math function

[lelepanz]

Hi, guys.
You would tell me what the mathematical function describes the amount of material that is on a film reel.
For example. I have a reel with internal diameter 100 and external 400, on which is wrapped a film (a film) that has a certain thickness s.

Thank you very much in advance.

I am now aware that the title is too generic, if anyone wants to change it, I would be grateful.

Hi.
emanuele

 

[cacciatorino]

Hi, guys.
You would tell me what the mathematical function describes the amount of material that is on a film reel.
For example. I have a reel with internal diameter 100 and external 400, on which is wrapped a film (a film) that has a certain thickness s.

Thank you very much in advance.

You mean the linear length?

We consider that the coil width is "h" (even if we don't need it because it's going to simplify), then the coil volume is:

v = (pi* (de^2-di^2)/4)

that will be the same volume of a rectangle (the tape performed) with volume:

v= h * s * l (where l is your length).

At this point you just have to do the equality between the two equations and you jump out there.

Bye.

 

[lelepanz]

You mean the linear length?

We consider that the coil width is "h" (even if we don't need it because it's going to simplify), then the coil volume is:

v = (pi* (de^2-di^2)/4)

that will be the same volume of a rectangle (the tape performed) with volume:

v= h * s * l (where l is your length).

At this point you just have to do the equality between the two equations and you jump out there.

Bye.

Thank you. In practice I wanted to know this, because I would have no doubt about knowing the de of this coil, as I shoot down the film....
... I had made a formula... but... I was framed... so it's much easier.

Thank you.

 

[cacciatorino]

In practice I wanted to know this

But consider that if the coil is not well wrapped (i.e., they didn't give it the right countertiro during the winding so it's not very "packed") in the reel you'll actually get some blanks for which the formula won't be exact.

 

[Fulvio Romano]

You mean the linear length?

We consider that the coil width is "h" (even if we don't need it because it's going to simplify), then the coil volume is:

v = (pi* (de^2-di^2)/4)

that will be the same volume of a rectangle (the tape performed) with volume:

v= h * s * l (where l is your length).

At this point you just have to do the equality between the two equations and you jump out there.

Bye.

elegant solution...I always made a series in excel, which is a rather brute method, as well as ugly!

 

[lelepanz]

But consider that if the coil is not well wrapped (i.e., they didn't give it the right countertiro during the winding so it's not very "packed") in the reel you'll actually get some blanks for which the formula won't be exact.

Yes. between the other in the formula we also set a k, a correction. because in theory between one layer and the other there would also be an interlayer film that is not used but must be recovered on another roll... It brings me the biggest mistake. Now we'll try. Thanks again.

elegant solution...I always made a series in excel, which is a rather brute method, as well as ugly!

It is what we were about to do.
It was enough to read the word volume, and even in me the springs... .
... and then to say... how did I not think about it before.... .

Thanks again hunter. +1 rep for you.

 

[cacciatorino]

Thanks again hunter. +1 rep for you.

Thank you! :smile:

I imagine there is also the relationship that binds the length of the spiral to its diameter, but we should go to some prontuary of mathematical formulas. . . .

 

[tecnico_plast]

excellent solution :finger:

I take the opportunity to see what you think of a similar problem that I have tried to solve (without losing too much time to tell the truth).

the function I would like to find I need to calculate the amount of wire (length) that it serves to fill the reel of a fishing reel (who goes fishing knows what I am talking about:tongue.

the problem is complicated by the fact that the wire is spiraled, from top to bottom and vice versa, also depending on the reels can be stuffed differently (crossed edges for example), but it would suffice to develop a formula also approximate, we say with an error margin of 10-15%.

 

[cacciatorino]

excellent solution :finger:

I take the opportunity to see what you think of a similar problem that I have tried to solve (without losing too much time to tell the truth).

the function I would like to find I need to calculate the amount of wire (length) that it serves to fill the reel of a fishing reel (who goes fishing knows what I am talking about:tongue.

the problem is complicated by the fact that the wire is spiraled, from top to bottom and vice versa, also depending on the reels can be stuffed differently (crossed edges for example), but it would suffice to develop a formula also approximate, we say with an error margin of 10-15%.

Perhaps the same system of the volume mentioned above works, with the kindness of considering your thread as if it were square section with square side equal to that of the square that circumscribes the section of your round wire. that is equal the volume of the coil to that of a rectangle of size hxhxl where h is the diameter of the wire and the length of the thread carried out... Just think about it, and then tell me if you agree.
I start from the assumption that being by force of things the coils cross the diametric difference between a turn of coils and the next cannot be lower than the diameter of the wire (can't be interlinking).

One thing that's going to make you kick your head off is that during the winding up of the aspo, the spires will never be exactly adjacent to each other but a little spaced, especially for thin wires.

 

[MBT]

where I worked a long time ago (we were dealing with, among other things, wire coilers ripped by building buildings) someone tried to make such a calculation.
the result was not the best.. .
In fact the unknown of the behavior of the wire leads you to kick the calculation, and not just....:redface:

 

[tecnico_plast]

the discourse to consider the square-section thread could function to simplify the calculation, but you should add a correction factor, since the square area is greater than the area of the actual round section of the wire. the percentage deviation is 21.5%, i.e. the round area is less than 21.5% compared to the square area, so found the length should be removed from the result 21.5%.

there is also to say that the height of the coil does not coincide with the side of the square (we are in the order of tenths of millimeter), so the formula should be this:

v [bobina] = pi*(r^2-r^2)*h (h = altezza bobina)

v [filo] = b*b*l = b^2*l (b = wire diameter = square side)

matching the two volumes and obtaining l, the formula should be this:

l = (pi*(r^2-r^2)*h) / b^2 (this quantity must be subtracted 21.5%)

Moreover, there will inevitably be spaces between the various layers of the wire that are wrapped one on the other, so you should add another correction factor.

the problem is the behaviour of the wire, in principle it could be said that the higher the diameter of the wire (one 0.70 mm is already a big thread), the more rigid and the less well it tends to envelop, leaving spaces between a coil and the other.

Anyway in the lunch break I see how much wire of a given diameter more being contained in a coil (data provided by the manufacturer), then I mix the coil and see what sorting is between the declared quantity and that calculated with the formula?
What do you think, it'll work? :smile:

 

[tecnico_plast]

I did the tests and it should work, with a single correction factor of 30-35%, but I tried with wires from 0.40 and 0.50 mm, I think with lower diameters something changes.

 

[cacciatorino]

I did the tests and it should work, with a single correction factor of 30-35%, but I tried with wires from 0.40 and 0.50 mm, I think with lower diameters something changes.

but a correction factor of 30-35% is no longer a correction factor, it becomes a magic number that makes the accounts square!!!! :biggrin::biggrin:

 

[Drag]

technical excuse_plast, I'm interested in the problem,
Could you tell me the results of your tests with wire 0.4 and 0.5mm
Thank you.

 

[tecnico_plast]

but a correction factor of 30-35% is no longer a correction factor, it becomes a magic number that makes the accounts square!!!! :biggrin::biggrin:

But if it works... :biggrin:

However, I would need more data to confirm everything, especially for thin diameters.

 

[tecnico_plast]

technical excuse_plast, I'm interested in the problem,
Could you tell me the results of your tests with wire 0.4 and 0.5mm
Thank you.

You can do it yourself, using that formula.

However on a coil with r=33, r= 28 and h=33, from the calculation comes out a length of about 197.5 m of 0.40 and 126 m of 0.50.
the declared capacity is instead of 140 m and 80 m respectively.
If we consider that the formula takes into account the square and non-round section, and that there are usually spaces between one end and the other, the correction factor comes out (also called magic number:biggrin of about 30-35% (of 21.5% is the discard due to the approximation of the section from round to square).

 

[Fulvio Romano]

Sorry... I didn't understand the difference between round and square thread... I thought the facts were like this: the thread is just round. to be able to consider the packing coefficient we imagine the square thread (perhaps hexagonal would have been better, but so much the packing is far from the perfect being) and so we consider this volume. But the length should not be affected, do you think?

 

[tecnico_plast]

but the square thread occupies more volume than the round, then at the same volume to fill it will take less turns to fill it with the round one.

 

[cacciatorino]

but the square thread occupies more volume than the round, then at the same volume to fill it will take less turns to fill it with the round one.

Actually, if you put two round wires close, the busy space is the square.

the speech of the hexagon that made Roman lightning according to me is not totally correct because the coils wrap a layer to the right and a left erso, so we can not soder that a thread of a higher layer lies in the "vallet" that exists between two adjacent threads of the lower layer.

 

[Fulvio Romano]

but the square thread occupies more volume than the round, then at the same volume to fill it will take less turns to fill it with the round one.

the volume to be considered is the same because you consider a volume "increased" in the case of the square not to consider the voids.

Actually, if you put two round wires close, the busy space is the square.

the speech of the hexagon that made Roman lightning according to me is not totally correct because the coils wrap a layer to the right and a left erso, so we can not soder that a thread of a higher layer lies in the "vallet" that exists between two adjacent threads of the lower layer.

In fact, I considered the hexagon only in case of maximum compaction. in the case of zigzag winding you should also consider the step...or perhaps it is the case of taking a spool, filling up to the edge a cup of water, dipping it slowly and finally finding the real volume. doing it with full and empty we can find the only volume of the wire, also because...for those who go fishing, often the lenze are conical and not cylindrical. . At least for towing rods.