Hi all
Quelequn to help me calculate the final winding diameter using Solidworks,
As an example, I would first like to know how to wind a kind of carpet or curtain of any thickness t and length L on a daimeter tube D with SW the fainal diameter will be d .
Then just measure the final diameter using the measure tool.
Thank you to everyone
So it's a bit empirical as a method but you can make a circle sketch of the diameter D (that of your roll) then use the helix/spiral function (which is located in "curves", tab "functions" then you select "spiral" in the part "defined by:" in the section not you enter your thickness, then you enter a number of revolutions at random (to find roughly you can do with your length to be wound and your diameter, 2ftradius all that)
Then you exit the spiral function, click on the spiral and read the arc length value at the bottom right of the solidworks window. This value corresponds to your length of fabrics to be rolled up. You then enter your spiral function and play with the number of revolutions (possible to put decimals) to get the desired length and your final winding diameter is displayed when you edit the Helix/Spiral function 
I hope this answers your question!
Otherwise there must be a mathematical formula to calculate this.
A mathematical formula?
Otherwise faster than SolidWorks you have this:
https://fr.planetcalc.com/9063/
Pierre32, it does exist so the formula, see on the site 
But I don't think I want to dissect it...
It's very simple, isn't it?
Hello
We can't say it's simple, but it's not very complicated.
You can find the explanation of the calculations in the file. doc and the spreadsheet to find out what you're looking for.
The sheet was made to calculate winding speeds, but in the graph you have the winding length as a function of the thickness of the material.
Kind regards
Winding and unwinding Version1-03.zip film (103.1 KB)
Winding and unwinding film Version1-03 2.zip (101.9 KB)
(d1+(e x 2)) x ntour = d2 coiled material and d2 x pi = unrolled length
This does not take into account the course of all the rounds, so the trick is not done!!
Hello
To avoid doliprane, here is a simple formula that assumes that the surface of the circular ring (circle ext - circle int) is equal to the surface L*t.
No need for SW for that.
The outer diameter is
d=√(4∗t∗L/π+D^2 )
(I used your annotations d for dim ext and D for diam int. I would have done the opposite)
Thank you all
Thank you too, learning and understanding are 2 good remedies for stupidity!!