to find the slant height of a conical frustum given its top radius, it's lateral surface area and it's common...
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I've searched high and low for a formula for finding the slant height of a conical frustum given it's top radius, it's lateral surface area and it's common central angle but have come up empty handed.
There are lots of formulae out there but none that I've seen with those inputs.
Any help would be greatly appreciated.
Thanks.
solid-geometry
asked Dec 22 '18 at 1:47
David WaldenDavid Walden
12
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add a comment |
$begingroup$
I've searched high and low for a formula for finding the slant height of a conical frustum given it's top radius, it's lateral surface area and it's common central angle but have come up empty handed.
There are lots of formulae out there but none that I've seen with those inputs.
Any help would be greatly appreciated.
Thanks.
solid-geometry
asked Dec 22 '18 at 1:47
David WaldenDavid Walden
12
$endgroup$
add a comment |
$begingroup$
I've searched high and low for a formula for finding the slant height of a conical frustum given it's top radius, it's lateral surface area and it's common central angle but have come up empty handed.
There are lots of formulae out there but none that I've seen with those inputs.
Any help would be greatly appreciated.
Thanks.
solid-geometry
asked Dec 22 '18 at 1:47
David WaldenDavid Walden
12
$endgroup$
I've searched high and low for a formula for finding the slant height of a conical frustum given it's top radius, it's lateral surface area and it's common central angle but have come up empty handed.
There are lots of formulae out there but none that I've seen with those inputs.
Any help would be greatly appreciated.
Thanks.
solid-geometry
solid-geometry
asked Dec 22 '18 at 1:47
David WaldenDavid Walden
12
asked Dec 22 '18 at 1:47
David WaldenDavid Walden
12
edited Dec 22 '18 at 1:55
David Walden
asked Dec 22 '18 at 1:47
David WaldenDavid Walden
12
asked Dec 22 '18 at 1:47
David WaldenDavid Walden
12
asked Dec 22 '18 at 1:47
David WaldenDavid Walden
12
12
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A frustum with top radius $r,$ bottom radius $R,$ and slant height $L$ has area
$$ A = pi (r + R) L.$$
You have $r,$ $A,$ and the common central angle.
Consider a plane through the axis of the cone from which the frustum is cut.
This plane takes a cross-section of the frustum;
the cross-section is an isosceles trapezoid with bases $2r$ and $2R$ whose non-parallel sides have length $L.$
Let the angle between the two non-parallel sides of that trapezoid be $2alpha.$
(I suppose your "common central angle" is either $alpha$ or $2alpha,$ depending on how you defined it.)
If we drop a perpendicular from one vertex of the smaller base of the trapezoid to the larger base, it cuts a right triangle off the trapezoid; the triangle has hypotenuse $L,$ one leg is $R - r,$ and the angle opposite that leg is $alpha.$
Therefore
$$
sinalpha = frac{R - r}{L},
$$
and solving for $R$ we find that
$$
R = r + L sinalpha.
$$
Plug this value of $R$ into the original equation for area:
begin{align}
A &= pi (r + (r + L sinalpha)) L \
&= pi (2r + L sinalpha) L. \
end{align}
This is an equation with one unknown that can be solved.
In order to find the solution, we can rearrange the equation above into
$$
(pisinalpha) L^2 + 2pi r L - A = 0
$$
and solve this quadratic equation in $L$:
$$
L = frac{-2pi r pm sqrt{4pi^2 r^2 + 4pi Asinalpha}}{2pisinalpha}.
$$
Assuming we measured lengths and angles in the usual way for such problems,
$L,$ $r,$ $A,$ and $sinalpha$ must all be positive, so we must take the solution
where $pm$ is $+$. We can simplify this to
$$
L = frac{-r + sqrt{r^2 + (A/pi)sinalpha}}{sinalpha}.
$$
answered Dec 22 '18 at 15:02
David KDavid K
55.7k345121
$endgroup$
$begingroup$
Thanks, David. You have made my day!
$endgroup$
– David Walden
Dec 22 '18 at 17:05
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
A frustum with top radius $r,$ bottom radius $R,$ and slant height $L$ has area
$$ A = pi (r + R) L.$$
You have $r,$ $A,$ and the common central angle.
Consider a plane through the axis of the cone from which the frustum is cut.
This plane takes a cross-section of the frustum;
the cross-section is an isosceles trapezoid with bases $2r$ and $2R$ whose non-parallel sides have length $L.$
Let the angle between the two non-parallel sides of that trapezoid be $2alpha.$
(I suppose your "common central angle" is either $alpha$ or $2alpha,$ depending on how you defined it.)
If we drop a perpendicular from one vertex of the smaller base of the trapezoid to the larger base, it cuts a right triangle off the trapezoid; the triangle has hypotenuse $L,$ one leg is $R - r,$ and the angle opposite that leg is $alpha.$
Therefore
$$
sinalpha = frac{R - r}{L},
$$
and solving for $R$ we find that
$$
R = r + L sinalpha.
$$
Plug this value of $R$ into the original equation for area:
begin{align}
A &= pi (r + (r + L sinalpha)) L \
&= pi (2r + L sinalpha) L. \
end{align}
This is an equation with one unknown that can be solved.
In order to find the solution, we can rearrange the equation above into
$$
(pisinalpha) L^2 + 2pi r L - A = 0
$$
and solve this quadratic equation in $L$:
$$
L = frac{-2pi r pm sqrt{4pi^2 r^2 + 4pi Asinalpha}}{2pisinalpha}.
$$
Assuming we measured lengths and angles in the usual way for such problems,
$L,$ $r,$ $A,$ and $sinalpha$ must all be positive, so we must take the solution
where $pm$ is $+$. We can simplify this to
$$
L = frac{-r + sqrt{r^2 + (A/pi)sinalpha}}{sinalpha}.
$$
answered Dec 22 '18 at 15:02
David KDavid K
55.7k345121
$endgroup$
$begingroup$
Thanks, David. You have made my day!
$endgroup$
– David Walden
Dec 22 '18 at 17:05
add a comment |
$begingroup$
A frustum with top radius $r,$ bottom radius $R,$ and slant height $L$ has area
$$ A = pi (r + R) L.$$
You have $r,$ $A,$ and the common central angle.
Consider a plane through the axis of the cone from which the frustum is cut.
This plane takes a cross-section of the frustum;
the cross-section is an isosceles trapezoid with bases $2r$ and $2R$ whose non-parallel sides have length $L.$
Let the angle between the two non-parallel sides of that trapezoid be $2alpha.$
(I suppose your "common central angle" is either $alpha$ or $2alpha,$ depending on how you defined it.)
If we drop a perpendicular from one vertex of the smaller base of the trapezoid to the larger base, it cuts a right triangle off the trapezoid; the triangle has hypotenuse $L,$ one leg is $R - r,$ and the angle opposite that leg is $alpha.$
Therefore
$$
sinalpha = frac{R - r}{L},
$$
and solving for $R$ we find that
$$
R = r + L sinalpha.
$$
Plug this value of $R$ into the original equation for area:
begin{align}
A &= pi (r + (r + L sinalpha)) L \
&= pi (2r + L sinalpha) L. \
end{align}
This is an equation with one unknown that can be solved.
In order to find the solution, we can rearrange the equation above into
$$
(pisinalpha) L^2 + 2pi r L - A = 0
$$
and solve this quadratic equation in $L$:
$$
L = frac{-2pi r pm sqrt{4pi^2 r^2 + 4pi Asinalpha}}{2pisinalpha}.
$$
Assuming we measured lengths and angles in the usual way for such problems,
$L,$ $r,$ $A,$ and $sinalpha$ must all be positive, so we must take the solution
where $pm$ is $+$. We can simplify this to
$$
L = frac{-r + sqrt{r^2 + (A/pi)sinalpha}}{sinalpha}.
$$
answered Dec 22 '18 at 15:02
David KDavid K
55.7k345121
$endgroup$
$begingroup$
Thanks, David. You have made my day!
$endgroup$
– David Walden
Dec 22 '18 at 17:05
add a comment |
$begingroup$
A frustum with top radius $r,$ bottom radius $R,$ and slant height $L$ has area
$$ A = pi (r + R) L.$$
You have $r,$ $A,$ and the common central angle.
Consider a plane through the axis of the cone from which the frustum is cut.
This plane takes a cross-section of the frustum;
the cross-section is an isosceles trapezoid with bases $2r$ and $2R$ whose non-parallel sides have length $L.$
Let the angle between the two non-parallel sides of that trapezoid be $2alpha.$
(I suppose your "common central angle" is either $alpha$ or $2alpha,$ depending on how you defined it.)
If we drop a perpendicular from one vertex of the smaller base of the trapezoid to the larger base, it cuts a right triangle off the trapezoid; the triangle has hypotenuse $L,$ one leg is $R - r,$ and the angle opposite that leg is $alpha.$
Therefore
$$
sinalpha = frac{R - r}{L},
$$
and solving for $R$ we find that
$$
R = r + L sinalpha.
$$
Plug this value of $R$ into the original equation for area:
begin{align}
A &= pi (r + (r + L sinalpha)) L \
&= pi (2r + L sinalpha) L. \
end{align}
This is an equation with one unknown that can be solved.
In order to find the solution, we can rearrange the equation above into
$$
(pisinalpha) L^2 + 2pi r L - A = 0
$$
and solve this quadratic equation in $L$:
$$
L = frac{-2pi r pm sqrt{4pi^2 r^2 + 4pi Asinalpha}}{2pisinalpha}.
$$
Assuming we measured lengths and angles in the usual way for such problems,
$L,$ $r,$ $A,$ and $sinalpha$ must all be positive, so we must take the solution
where $pm$ is $+$. We can simplify this to
$$
L = frac{-r + sqrt{r^2 + (A/pi)sinalpha}}{sinalpha}.
$$
answered Dec 22 '18 at 15:02
David KDavid K
55.7k345121
$endgroup$
A frustum with top radius $r,$ bottom radius $R,$ and slant height $L$ has area
$$ A = pi (r + R) L.$$
You have $r,$ $A,$ and the common central angle.
Consider a plane through the axis of the cone from which the frustum is cut.
This plane takes a cross-section of the frustum;
the cross-section is an isosceles trapezoid with bases $2r$ and $2R$ whose non-parallel sides have length $L.$
Let the angle between the two non-parallel sides of that trapezoid be $2alpha.$
(I suppose your "common central angle" is either $alpha$ or $2alpha,$ depending on how you defined it.)
If we drop a perpendicular from one vertex of the smaller base of the trapezoid to the larger base, it cuts a right triangle off the trapezoid; the triangle has hypotenuse $L,$ one leg is $R - r,$ and the angle opposite that leg is $alpha.$
Therefore
$$
sinalpha = frac{R - r}{L},
$$
and solving for $R$ we find that
$$
R = r + L sinalpha.
$$
Plug this value of $R$ into the original equation for area:
begin{align}
A &= pi (r + (r + L sinalpha)) L \
&= pi (2r + L sinalpha) L. \
end{align}
This is an equation with one unknown that can be solved.
In order to find the solution, we can rearrange the equation above into
$$
(pisinalpha) L^2 + 2pi r L - A = 0
$$
and solve this quadratic equation in $L$:
$$
L = frac{-2pi r pm sqrt{4pi^2 r^2 + 4pi Asinalpha}}{2pisinalpha}.
$$
Assuming we measured lengths and angles in the usual way for such problems,
$L,$ $r,$ $A,$ and $sinalpha$ must all be positive, so we must take the solution
where $pm$ is $+$. We can simplify this to
$$
L = frac{-r + sqrt{r^2 + (A/pi)sinalpha}}{sinalpha}.
$$
answered Dec 22 '18 at 15:02
David KDavid K
55.7k345121
answered Dec 22 '18 at 15:02
David KDavid K
55.7k345121
answered Dec 22 '18 at 15:02
David KDavid K
55.7k345121
answered Dec 22 '18 at 15:02
David KDavid K
55.7k345121
55.7k345121
$begingroup$
Thanks, David. You have made my day!
$endgroup$
– David Walden
Dec 22 '18 at 17:05
add a comment |
$begingroup$
Thanks, David. You have made my day!
$endgroup$
– David Walden
Dec 22 '18 at 17:05
$begingroup$
Thanks, David. You have made my day!
$endgroup$
– David Walden
Dec 22 '18 at 17:05
$begingroup$
Thanks, David. You have made my day!
$endgroup$
– David Walden
Dec 22 '18 at 17:05
add a comment |
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