calculating the volume of a room with a lopsided ceiling
$begingroup$
Part of my job description requires that I find the volume of a room for calculating air leakage. Normally no problem, but this is an unusual house for many reasons. The main issue I'm having a problem with is that I don't know how to calculate the volume for a room where the ceiling slopes in two directions.
I knew that I'd have a problem with describing the room so I included a small example to illustrate the room i'm trying to find the volume of.
Would you also show the work, and/or at least provide a formula so that I can apply it on the job?? Thank you
volume
asked Jun 22 '15 at 19:55
Jacob HooperJacob Hooper
455
$endgroup$
|
show 8 more comments
$begingroup$
Part of my job description requires that I find the volume of a room for calculating air leakage. Normally no problem, but this is an unusual house for many reasons. The main issue I'm having a problem with is that I don't know how to calculate the volume for a room where the ceiling slopes in two directions.
I knew that I'd have a problem with describing the room so I included a small example to illustrate the room i'm trying to find the volume of.
Would you also show the work, and/or at least provide a formula so that I can apply it on the job?? Thank you
volume
asked Jun 22 '15 at 19:55
Jacob HooperJacob Hooper
455
$endgroup$
3
$begingroup$
Is the ceiling flat or curved? In the first case it seems the room isn't even rectangular ...
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 19:57
2
$begingroup$
Are the walls orthogonal to the floor? And I think we're missing a measure on the depth of the room.
$endgroup$
– Arthur
Jun 22 '15 at 19:59
$begingroup$
right. sorry guys, this is just an example. assume the depth is the same as the width. and yes, the ceiling is flat, and the walls are orthogonal.
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:01
2
$begingroup$
Without going too deep into massive measurements and modeling, I suppose a good approximation can be obtained by taking the average $frac{9+9+10+11}4=9.75$ as height ... (And if the base assumptions - rectangular floor, vertical walls, flat ceiling - are correct, this gives even the exact result)
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 20:02
$begingroup$
would an average actually work on this?? I had not thought of that...
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:03
|
show 8 more comments
$begingroup$
Part of my job description requires that I find the volume of a room for calculating air leakage. Normally no problem, but this is an unusual house for many reasons. The main issue I'm having a problem with is that I don't know how to calculate the volume for a room where the ceiling slopes in two directions.
I knew that I'd have a problem with describing the room so I included a small example to illustrate the room i'm trying to find the volume of.
Would you also show the work, and/or at least provide a formula so that I can apply it on the job?? Thank you
volume
asked Jun 22 '15 at 19:55
Jacob HooperJacob Hooper
455
$endgroup$
Part of my job description requires that I find the volume of a room for calculating air leakage. Normally no problem, but this is an unusual house for many reasons. The main issue I'm having a problem with is that I don't know how to calculate the volume for a room where the ceiling slopes in two directions.
I knew that I'd have a problem with describing the room so I included a small example to illustrate the room i'm trying to find the volume of.
Would you also show the work, and/or at least provide a formula so that I can apply it on the job?? Thank you
volume
volume
asked Jun 22 '15 at 19:55
Jacob HooperJacob Hooper
455
asked Jun 22 '15 at 19:55
Jacob HooperJacob Hooper
455
edited Jun 22 '15 at 20:17
Jacob Hooper
asked Jun 22 '15 at 19:55
Jacob HooperJacob Hooper
455
asked Jun 22 '15 at 19:55
Jacob HooperJacob Hooper
455
asked Jun 22 '15 at 19:55
Jacob HooperJacob Hooper
455
455
3
$begingroup$
Is the ceiling flat or curved? In the first case it seems the room isn't even rectangular ...
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 19:57
2
$begingroup$
Are the walls orthogonal to the floor? And I think we're missing a measure on the depth of the room.
$endgroup$
– Arthur
Jun 22 '15 at 19:59
$begingroup$
right. sorry guys, this is just an example. assume the depth is the same as the width. and yes, the ceiling is flat, and the walls are orthogonal.
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:01
2
$begingroup$
Without going too deep into massive measurements and modeling, I suppose a good approximation can be obtained by taking the average $frac{9+9+10+11}4=9.75$ as height ... (And if the base assumptions - rectangular floor, vertical walls, flat ceiling - are correct, this gives even the exact result)
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 20:02
$begingroup$
would an average actually work on this?? I had not thought of that...
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:03
|
show 8 more comments
3
$begingroup$
Is the ceiling flat or curved? In the first case it seems the room isn't even rectangular ...
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 19:57
2
$begingroup$
Are the walls orthogonal to the floor? And I think we're missing a measure on the depth of the room.
$endgroup$
– Arthur
Jun 22 '15 at 19:59
$begingroup$
right. sorry guys, this is just an example. assume the depth is the same as the width. and yes, the ceiling is flat, and the walls are orthogonal.
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:01
2
$begingroup$
Without going too deep into massive measurements and modeling, I suppose a good approximation can be obtained by taking the average $frac{9+9+10+11}4=9.75$ as height ... (And if the base assumptions - rectangular floor, vertical walls, flat ceiling - are correct, this gives even the exact result)
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 20:02
$begingroup$
would an average actually work on this?? I had not thought of that...
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:03
3
3
$begingroup$
Is the ceiling flat or curved? In the first case it seems the room isn't even rectangular ...
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 19:57
$begingroup$
Is the ceiling flat or curved? In the first case it seems the room isn't even rectangular ...
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 19:57
2
2
$begingroup$
Are the walls orthogonal to the floor? And I think we're missing a measure on the depth of the room.
$endgroup$
– Arthur
Jun 22 '15 at 19:59
$begingroup$
Are the walls orthogonal to the floor? And I think we're missing a measure on the depth of the room.
$endgroup$
– Arthur
Jun 22 '15 at 19:59
$begingroup$
right. sorry guys, this is just an example. assume the depth is the same as the width. and yes, the ceiling is flat, and the walls are orthogonal.
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:01
$begingroup$
right. sorry guys, this is just an example. assume the depth is the same as the width. and yes, the ceiling is flat, and the walls are orthogonal.
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:01
2
2
$begingroup$
Without going too deep into massive measurements and modeling, I suppose a good approximation can be obtained by taking the average $frac{9+9+10+11}4=9.75$ as height ... (And if the base assumptions - rectangular floor, vertical walls, flat ceiling - are correct, this gives even the exact result)
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 20:02
$begingroup$
Without going too deep into massive measurements and modeling, I suppose a good approximation can be obtained by taking the average $frac{9+9+10+11}4=9.75$ as height ... (And if the base assumptions - rectangular floor, vertical walls, flat ceiling - are correct, this gives even the exact result)
$endgroup$
– Hagen von Eitzen
Jun 22 '15 at 20:02
$begingroup$
would an average actually work on this?? I had not thought of that...
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:03
$begingroup$
would an average actually work on this?? I had not thought of that...
$endgroup$
– Jacob Hooper
Jun 22 '15 at 20:03
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I get a total volume of around 16,654 cubit feet.
See my schematic:
Hopefully it's legible enough.
Basically, we put the floor onto a coordinate plane of our choosing. Since the upper right corner is the only right angle at an extreme of the room, I chose that as the origin.
We then find the coordinates of all the corners of the room, and express the height of the ceiling as a plane; the volume of the room is then the double integral over your floor plan of that plane.
The double integral is simplest when splitting into trapezoidal regions; I couldn't see how to use less than 4 (labeled $A,B,C,D$ on the graph).
We need to use the Pythagorean theorem to find the coordinates of two points: the bottom of region $B$ and the bottom of region $C$. The equations used are to the right of the schematic.
We solve for the plane describing the height of your ceiling by using the general formula for a plane, $f(x,y)=ax+by+c$ and plugging the three points whose heights we know to find $(a,b,c)$. If we use three points you told us: $(0,0,0), (-25,-35.5,11), (-11,-38.15,10)$, we get $(a,b,c)approx(-.072,-.005,9)$, but this is inconsistent with your saying that we also have $(-25,0,9)$ (because $f(-25,0)approx 11$). I'll assume the 9 is wrong, for now; the procedure from here is the same once we've figured out the plane for the height of the ceiling:
Finally, we integrate:
$$V=intopintop_A f(x,y)dydx+intopintop_{B}f(x,y)dydx+intopintop_{C}f(x,y)dydx+intopintop_D f(x,y)dydx$$
$$V=intop_{-25}^0intop_{-17}^0f(x,y)dydx+intop_{-31}^{-25}intop_{-.187x-34.29}^{-16.5}f(x,y)dydx+intop_{-25}^{-11}intop_{-40.23-.189x}^{-17}f(x,y)dydx+intop_{-11}^{-7}intop_{20.01+5.29x}^{-17}f(x,y)dydx$$
I plugged those integrals into Wolfram Alpha (e.g.) to get the final approximations and added them.
EDIT:
In fact, with some care taken we can get WolframAlpha to give us the answer in general in terms of the $(a,b,c)$ which describe your roof by using $max$ and $min$ to describe the lower boundary (otherwise the query is too long for the website):
$$V(a,b,c)=-12781.8a-13824.9b+820.356c$$
So once you settle on which heights to use to describe your roof, solve for $(a,b,c)$ and plug it in to that simple equation.
answered Jun 22 '15 at 21:36
MichaelChiricoMichaelChirico
3,5381126
$endgroup$
$begingroup$
But the ceiling can't be planar, given the dimensions in the diagram. It must be curved, and we don't know how it's curved. So we can't possibly calculate the volume of the room.
$endgroup$
– TonyK
Jun 22 '15 at 21:50
$begingroup$
OP says "roof is flat" in the comments
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:50
$begingroup$
Yes. But that is impossible, given the heights in the diagram.
$endgroup$
– TonyK
Jun 22 '15 at 21:51
$begingroup$
He's saying 11' and 10' "assumed". I'll leave it to OP to say which of the measurements we should trust more.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:53
$begingroup$
But how did you get your answer, if the input data were inconsistent?
$endgroup$
– TonyK
Jun 22 '15 at 21:54
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
I get a total volume of around 16,654 cubit feet.
See my schematic:
Hopefully it's legible enough.
Basically, we put the floor onto a coordinate plane of our choosing. Since the upper right corner is the only right angle at an extreme of the room, I chose that as the origin.
We then find the coordinates of all the corners of the room, and express the height of the ceiling as a plane; the volume of the room is then the double integral over your floor plan of that plane.
The double integral is simplest when splitting into trapezoidal regions; I couldn't see how to use less than 4 (labeled $A,B,C,D$ on the graph).
We need to use the Pythagorean theorem to find the coordinates of two points: the bottom of region $B$ and the bottom of region $C$. The equations used are to the right of the schematic.
We solve for the plane describing the height of your ceiling by using the general formula for a plane, $f(x,y)=ax+by+c$ and plugging the three points whose heights we know to find $(a,b,c)$. If we use three points you told us: $(0,0,0), (-25,-35.5,11), (-11,-38.15,10)$, we get $(a,b,c)approx(-.072,-.005,9)$, but this is inconsistent with your saying that we also have $(-25,0,9)$ (because $f(-25,0)approx 11$). I'll assume the 9 is wrong, for now; the procedure from here is the same once we've figured out the plane for the height of the ceiling:
Finally, we integrate:
$$V=intopintop_A f(x,y)dydx+intopintop_{B}f(x,y)dydx+intopintop_{C}f(x,y)dydx+intopintop_D f(x,y)dydx$$
$$V=intop_{-25}^0intop_{-17}^0f(x,y)dydx+intop_{-31}^{-25}intop_{-.187x-34.29}^{-16.5}f(x,y)dydx+intop_{-25}^{-11}intop_{-40.23-.189x}^{-17}f(x,y)dydx+intop_{-11}^{-7}intop_{20.01+5.29x}^{-17}f(x,y)dydx$$
I plugged those integrals into Wolfram Alpha (e.g.) to get the final approximations and added them.
EDIT:
In fact, with some care taken we can get WolframAlpha to give us the answer in general in terms of the $(a,b,c)$ which describe your roof by using $max$ and $min$ to describe the lower boundary (otherwise the query is too long for the website):
$$V(a,b,c)=-12781.8a-13824.9b+820.356c$$
So once you settle on which heights to use to describe your roof, solve for $(a,b,c)$ and plug it in to that simple equation.
answered Jun 22 '15 at 21:36
MichaelChiricoMichaelChirico
3,5381126
$endgroup$
$begingroup$
But the ceiling can't be planar, given the dimensions in the diagram. It must be curved, and we don't know how it's curved. So we can't possibly calculate the volume of the room.
$endgroup$
– TonyK
Jun 22 '15 at 21:50
$begingroup$
OP says "roof is flat" in the comments
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:50
$begingroup$
Yes. But that is impossible, given the heights in the diagram.
$endgroup$
– TonyK
Jun 22 '15 at 21:51
$begingroup$
He's saying 11' and 10' "assumed". I'll leave it to OP to say which of the measurements we should trust more.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:53
$begingroup$
But how did you get your answer, if the input data were inconsistent?
$endgroup$
– TonyK
Jun 22 '15 at 21:54
|
show 3 more comments
$begingroup$
I get a total volume of around 16,654 cubit feet.
See my schematic:
Hopefully it's legible enough.
Basically, we put the floor onto a coordinate plane of our choosing. Since the upper right corner is the only right angle at an extreme of the room, I chose that as the origin.
We then find the coordinates of all the corners of the room, and express the height of the ceiling as a plane; the volume of the room is then the double integral over your floor plan of that plane.
The double integral is simplest when splitting into trapezoidal regions; I couldn't see how to use less than 4 (labeled $A,B,C,D$ on the graph).
We need to use the Pythagorean theorem to find the coordinates of two points: the bottom of region $B$ and the bottom of region $C$. The equations used are to the right of the schematic.
We solve for the plane describing the height of your ceiling by using the general formula for a plane, $f(x,y)=ax+by+c$ and plugging the three points whose heights we know to find $(a,b,c)$. If we use three points you told us: $(0,0,0), (-25,-35.5,11), (-11,-38.15,10)$, we get $(a,b,c)approx(-.072,-.005,9)$, but this is inconsistent with your saying that we also have $(-25,0,9)$ (because $f(-25,0)approx 11$). I'll assume the 9 is wrong, for now; the procedure from here is the same once we've figured out the plane for the height of the ceiling:
Finally, we integrate:
$$V=intopintop_A f(x,y)dydx+intopintop_{B}f(x,y)dydx+intopintop_{C}f(x,y)dydx+intopintop_D f(x,y)dydx$$
$$V=intop_{-25}^0intop_{-17}^0f(x,y)dydx+intop_{-31}^{-25}intop_{-.187x-34.29}^{-16.5}f(x,y)dydx+intop_{-25}^{-11}intop_{-40.23-.189x}^{-17}f(x,y)dydx+intop_{-11}^{-7}intop_{20.01+5.29x}^{-17}f(x,y)dydx$$
I plugged those integrals into Wolfram Alpha (e.g.) to get the final approximations and added them.
EDIT:
In fact, with some care taken we can get WolframAlpha to give us the answer in general in terms of the $(a,b,c)$ which describe your roof by using $max$ and $min$ to describe the lower boundary (otherwise the query is too long for the website):
$$V(a,b,c)=-12781.8a-13824.9b+820.356c$$
So once you settle on which heights to use to describe your roof, solve for $(a,b,c)$ and plug it in to that simple equation.
answered Jun 22 '15 at 21:36
MichaelChiricoMichaelChirico
3,5381126
$endgroup$
$begingroup$
But the ceiling can't be planar, given the dimensions in the diagram. It must be curved, and we don't know how it's curved. So we can't possibly calculate the volume of the room.
$endgroup$
– TonyK
Jun 22 '15 at 21:50
$begingroup$
OP says "roof is flat" in the comments
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:50
$begingroup$
Yes. But that is impossible, given the heights in the diagram.
$endgroup$
– TonyK
Jun 22 '15 at 21:51
$begingroup$
He's saying 11' and 10' "assumed". I'll leave it to OP to say which of the measurements we should trust more.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:53
$begingroup$
But how did you get your answer, if the input data were inconsistent?
$endgroup$
– TonyK
Jun 22 '15 at 21:54
|
show 3 more comments
$begingroup$
I get a total volume of around 16,654 cubit feet.
See my schematic:
Hopefully it's legible enough.
Basically, we put the floor onto a coordinate plane of our choosing. Since the upper right corner is the only right angle at an extreme of the room, I chose that as the origin.
We then find the coordinates of all the corners of the room, and express the height of the ceiling as a plane; the volume of the room is then the double integral over your floor plan of that plane.
The double integral is simplest when splitting into trapezoidal regions; I couldn't see how to use less than 4 (labeled $A,B,C,D$ on the graph).
We need to use the Pythagorean theorem to find the coordinates of two points: the bottom of region $B$ and the bottom of region $C$. The equations used are to the right of the schematic.
We solve for the plane describing the height of your ceiling by using the general formula for a plane, $f(x,y)=ax+by+c$ and plugging the three points whose heights we know to find $(a,b,c)$. If we use three points you told us: $(0,0,0), (-25,-35.5,11), (-11,-38.15,10)$, we get $(a,b,c)approx(-.072,-.005,9)$, but this is inconsistent with your saying that we also have $(-25,0,9)$ (because $f(-25,0)approx 11$). I'll assume the 9 is wrong, for now; the procedure from here is the same once we've figured out the plane for the height of the ceiling:
Finally, we integrate:
$$V=intopintop_A f(x,y)dydx+intopintop_{B}f(x,y)dydx+intopintop_{C}f(x,y)dydx+intopintop_D f(x,y)dydx$$
$$V=intop_{-25}^0intop_{-17}^0f(x,y)dydx+intop_{-31}^{-25}intop_{-.187x-34.29}^{-16.5}f(x,y)dydx+intop_{-25}^{-11}intop_{-40.23-.189x}^{-17}f(x,y)dydx+intop_{-11}^{-7}intop_{20.01+5.29x}^{-17}f(x,y)dydx$$
I plugged those integrals into Wolfram Alpha (e.g.) to get the final approximations and added them.
EDIT:
In fact, with some care taken we can get WolframAlpha to give us the answer in general in terms of the $(a,b,c)$ which describe your roof by using $max$ and $min$ to describe the lower boundary (otherwise the query is too long for the website):
$$V(a,b,c)=-12781.8a-13824.9b+820.356c$$
So once you settle on which heights to use to describe your roof, solve for $(a,b,c)$ and plug it in to that simple equation.
answered Jun 22 '15 at 21:36
MichaelChiricoMichaelChirico
3,5381126
$endgroup$
I get a total volume of around 16,654 cubit feet.
See my schematic:
Hopefully it's legible enough.
Basically, we put the floor onto a coordinate plane of our choosing. Since the upper right corner is the only right angle at an extreme of the room, I chose that as the origin.
We then find the coordinates of all the corners of the room, and express the height of the ceiling as a plane; the volume of the room is then the double integral over your floor plan of that plane.
The double integral is simplest when splitting into trapezoidal regions; I couldn't see how to use less than 4 (labeled $A,B,C,D$ on the graph).
We need to use the Pythagorean theorem to find the coordinates of two points: the bottom of region $B$ and the bottom of region $C$. The equations used are to the right of the schematic.
We solve for the plane describing the height of your ceiling by using the general formula for a plane, $f(x,y)=ax+by+c$ and plugging the three points whose heights we know to find $(a,b,c)$. If we use three points you told us: $(0,0,0), (-25,-35.5,11), (-11,-38.15,10)$, we get $(a,b,c)approx(-.072,-.005,9)$, but this is inconsistent with your saying that we also have $(-25,0,9)$ (because $f(-25,0)approx 11$). I'll assume the 9 is wrong, for now; the procedure from here is the same once we've figured out the plane for the height of the ceiling:
Finally, we integrate:
$$V=intopintop_A f(x,y)dydx+intopintop_{B}f(x,y)dydx+intopintop_{C}f(x,y)dydx+intopintop_D f(x,y)dydx$$
$$V=intop_{-25}^0intop_{-17}^0f(x,y)dydx+intop_{-31}^{-25}intop_{-.187x-34.29}^{-16.5}f(x,y)dydx+intop_{-25}^{-11}intop_{-40.23-.189x}^{-17}f(x,y)dydx+intop_{-11}^{-7}intop_{20.01+5.29x}^{-17}f(x,y)dydx$$
I plugged those integrals into Wolfram Alpha (e.g.) to get the final approximations and added them.
EDIT:
In fact, with some care taken we can get WolframAlpha to give us the answer in general in terms of the $(a,b,c)$ which describe your roof by using $max$ and $min$ to describe the lower boundary (otherwise the query is too long for the website):
$$V(a,b,c)=-12781.8a-13824.9b+820.356c$$
So once you settle on which heights to use to describe your roof, solve for $(a,b,c)$ and plug it in to that simple equation.
answered Jun 22 '15 at 21:36
MichaelChiricoMichaelChirico
3,5381126
edited Jun 22 '15 at 23:40
answered Jun 22 '15 at 21:36
MichaelChiricoMichaelChirico
3,5381126
answered Jun 22 '15 at 21:36
MichaelChiricoMichaelChirico
3,5381126
answered Jun 22 '15 at 21:36
MichaelChiricoMichaelChirico
3,5381126
3,5381126
$begingroup$
But the ceiling can't be planar, given the dimensions in the diagram. It must be curved, and we don't know how it's curved. So we can't possibly calculate the volume of the room.
$endgroup$
– TonyK
Jun 22 '15 at 21:50
$begingroup$
OP says "roof is flat" in the comments
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:50
$begingroup$
Yes. But that is impossible, given the heights in the diagram.
$endgroup$
– TonyK
Jun 22 '15 at 21:51
$begingroup$
He's saying 11' and 10' "assumed". I'll leave it to OP to say which of the measurements we should trust more.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:53
$begingroup$
But how did you get your answer, if the input data were inconsistent?
$endgroup$
– TonyK
Jun 22 '15 at 21:54
|
show 3 more comments
$begingroup$
But the ceiling can't be planar, given the dimensions in the diagram. It must be curved, and we don't know how it's curved. So we can't possibly calculate the volume of the room.
$endgroup$
– TonyK
Jun 22 '15 at 21:50
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OP says "roof is flat" in the comments
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– MichaelChirico
Jun 22 '15 at 21:50
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Yes. But that is impossible, given the heights in the diagram.
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– TonyK
Jun 22 '15 at 21:51
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He's saying 11' and 10' "assumed". I'll leave it to OP to say which of the measurements we should trust more.
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– MichaelChirico
Jun 22 '15 at 21:53
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But how did you get your answer, if the input data were inconsistent?
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– TonyK
Jun 22 '15 at 21:54
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But the ceiling can't be planar, given the dimensions in the diagram. It must be curved, and we don't know how it's curved. So we can't possibly calculate the volume of the room.
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– TonyK
Jun 22 '15 at 21:50
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But the ceiling can't be planar, given the dimensions in the diagram. It must be curved, and we don't know how it's curved. So we can't possibly calculate the volume of the room.
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– TonyK
Jun 22 '15 at 21:50
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OP says "roof is flat" in the comments
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– MichaelChirico
Jun 22 '15 at 21:50
$begingroup$
OP says "roof is flat" in the comments
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:50
$begingroup$
Yes. But that is impossible, given the heights in the diagram.
$endgroup$
– TonyK
Jun 22 '15 at 21:51
$begingroup$
Yes. But that is impossible, given the heights in the diagram.
$endgroup$
– TonyK
Jun 22 '15 at 21:51
$begingroup$
He's saying 11' and 10' "assumed". I'll leave it to OP to say which of the measurements we should trust more.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:53
$begingroup$
He's saying 11' and 10' "assumed". I'll leave it to OP to say which of the measurements we should trust more.
$endgroup$
– MichaelChirico
Jun 22 '15 at 21:53
$begingroup$
But how did you get your answer, if the input data were inconsistent?
$endgroup$
– TonyK
Jun 22 '15 at 21:54
$begingroup$
But how did you get your answer, if the input data were inconsistent?
$endgroup$
– TonyK
Jun 22 '15 at 21:54
|
show 3 more comments
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3
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Is the ceiling flat or curved? In the first case it seems the room isn't even rectangular ...
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– Hagen von Eitzen
Jun 22 '15 at 19:57
2
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Are the walls orthogonal to the floor? And I think we're missing a measure on the depth of the room.
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– Arthur
Jun 22 '15 at 19:59
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right. sorry guys, this is just an example. assume the depth is the same as the width. and yes, the ceiling is flat, and the walls are orthogonal.
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– Jacob Hooper
Jun 22 '15 at 20:01
2
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Without going too deep into massive measurements and modeling, I suppose a good approximation can be obtained by taking the average $frac{9+9+10+11}4=9.75$ as height ... (And if the base assumptions - rectangular floor, vertical walls, flat ceiling - are correct, this gives even the exact result)
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– Hagen von Eitzen
Jun 22 '15 at 20:02
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would an average actually work on this?? I had not thought of that...
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– Jacob Hooper
Jun 22 '15 at 20:03