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Imagine that we have a regular plane defined in three-dimensional Euclidean space ("world space").
On this plane, we trace an arc from point A to point B, like a bullet trajectory, in the plane's space ("local space"). That is to say, if the plane was transformed in "world space" using for example rotation, the arc would still remain relative to the plane as it is defined in the plane's space.
We now apply an arbitrary deformation to the plane. It follows that the arc would likewise be deformed in world space.
Which branch of geometry (or mathematics in general) would be used to approach calculating the path of this arc in world space; and for problems like these in general; and how would this problem be approached in software?
Update:
This falls within the domain of Differential geometry.
I believe you are looking for affine transformations, though if your arbitrary deformations to the plane are "strong enough" you might be working with Projective geometry instead.
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I have a 3D mesh I'm calculating vertex normals for, from the mesh's face normals. Each vertex normal is computed as an average of all the normals of the faces that share that vertex.
All that works fine, except in cases where -- due to uneven subdivision of faces over the surface of the mesh -- vertex normals can be skewed. Here is an example of an object with a skewed vertex normal, in one of its corners:
In this image you can see the various face normals (blue), the ideal vertex normals (yellow), and the problematic vertex normal (red) which is being skewed by the many face normals on the heavily subdivided side of the mesh which all share the vertex.
So my question is: is it possible to calculate a vector median, instead of an average? Here's an extra image to further illustrate:
The idea is to weight the face normals by the angle of the edges that connect to the vertex in question, before adding them to that vertex's normal.
So in my first image, the large grouping of normals on the right would have the same cumulative weight as the single normal on the top/left....resulting in an overall ideal normal for that corner vertex.
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I am looking for a library or other source code that will help me display non-intersecting spheres or spheroids, not necessarily a bin packing issue, but it could be. Each of these spheres can also contain a further set of much smaller spheres within. I would then like to take a slice, i.e. two parallel planes a distance apart or a cuboid, through this "world", and calculate how many larger as well as smaller spheres are "captured" by the given slice, and to generate a 3D graphic.
Although I have plenty of experience in programming, etc. 3D spaces are new to me, and my geometry a bit rusty. I've looked into the rgl package for R+ as well as D3.js, but don't see an obvious solution. I've searched through StackOverflow and there is the "Cube sphere intersection test?" post, but it doesn't quite get me what I'm looking for. Not sure if this falls under a "collision" problem or more of a stereology problem perhaps, but the CRAN package StereoMorph also didn't yield any answers.
Any suggestions as to where I should look are welcome. I'm also not wed to any particular computer language.
Thanks!
A sphere is defined as:
[centerX,centerY,centerZ,radius]
and a plane is defined as
[normalVectorX,normalVectorY,normalVectorZ,distance]
Distance is the distance from the origin to the plane if you were to travel along the direction of the normal vector. In other words, if you multiply each component of the normal vector by distance you will have a point somewhere on the plane.
To test if a sphere is on one side of the plane or the other:
Project the sphere center onto the plane normal, add the radius of the sphere, and if that sum is less than the plane distance, then it has not intersected.
CENTER = [centerX,centerY,centerZ]
NORMAL = [normalVectorX,normalVectorY,normalVectorZ]
isBelowHighPlane = dotProduct(CENTER,NORMAL)+radius < distance
For the other plane, lets say the "lower" plane, subtract the radius and test if the sum is greater:
isAboveLowPlane = dotProduct(CENTER,NORMAL)-radius > distance
Therefore, if both the above tests are true the sphere is contained.
NOTE: If your plane is defined some other way, for example by three points in space A,B,C, you can find the normal by taking the cross product of any two sides:
normalVector = normalize(crossProduct((B-A),(B-C)))
And once you have that normal vector, you can find the distance by projecting the vector from the origin to any of those points onto the normal:
distance = dotProduct(B,normalVector)
Note, distance may be negative.
So the 3d functions you need in whatever language you choose are:
dotProduct(), crossProduct(), and normalize()
If you have some connection from small spheres to their containing big sphere, and that big sphere is either completely in or completely out, then you can instantly know the state of all the small spheres. However, to test if a big sphere is completely out, you'll need extra variations on the test:
isAboveHighPlane = dotProduct(CENTER,NORMAL)-radius > distance
isBelowLowPlane = dotProduct(CENTER,NORMAL)+radius < distance
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I have searched a lot of google, reviewed my old mathematics books, But I am stuck why can't we find area of an irregular shape from its perimeter? If we have find perimeter from area, Is that possible to reverse the process and find area from perimeter.
This process is working fine for square and rectangle but not for totally irregular shapes.
I'm asking this question because we have one field with irregular shape, and we want to know its area, We can measure its 5 sides.
Don't we have any method to find area from perimeter for irregular shapes.
Any help will be appreciated.
Consider a rope and form an irregular shape. Measure it's area. Now change the shape you just formed with the rope. Measure the area again. Is it the same? Not necessarily.
Or consider a square and a rhomb (which is not a square) with the same perimeter. Do they have the same area? No. Therefore, how do you think could you find out what its area is if you don't have the information of the shape?
You cannot get the area if you just have the information of the perimeter, but don't know anything else.
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I would realize the papercraft of a geodesic sphere like THIS .
Is the dual of the one discussed in THIS OTHER QUESTION .
Where can I find the printable nets, or the online resources to create them?
In the other discussion I learned that there are 3 classes of possible tessellations.
One of these classes needs less pentagons to tessellate the sphere?
I don't exactly know how much faces I need, indicatively between 100 and 200.
Which chord factors I should consider?
Thanks to all
genna,
The link that I provided in the former discussion...
http://thomson.phy.syr.edu/thomsonapplet.php
...provides for display of a dual. Unfortunately, the java applet will only export a data set for the geodesic triangular tessellation; not the dual.
The dual can be constructed from the triangular tessellation, by importing the vertex (point) data set (from the java applet) into a 3D-graphics program, and then following the rules of the definition of a dual. (Each pent/hex face will use the triangular vertex point as the centerpoint of a circle, being equidistant from the sphere center and normal to the central vector. The intersections of these circles will produce the linear edges of the dual faces.)
All tessellated duals, based on the icosahedron, will have the same number of pentagonal faces (12), regardless of how many hex faces are produced.
-Taff
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I'm working on a problem, wherein a sphere of known radius is dropped vertically and strikes the edge of a cup. I need to figure out the angle of deflection, which will be a function of where along the sphere contact occurs. I.e., if the contact point is dead-center (the pole), the angle of deflection is 0. The further from the pole that contact is made, the larger the deflection angle.
In the formulation of the entire problem (this is just part of it), I'm strictly using the projection onto the x-y plane, so the only information I have is about the projected circles. I know the location of the center of the ball-circle and cup-circle, their radii, and the size of the (downward projected) overlap.
I can compute the angle of deflection, if I only have the length of the chord from the pole to the contact point. How can I get the length of this chord, working from the x-y projection givens? The attached image shows some of how the problem is formulated.
Answer duplicated from my correct repost on math.stackexchange.com
Thank you very much to my friend Dr. Andrew McHugh for helping me see the solution. I overlooked that I can compute the angle theta by knowing the horizontal distance (X) from the lip of the cup to the ball's vertical axis. Theta is then the arcsin of the ratio of X to the radius (r). In the x-y plane projection, X is the radius - the line connecting the cusps of the overlapping lenticular area (d). Hence, theta = arcsin((r-d)/r).