More a Maths question than a FreeCAD one.
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More a Maths question than a FreeCAD one.
Sorry for the Maths question but if a have a sphere and at its centre the apex of a circlular cone. I would like to know how to work out the angle for the cone such that its intersection with the sphere is to take up x% of the surface area of a hemisphere of the sphere.
Thanks in anticipation
Thanks in anticipation
Re: More a Maths question than a FreeCAD one.
The angle between the cone axis and a line on its surface should be arccos(1-p), if I am not mistaken.
(p is your percentage expressed as a ratio, so 40% is 0.4)
(p is your percentage expressed as a ratio, so 40% is 0.4)
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Re: More a Maths question than a FreeCAD one.
It will be easier if you - mentally - work with a quarter of circle and a line segment going through the center of the arc of circle. So instead of surface, you work with a perimeter.
The perimeter above the line is proportinal to the angle: p_above = r * angle, where r is the radius and angle is the angle between the line segment and the revolution axis. If you normalize the radius with r = 2 / pi, so that the perimeter p_above goes from 0 to 1, you get the formula p_above = angle / (pi / 2). The final answer for the planar case is angle = p_above * pi / 2.
The revolution doesn't change the result because the surface is proportional to the revolution angle, so the final answer is
angle = p * pi / 2, with p in [0, 1].
Be aware, that angle here is the half cone angle.
Does it sound ok?
Gaël
N.B. what did you mean with "circular" cone?
The perimeter above the line is proportinal to the angle: p_above = r * angle, where r is the radius and angle is the angle between the line segment and the revolution axis. If you normalize the radius with r = 2 / pi, so that the perimeter p_above goes from 0 to 1, you get the formula p_above = angle / (pi / 2). The final answer for the planar case is angle = p_above * pi / 2.
The revolution doesn't change the result because the surface is proportional to the revolution angle, so the final answer is
angle = p * pi / 2, with p in [0, 1].
Be aware, that angle here is the half cone angle.
Does it sound ok?
Gaël
N.B. what did you mean with "circular" cone?
Re: More a Maths question than a FreeCAD one.
This is wrong. The result is skewed by the position of each infinitesimal line segment.galou_breizh wrote: The revolution doesn't change the result because the surface is proportional to the revolution angle, so the final answer is
angle = p * pi / 2, with p in [0, 1].
Take a look at the picture of a spherical cap at Wikipedia:
http://en.wikipedia.org/wiki/Spherical_cap
The surface area of the cap is 2(pi)rh. The area of the upper hemisphere is 2(pi)r^2. So you need 2(pi)rh = 2p(pi)r^2, which gives you h = pr. You can also see, that cosine of the angle is (r-h)/r = (r-pr)/r = 1 - p. This yealds the result in my previous post.
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Re: More a Maths question than a FreeCAD one.
I agree galou's approach does not seem right to me. Area's are related to powers of 2, so I don't see how you can relate to non powers of 2.vrytar wrote:This is wrong. The result is skewed by the position of each infinitesimal line segment.galou_breizh wrote: The revolution doesn't change the result because the surface is proportional to the revolution angle, so the final answer is
angle = p * pi / 2, with p in [0, 1].
Take a look at the picture of a spherical cap at Wikipedia:
http://en.wikipedia.org/wiki/Spherical_cap
The surface area of the cap is 2(pi)rh. The area of the upper hemisphere is 2(pi)r^2. So you need 2(pi)rh = 2p(pi)r^2, which gives you h = pr. You can also see, that cosine of the angle is (r-h)/r = (r-pr)/r = 1 - p. This yealds the result in my previous post.
The wiki is very helpful it gives the the Area as A= 2(pi) rh and given h= r - r cos(theta) where theta is angle I am concerned about.
Percentage P = 2(pi) r ( r - r cos(theta))/ 2 (pi) r ^2 so gives P = 1 - cos(theta) => theta = cos^-1(1-P)
So I think vrytar is correct and thanks.
Re: More a Maths question than a FreeCAD one.
There is an amazing theorem called Archimedes' Hat-Box Theorem, which you might find instructive for this sort of problems. It says that the area of a spherical segment equals the area of its projection on a surrounding cylinder. See the picture at
http://mathworld.wolfram.com/Archimedes ... eorem.html
This shows you clearly you only need to compare the hights of various spherical segments, including the special cases of a spherical cap or a hemisphere.
http://mathworld.wolfram.com/Archimedes ... eorem.html
This shows you clearly you only need to compare the hights of various spherical segments, including the special cases of a spherical cap or a hemisphere.
Re: More a Maths question than a FreeCAD one.
Because what we usually call a cone is a very special case of the general definition: http://en.wikipedia.org/wiki/Conical_surfaceN.B. what did you mean with "circular" cone?
So, other special cases of a cone are e.g. all kind of pyramids where the base curve is a polygon.
Re: More a Maths question than a FreeCAD one.
Wow, I've learned something new.wmayer wrote:
So, other special cases of a cone are e.g. all kind of pyramids where the base curve is a polygon.
"A cone is a three-dimensional geometric shape that tapers smoothly from a flat base (frequently, though not necessarily, circular) to a point called the apex or vertex."
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Re: More a Maths question than a FreeCAD one.
I think this an even clearer wikipedia entry http://en.wikipedia.org/wiki/Cone_%28geometry%29wmayer wrote:Because what we usually call a cone is a very special case of the general definition: http://en.wikipedia.org/wiki/Conical_surfaceN.B. what did you mean with "circular" cone?
So, other special cases of a cone are e.g. all kind of pyramids where the base curve is a polygon.
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Re: More a Maths question than a FreeCAD one.
I just wanted to say that I'm sorry about having given a wrong answer. I was too quick in my conclusion about the relation arc of circle to surface.
It looks that you found a answer. I'm glad about it.
Gaël
It looks that you found a answer. I'm glad about it.
Gaël