Optimal number of points to plot a sphere

hi all,

I have thousands of data points in (x y z) format to represent the human eye. Since the human eye is modeled like a sphere, is there any way where i can reduce the number of data points to plot the human eye without any distortion of the shape?

Best Regards, natur3

Answer 1

[Walter Roberson]

Calculate the linear distance in pixel-widths between the center of the sphere and the edge. Using that as a radius, calculate the surface area in pixels. If you have more points than that surface area, you can prune back and still have full graphic representation at that drawing scale.

You can probably prune even further than that, but the calculation gets more difficult.

[Rick]

Hi,

What do you mean by prune back?

Best Regards, natur3

[Walter Roberson]

"prune back" is to remove excess points not needed for the graphical representation.

[Walter Roberson]

My calculation is that a view of a sphere with a 16 pixel radius corresponds to a surface area of over 3141 pixels. Therefore to the first approximation, you cannot reduce the number of data points from your "thousands" without distortion unless you are displaying the sphere as less than about 30 pixels wide. This is a first approximation only.

[Rick]

wow, < 30 pixels wide is very small even though its an approximation.

seems like I can't afford to reduce the data points if i want to avoid distortion.

[Walter Roberson]

A brief test with radius 15.5 <= R < 16.5 pixel-widths shows 3338 pixels on the surface, slightly more than 4*Pi*r^2 would predict.

[Rick]

Hi,

sorry for the late reply. Have been trying out and searching for info. I have some questions.

That is, how do I calculate in pixels width?

Whats the difference between "R" and "r"?

By having slightly more pixels on the surface as compared to 4*Pi*r^2, what does it imply?

Best regards, natur3

[Walter Roberson]

[X, Y, Z] = ndgrid(-18:18);  %pixels
R = sqrt(X.^2 + Y.^2 + Z.^2);
onsurface = 15.5 <= R & R < 16.5;
nnz(onsurface)

gives 3338.

Here, "R" is the actual radius at each integer coordinate triple, whereas in 4*Pi*r^2 the "r" is the desired theoretical radius.

There are going to be only 6 integer coordinate triples whose distance is exactly 16: (-16,0,0), (16,0,0), (0,-16,0), (0,16,0), (0,0,-16), (0,0,16). In order to fill out the sphere, we need to select which integer triples are "effectively" at distance 16. I used the semi-open radius range [15.5, 16.5) in my test, selecting the points whose center lies in a thin shell between 15.5 (inclusive) and 16.5 (exclusive) away from the origin. This slightly over-selects points compared to the ideal 4*Pi*(16)^2 as a larger radius is being included and reachable points within a given distance expand with the cube of the radius. The best matching spherical shell should likely be a little thinner than 16.5 at maximum, possibly around 16.2

[Rick]

I see.

So we can cut down the number of points where we take only the "R" which gives us the best matching surface area compared to the desired surface area?

If correct, the range that i used will be more accurate if i used smaller range?

Best regards, natur3