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Dimensions of paper needed to roll a cone (Updated with clarifications)

Calulating length of cable running along exterior of an axleArea of equilateral triangle around circles placed in equilateral triangleProjecting a surface segment of a cone onto a 2D plane?Roll Weight calculationWas this a correct approach? Is there a simpler one?Dihedral Calculation.What 'uniform' shapes can be used to build an approximated spherical object?Find diameter of partially filled coneHow to label the intersection of edges in arbitrary four-sided shapes?Is the right intersection of an oblique circular cone an ellipse?

2

0

$begingroup$

I'm looking for a way to calculate the dimensions of a piece of paper needed to roll up into a cone shape. Please consider the following diagram I created (nothing is drawn to scale):

In this example, I'm trying to create a cone shape, that is 100 mm tall and has a top diameter of 3 mm, and a bottom diameter of 12 mm (FIG 1). FIG 2 shows a piece of paper that is 0.1 mm thick shaped like what I would expect the final piece to look like. However, sides A and C would be much longer than illustrated.

Essentially, what I need is a way to figure out the lengths of sides A, B, and C in FIG 2, such that when the paper is rolled tightly, I could achieve the cone in FIG 1.

NOTE: I'm not trying to just create a "shell" of the cone as shown in FIG 3. I would like to roll the paper shape tightly into the cone to provide a solid structure (as shown in FIG 4).

---

EDIT: I'm adding some clarification. I did a "real world" example to try to illustrate what I'm trying to accomplish.

1. I cut out a shape similar to the one is FIG 2 above.
   image 1




2. I roll it as shown here.
   image 2




3. This is what the cone looks like when done rolling.
   image 3




4. The top of the cone shows a diameter of 2.1 mm
   image 4




5. And the bottom of the cone shows a diameter of 3.4 mm
   image 5

I hope this makes things more clear.

EDIT 2: Maybe the numbers in the initial example are just not realistic. So how about something like 2 mm for the top diameter and 5 mm for the bottom one.

geometry

share|cite|improve this question

edited Mar 20 at 22:48

Alarik

asked Mar 20 at 19:33

AlarikAlarik

133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In what direction will you be rolling. Also: what have you tried?
  $endgroup$
  – clathratus
  Mar 20 at 20:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are several problems with your description. First of all, to get a conic surface you must fold a circular sector, and not a trapezoid as in your picture. Moreover, you cannot get a uniformly tight structure: to get 12 mm at the base you should roll 120 layers of paper, but for 3 mm on top you need only 30 layers.
  $endgroup$
  – Aretino
  Mar 20 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the responses. @ Aretino: I added a "real-world" example to show what I'm trying to accomplish.
  $endgroup$
  – Alarik
  Mar 20 at 21:52
  

  
  
  
  

add a comment | 

2

0

$begingroup$

I'm looking for a way to calculate the dimensions of a piece of paper needed to roll up into a cone shape. Please consider the following diagram I created (nothing is drawn to scale):

In this example, I'm trying to create a cone shape, that is 100 mm tall and has a top diameter of 3 mm, and a bottom diameter of 12 mm (FIG 1). FIG 2 shows a piece of paper that is 0.1 mm thick shaped like what I would expect the final piece to look like. However, sides A and C would be much longer than illustrated.

Essentially, what I need is a way to figure out the lengths of sides A, B, and C in FIG 2, such that when the paper is rolled tightly, I could achieve the cone in FIG 1.

NOTE: I'm not trying to just create a "shell" of the cone as shown in FIG 3. I would like to roll the paper shape tightly into the cone to provide a solid structure (as shown in FIG 4).

---

EDIT: I'm adding some clarification. I did a "real world" example to try to illustrate what I'm trying to accomplish.

1. I cut out a shape similar to the one is FIG 2 above.
   image 1




2. I roll it as shown here.
   image 2




3. This is what the cone looks like when done rolling.
   image 3




4. The top of the cone shows a diameter of 2.1 mm
   image 4




5. And the bottom of the cone shows a diameter of 3.4 mm
   image 5

I hope this makes things more clear.

EDIT 2: Maybe the numbers in the initial example are just not realistic. So how about something like 2 mm for the top diameter and 5 mm for the bottom one.

geometry

share|cite|improve this question

edited Mar 20 at 22:48

Alarik

asked Mar 20 at 19:33

AlarikAlarik

133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In what direction will you be rolling. Also: what have you tried?
  $endgroup$
  – clathratus
  Mar 20 at 20:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are several problems with your description. First of all, to get a conic surface you must fold a circular sector, and not a trapezoid as in your picture. Moreover, you cannot get a uniformly tight structure: to get 12 mm at the base you should roll 120 layers of paper, but for 3 mm on top you need only 30 layers.
  $endgroup$
  – Aretino
  Mar 20 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the responses. @ Aretino: I added a "real-world" example to show what I'm trying to accomplish.
  $endgroup$
  – Alarik
  Mar 20 at 21:52
  

  
  
  
  

add a comment | 

$begingroup$

I'm looking for a way to calculate the dimensions of a piece of paper needed to roll up into a cone shape. Please consider the following diagram I created (nothing is drawn to scale):

In this example, I'm trying to create a cone shape, that is 100 mm tall and has a top diameter of 3 mm, and a bottom diameter of 12 mm (FIG 1). FIG 2 shows a piece of paper that is 0.1 mm thick shaped like what I would expect the final piece to look like. However, sides A and C would be much longer than illustrated.

Essentially, what I need is a way to figure out the lengths of sides A, B, and C in FIG 2, such that when the paper is rolled tightly, I could achieve the cone in FIG 1.

NOTE: I'm not trying to just create a "shell" of the cone as shown in FIG 3. I would like to roll the paper shape tightly into the cone to provide a solid structure (as shown in FIG 4).

---

EDIT: I'm adding some clarification. I did a "real world" example to try to illustrate what I'm trying to accomplish.

1. I cut out a shape similar to the one is FIG 2 above.
   image 1




2. I roll it as shown here.
   image 2




3. This is what the cone looks like when done rolling.
   image 3




4. The top of the cone shows a diameter of 2.1 mm
   image 4




5. And the bottom of the cone shows a diameter of 3.4 mm
   image 5

I hope this makes things more clear.

EDIT 2: Maybe the numbers in the initial example are just not realistic. So how about something like 2 mm for the top diameter and 5 mm for the bottom one.

geometry

share|cite|improve this question

edited Mar 20 at 22:48

Alarik

asked Mar 20 at 19:33

AlarikAlarik

133

$endgroup$

share|cite|improve this question

edited Mar 20 at 22:48

Alarik

asked Mar 20 at 19:33

AlarikAlarik

133

share|cite|improve this question

edited Mar 20 at 22:48

Alarik

asked Mar 20 at 19:33

AlarikAlarik

133

asked Mar 20 at 19:33

AlarikAlarik

133

asked Mar 20 at 19:33

AlarikAlarik

133

1 Answer
1

active

oldest

votes

1

$begingroup$

If we slightly simplify the situation, we can say that at the top (and at the bottom) the paper is arranged in concentric circles. If $t=0.1$mm is the thickness of the paper, we get a radius $r_n = n cdot t$ for the $n$th circle. The circumference of the $n$th circle is $2 pi r_n$. The total length of the paper (for $N$ circles) is the sum of the circumferences
$$
l = sum_{n=1}^N 2 pi r_n = 2 pi cdot t cdot sum_{n=1}^N n = 2 pi cdot t cdot frac{N(N+1)}{2}
$$
and the diameter is $d = 2 N t$.

So if you want a diameter $d$, then you need $N=frac{d}{2t}$ many layers resulting in a length of
$$l = 2 pi cdot t cdot frac{frac{d}{2t}left(frac{d}{2t}+1right)}{2} = pi cdot frac{d}{2} left(frac{d}{2t}+1right).$$

According to the pictures in your clarification, $A$ and $C$ are independent of each other, thus the above calculation can be done seperatly for the top and the bottom.

At the top, you want the diameter to be $d=3$mm, resulting in a length of $Acong 65$mm.
At the bottom, you want the diameter to be $d=12$mm, thus $C cong 1100$mm would do the trick.

However, rolling up a meter of paper perfectly tight is probably not physically feasible, thus the calculation might not agree with experimental results.

share|cite|improve this answer

edited Mar 21 at 11:54

answered Mar 21 at 1:03

StrichcoderStrichcoder

3977

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The sum is $sum_{n=1}^N n=N(N+1)/2$.
  $endgroup$
  – Aretino
  Mar 21 at 8:16
  

  
  
  
  

add a comment | 

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1

$begingroup$

If we slightly simplify the situation, we can say that at the top (and at the bottom) the paper is arranged in concentric circles. If $t=0.1$mm is the thickness of the paper, we get a radius $r_n = n cdot t$ for the $n$th circle. The circumference of the $n$th circle is $2 pi r_n$. The total length of the paper (for $N$ circles) is the sum of the circumferences
$$
l = sum_{n=1}^N 2 pi r_n = 2 pi cdot t cdot sum_{n=1}^N n = 2 pi cdot t cdot frac{N(N+1)}{2}
$$
and the diameter is $d = 2 N t$.

So if you want a diameter $d$, then you need $N=frac{d}{2t}$ many layers resulting in a length of
$$l = 2 pi cdot t cdot frac{frac{d}{2t}left(frac{d}{2t}+1right)}{2} = pi cdot frac{d}{2} left(frac{d}{2t}+1right).$$

According to the pictures in your clarification, $A$ and $C$ are independent of each other, thus the above calculation can be done seperatly for the top and the bottom.

At the top, you want the diameter to be $d=3$mm, resulting in a length of $Acong 65$mm.
At the bottom, you want the diameter to be $d=12$mm, thus $C cong 1100$mm would do the trick.

However, rolling up a meter of paper perfectly tight is probably not physically feasible, thus the calculation might not agree with experimental results.

share|cite|improve this answer

edited Mar 21 at 11:54

answered Mar 21 at 1:03

StrichcoderStrichcoder

3977

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The sum is $sum_{n=1}^N n=N(N+1)/2$.
  $endgroup$
  – Aretino
  Mar 21 at 8:16
  

  
  
  
  

add a comment | 

1

$begingroup$

If we slightly simplify the situation, we can say that at the top (and at the bottom) the paper is arranged in concentric circles. If $t=0.1$mm is the thickness of the paper, we get a radius $r_n = n cdot t$ for the $n$th circle. The circumference of the $n$th circle is $2 pi r_n$. The total length of the paper (for $N$ circles) is the sum of the circumferences
$$
l = sum_{n=1}^N 2 pi r_n = 2 pi cdot t cdot sum_{n=1}^N n = 2 pi cdot t cdot frac{N(N+1)}{2}
$$
and the diameter is $d = 2 N t$.

So if you want a diameter $d$, then you need $N=frac{d}{2t}$ many layers resulting in a length of
$$l = 2 pi cdot t cdot frac{frac{d}{2t}left(frac{d}{2t}+1right)}{2} = pi cdot frac{d}{2} left(frac{d}{2t}+1right).$$

According to the pictures in your clarification, $A$ and $C$ are independent of each other, thus the above calculation can be done seperatly for the top and the bottom.

At the top, you want the diameter to be $d=3$mm, resulting in a length of $Acong 65$mm.
At the bottom, you want the diameter to be $d=12$mm, thus $C cong 1100$mm would do the trick.

However, rolling up a meter of paper perfectly tight is probably not physically feasible, thus the calculation might not agree with experimental results.

share|cite|improve this answer

edited Mar 21 at 11:54

answered Mar 21 at 1:03

StrichcoderStrichcoder

3977

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The sum is $sum_{n=1}^N n=N(N+1)/2$.
  $endgroup$
  – Aretino
  Mar 21 at 8:16
  

  
  
  
  

add a comment | 

$begingroup$

If we slightly simplify the situation, we can say that at the top (and at the bottom) the paper is arranged in concentric circles. If $t=0.1$mm is the thickness of the paper, we get a radius $r_n = n cdot t$ for the $n$th circle. The circumference of the $n$th circle is $2 pi r_n$. The total length of the paper (for $N$ circles) is the sum of the circumferences
$$
l = sum_{n=1}^N 2 pi r_n = 2 pi cdot t cdot sum_{n=1}^N n = 2 pi cdot t cdot frac{N(N+1)}{2}
$$
and the diameter is $d = 2 N t$.

So if you want a diameter $d$, then you need $N=frac{d}{2t}$ many layers resulting in a length of
$$l = 2 pi cdot t cdot frac{frac{d}{2t}left(frac{d}{2t}+1right)}{2} = pi cdot frac{d}{2} left(frac{d}{2t}+1right).$$

According to the pictures in your clarification, $A$ and $C$ are independent of each other, thus the above calculation can be done seperatly for the top and the bottom.

At the top, you want the diameter to be $d=3$mm, resulting in a length of $Acong 65$mm.
At the bottom, you want the diameter to be $d=12$mm, thus $C cong 1100$mm would do the trick.

However, rolling up a meter of paper perfectly tight is probably not physically feasible, thus the calculation might not agree with experimental results.

share|cite|improve this answer

edited Mar 21 at 11:54

answered Mar 21 at 1:03

StrichcoderStrichcoder

3977

$endgroup$

share|cite|improve this answer

edited Mar 21 at 11:54

answered Mar 21 at 1:03

StrichcoderStrichcoder

3977

answered Mar 21 at 1:03

StrichcoderStrichcoder

3977

answered Mar 21 at 1:03

StrichcoderStrichcoder

3977