Mixing
Height of a part volume in a cone
$begingroup$
Hi,
I know this is high school math but i feel kinda stupid right now so I am asking it here:
Given is a truncated cone (upside down if relevant). Given is the volume as well as the upper and lower radius.
Now the cone got filled with a given volume. How do i determine the height of the filled liquid?
As you see in the picture, there is a cone (not a triangle). Given is r1w, r2w, and VW and VA
With that at least I am able to calculate everything else, but not hA. Can someone of you please help me?
geometry
edited Jan 29 at 9:46
Larry
2,53031131
asked Jan 29 at 9:40
FrostyFrosty
32
$endgroup$
add a comment |
$begingroup$
Hi,
I know this is high school math but i feel kinda stupid right now so I am asking it here:
Given is a truncated cone (upside down if relevant). Given is the volume as well as the upper and lower radius.
Now the cone got filled with a given volume. How do i determine the height of the filled liquid?
As you see in the picture, there is a cone (not a triangle). Given is r1w, r2w, and VW and VA
With that at least I am able to calculate everything else, but not hA. Can someone of you please help me?
geometry
edited Jan 29 at 9:46
Larry
2,53031131
asked Jan 29 at 9:40
FrostyFrosty
32
$endgroup$
add a comment |
$begingroup$
Hi,
I know this is high school math but i feel kinda stupid right now so I am asking it here:
Given is a truncated cone (upside down if relevant). Given is the volume as well as the upper and lower radius.
Now the cone got filled with a given volume. How do i determine the height of the filled liquid?
As you see in the picture, there is a cone (not a triangle). Given is r1w, r2w, and VW and VA
With that at least I am able to calculate everything else, but not hA. Can someone of you please help me?
geometry
edited Jan 29 at 9:46
Larry
2,53031131
asked Jan 29 at 9:40
FrostyFrosty
32
$endgroup$
Hi,
I know this is high school math but i feel kinda stupid right now so I am asking it here:
Given is a truncated cone (upside down if relevant). Given is the volume as well as the upper and lower radius.
Now the cone got filled with a given volume. How do i determine the height of the filled liquid?
As you see in the picture, there is a cone (not a triangle). Given is r1w, r2w, and VW and VA
With that at least I am able to calculate everything else, but not hA. Can someone of you please help me?
geometry
geometry
edited Jan 29 at 9:46
Larry
2,53031131
asked Jan 29 at 9:40
FrostyFrosty
32
edited Jan 29 at 9:46
Larry
2,53031131
asked Jan 29 at 9:40
FrostyFrosty
32
edited Jan 29 at 9:46
Larry
2,53031131
edited Jan 29 at 9:46
Larry
2,53031131
edited Jan 29 at 9:46
Larry
2,53031131
2,53031131
asked Jan 29 at 9:40
FrostyFrosty
32
asked Jan 29 at 9:40
FrostyFrosty
32
asked Jan 29 at 9:40
FrostyFrosty
32
32
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $alpha$ is the half-aperture angle of a cone, $r$ its base radius and $h$ its height, then $h=rcotalpha$ and the volume of the cone can be written as $V={piover3}r^3cotalpha$.
Let then $r_A$ be the radius of the surface of the liquid. We have:
$$
V_W={piover3}cotalpha(r_{2W}^3-r_{1W}^3),
quad
V_A={piover3}cotalpha(r_{A}^3-r_{1W}^3).
$$
From the first equation we get
$$
cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
$$
and substituting that into the second equation we can solve for $r_A$:
$$
r_A=root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}
$$
and finally:
$$
h_A=(r_A-r_{1W})cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
left(root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}-r_{1W}right).
$$
answered Jan 29 at 23:00
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
Thank you so far. I now get how you got to $r_A$ . But the next line is not really clear to me. You said $h=rcotalpha$, so why did you said in the last equation $h_A=(r_A-r_{1W})cotalpha$? Applying the volume formula of a a conial frustum to $V_A$: $$ {V_A={1over3}pi(r_A^2+r_Ar_{1W}+r_{1W}^2)*h_A} $$ Solve for $h_A$: $$ {h_A = {V_Aover{1over3}pi(r_A^2+r_Ar_{1w}+r_{1w}^2)}} $$ Now comparing your $h_A$ solution to mine, there are differences. Where is my mistake?
$endgroup$
– Frosty
Jan 30 at 9:08
$begingroup$
Height $h$ in $h=rcotalpha$ is referred to the whole cone, i.e. it is the height from the cone vertex to the base. The height of a frustum is the difference of the heights of two cones.
$endgroup$
– Aretino
Jan 30 at 9:54
$begingroup$
I think your formula (it isn't easily readable in comment above) should give the same results as mine: please check for some numerical values of the variables.
$endgroup$
– Aretino
Jan 30 at 9:57
$begingroup$
Sorry for the weird formula. I typed it in many other mathjax-previews and they were shown without failours. I went wrong. Maybe now its going to work. I got a Volume formula for a truncated cone and solved it for $h_A$: $h_A={V_Aover{1over3}pi(r_A^2+r_A*r_{1w}+r_{1w}^2)}$.
$endgroup$
– Frosty
Jan 30 at 12:08
$begingroup$
Now if i type i try your take your $r_A$ and $h_A$ i get other results as with my $h_A$ calculation.
$endgroup$
– Frosty
Jan 30 at 12:13
|
show 2 more comments
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1 Answer
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1 Answer
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$begingroup$
If $alpha$ is the half-aperture angle of a cone, $r$ its base radius and $h$ its height, then $h=rcotalpha$ and the volume of the cone can be written as $V={piover3}r^3cotalpha$.
Let then $r_A$ be the radius of the surface of the liquid. We have:
$$
V_W={piover3}cotalpha(r_{2W}^3-r_{1W}^3),
quad
V_A={piover3}cotalpha(r_{A}^3-r_{1W}^3).
$$
From the first equation we get
$$
cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
$$
and substituting that into the second equation we can solve for $r_A$:
$$
r_A=root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}
$$
and finally:
$$
h_A=(r_A-r_{1W})cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
left(root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}-r_{1W}right).
$$
answered Jan 29 at 23:00
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
Thank you so far. I now get how you got to $r_A$ . But the next line is not really clear to me. You said $h=rcotalpha$, so why did you said in the last equation $h_A=(r_A-r_{1W})cotalpha$? Applying the volume formula of a a conial frustum to $V_A$: $$ {V_A={1over3}pi(r_A^2+r_Ar_{1W}+r_{1W}^2)*h_A} $$ Solve for $h_A$: $$ {h_A = {V_Aover{1over3}pi(r_A^2+r_Ar_{1w}+r_{1w}^2)}} $$ Now comparing your $h_A$ solution to mine, there are differences. Where is my mistake?
$endgroup$
– Frosty
Jan 30 at 9:08
$begingroup$
Height $h$ in $h=rcotalpha$ is referred to the whole cone, i.e. it is the height from the cone vertex to the base. The height of a frustum is the difference of the heights of two cones.
$endgroup$
– Aretino
Jan 30 at 9:54
$begingroup$
I think your formula (it isn't easily readable in comment above) should give the same results as mine: please check for some numerical values of the variables.
$endgroup$
– Aretino
Jan 30 at 9:57
$begingroup$
Sorry for the weird formula. I typed it in many other mathjax-previews and they were shown without failours. I went wrong. Maybe now its going to work. I got a Volume formula for a truncated cone and solved it for $h_A$: $h_A={V_Aover{1over3}pi(r_A^2+r_A*r_{1w}+r_{1w}^2)}$.
$endgroup$
– Frosty
Jan 30 at 12:08
$begingroup$
Now if i type i try your take your $r_A$ and $h_A$ i get other results as with my $h_A$ calculation.
$endgroup$
– Frosty
Jan 30 at 12:13
|
show 2 more comments
$begingroup$
If $alpha$ is the half-aperture angle of a cone, $r$ its base radius and $h$ its height, then $h=rcotalpha$ and the volume of the cone can be written as $V={piover3}r^3cotalpha$.
Let then $r_A$ be the radius of the surface of the liquid. We have:
$$
V_W={piover3}cotalpha(r_{2W}^3-r_{1W}^3),
quad
V_A={piover3}cotalpha(r_{A}^3-r_{1W}^3).
$$
From the first equation we get
$$
cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
$$
and substituting that into the second equation we can solve for $r_A$:
$$
r_A=root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}
$$
and finally:
$$
h_A=(r_A-r_{1W})cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
left(root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}-r_{1W}right).
$$
answered Jan 29 at 23:00
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
Thank you so far. I now get how you got to $r_A$ . But the next line is not really clear to me. You said $h=rcotalpha$, so why did you said in the last equation $h_A=(r_A-r_{1W})cotalpha$? Applying the volume formula of a a conial frustum to $V_A$: $$ {V_A={1over3}pi(r_A^2+r_Ar_{1W}+r_{1W}^2)*h_A} $$ Solve for $h_A$: $$ {h_A = {V_Aover{1over3}pi(r_A^2+r_Ar_{1w}+r_{1w}^2)}} $$ Now comparing your $h_A$ solution to mine, there are differences. Where is my mistake?
$endgroup$
– Frosty
Jan 30 at 9:08
$begingroup$
Height $h$ in $h=rcotalpha$ is referred to the whole cone, i.e. it is the height from the cone vertex to the base. The height of a frustum is the difference of the heights of two cones.
$endgroup$
– Aretino
Jan 30 at 9:54
$begingroup$
I think your formula (it isn't easily readable in comment above) should give the same results as mine: please check for some numerical values of the variables.
$endgroup$
– Aretino
Jan 30 at 9:57
$begingroup$
Sorry for the weird formula. I typed it in many other mathjax-previews and they were shown without failours. I went wrong. Maybe now its going to work. I got a Volume formula for a truncated cone and solved it for $h_A$: $h_A={V_Aover{1over3}pi(r_A^2+r_A*r_{1w}+r_{1w}^2)}$.
$endgroup$
– Frosty
Jan 30 at 12:08
$begingroup$
Now if i type i try your take your $r_A$ and $h_A$ i get other results as with my $h_A$ calculation.
$endgroup$
– Frosty
Jan 30 at 12:13
|
show 2 more comments
$begingroup$
If $alpha$ is the half-aperture angle of a cone, $r$ its base radius and $h$ its height, then $h=rcotalpha$ and the volume of the cone can be written as $V={piover3}r^3cotalpha$.
Let then $r_A$ be the radius of the surface of the liquid. We have:
$$
V_W={piover3}cotalpha(r_{2W}^3-r_{1W}^3),
quad
V_A={piover3}cotalpha(r_{A}^3-r_{1W}^3).
$$
From the first equation we get
$$
cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
$$
and substituting that into the second equation we can solve for $r_A$:
$$
r_A=root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}
$$
and finally:
$$
h_A=(r_A-r_{1W})cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
left(root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}-r_{1W}right).
$$
answered Jan 29 at 23:00
AretinoAretino
25.8k31545
$endgroup$
If $alpha$ is the half-aperture angle of a cone, $r$ its base radius and $h$ its height, then $h=rcotalpha$ and the volume of the cone can be written as $V={piover3}r^3cotalpha$.
Let then $r_A$ be the radius of the surface of the liquid. We have:
$$
V_W={piover3}cotalpha(r_{2W}^3-r_{1W}^3),
quad
V_A={piover3}cotalpha(r_{A}^3-r_{1W}^3).
$$
From the first equation we get
$$
cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
$$
and substituting that into the second equation we can solve for $r_A$:
$$
r_A=root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}
$$
and finally:
$$
h_A=(r_A-r_{1W})cotalpha={3V_Wover{pi}(r_{2W}^3-r_{1W}^3)}
left(root3of{{V_Aover V_W}(r_{2W}^3-r_{1W}^3)+r_{1W}^3}-r_{1W}right).
$$
answered Jan 29 at 23:00
AretinoAretino
25.8k31545
answered Jan 29 at 23:00
AretinoAretino
25.8k31545
answered Jan 29 at 23:00
AretinoAretino
25.8k31545
answered Jan 29 at 23:00
AretinoAretino
25.8k31545
25.8k31545
$begingroup$
Thank you so far. I now get how you got to $r_A$ . But the next line is not really clear to me. You said $h=rcotalpha$, so why did you said in the last equation $h_A=(r_A-r_{1W})cotalpha$? Applying the volume formula of a a conial frustum to $V_A$: $$ {V_A={1over3}pi(r_A^2+r_Ar_{1W}+r_{1W}^2)*h_A} $$ Solve for $h_A$: $$ {h_A = {V_Aover{1over3}pi(r_A^2+r_Ar_{1w}+r_{1w}^2)}} $$ Now comparing your $h_A$ solution to mine, there are differences. Where is my mistake?
$endgroup$
– Frosty
Jan 30 at 9:08
$begingroup$
Height $h$ in $h=rcotalpha$ is referred to the whole cone, i.e. it is the height from the cone vertex to the base. The height of a frustum is the difference of the heights of two cones.
$endgroup$
– Aretino
Jan 30 at 9:54
$begingroup$
I think your formula (it isn't easily readable in comment above) should give the same results as mine: please check for some numerical values of the variables.
$endgroup$
– Aretino
Jan 30 at 9:57
$begingroup$
Sorry for the weird formula. I typed it in many other mathjax-previews and they were shown without failours. I went wrong. Maybe now its going to work. I got a Volume formula for a truncated cone and solved it for $h_A$: $h_A={V_Aover{1over3}pi(r_A^2+r_A*r_{1w}+r_{1w}^2)}$.
$endgroup$
– Frosty
Jan 30 at 12:08
$begingroup$
Now if i type i try your take your $r_A$ and $h_A$ i get other results as with my $h_A$ calculation.
$endgroup$
– Frosty
Jan 30 at 12:13
|
show 2 more comments
$begingroup$
Thank you so far. I now get how you got to $r_A$ . But the next line is not really clear to me. You said $h=rcotalpha$, so why did you said in the last equation $h_A=(r_A-r_{1W})cotalpha$? Applying the volume formula of a a conial frustum to $V_A$: $$ {V_A={1over3}pi(r_A^2+r_Ar_{1W}+r_{1W}^2)*h_A} $$ Solve for $h_A$: $$ {h_A = {V_Aover{1over3}pi(r_A^2+r_Ar_{1w}+r_{1w}^2)}} $$ Now comparing your $h_A$ solution to mine, there are differences. Where is my mistake?
$endgroup$
– Frosty
Jan 30 at 9:08
$begingroup$
Height $h$ in $h=rcotalpha$ is referred to the whole cone, i.e. it is the height from the cone vertex to the base. The height of a frustum is the difference of the heights of two cones.
$endgroup$
– Aretino
Jan 30 at 9:54
$begingroup$
I think your formula (it isn't easily readable in comment above) should give the same results as mine: please check for some numerical values of the variables.
$endgroup$
– Aretino
Jan 30 at 9:57
$begingroup$
Sorry for the weird formula. I typed it in many other mathjax-previews and they were shown without failours. I went wrong. Maybe now its going to work. I got a Volume formula for a truncated cone and solved it for $h_A$: $h_A={V_Aover{1over3}pi(r_A^2+r_A*r_{1w}+r_{1w}^2)}$.
$endgroup$
– Frosty
Jan 30 at 12:08
$begingroup$
Now if i type i try your take your $r_A$ and $h_A$ i get other results as with my $h_A$ calculation.
$endgroup$
– Frosty
Jan 30 at 12:13
$begingroup$
Thank you so far. I now get how you got to $r_A$ . But the next line is not really clear to me. You said $h=rcotalpha$, so why did you said in the last equation $h_A=(r_A-r_{1W})cotalpha$? Applying the volume formula of a a conial frustum to $V_A$: $$ {V_A={1over3}pi(r_A^2+r_Ar_{1W}+r_{1W}^2)*h_A} $$ Solve for $h_A$: $$ {h_A = {V_Aover{1over3}pi(r_A^2+r_Ar_{1w}+r_{1w}^2)}} $$ Now comparing your $h_A$ solution to mine, there are differences. Where is my mistake?
$endgroup$
– Frosty
Jan 30 at 9:08
$begingroup$
Thank you so far. I now get how you got to $r_A$ . But the next line is not really clear to me. You said $h=rcotalpha$, so why did you said in the last equation $h_A=(r_A-r_{1W})cotalpha$? Applying the volume formula of a a conial frustum to $V_A$: $$ {V_A={1over3}pi(r_A^2+r_Ar_{1W}+r_{1W}^2)*h_A} $$ Solve for $h_A$: $$ {h_A = {V_Aover{1over3}pi(r_A^2+r_Ar_{1w}+r_{1w}^2)}} $$ Now comparing your $h_A$ solution to mine, there are differences. Where is my mistake?
$endgroup$
– Frosty
Jan 30 at 9:08
$begingroup$
Height $h$ in $h=rcotalpha$ is referred to the whole cone, i.e. it is the height from the cone vertex to the base. The height of a frustum is the difference of the heights of two cones.
$endgroup$
– Aretino
Jan 30 at 9:54
$begingroup$
Height $h$ in $h=rcotalpha$ is referred to the whole cone, i.e. it is the height from the cone vertex to the base. The height of a frustum is the difference of the heights of two cones.
$endgroup$
– Aretino
Jan 30 at 9:54
$begingroup$
I think your formula (it isn't easily readable in comment above) should give the same results as mine: please check for some numerical values of the variables.
$endgroup$
– Aretino
Jan 30 at 9:57
$begingroup$
I think your formula (it isn't easily readable in comment above) should give the same results as mine: please check for some numerical values of the variables.
$endgroup$
– Aretino
Jan 30 at 9:57
$begingroup$
Sorry for the weird formula. I typed it in many other mathjax-previews and they were shown without failours. I went wrong. Maybe now its going to work. I got a Volume formula for a truncated cone and solved it for $h_A$: $h_A={V_Aover{1over3}pi(r_A^2+r_A*r_{1w}+r_{1w}^2)}$.
$endgroup$
– Frosty
Jan 30 at 12:08
$begingroup$
Sorry for the weird formula. I typed it in many other mathjax-previews and they were shown without failours. I went wrong. Maybe now its going to work. I got a Volume formula for a truncated cone and solved it for $h_A$: $h_A={V_Aover{1over3}pi(r_A^2+r_A*r_{1w}+r_{1w}^2)}$.
$endgroup$
– Frosty
Jan 30 at 12:08
$begingroup$
Now if i type i try your take your $r_A$ and $h_A$ i get other results as with my $h_A$ calculation.
$endgroup$
– Frosty
Jan 30 at 12:13
$begingroup$
Now if i type i try your take your $r_A$ and $h_A$ i get other results as with my $h_A$ calculation.
$endgroup$
– Frosty
Jan 30 at 12:13
|
show 2 more comments
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