Spherical cap problem - trigonometry / circle theorems problem / surface area
$begingroup$
Graph here
I am trying to derive the following equation from a paper I am studying, which the author has derived from the graph above. The two slightly curved lines here are modelled as the surfaces of spherical caps, with surface area S (known).
The equation is
$$frac{X}{4sqrt{S/pi}} = sin( phi_0) + frac{sqrt{S/ pi}}{R}cos( phi_0).$$
I am thinking that if the spherical cap is flattened out, the curved line can be seen as the diameter (D) of this flattened circle, hence $D=2sqrt{S/pi}$. I believe this is where the $sqrt{S/pi}$ term comes from.
As for the rest, I have managed to get an equation for $D$ in terms of $phi$, $alpha$ and $D$, using that $D=Rtheta$ (as the triangle in the diagram can be seen as a segment of a circle).
My solutions however all seem pretty complicated and I cannot manage to get them to match up to the correct one. Where am I going wrong? Do I need to use a different approach? (I thought about using the volume of revolution but I can't work out how to use that here).
Can anyone help?
geometry trigonometry spheres spherical-geometry spherical-trigonometry
asked Dec 17 '18 at 14:33
maria1991maria1991
104
$endgroup$
add a comment |
$begingroup$
Graph here
I am trying to derive the following equation from a paper I am studying, which the author has derived from the graph above. The two slightly curved lines here are modelled as the surfaces of spherical caps, with surface area S (known).
The equation is
$$frac{X}{4sqrt{S/pi}} = sin( phi_0) + frac{sqrt{S/ pi}}{R}cos( phi_0).$$
I am thinking that if the spherical cap is flattened out, the curved line can be seen as the diameter (D) of this flattened circle, hence $D=2sqrt{S/pi}$. I believe this is where the $sqrt{S/pi}$ term comes from.
As for the rest, I have managed to get an equation for $D$ in terms of $phi$, $alpha$ and $D$, using that $D=Rtheta$ (as the triangle in the diagram can be seen as a segment of a circle).
My solutions however all seem pretty complicated and I cannot manage to get them to match up to the correct one. Where am I going wrong? Do I need to use a different approach? (I thought about using the volume of revolution but I can't work out how to use that here).
Can anyone help?
geometry trigonometry spheres spherical-geometry spherical-trigonometry
asked Dec 17 '18 at 14:33
maria1991maria1991
104
$endgroup$
add a comment |
$begingroup$
Graph here
I am trying to derive the following equation from a paper I am studying, which the author has derived from the graph above. The two slightly curved lines here are modelled as the surfaces of spherical caps, with surface area S (known).
The equation is
$$frac{X}{4sqrt{S/pi}} = sin( phi_0) + frac{sqrt{S/ pi}}{R}cos( phi_0).$$
I am thinking that if the spherical cap is flattened out, the curved line can be seen as the diameter (D) of this flattened circle, hence $D=2sqrt{S/pi}$. I believe this is where the $sqrt{S/pi}$ term comes from.
As for the rest, I have managed to get an equation for $D$ in terms of $phi$, $alpha$ and $D$, using that $D=Rtheta$ (as the triangle in the diagram can be seen as a segment of a circle).
My solutions however all seem pretty complicated and I cannot manage to get them to match up to the correct one. Where am I going wrong? Do I need to use a different approach? (I thought about using the volume of revolution but I can't work out how to use that here).
Can anyone help?
geometry trigonometry spheres spherical-geometry spherical-trigonometry
asked Dec 17 '18 at 14:33
maria1991maria1991
104
$endgroup$
Graph here
I am trying to derive the following equation from a paper I am studying, which the author has derived from the graph above. The two slightly curved lines here are modelled as the surfaces of spherical caps, with surface area S (known).
The equation is
$$frac{X}{4sqrt{S/pi}} = sin( phi_0) + frac{sqrt{S/ pi}}{R}cos( phi_0).$$
I am thinking that if the spherical cap is flattened out, the curved line can be seen as the diameter (D) of this flattened circle, hence $D=2sqrt{S/pi}$. I believe this is where the $sqrt{S/pi}$ term comes from.
As for the rest, I have managed to get an equation for $D$ in terms of $phi$, $alpha$ and $D$, using that $D=Rtheta$ (as the triangle in the diagram can be seen as a segment of a circle).
My solutions however all seem pretty complicated and I cannot manage to get them to match up to the correct one. Where am I going wrong? Do I need to use a different approach? (I thought about using the volume of revolution but I can't work out how to use that here).
Can anyone help?
geometry trigonometry spheres spherical-geometry spherical-trigonometry
geometry trigonometry spheres spherical-geometry spherical-trigonometry
asked Dec 17 '18 at 14:33
maria1991maria1991
104
asked Dec 17 '18 at 14:33
maria1991maria1991
104
edited Dec 20 '18 at 0:48
maria1991
asked Dec 17 '18 at 14:33
maria1991maria1991
104
asked Dec 17 '18 at 14:33
maria1991maria1991
104
asked Dec 17 '18 at 14:33
maria1991maria1991
104
104
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This doesn't seem quite right to me. If I take $S$ to be half the surface area (i.e., the surface area of one of the "arcs"), then I have $S=4pi R^2sin^2alpha$, so $sqrt{frac Spi} = 2Rsinalpha$. By similar triangles, the angle between the arc of the circle and the chord joining the endpoints of that arc is $alpha$ as well, and so
$$frac x2 = 2Rsinalphasin(phi+alpha) = 2Rsinalphabig(sinphicosalpha + cosphisinalphabig).$$
Since $alpha$ is presumably small, we take $cosalphaapprox 1$, and this becomes
$$frac x{2sqrt{S/pi}} approx sinphi + cosphisinalpha = sinphi+frac{sqrt{S/pi}}{2R}cosphi.$$
This is "close" to what you posted, but certainly different. At this point, I'm not sure the author is correct.
answered Dec 20 '18 at 1:29
Ted ShifrinTed Shifrin
63.9k44591
$endgroup$
$begingroup$
Could you please explain where that equation of S comes from? (This is the first time I've encountered spherical caps so I'm still not sure on the relevant equations.) Also, could you please explain the similar triangles part? I'm struggling to see how they are similar. Though I think you're right, it is very possible that the author has made an error!
$endgroup$
– maria1991
Dec 20 '18 at 2:10
$begingroup$
You can find the surface area of the cap by doing the surface integral $int_0^{2pi}int_0^{2alpha} R^2sinphi,dphi = 2pi R^2(1-cos 2alpha) = 4pi R^2 sin^2alpha$. For the similar triangles, you have a right triangle (because the tangent to a circle is perpendicular to the radius) with one leg the radius and angle $alpha$ at the center of the circle (so I'm doing half the chord); now draw the chord, which will be perpendicular to the hypotenuse of the original right triangle. That gives you similar triangles. The angle between the chord and the tangent is $alpha$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 2:23
$begingroup$
Ah, I understand now! Thank you very much for your help!
$endgroup$
– maria1991
Dec 20 '18 at 2:45
$begingroup$
Sorry, I have a final question! How did you work out that $frac{x}{2}=2Rsinalpha sin(phi +alpha )$?
$endgroup$
– maria1991
Dec 20 '18 at 3:00
$begingroup$
Make a right triangle with one leg $x/2$ and hypotenuse the chord joining the endpoints of the arc.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 3:04
|
show 1 more comment
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$begingroup$
This doesn't seem quite right to me. If I take $S$ to be half the surface area (i.e., the surface area of one of the "arcs"), then I have $S=4pi R^2sin^2alpha$, so $sqrt{frac Spi} = 2Rsinalpha$. By similar triangles, the angle between the arc of the circle and the chord joining the endpoints of that arc is $alpha$ as well, and so
$$frac x2 = 2Rsinalphasin(phi+alpha) = 2Rsinalphabig(sinphicosalpha + cosphisinalphabig).$$
Since $alpha$ is presumably small, we take $cosalphaapprox 1$, and this becomes
$$frac x{2sqrt{S/pi}} approx sinphi + cosphisinalpha = sinphi+frac{sqrt{S/pi}}{2R}cosphi.$$
This is "close" to what you posted, but certainly different. At this point, I'm not sure the author is correct.
answered Dec 20 '18 at 1:29
Ted ShifrinTed Shifrin
63.9k44591
$endgroup$
$begingroup$
Could you please explain where that equation of S comes from? (This is the first time I've encountered spherical caps so I'm still not sure on the relevant equations.) Also, could you please explain the similar triangles part? I'm struggling to see how they are similar. Though I think you're right, it is very possible that the author has made an error!
$endgroup$
– maria1991
Dec 20 '18 at 2:10
$begingroup$
You can find the surface area of the cap by doing the surface integral $int_0^{2pi}int_0^{2alpha} R^2sinphi,dphi = 2pi R^2(1-cos 2alpha) = 4pi R^2 sin^2alpha$. For the similar triangles, you have a right triangle (because the tangent to a circle is perpendicular to the radius) with one leg the radius and angle $alpha$ at the center of the circle (so I'm doing half the chord); now draw the chord, which will be perpendicular to the hypotenuse of the original right triangle. That gives you similar triangles. The angle between the chord and the tangent is $alpha$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 2:23
$begingroup$
Ah, I understand now! Thank you very much for your help!
$endgroup$
– maria1991
Dec 20 '18 at 2:45
$begingroup$
Sorry, I have a final question! How did you work out that $frac{x}{2}=2Rsinalpha sin(phi +alpha )$?
$endgroup$
– maria1991
Dec 20 '18 at 3:00
$begingroup$
Make a right triangle with one leg $x/2$ and hypotenuse the chord joining the endpoints of the arc.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 3:04
|
show 1 more comment
$begingroup$
This doesn't seem quite right to me. If I take $S$ to be half the surface area (i.e., the surface area of one of the "arcs"), then I have $S=4pi R^2sin^2alpha$, so $sqrt{frac Spi} = 2Rsinalpha$. By similar triangles, the angle between the arc of the circle and the chord joining the endpoints of that arc is $alpha$ as well, and so
$$frac x2 = 2Rsinalphasin(phi+alpha) = 2Rsinalphabig(sinphicosalpha + cosphisinalphabig).$$
Since $alpha$ is presumably small, we take $cosalphaapprox 1$, and this becomes
$$frac x{2sqrt{S/pi}} approx sinphi + cosphisinalpha = sinphi+frac{sqrt{S/pi}}{2R}cosphi.$$
This is "close" to what you posted, but certainly different. At this point, I'm not sure the author is correct.
answered Dec 20 '18 at 1:29
Ted ShifrinTed Shifrin
63.9k44591
$endgroup$
$begingroup$
Could you please explain where that equation of S comes from? (This is the first time I've encountered spherical caps so I'm still not sure on the relevant equations.) Also, could you please explain the similar triangles part? I'm struggling to see how they are similar. Though I think you're right, it is very possible that the author has made an error!
$endgroup$
– maria1991
Dec 20 '18 at 2:10
$begingroup$
You can find the surface area of the cap by doing the surface integral $int_0^{2pi}int_0^{2alpha} R^2sinphi,dphi = 2pi R^2(1-cos 2alpha) = 4pi R^2 sin^2alpha$. For the similar triangles, you have a right triangle (because the tangent to a circle is perpendicular to the radius) with one leg the radius and angle $alpha$ at the center of the circle (so I'm doing half the chord); now draw the chord, which will be perpendicular to the hypotenuse of the original right triangle. That gives you similar triangles. The angle between the chord and the tangent is $alpha$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 2:23
$begingroup$
Ah, I understand now! Thank you very much for your help!
$endgroup$
– maria1991
Dec 20 '18 at 2:45
$begingroup$
Sorry, I have a final question! How did you work out that $frac{x}{2}=2Rsinalpha sin(phi +alpha )$?
$endgroup$
– maria1991
Dec 20 '18 at 3:00
$begingroup$
Make a right triangle with one leg $x/2$ and hypotenuse the chord joining the endpoints of the arc.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 3:04
|
show 1 more comment
$begingroup$
This doesn't seem quite right to me. If I take $S$ to be half the surface area (i.e., the surface area of one of the "arcs"), then I have $S=4pi R^2sin^2alpha$, so $sqrt{frac Spi} = 2Rsinalpha$. By similar triangles, the angle between the arc of the circle and the chord joining the endpoints of that arc is $alpha$ as well, and so
$$frac x2 = 2Rsinalphasin(phi+alpha) = 2Rsinalphabig(sinphicosalpha + cosphisinalphabig).$$
Since $alpha$ is presumably small, we take $cosalphaapprox 1$, and this becomes
$$frac x{2sqrt{S/pi}} approx sinphi + cosphisinalpha = sinphi+frac{sqrt{S/pi}}{2R}cosphi.$$
This is "close" to what you posted, but certainly different. At this point, I'm not sure the author is correct.
answered Dec 20 '18 at 1:29
Ted ShifrinTed Shifrin
63.9k44591
$endgroup$
This doesn't seem quite right to me. If I take $S$ to be half the surface area (i.e., the surface area of one of the "arcs"), then I have $S=4pi R^2sin^2alpha$, so $sqrt{frac Spi} = 2Rsinalpha$. By similar triangles, the angle between the arc of the circle and the chord joining the endpoints of that arc is $alpha$ as well, and so
$$frac x2 = 2Rsinalphasin(phi+alpha) = 2Rsinalphabig(sinphicosalpha + cosphisinalphabig).$$
Since $alpha$ is presumably small, we take $cosalphaapprox 1$, and this becomes
$$frac x{2sqrt{S/pi}} approx sinphi + cosphisinalpha = sinphi+frac{sqrt{S/pi}}{2R}cosphi.$$
This is "close" to what you posted, but certainly different. At this point, I'm not sure the author is correct.
answered Dec 20 '18 at 1:29
Ted ShifrinTed Shifrin
63.9k44591
answered Dec 20 '18 at 1:29
Ted ShifrinTed Shifrin
63.9k44591
answered Dec 20 '18 at 1:29
Ted ShifrinTed Shifrin
63.9k44591
answered Dec 20 '18 at 1:29
Ted ShifrinTed Shifrin
63.9k44591
63.9k44591
$begingroup$
Could you please explain where that equation of S comes from? (This is the first time I've encountered spherical caps so I'm still not sure on the relevant equations.) Also, could you please explain the similar triangles part? I'm struggling to see how they are similar. Though I think you're right, it is very possible that the author has made an error!
$endgroup$
– maria1991
Dec 20 '18 at 2:10
$begingroup$
You can find the surface area of the cap by doing the surface integral $int_0^{2pi}int_0^{2alpha} R^2sinphi,dphi = 2pi R^2(1-cos 2alpha) = 4pi R^2 sin^2alpha$. For the similar triangles, you have a right triangle (because the tangent to a circle is perpendicular to the radius) with one leg the radius and angle $alpha$ at the center of the circle (so I'm doing half the chord); now draw the chord, which will be perpendicular to the hypotenuse of the original right triangle. That gives you similar triangles. The angle between the chord and the tangent is $alpha$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 2:23
$begingroup$
Ah, I understand now! Thank you very much for your help!
$endgroup$
– maria1991
Dec 20 '18 at 2:45
$begingroup$
Sorry, I have a final question! How did you work out that $frac{x}{2}=2Rsinalpha sin(phi +alpha )$?
$endgroup$
– maria1991
Dec 20 '18 at 3:00
$begingroup$
Make a right triangle with one leg $x/2$ and hypotenuse the chord joining the endpoints of the arc.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 3:04
|
show 1 more comment
$begingroup$
Could you please explain where that equation of S comes from? (This is the first time I've encountered spherical caps so I'm still not sure on the relevant equations.) Also, could you please explain the similar triangles part? I'm struggling to see how they are similar. Though I think you're right, it is very possible that the author has made an error!
$endgroup$
– maria1991
Dec 20 '18 at 2:10
$begingroup$
You can find the surface area of the cap by doing the surface integral $int_0^{2pi}int_0^{2alpha} R^2sinphi,dphi = 2pi R^2(1-cos 2alpha) = 4pi R^2 sin^2alpha$. For the similar triangles, you have a right triangle (because the tangent to a circle is perpendicular to the radius) with one leg the radius and angle $alpha$ at the center of the circle (so I'm doing half the chord); now draw the chord, which will be perpendicular to the hypotenuse of the original right triangle. That gives you similar triangles. The angle between the chord and the tangent is $alpha$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 2:23
$begingroup$
Ah, I understand now! Thank you very much for your help!
$endgroup$
– maria1991
Dec 20 '18 at 2:45
$begingroup$
Sorry, I have a final question! How did you work out that $frac{x}{2}=2Rsinalpha sin(phi +alpha )$?
$endgroup$
– maria1991
Dec 20 '18 at 3:00
$begingroup$
Make a right triangle with one leg $x/2$ and hypotenuse the chord joining the endpoints of the arc.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 3:04
$begingroup$
Could you please explain where that equation of S comes from? (This is the first time I've encountered spherical caps so I'm still not sure on the relevant equations.) Also, could you please explain the similar triangles part? I'm struggling to see how they are similar. Though I think you're right, it is very possible that the author has made an error!
$endgroup$
– maria1991
Dec 20 '18 at 2:10
$begingroup$
Could you please explain where that equation of S comes from? (This is the first time I've encountered spherical caps so I'm still not sure on the relevant equations.) Also, could you please explain the similar triangles part? I'm struggling to see how they are similar. Though I think you're right, it is very possible that the author has made an error!
$endgroup$
– maria1991
Dec 20 '18 at 2:10
$begingroup$
You can find the surface area of the cap by doing the surface integral $int_0^{2pi}int_0^{2alpha} R^2sinphi,dphi = 2pi R^2(1-cos 2alpha) = 4pi R^2 sin^2alpha$. For the similar triangles, you have a right triangle (because the tangent to a circle is perpendicular to the radius) with one leg the radius and angle $alpha$ at the center of the circle (so I'm doing half the chord); now draw the chord, which will be perpendicular to the hypotenuse of the original right triangle. That gives you similar triangles. The angle between the chord and the tangent is $alpha$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 2:23
$begingroup$
You can find the surface area of the cap by doing the surface integral $int_0^{2pi}int_0^{2alpha} R^2sinphi,dphi = 2pi R^2(1-cos 2alpha) = 4pi R^2 sin^2alpha$. For the similar triangles, you have a right triangle (because the tangent to a circle is perpendicular to the radius) with one leg the radius and angle $alpha$ at the center of the circle (so I'm doing half the chord); now draw the chord, which will be perpendicular to the hypotenuse of the original right triangle. That gives you similar triangles. The angle between the chord and the tangent is $alpha$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 2:23
$begingroup$
Ah, I understand now! Thank you very much for your help!
$endgroup$
– maria1991
Dec 20 '18 at 2:45
$begingroup$
Ah, I understand now! Thank you very much for your help!
$endgroup$
– maria1991
Dec 20 '18 at 2:45
$begingroup$
Sorry, I have a final question! How did you work out that $frac{x}{2}=2Rsinalpha sin(phi +alpha )$?
$endgroup$
– maria1991
Dec 20 '18 at 3:00
$begingroup$
Sorry, I have a final question! How did you work out that $frac{x}{2}=2Rsinalpha sin(phi +alpha )$?
$endgroup$
– maria1991
Dec 20 '18 at 3:00
$begingroup$
Make a right triangle with one leg $x/2$ and hypotenuse the chord joining the endpoints of the arc.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 3:04
$begingroup$
Make a right triangle with one leg $x/2$ and hypotenuse the chord joining the endpoints of the arc.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 3:04
|
show 1 more comment
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