Posted by Dark Alchemist

Weird question I think but I Googled... July 21, 2012 05:24AM |
Registered: 9 years ago Posts: 1,277 |

Re: Weird question I think but I Googled... July 21, 2012 06:36AM |
AdminRegistered: 14 years ago Posts: 7,857 |

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.

[www.hydraraptor.blogspot.com]

[www.hydraraptor.blogspot.com]

Re: Weird question I think but I Googled... July 21, 2012 06:37AM |
Registered: 9 years ago Posts: 120 |

Re: Weird question I think but I Googled... July 21, 2012 07:01AM |
Registered: 9 years ago Posts: 1,277 |

Lost me.

Density for PLA is 1.23 - 1.25 g/cmÂ³ 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.

Density for PLA is 1.23 - 1.25 g/cmÂ³ 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.

Re: Weird question I think but I Googled... July 21, 2012 07:10AM |
Registered: 9 years ago Posts: 6 |

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.

Also understanding that

The equation becomes:

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.

Re: Weird question I think but I Googled... July 21, 2012 07:15AM |
Registered: 9 years ago Posts: 1,277 |

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.

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.

Re: Weird question I think but I Googled... July 21, 2012 12:58PM |
Registered: 9 years ago Posts: 120 |

Re: Weird question I think but I Googled... July 21, 2012 02:06PM |
Registered: 9 years ago Posts: 6 |

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).

Reel Cross Section by MacAleJam, on Flickr

For demonstration purposes assume that:

**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.

**Notes**

Hope this helps,

Mac

mass = 1kg

Diameter = 0.003 m

Density = 1240 kg / m^3

The formula would become

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

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

We know the length of tube we have to wrap the filament around,

For this example that would be N

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:

Multiplying this length by the total number of loops gives the overall length of the first layer (

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

Continuing the example, the length of filament for the 2nd layer of the spool would be L

This is essentially the same as calculating the number of coils, except this time

Hence the maximum number of layers for the example are N

- 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

Re: Weird question I think but I Googled... July 21, 2012 04:24PM |
AdminRegistered: 14 years ago Posts: 7,857 |

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

[www.hydraraptor.blogspot.com]

[www.hydraraptor.blogspot.com]

Re: Weird question I think but I Googled... July 21, 2012 05:35PM |
Registered: 9 years ago Posts: 1,277 |

Re: Weird question I think but I Googled... July 21, 2012 05:36PM |
Registered: 9 years ago Posts: 1,277 |

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.

-------------------------------------------------------

> 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.

Re: Weird question I think but I Googled... July 21, 2012 06:22PM |
AdminRegistered: 12 years ago Posts: 1,461 |

nophead Wrote:

-------------------------------------------------------

> What has the spool got to do with how long a

> certain weight of plastic is?

That's what I was wondering...

-------------------------------------------------------

> What has the spool got to do with how long a

> certain weight of plastic is?

That's what I was wondering...

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

Re: Weird question I think but I Googled... July 22, 2012 05:54AM |
Registered: 9 years ago Posts: 120 |

Re: Weird question I think but I Googled... July 22, 2012 06:02AM |
Registered: 9 years ago Posts: 6 |

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.

-------------------------------------------------------

> 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.

Re: Weird question I think but I Googled... July 22, 2012 07:09AM |
AdminRegistered: 14 years ago Posts: 7,857 |

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

[www.hydraraptor.blogspot.com]

[www.hydraraptor.blogspot.com]

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