Calculating height in a truncated cone using volume
$begingroup$
I need to measure the height liquid would come up to in a jug I have. I need to determin how far up the jug the liquid would go (similar to markings on a measuring jug).
Using the volume and lower radius and slant angles to Im looking for a formula I could use. (After translating ml in mmcubed to make it possible)
As an example, if i know the dimensions of the vessel, I would want to know how far up 200ml of liquid would go.
Thanks in advance people!
geometry
asked Dec 21 '15 at 11:13
WillWill
62
$endgroup$
add a comment |
$begingroup$
I need to measure the height liquid would come up to in a jug I have. I need to determin how far up the jug the liquid would go (similar to markings on a measuring jug).
Using the volume and lower radius and slant angles to Im looking for a formula I could use. (After translating ml in mmcubed to make it possible)
As an example, if i know the dimensions of the vessel, I would want to know how far up 200ml of liquid would go.
Thanks in advance people!
geometry
asked Dec 21 '15 at 11:13
WillWill
62
$endgroup$
add a comment |
$begingroup$
I need to measure the height liquid would come up to in a jug I have. I need to determin how far up the jug the liquid would go (similar to markings on a measuring jug).
Using the volume and lower radius and slant angles to Im looking for a formula I could use. (After translating ml in mmcubed to make it possible)
As an example, if i know the dimensions of the vessel, I would want to know how far up 200ml of liquid would go.
Thanks in advance people!
geometry
asked Dec 21 '15 at 11:13
WillWill
62
$endgroup$
I need to measure the height liquid would come up to in a jug I have. I need to determin how far up the jug the liquid would go (similar to markings on a measuring jug).
Using the volume and lower radius and slant angles to Im looking for a formula I could use. (After translating ml in mmcubed to make it possible)
As an example, if i know the dimensions of the vessel, I would want to know how far up 200ml of liquid would go.
Thanks in advance people!
geometry
geometry
asked Dec 21 '15 at 11:13
WillWill
62
asked Dec 21 '15 at 11:13
WillWill
62
asked Dec 21 '15 at 11:13
WillWill
62
asked Dec 21 '15 at 11:13
WillWill
62
asked Dec 21 '15 at 11:13
WillWill
62
62
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Ok, so call $r$ the radius of the basis, $theta$ the angle between the vertical axis of the cone and the surface.
$$color{red}{h}=frac{r}{tan theta}-sqrt[3]{frac{r^3}{tan^3 theta}-frac{3V}{pi tan^2 theta}}.$$
Of course you have to plug in the measures with the right units, for example: $r$ in $mm$ and $V$ in $mm^3$.
EXAMPLE: you have that $r=45mm$, the graph below for computing $tan theta$ remains valid, so $tan theta= frac{6.5}{110}$. Let's say you want $V=200ml=200000mm^3$, the formula gives
$$h=frac{45cdot 110}{6.5}-sqrt[3]{frac{45^3cdot 6.5^3}{110^3}-frac{3cdot 200000cdot 110^2}{picdot 6.5^2}}=sim 32.8 mm$$
EDIT: Since $tan theta$ can be difficulte to find, this is an easier way you can calculate it: here is a section of the truncated cone, you have to measure the lenghts $a$ and $b$. Then
$$tan theta =frac{a}{b}$$
Once you have found $tan theta$, it should be easy to compute $tan^3 theta$ with a calculator. The quantity $r$ can be measured directly and $V$ is known.
answered Dec 21 '15 at 12:44
mrprottolomrprottolo
3,2291718
$endgroup$
$begingroup$
Nice one. Thanks, I'll give it a go
$endgroup$
– Will
Dec 24 '15 at 10:15
$begingroup$
Ive tried to work this formula out and cant seem to make sense of the second half of the formula to calculate height from volume, and also the part 'h=x-r/tan...'. I dont seem to be able to calculate tan squared and tan cubed on my calculator which may be a problem. Im getting a figure buts its quite a bit out.
$endgroup$
– Will
Dec 29 '15 at 23:09
$begingroup$
Please help me firgure this out if you dont mind. Many thanks
$endgroup$
– Will
Dec 29 '15 at 23:12
$begingroup$
@will See if the edit can help you. Also the only formula you need is the last one, merked in red, which gives you the height given the volume of liquid.
$endgroup$
– mrprottolo
Dec 29 '15 at 23:33
$begingroup$
Hi, thanks for the help but im still not there. Would it matter if the truncated cone I am working with is the other way up, with the wide part at the bottom rather than the top? Because the height equation I am using still doesnt give me the correct results. I am using a base radius of 45mm and a vertical height of 110m with an upper radius of 38.5mm (if you can envisage the shape with that in mind). Should I be using Cos instead? Etc.
$endgroup$
– Will
Jan 2 '16 at 6:11
|
show 4 more comments
$begingroup$
Assume the container has shape of a truncated cone .. $r$ is minimum radius at bottom, $R$ at top, the cone has a flare up with $alpha$, $h$ is height reached. Using well known relation of truncated cone volume we work up the algebra, defining $H,a^3$ as
$$ V = frac{pi h}{3} ( R^2 + r^2 + R r ), quad tan alpha = frac{R-r}{h}
,quad H = h tan alpha,, , a^3= frac{V tan alpha}{pi} tag1 $$
When $R$ is eliminated to make a cubic in $H$,
$$ H^3/3 + H^2 r + H r^2 - a^3 = 0 tag2 $$
whose only real root is found by Mathematica as
$$ H = (3 a^3 +r^3)^{frac13} - r $$
answered Apr 5 '17 at 12:02
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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active
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$begingroup$
Ok, so call $r$ the radius of the basis, $theta$ the angle between the vertical axis of the cone and the surface.
$$color{red}{h}=frac{r}{tan theta}-sqrt[3]{frac{r^3}{tan^3 theta}-frac{3V}{pi tan^2 theta}}.$$
Of course you have to plug in the measures with the right units, for example: $r$ in $mm$ and $V$ in $mm^3$.
EXAMPLE: you have that $r=45mm$, the graph below for computing $tan theta$ remains valid, so $tan theta= frac{6.5}{110}$. Let's say you want $V=200ml=200000mm^3$, the formula gives
$$h=frac{45cdot 110}{6.5}-sqrt[3]{frac{45^3cdot 6.5^3}{110^3}-frac{3cdot 200000cdot 110^2}{picdot 6.5^2}}=sim 32.8 mm$$
EDIT: Since $tan theta$ can be difficulte to find, this is an easier way you can calculate it: here is a section of the truncated cone, you have to measure the lenghts $a$ and $b$. Then
$$tan theta =frac{a}{b}$$
Once you have found $tan theta$, it should be easy to compute $tan^3 theta$ with a calculator. The quantity $r$ can be measured directly and $V$ is known.
answered Dec 21 '15 at 12:44
mrprottolomrprottolo
3,2291718
$endgroup$
$begingroup$
Nice one. Thanks, I'll give it a go
$endgroup$
– Will
Dec 24 '15 at 10:15
$begingroup$
Ive tried to work this formula out and cant seem to make sense of the second half of the formula to calculate height from volume, and also the part 'h=x-r/tan...'. I dont seem to be able to calculate tan squared and tan cubed on my calculator which may be a problem. Im getting a figure buts its quite a bit out.
$endgroup$
– Will
Dec 29 '15 at 23:09
$begingroup$
Please help me firgure this out if you dont mind. Many thanks
$endgroup$
– Will
Dec 29 '15 at 23:12
$begingroup$
@will See if the edit can help you. Also the only formula you need is the last one, merked in red, which gives you the height given the volume of liquid.
$endgroup$
– mrprottolo
Dec 29 '15 at 23:33
$begingroup$
Hi, thanks for the help but im still not there. Would it matter if the truncated cone I am working with is the other way up, with the wide part at the bottom rather than the top? Because the height equation I am using still doesnt give me the correct results. I am using a base radius of 45mm and a vertical height of 110m with an upper radius of 38.5mm (if you can envisage the shape with that in mind). Should I be using Cos instead? Etc.
$endgroup$
– Will
Jan 2 '16 at 6:11
|
show 4 more comments
$begingroup$
Ok, so call $r$ the radius of the basis, $theta$ the angle between the vertical axis of the cone and the surface.
$$color{red}{h}=frac{r}{tan theta}-sqrt[3]{frac{r^3}{tan^3 theta}-frac{3V}{pi tan^2 theta}}.$$
Of course you have to plug in the measures with the right units, for example: $r$ in $mm$ and $V$ in $mm^3$.
EXAMPLE: you have that $r=45mm$, the graph below for computing $tan theta$ remains valid, so $tan theta= frac{6.5}{110}$. Let's say you want $V=200ml=200000mm^3$, the formula gives
$$h=frac{45cdot 110}{6.5}-sqrt[3]{frac{45^3cdot 6.5^3}{110^3}-frac{3cdot 200000cdot 110^2}{picdot 6.5^2}}=sim 32.8 mm$$
EDIT: Since $tan theta$ can be difficulte to find, this is an easier way you can calculate it: here is a section of the truncated cone, you have to measure the lenghts $a$ and $b$. Then
$$tan theta =frac{a}{b}$$
Once you have found $tan theta$, it should be easy to compute $tan^3 theta$ with a calculator. The quantity $r$ can be measured directly and $V$ is known.
answered Dec 21 '15 at 12:44
mrprottolomrprottolo
3,2291718
$endgroup$
$begingroup$
Nice one. Thanks, I'll give it a go
$endgroup$
– Will
Dec 24 '15 at 10:15
$begingroup$
Ive tried to work this formula out and cant seem to make sense of the second half of the formula to calculate height from volume, and also the part 'h=x-r/tan...'. I dont seem to be able to calculate tan squared and tan cubed on my calculator which may be a problem. Im getting a figure buts its quite a bit out.
$endgroup$
– Will
Dec 29 '15 at 23:09
$begingroup$
Please help me firgure this out if you dont mind. Many thanks
$endgroup$
– Will
Dec 29 '15 at 23:12
$begingroup$
@will See if the edit can help you. Also the only formula you need is the last one, merked in red, which gives you the height given the volume of liquid.
$endgroup$
– mrprottolo
Dec 29 '15 at 23:33
$begingroup$
Hi, thanks for the help but im still not there. Would it matter if the truncated cone I am working with is the other way up, with the wide part at the bottom rather than the top? Because the height equation I am using still doesnt give me the correct results. I am using a base radius of 45mm and a vertical height of 110m with an upper radius of 38.5mm (if you can envisage the shape with that in mind). Should I be using Cos instead? Etc.
$endgroup$
– Will
Jan 2 '16 at 6:11
|
show 4 more comments
$begingroup$
Ok, so call $r$ the radius of the basis, $theta$ the angle between the vertical axis of the cone and the surface.
$$color{red}{h}=frac{r}{tan theta}-sqrt[3]{frac{r^3}{tan^3 theta}-frac{3V}{pi tan^2 theta}}.$$
Of course you have to plug in the measures with the right units, for example: $r$ in $mm$ and $V$ in $mm^3$.
EXAMPLE: you have that $r=45mm$, the graph below for computing $tan theta$ remains valid, so $tan theta= frac{6.5}{110}$. Let's say you want $V=200ml=200000mm^3$, the formula gives
$$h=frac{45cdot 110}{6.5}-sqrt[3]{frac{45^3cdot 6.5^3}{110^3}-frac{3cdot 200000cdot 110^2}{picdot 6.5^2}}=sim 32.8 mm$$
EDIT: Since $tan theta$ can be difficulte to find, this is an easier way you can calculate it: here is a section of the truncated cone, you have to measure the lenghts $a$ and $b$. Then
$$tan theta =frac{a}{b}$$
Once you have found $tan theta$, it should be easy to compute $tan^3 theta$ with a calculator. The quantity $r$ can be measured directly and $V$ is known.
answered Dec 21 '15 at 12:44
mrprottolomrprottolo
3,2291718
$endgroup$
Ok, so call $r$ the radius of the basis, $theta$ the angle between the vertical axis of the cone and the surface.
$$color{red}{h}=frac{r}{tan theta}-sqrt[3]{frac{r^3}{tan^3 theta}-frac{3V}{pi tan^2 theta}}.$$
Of course you have to plug in the measures with the right units, for example: $r$ in $mm$ and $V$ in $mm^3$.
EXAMPLE: you have that $r=45mm$, the graph below for computing $tan theta$ remains valid, so $tan theta= frac{6.5}{110}$. Let's say you want $V=200ml=200000mm^3$, the formula gives
$$h=frac{45cdot 110}{6.5}-sqrt[3]{frac{45^3cdot 6.5^3}{110^3}-frac{3cdot 200000cdot 110^2}{picdot 6.5^2}}=sim 32.8 mm$$
EDIT: Since $tan theta$ can be difficulte to find, this is an easier way you can calculate it: here is a section of the truncated cone, you have to measure the lenghts $a$ and $b$. Then
$$tan theta =frac{a}{b}$$
Once you have found $tan theta$, it should be easy to compute $tan^3 theta$ with a calculator. The quantity $r$ can be measured directly and $V$ is known.
answered Dec 21 '15 at 12:44
mrprottolomrprottolo
3,2291718
edited Jan 2 '16 at 11:03
answered Dec 21 '15 at 12:44
mrprottolomrprottolo
3,2291718
answered Dec 21 '15 at 12:44
mrprottolomrprottolo
3,2291718
answered Dec 21 '15 at 12:44
mrprottolomrprottolo
3,2291718
3,2291718
$begingroup$
Nice one. Thanks, I'll give it a go
$endgroup$
– Will
Dec 24 '15 at 10:15
$begingroup$
Ive tried to work this formula out and cant seem to make sense of the second half of the formula to calculate height from volume, and also the part 'h=x-r/tan...'. I dont seem to be able to calculate tan squared and tan cubed on my calculator which may be a problem. Im getting a figure buts its quite a bit out.
$endgroup$
– Will
Dec 29 '15 at 23:09
$begingroup$
Please help me firgure this out if you dont mind. Many thanks
$endgroup$
– Will
Dec 29 '15 at 23:12
$begingroup$
@will See if the edit can help you. Also the only formula you need is the last one, merked in red, which gives you the height given the volume of liquid.
$endgroup$
– mrprottolo
Dec 29 '15 at 23:33
$begingroup$
Hi, thanks for the help but im still not there. Would it matter if the truncated cone I am working with is the other way up, with the wide part at the bottom rather than the top? Because the height equation I am using still doesnt give me the correct results. I am using a base radius of 45mm and a vertical height of 110m with an upper radius of 38.5mm (if you can envisage the shape with that in mind). Should I be using Cos instead? Etc.
$endgroup$
– Will
Jan 2 '16 at 6:11
|
show 4 more comments
$begingroup$
Nice one. Thanks, I'll give it a go
$endgroup$
– Will
Dec 24 '15 at 10:15
$begingroup$
Ive tried to work this formula out and cant seem to make sense of the second half of the formula to calculate height from volume, and also the part 'h=x-r/tan...'. I dont seem to be able to calculate tan squared and tan cubed on my calculator which may be a problem. Im getting a figure buts its quite a bit out.
$endgroup$
– Will
Dec 29 '15 at 23:09
$begingroup$
Please help me firgure this out if you dont mind. Many thanks
$endgroup$
– Will
Dec 29 '15 at 23:12
$begingroup$
@will See if the edit can help you. Also the only formula you need is the last one, merked in red, which gives you the height given the volume of liquid.
$endgroup$
– mrprottolo
Dec 29 '15 at 23:33
$begingroup$
Hi, thanks for the help but im still not there. Would it matter if the truncated cone I am working with is the other way up, with the wide part at the bottom rather than the top? Because the height equation I am using still doesnt give me the correct results. I am using a base radius of 45mm and a vertical height of 110m with an upper radius of 38.5mm (if you can envisage the shape with that in mind). Should I be using Cos instead? Etc.
$endgroup$
– Will
Jan 2 '16 at 6:11
$begingroup$
Nice one. Thanks, I'll give it a go
$endgroup$
– Will
Dec 24 '15 at 10:15
$begingroup$
Nice one. Thanks, I'll give it a go
$endgroup$
– Will
Dec 24 '15 at 10:15
$begingroup$
Ive tried to work this formula out and cant seem to make sense of the second half of the formula to calculate height from volume, and also the part 'h=x-r/tan...'. I dont seem to be able to calculate tan squared and tan cubed on my calculator which may be a problem. Im getting a figure buts its quite a bit out.
$endgroup$
– Will
Dec 29 '15 at 23:09
$begingroup$
Ive tried to work this formula out and cant seem to make sense of the second half of the formula to calculate height from volume, and also the part 'h=x-r/tan...'. I dont seem to be able to calculate tan squared and tan cubed on my calculator which may be a problem. Im getting a figure buts its quite a bit out.
$endgroup$
– Will
Dec 29 '15 at 23:09
$begingroup$
Please help me firgure this out if you dont mind. Many thanks
$endgroup$
– Will
Dec 29 '15 at 23:12
$begingroup$
Please help me firgure this out if you dont mind. Many thanks
$endgroup$
– Will
Dec 29 '15 at 23:12
$begingroup$
@will See if the edit can help you. Also the only formula you need is the last one, merked in red, which gives you the height given the volume of liquid.
$endgroup$
– mrprottolo
Dec 29 '15 at 23:33
$begingroup$
@will See if the edit can help you. Also the only formula you need is the last one, merked in red, which gives you the height given the volume of liquid.
$endgroup$
– mrprottolo
Dec 29 '15 at 23:33
$begingroup$
Hi, thanks for the help but im still not there. Would it matter if the truncated cone I am working with is the other way up, with the wide part at the bottom rather than the top? Because the height equation I am using still doesnt give me the correct results. I am using a base radius of 45mm and a vertical height of 110m with an upper radius of 38.5mm (if you can envisage the shape with that in mind). Should I be using Cos instead? Etc.
$endgroup$
– Will
Jan 2 '16 at 6:11
$begingroup$
Hi, thanks for the help but im still not there. Would it matter if the truncated cone I am working with is the other way up, with the wide part at the bottom rather than the top? Because the height equation I am using still doesnt give me the correct results. I am using a base radius of 45mm and a vertical height of 110m with an upper radius of 38.5mm (if you can envisage the shape with that in mind). Should I be using Cos instead? Etc.
$endgroup$
– Will
Jan 2 '16 at 6:11
|
show 4 more comments
$begingroup$
Assume the container has shape of a truncated cone .. $r$ is minimum radius at bottom, $R$ at top, the cone has a flare up with $alpha$, $h$ is height reached. Using well known relation of truncated cone volume we work up the algebra, defining $H,a^3$ as
$$ V = frac{pi h}{3} ( R^2 + r^2 + R r ), quad tan alpha = frac{R-r}{h}
,quad H = h tan alpha,, , a^3= frac{V tan alpha}{pi} tag1 $$
When $R$ is eliminated to make a cubic in $H$,
$$ H^3/3 + H^2 r + H r^2 - a^3 = 0 tag2 $$
whose only real root is found by Mathematica as
$$ H = (3 a^3 +r^3)^{frac13} - r $$
answered Apr 5 '17 at 12:02
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
Assume the container has shape of a truncated cone .. $r$ is minimum radius at bottom, $R$ at top, the cone has a flare up with $alpha$, $h$ is height reached. Using well known relation of truncated cone volume we work up the algebra, defining $H,a^3$ as
$$ V = frac{pi h}{3} ( R^2 + r^2 + R r ), quad tan alpha = frac{R-r}{h}
,quad H = h tan alpha,, , a^3= frac{V tan alpha}{pi} tag1 $$
When $R$ is eliminated to make a cubic in $H$,
$$ H^3/3 + H^2 r + H r^2 - a^3 = 0 tag2 $$
whose only real root is found by Mathematica as
$$ H = (3 a^3 +r^3)^{frac13} - r $$
answered Apr 5 '17 at 12:02
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
Assume the container has shape of a truncated cone .. $r$ is minimum radius at bottom, $R$ at top, the cone has a flare up with $alpha$, $h$ is height reached. Using well known relation of truncated cone volume we work up the algebra, defining $H,a^3$ as
$$ V = frac{pi h}{3} ( R^2 + r^2 + R r ), quad tan alpha = frac{R-r}{h}
,quad H = h tan alpha,, , a^3= frac{V tan alpha}{pi} tag1 $$
When $R$ is eliminated to make a cubic in $H$,
$$ H^3/3 + H^2 r + H r^2 - a^3 = 0 tag2 $$
whose only real root is found by Mathematica as
$$ H = (3 a^3 +r^3)^{frac13} - r $$
answered Apr 5 '17 at 12:02
NarasimhamNarasimham
20.8k52158
$endgroup$
Assume the container has shape of a truncated cone .. $r$ is minimum radius at bottom, $R$ at top, the cone has a flare up with $alpha$, $h$ is height reached. Using well known relation of truncated cone volume we work up the algebra, defining $H,a^3$ as
$$ V = frac{pi h}{3} ( R^2 + r^2 + R r ), quad tan alpha = frac{R-r}{h}
,quad H = h tan alpha,, , a^3= frac{V tan alpha}{pi} tag1 $$
When $R$ is eliminated to make a cubic in $H$,
$$ H^3/3 + H^2 r + H r^2 - a^3 = 0 tag2 $$
whose only real root is found by Mathematica as
$$ H = (3 a^3 +r^3)^{frac13} - r $$
answered Apr 5 '17 at 12:02
NarasimhamNarasimham
20.8k52158
edited Jul 9 '17 at 14:24
answered Apr 5 '17 at 12:02
NarasimhamNarasimham
20.8k52158
answered Apr 5 '17 at 12:02
NarasimhamNarasimham
20.8k52158
answered Apr 5 '17 at 12:02
NarasimhamNarasimham
20.8k52158
20.8k52158
add a comment |
add a comment |
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Required, but never shown
Required, but never shown
Required, but never shown
