to wrap the
> filament around, HSPOOL, and we know the diameter
> of the filament, DFIL, so it is possible to work
> out the number of times the filament can lay side
> by side along the length of tube using:
>
> NLOOPS = HSPOOL / DFIL
>
> For this example that would be NLOOPS = 0.05 /
> 0.003 = 16 times (roughly). As HSPOOL should not
> change as you add more layers the number of coils
> (or loops dependant on what you would rather call
> a revolution of filament) should stay the same on
> each layer.
>
> Finding the Length of the First Layer
> Now that we know the number of coils it is
> possible to find the length of filament used to
> create the first layer of coils. To do this the
> distance travelled by the filament for one coil
> must be found -- and it can as one coil of
> filament roughly follows the circumference of the
> core. Hence the length of one coil can be found
> using:
>
> LLOOP = pi * WCORE
>
> Multiplying this length by the total number of
> loops gives the overall length of the first layer
> ( LLAYER1 = LLOOP * NLOOPS ). For the current
> example this would be LLAYER1 = 16 * pi * 0.01 =
> 0.5 metres.
>
> Finding the Length of Subsequent Layers
> If you look at the image, specifically where the
> second layer starts, you can see that each layer
> is added over the previous layer. This essentially
> means that each completed layer adds to WCORE by
> twice the diameter of the filament, 2 * DFIL. This
> can be seen on the image cross section by the
> filament of the first layer on either side of the
> tube. Hence to find the length of filament
> required to complete a certain layer, NLAYER you
> can use:
>
> LLAYERN = ( 2 * NLAYER + WCORE ) * pi * NLOOPS
>
> Continuing the example, the length of filament for
> the 2nd layer of the spool would be LLAYERN = ( 2
> * 2 + 0.01 ) * pi * 16 = 0.8042477193 metres.
>
> Calculating the Maximum Number of Layers
> This is essentially the same as calculating the
> number of coils, except this time WOUTER is used
> as the ratioing edge:
>
> NLAYERMAX = WOUTER / DFIL
>
> Hence the maximum number of layers for the example
> are NLAYERMAX = 0.05 / 0.003 = 16 approximately.
> From here it would be possible to work out the
> total amount of filament the spool will hold or to
> create an algorithm to determine how many layers a
> certain length of filament would produce.
>
> Notes
>
> [*] This information is a rewrite of This Link
> changing imperial to metric and, hopefully, being
> a little more descriptive.
> [*] These equations are generalised, for example
> the one working out the length of filament per
> coil does not take into account the extra small
> amount required to create the coil shape rather
> than a circle.
> [*] When I have a spare moment I shall try and
> create a spool calculator if desired.
>
>
> Hope this helps,
> Mac

WOW, so detailed. Thank you because this will help me for some other stuff later that has zero to do with 3d printing.

nophead Wrote:
-------------------------------------------------------
> What has the spool got to do with how long a
> certain weight of plastic is?

That's what I was wondering...

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Help improve the RepRap wiki! Just click "Edit" in the top-right corner of the page and start typing. Anyone can edit the wiki!

maybe he wants to know how wide a spoolholder has to be for a certain amount (weight ) of filament......

NewPerfection Wrote:
-------------------------------------------------------
> nophead Wrote:
> --------------------------------------------------
> -----
> > What has the spool got to do with how long a
> > certain weight of plastic is?
>
> That's what I was wondering...

Nothing to do with calculating a certain weight of plastic -- and sorry if it is too far off topic. In one of the posts above Dark Alchemist mentions having had difficulties with spool calculations. I'll happily delete the post if you all wish?

Edited 1 time(s). Last edit at 07/22/2012 06:42AM by MacAleJam.

No its useful information, I am just confused about what the original question was.

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[www.hydraraptor.blogspot.com]