The area of a truncated pyramid with irregular top and bottom surface, given the height $h$
0
$begingroup$
This question is inspired by the question The volume for truncated pyramid with irregular base.
Given that we have the top and bottom surface area ($A_1$, $A_2$) of a pyramid, ad the height of the truncated pyramid is given by $H$, how can we find the surface area of the pyramid, given the height from the bottom $h$? Without a loss of generality we can assume that $A_2$ is always bigger than $A_1$.
I can prove that in two limiting cases where the surface is circle and rectangle, the area $A_h$ at height $h$ is somehow proportional to the square of $h$, ie: $A_h
varpropto h^2$. But is there a general formula connecting $A_h$, $h$, $H$, $A_1$ and $A_2$?
geometry volume
asked Dec 24 '18 at 9:58
GravitonGraviton
1,09622044
$endgroup$
|
show 1 more comment
0
$begingroup$
This question is inspired by the question The volume for truncated pyramid with irregular base.
Given that we have the top and bottom surface area ($A_1$, $A_2$) of a pyramid, ad the height of the truncated pyramid is given by $H$, how can we find the surface area of the pyramid, given the height from the bottom $h$? Without a loss of generality we can assume that $A_2$ is always bigger than $A_1$.
I can prove that in two limiting cases where the surface is circle and rectangle, the area $A_h$ at height $h$ is somehow proportional to the square of $h$, ie: $A_h
varpropto h^2$. But is there a general formula connecting $A_h$, $h$, $H$, $A_1$ and $A_2$?
geometry volume
asked Dec 24 '18 at 9:58
GravitonGraviton
1,09622044
$endgroup$
$begingroup$
The combination of the bottom surface and a horizontally shifted copy of the top surface describes a new truncated pyramid with a generally different area.
$endgroup$
– random
Dec 24 '18 at 11:57
$begingroup$
@random, what does your comment has to do with my question?
$endgroup$
– Graviton
Dec 25 '18 at 9:13
$begingroup$
You are asking for the area of a truncated pyramid, which is to be derived from only its height and the areas of the upper and lower surface. Such a derivation is impossible if there are truncated pyramids that fit that description, but have different total areas. For the class of truncated pyramids described in the comment, the area of their vertical projection on the base plane can be arbitrarily large, with the real area being at least twice as large.
$endgroup$
– random
Dec 25 '18 at 10:34
$begingroup$
If your real question is described in its last paragraph instead of in the header then the answer would be that the computation additionally requires the height of the not yet truncated pyramid or the distance between the top and bottom surface.
$endgroup$
– random
Dec 25 '18 at 10:43
$begingroup$
@random you are right. I have modified the question , we do know the $H$ after all
$endgroup$
– Graviton
Dec 25 '18 at 11:21
|
show 1 more comment
0
0
0
$begingroup$
This question is inspired by the question The volume for truncated pyramid with irregular base.
Given that we have the top and bottom surface area ($A_1$, $A_2$) of a pyramid, ad the height of the truncated pyramid is given by $H$, how can we find the surface area of the pyramid, given the height from the bottom $h$? Without a loss of generality we can assume that $A_2$ is always bigger than $A_1$.
I can prove that in two limiting cases where the surface is circle and rectangle, the area $A_h$ at height $h$ is somehow proportional to the square of $h$, ie: $A_h
varpropto h^2$. But is there a general formula connecting $A_h$, $h$, $H$, $A_1$ and $A_2$?
geometry volume
asked Dec 24 '18 at 9:58
GravitonGraviton
1,09622044
$endgroup$
This question is inspired by the question The volume for truncated pyramid with irregular base.
Given that we have the top and bottom surface area ($A_1$, $A_2$) of a pyramid, ad the height of the truncated pyramid is given by $H$, how can we find the surface area of the pyramid, given the height from the bottom $h$? Without a loss of generality we can assume that $A_2$ is always bigger than $A_1$.
I can prove that in two limiting cases where the surface is circle and rectangle, the area $A_h$ at height $h$ is somehow proportional to the square of $h$, ie: $A_h
varpropto h^2$. But is there a general formula connecting $A_h$, $h$, $H$, $A_1$ and $A_2$?
geometry volume
geometry volume
asked Dec 24 '18 at 9:58
GravitonGraviton
1,09622044
asked Dec 24 '18 at 9:58
GravitonGraviton
1,09622044
edited Dec 25 '18 at 11:20
Graviton
asked Dec 24 '18 at 9:58
GravitonGraviton
1,09622044
asked Dec 24 '18 at 9:58
GravitonGraviton
1,09622044
asked Dec 24 '18 at 9:58
GravitonGraviton
1,09622044
1,09622044
$begingroup$
The combination of the bottom surface and a horizontally shifted copy of the top surface describes a new truncated pyramid with a generally different area.
$endgroup$
– random
Dec 24 '18 at 11:57
$begingroup$
@random, what does your comment has to do with my question?
$endgroup$
– Graviton
Dec 25 '18 at 9:13
$begingroup$
You are asking for the area of a truncated pyramid, which is to be derived from only its height and the areas of the upper and lower surface. Such a derivation is impossible if there are truncated pyramids that fit that description, but have different total areas. For the class of truncated pyramids described in the comment, the area of their vertical projection on the base plane can be arbitrarily large, with the real area being at least twice as large.
$endgroup$
– random
Dec 25 '18 at 10:34
$begingroup$
If your real question is described in its last paragraph instead of in the header then the answer would be that the computation additionally requires the height of the not yet truncated pyramid or the distance between the top and bottom surface.
$endgroup$
– random
Dec 25 '18 at 10:43
$begingroup$
@random you are right. I have modified the question , we do know the $H$ after all
$endgroup$
– Graviton
Dec 25 '18 at 11:21
|
show 1 more comment
$begingroup$
The combination of the bottom surface and a horizontally shifted copy of the top surface describes a new truncated pyramid with a generally different area.
$endgroup$
– random
Dec 24 '18 at 11:57
$begingroup$
@random, what does your comment has to do with my question?
$endgroup$
– Graviton
Dec 25 '18 at 9:13
$begingroup$
You are asking for the area of a truncated pyramid, which is to be derived from only its height and the areas of the upper and lower surface. Such a derivation is impossible if there are truncated pyramids that fit that description, but have different total areas. For the class of truncated pyramids described in the comment, the area of their vertical projection on the base plane can be arbitrarily large, with the real area being at least twice as large.
$endgroup$
– random
Dec 25 '18 at 10:34
$begingroup$
If your real question is described in its last paragraph instead of in the header then the answer would be that the computation additionally requires the height of the not yet truncated pyramid or the distance between the top and bottom surface.
$endgroup$
– random
Dec 25 '18 at 10:43
$begingroup$
@random you are right. I have modified the question , we do know the $H$ after all
$endgroup$
– Graviton
Dec 25 '18 at 11:21
$begingroup$
The combination of the bottom surface and a horizontally shifted copy of the top surface describes a new truncated pyramid with a generally different area.
$endgroup$
– random
Dec 24 '18 at 11:57
$begingroup$
The combination of the bottom surface and a horizontally shifted copy of the top surface describes a new truncated pyramid with a generally different area.
$endgroup$
– random
Dec 24 '18 at 11:57
$begingroup$
@random, what does your comment has to do with my question?
$endgroup$
– Graviton
Dec 25 '18 at 9:13
$begingroup$
@random, what does your comment has to do with my question?
$endgroup$
– Graviton
Dec 25 '18 at 9:13
$begingroup$
You are asking for the area of a truncated pyramid, which is to be derived from only its height and the areas of the upper and lower surface. Such a derivation is impossible if there are truncated pyramids that fit that description, but have different total areas. For the class of truncated pyramids described in the comment, the area of their vertical projection on the base plane can be arbitrarily large, with the real area being at least twice as large.
$endgroup$
– random
Dec 25 '18 at 10:34
$begingroup$
You are asking for the area of a truncated pyramid, which is to be derived from only its height and the areas of the upper and lower surface. Such a derivation is impossible if there are truncated pyramids that fit that description, but have different total areas. For the class of truncated pyramids described in the comment, the area of their vertical projection on the base plane can be arbitrarily large, with the real area being at least twice as large.
$endgroup$
– random
Dec 25 '18 at 10:34
$begingroup$
If your real question is described in its last paragraph instead of in the header then the answer would be that the computation additionally requires the height of the not yet truncated pyramid or the distance between the top and bottom surface.
$endgroup$
– random
Dec 25 '18 at 10:43
$begingroup$
If your real question is described in its last paragraph instead of in the header then the answer would be that the computation additionally requires the height of the not yet truncated pyramid or the distance between the top and bottom surface.
$endgroup$
– random
Dec 25 '18 at 10:43
$begingroup$
@random you are right. I have modified the question , we do know the $H$ after all
$endgroup$
– Graviton
Dec 25 '18 at 11:21
$begingroup$
@random you are right. I have modified the question , we do know the $H$ after all
$endgroup$
– Graviton
Dec 25 '18 at 11:21
|
show 1 more comment
1 Answer
1
active
oldest
votes
2
$begingroup$
$A_h$ is proportional with the square of the distance from its plane to the virtual top of the pyramid at height $H_{top}$, so from $(frac{H_{top}-H}{H_{top}})^2=frac{A_1}{A_2}$ it follows that $H_{top}=frac H{1-sqrt{frac{A_1}{A_2}}}$ and $A_h=(1-frac h{H_{top}})^2A_2$.
answered Dec 25 '18 at 12:25
randomrandom
57626
$endgroup$
$begingroup$
$A_h$ is proportional with the square of the distance... but how am I going to know for sure that this is true for any shape $A$? Any proof for that?
$endgroup$
– Graviton
Dec 25 '18 at 23:46
$begingroup$
The basic idea of "any shape" even having a well defined area is that it contains and is contained in shapes for which the area definition has already been accepted (such as combinations of many small rectangles) where the difference between the areas of the inner and outer shapes can be made arbitrarily small. Then the upper bound of the inner area equals the lower bound of the outer area and that value defines the area of the "any shape". The proportionality of the areas of the "any shape"s then follows from the proportionality of the areas of rectangles etc.
$endgroup$
– random
Dec 26 '18 at 0:16
$begingroup$
@Graviton Not every shape has an area. But you say the top and bottom shapes do. So the question is, how did you measure the area of the shape on the top of the frustum? Whatever the answer to that question is, scale it up by the appropriate factor and it will measure the area of any other parallel section of the frustum in the same way.
$endgroup$
– David K
Dec 26 '18 at 3:58
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
2
$begingroup$
$A_h$ is proportional with the square of the distance from its plane to the virtual top of the pyramid at height $H_{top}$, so from $(frac{H_{top}-H}{H_{top}})^2=frac{A_1}{A_2}$ it follows that $H_{top}=frac H{1-sqrt{frac{A_1}{A_2}}}$ and $A_h=(1-frac h{H_{top}})^2A_2$.
answered Dec 25 '18 at 12:25
randomrandom
57626
$endgroup$
$begingroup$
$A_h$ is proportional with the square of the distance... but how am I going to know for sure that this is true for any shape $A$? Any proof for that?
$endgroup$
– Graviton
Dec 25 '18 at 23:46
$begingroup$
The basic idea of "any shape" even having a well defined area is that it contains and is contained in shapes for which the area definition has already been accepted (such as combinations of many small rectangles) where the difference between the areas of the inner and outer shapes can be made arbitrarily small. Then the upper bound of the inner area equals the lower bound of the outer area and that value defines the area of the "any shape". The proportionality of the areas of the "any shape"s then follows from the proportionality of the areas of rectangles etc.
$endgroup$
– random
Dec 26 '18 at 0:16
$begingroup$
@Graviton Not every shape has an area. But you say the top and bottom shapes do. So the question is, how did you measure the area of the shape on the top of the frustum? Whatever the answer to that question is, scale it up by the appropriate factor and it will measure the area of any other parallel section of the frustum in the same way.
$endgroup$
– David K
Dec 26 '18 at 3:58
add a comment |
2
$begingroup$
$A_h$ is proportional with the square of the distance from its plane to the virtual top of the pyramid at height $H_{top}$, so from $(frac{H_{top}-H}{H_{top}})^2=frac{A_1}{A_2}$ it follows that $H_{top}=frac H{1-sqrt{frac{A_1}{A_2}}}$ and $A_h=(1-frac h{H_{top}})^2A_2$.
answered Dec 25 '18 at 12:25
randomrandom
57626
$endgroup$
$begingroup$
$A_h$ is proportional with the square of the distance... but how am I going to know for sure that this is true for any shape $A$? Any proof for that?
$endgroup$
– Graviton
Dec 25 '18 at 23:46
$begingroup$
The basic idea of "any shape" even having a well defined area is that it contains and is contained in shapes for which the area definition has already been accepted (such as combinations of many small rectangles) where the difference between the areas of the inner and outer shapes can be made arbitrarily small. Then the upper bound of the inner area equals the lower bound of the outer area and that value defines the area of the "any shape". The proportionality of the areas of the "any shape"s then follows from the proportionality of the areas of rectangles etc.
$endgroup$
– random
Dec 26 '18 at 0:16
$begingroup$
@Graviton Not every shape has an area. But you say the top and bottom shapes do. So the question is, how did you measure the area of the shape on the top of the frustum? Whatever the answer to that question is, scale it up by the appropriate factor and it will measure the area of any other parallel section of the frustum in the same way.
$endgroup$
– David K
Dec 26 '18 at 3:58
add a comment |
2
2
2
$begingroup$
$A_h$ is proportional with the square of the distance from its plane to the virtual top of the pyramid at height $H_{top}$, so from $(frac{H_{top}-H}{H_{top}})^2=frac{A_1}{A_2}$ it follows that $H_{top}=frac H{1-sqrt{frac{A_1}{A_2}}}$ and $A_h=(1-frac h{H_{top}})^2A_2$.
answered Dec 25 '18 at 12:25
randomrandom
57626
$endgroup$
$A_h$ is proportional with the square of the distance from its plane to the virtual top of the pyramid at height $H_{top}$, so from $(frac{H_{top}-H}{H_{top}})^2=frac{A_1}{A_2}$ it follows that $H_{top}=frac H{1-sqrt{frac{A_1}{A_2}}}$ and $A_h=(1-frac h{H_{top}})^2A_2$.
answered Dec 25 '18 at 12:25
randomrandom
57626
answered Dec 25 '18 at 12:25
randomrandom
57626
answered Dec 25 '18 at 12:25
randomrandom
57626
answered Dec 25 '18 at 12:25
randomrandom
57626
57626
$begingroup$
$A_h$ is proportional with the square of the distance... but how am I going to know for sure that this is true for any shape $A$? Any proof for that?
$endgroup$
– Graviton
Dec 25 '18 at 23:46
$begingroup$
The basic idea of "any shape" even having a well defined area is that it contains and is contained in shapes for which the area definition has already been accepted (such as combinations of many small rectangles) where the difference between the areas of the inner and outer shapes can be made arbitrarily small. Then the upper bound of the inner area equals the lower bound of the outer area and that value defines the area of the "any shape". The proportionality of the areas of the "any shape"s then follows from the proportionality of the areas of rectangles etc.
$endgroup$
– random
Dec 26 '18 at 0:16
$begingroup$
@Graviton Not every shape has an area. But you say the top and bottom shapes do. So the question is, how did you measure the area of the shape on the top of the frustum? Whatever the answer to that question is, scale it up by the appropriate factor and it will measure the area of any other parallel section of the frustum in the same way.
$endgroup$
– David K
Dec 26 '18 at 3:58
add a comment |
$begingroup$
$A_h$ is proportional with the square of the distance... but how am I going to know for sure that this is true for any shape $A$? Any proof for that?
$endgroup$
– Graviton
Dec 25 '18 at 23:46
$begingroup$
The basic idea of "any shape" even having a well defined area is that it contains and is contained in shapes for which the area definition has already been accepted (such as combinations of many small rectangles) where the difference between the areas of the inner and outer shapes can be made arbitrarily small. Then the upper bound of the inner area equals the lower bound of the outer area and that value defines the area of the "any shape". The proportionality of the areas of the "any shape"s then follows from the proportionality of the areas of rectangles etc.
$endgroup$
– random
Dec 26 '18 at 0:16
$begingroup$
@Graviton Not every shape has an area. But you say the top and bottom shapes do. So the question is, how did you measure the area of the shape on the top of the frustum? Whatever the answer to that question is, scale it up by the appropriate factor and it will measure the area of any other parallel section of the frustum in the same way.
$endgroup$
– David K
Dec 26 '18 at 3:58
$begingroup$
$A_h$ is proportional with the square of the distance... but how am I going to know for sure that this is true for any shape $A$? Any proof for that?
$endgroup$
– Graviton
Dec 25 '18 at 23:46
$begingroup$
$A_h$ is proportional with the square of the distance... but how am I going to know for sure that this is true for any shape $A$? Any proof for that?
$endgroup$
– Graviton
Dec 25 '18 at 23:46
$begingroup$
The basic idea of "any shape" even having a well defined area is that it contains and is contained in shapes for which the area definition has already been accepted (such as combinations of many small rectangles) where the difference between the areas of the inner and outer shapes can be made arbitrarily small. Then the upper bound of the inner area equals the lower bound of the outer area and that value defines the area of the "any shape". The proportionality of the areas of the "any shape"s then follows from the proportionality of the areas of rectangles etc.
$endgroup$
– random
Dec 26 '18 at 0:16
$begingroup$
The basic idea of "any shape" even having a well defined area is that it contains and is contained in shapes for which the area definition has already been accepted (such as combinations of many small rectangles) where the difference between the areas of the inner and outer shapes can be made arbitrarily small. Then the upper bound of the inner area equals the lower bound of the outer area and that value defines the area of the "any shape". The proportionality of the areas of the "any shape"s then follows from the proportionality of the areas of rectangles etc.
$endgroup$
– random
Dec 26 '18 at 0:16
$begingroup$
@Graviton Not every shape has an area. But you say the top and bottom shapes do. So the question is, how did you measure the area of the shape on the top of the frustum? Whatever the answer to that question is, scale it up by the appropriate factor and it will measure the area of any other parallel section of the frustum in the same way.
$endgroup$
– David K
Dec 26 '18 at 3:58
$begingroup$
@Graviton Not every shape has an area. But you say the top and bottom shapes do. So the question is, how did you measure the area of the shape on the top of the frustum? Whatever the answer to that question is, scale it up by the appropriate factor and it will measure the area of any other parallel section of the frustum in the same way.
$endgroup$
– David K
Dec 26 '18 at 3:58
add a comment |
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$begingroup$
The combination of the bottom surface and a horizontally shifted copy of the top surface describes a new truncated pyramid with a generally different area.
$endgroup$
– random
Dec 24 '18 at 11:57
$begingroup$
@random, what does your comment has to do with my question?
$endgroup$
– Graviton
Dec 25 '18 at 9:13
$begingroup$
You are asking for the area of a truncated pyramid, which is to be derived from only its height and the areas of the upper and lower surface. Such a derivation is impossible if there are truncated pyramids that fit that description, but have different total areas. For the class of truncated pyramids described in the comment, the area of their vertical projection on the base plane can be arbitrarily large, with the real area being at least twice as large.
$endgroup$
– random
Dec 25 '18 at 10:34
$begingroup$
If your real question is described in its last paragraph instead of in the header then the answer would be that the computation additionally requires the height of the not yet truncated pyramid or the distance between the top and bottom surface.
$endgroup$
– random
Dec 25 '18 at 10:43
$begingroup$
@random you are right. I have modified the question , we do know the $H$ after all
$endgroup$
– Graviton
Dec 25 '18 at 11:21