Questions tagged [analytic-geometry]

6 questions
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How to write an algorithm to calculate the occlusion percentage of a sphere by another sphere when viewed from a point?

Language doesn't matter all that much since this is an algorithm problem, but let's say I'm using Python 3. Imagine there are two spheres in a 3D space. The spheres have a center position [x, y, z] and a radius r. I also have a point in the same…
2
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1 answer

How to calculate intersection over union for more than two bounding boxes?

I have more than two bounding boxes where I want to calculate the intersection over union. I have done it for two bounding boxes but is there a way to accommodate more than two bounding boxes? I have tried this with two bounding boxes def…
sam
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volume of solid bounded by the graph of y = cos(x), the x−axis, and the y− axis over the interval [0, π/4] is revolved about the x-axis

I need to know what formula to use I know how to solve integrals so I don't need the whole thing done but i just need to know how to set it up. this is what I did: the integral of 2πx(cosx)dx from 0,π/4 Thanks!
blackdiamondx
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Quadratic Bezier curve arc length

I am trying to calculate the length of a quadratic Bezier curve. Found a solution and a formula to calculate the length of the entire curve in a "closed form" - without having to do numeric methods. The solution is covered at: quadratic Bezier curve…
Igor Shmukler
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Drawing markers on a quadratic curve

I am trying to place evenly-spaced markers/dots on a quadratic curve drawn with HTML Canvas API. Found a nice article explaining how the paths are calculated in the first place, at determining coordinates on canvas curve. There is a formula, at the…
0
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2 answers

Analytic geometry, ordering vertices of triangle to capture shortest and second shortest sides

If I have x and y coordinates for triangle corners A, B, and C, I want to know which of the six orderings of {A, B, C} put the shortest side of the triangle between the first two vertices in the ordering, and the second shortest side between the…
Lori
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