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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












$begingroup$


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):



enter image description here



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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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















2












$begingroup$


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):



enter image description here



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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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













2












2








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):



enter image description here



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.










share|cite|improve this question











$endgroup$




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):



enter image description here



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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 20 at 22:48







Alarik

















asked Mar 20 at 19:33









AlarikAlarik

133




133











  • $begingroup$
    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
















  • $begingroup$
    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















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




$begingroup$
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




$begingroup$
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




$begingroup$
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










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
$$
l = sum_n=1^N 2 pi r_n = 2 pi cdot t cdot sum_n=1^N n = 2 pi cdot t cdot fracN(N+1)2
$$

and the diameter is $d = 2 N t$.



So if you want a diameter $d$, then you need $N=fracd2t$ many layers resulting in a length of
$$l = 2 pi cdot t cdot fracfracd2tleft(fracd2t+1right)2 = pi cdot fracd2 left(fracd2t+1right).$$



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











$endgroup$








  • 1




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











Your Answer





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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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
$$
l = sum_n=1^N 2 pi r_n = 2 pi cdot t cdot sum_n=1^N n = 2 pi cdot t cdot fracN(N+1)2
$$

and the diameter is $d = 2 N t$.



So if you want a diameter $d$, then you need $N=fracd2t$ many layers resulting in a length of
$$l = 2 pi cdot t cdot fracfracd2tleft(fracd2t+1right)2 = pi cdot fracd2 left(fracd2t+1right).$$



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











$endgroup$








  • 1




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















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
$$
l = sum_n=1^N 2 pi r_n = 2 pi cdot t cdot sum_n=1^N n = 2 pi cdot t cdot fracN(N+1)2
$$

and the diameter is $d = 2 N t$.



So if you want a diameter $d$, then you need $N=fracd2t$ many layers resulting in a length of
$$l = 2 pi cdot t cdot fracfracd2tleft(fracd2t+1right)2 = pi cdot fracd2 left(fracd2t+1right).$$



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











$endgroup$








  • 1




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













1












1








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
$$
l = sum_n=1^N 2 pi r_n = 2 pi cdot t cdot sum_n=1^N n = 2 pi cdot t cdot fracN(N+1)2
$$

and the diameter is $d = 2 N t$.



So if you want a diameter $d$, then you need $N=fracd2t$ many layers resulting in a length of
$$l = 2 pi cdot t cdot fracfracd2tleft(fracd2t+1right)2 = pi cdot fracd2 left(fracd2t+1right).$$



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











$endgroup$



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
$$
l = sum_n=1^N 2 pi r_n = 2 pi cdot t cdot sum_n=1^N n = 2 pi cdot t cdot fracN(N+1)2
$$

and the diameter is $d = 2 N t$.



So if you want a diameter $d$, then you need $N=fracd2t$ many layers resulting in a length of
$$l = 2 pi cdot t cdot fracfracd2tleft(fracd2t+1right)2 = pi cdot fracd2 left(fracd2t+1right).$$



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














share|cite|improve this answer



share|cite|improve this answer








edited Mar 21 at 11:54

























answered Mar 21 at 1:03









StrichcoderStrichcoder

1866




1866







  • 1




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












  • 1




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







1




1




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




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

















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