More a Maths question than a FreeCAD one.

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keithsloan52
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More a Maths question than a FreeCAD one.

Post by keithsloan52 »

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
vrytar
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Re: More a Maths question than a FreeCAD one.

Post by vrytar »

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)
galou_breizh
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Re: More a Maths question than a FreeCAD one.

Post by galou_breizh »

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?
vrytar
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Re: More a Maths question than a FreeCAD one.

Post by vrytar »

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].
This is wrong. The result is skewed by the position of each infinitesimal line segment.

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.
keithsloan52
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Re: More a Maths question than a FreeCAD one.

Post by keithsloan52 »

vrytar wrote:
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].
This is wrong. The result is skewed by the position of each infinitesimal line segment.

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.
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.

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.
vrytar
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Re: More a Maths question than a FreeCAD one.

Post by vrytar »

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.
wmayer
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Re: More a Maths question than a FreeCAD one.

Post by wmayer »

N.B. what did you mean with "circular" cone?
Because what we usually call a cone is a very special case of the general definition: http://en.wikipedia.org/wiki/Conical_surface

So, other special cases of a cone are e.g. all kind of pyramids where the base curve is a polygon.
keithsloan52
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Re: More a Maths question than a FreeCAD one.

Post by keithsloan52 »

wmayer wrote:
N.B. what did you mean with "circular" cone?
Because what we usually call a cone is a very special case of the general definition: http://en.wikipedia.org/wiki/Conical_surface

So, other special cases of a cone are e.g. all kind of pyramids where the base curve is a polygon.
I think this an even clearer wikipedia entry http://en.wikipedia.org/wiki/Cone_%28geometry%29
galou_breizh
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Re: More a Maths question than a FreeCAD one.

Post by galou_breizh »

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
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