rope winding on disc

[eugenio_78]

Bye to all,
today I have met a very interesting problem that I would like to share with you:
I have to wrap a rope on a disk forming a kind of pizza, the rope has a length of 100 mt. and the wrapping on the disk starts from an initial diameter of 90mm.
I have to figure out how many laps it has to make the disk and what will be the outer diameter of the pizza.
I've been sitting on it for a while, but having lost confidence with the entirety doesn't help me.
I tried to use this formula but it is not correct:
N = 100,000 / (90+5*n)*pi

If I find the solution, I will

 

[marcof]

Bye to all,
today I have met a very interesting problem that I would like to share with you:
I have to wrap a rope on a disk forming a kind of pizza, the rope has a length of 100 mt. and the wrapping on the disk starts from an initial diameter of 90mm.
I have to figure out how many laps it has to make the disk and what will be the outer diameter of the pizza.
I've been sitting on it for a while, but having lost confidence with the entirety doesn't help me.
I tried to use this formula but it is not correct:
N = 100,000 / (90+5*n)*pi

If I find the solution, I will

You're considering circumferences. I don't know what degree of caxxi :smile teperation: you need to arrive in this calculation, but the approach, at least after the first turns, is much more approximate to a spiral of Archimedes than to a overlap of circumferences. I, known the step that is equal to the diameter of the rope, would make calculations of the length of the spiral. Maybe I'm wrong and it really needs to shake the entirety, even if the one above still seems to me a second degree equation with a, b and c that are known.

 

[Fulvio Romano]

do not need to shake the whole pieces, just equal the volume of the rope to that of the pizza.
if the rope has diameter d, imagine wrapping a square section rope on the side d.
the volume of the rope (length per d^2) must be equal to the volume of the pizza (pi*rp^2*d) where r is the radius of the pizza.

you'll have a approximation on the first round that's "empty." if numerically relevant you can take account of the second member.

A beer is good, too.

 

[SOLID]

made with excel considering the sum of the lengths of many 5mm step circumferences with departure from an initial diameter of 90 turns out to be 150 laps with for a final diament of about 850mm. .

 

[Focus]

Save. qui, should help you.

Hi.

 

[eugenio_78]

do not need to shake the whole pieces, just equal the volume of the rope to that of the pizza.
if the rope has diameter d, imagine wrapping a square section rope on the side d.
the volume of the rope (length per d^2) must be equal to the volume of the pizza (pi*rp^2*d) where r is the radius of the pizza.

you'll have a approximation on the first round that's "empty." if numerically relevant you can take account of the second member.

A beer is good, too.

interesting...

I also found another similar way:
(d2-d2)

where l is the length s the thickness I can calculate the outer diameter and then

(d-d)/2/s = number of turns to wrap the date length

 

[marcof]

do not need to shake the whole pieces, just equal the volume of the rope to that of the pizza.
if the rope has diameter d, imagine wrapping a square section rope on the side d.
the volume of the rope (length per d^2) must be equal to the volume of the pizza (pi*rp^2*d) where r is the radius of the pizza.

chapeau for the idea like colombo egg!
that online computer returns a diameter of 1129 and with your method, removed the volume of pizza hazelnut vien out 1132 mm

 

[eugenio_78]

try to copy this code on a broser console and run it:

function mia(){
d=90;
n=0;
l=0;

while(l<100000){
l=l+d*3.14;
d += 0.05;
n++
?
alert('final diameter = ' + d + '\nnumero diri = ' + n)

?
my();

 

[meccanicamg]

you have to use the correct string formula of the rope thickness. it's the same one used for sheet coils or find the explanation qui