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There will be an intersection if 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1. The intersection point falls within the first line segment if 0 ≤ t ≤ 1, and it falls within the second line segment if 0 ≤ u ≤ 1. These inequalities can be tested without the need for division, allowing rapid determination of the existence of any line segment ...
The formula for calculating it can be derived and expressed in several ways. Knowing the shortest distance from a point to a line can be useful in various situations—for example, finding the shortest distance to reach a road, quantifying the scatter on a graph, etc.
Cyan line has a single point of intersection. Green line has two intersections. Yellow line lies tangent to the cylinder, so has infinitely many points of intersection. Line-cylinder intersection is the calculation of any points of intersection, given an analytic geometry description of a line and a cylinder in 3d space.
It is an easy task to determine the intersection points of a line with a quadric (i.e. line-sphere); one only has to solve a quadratic equation. So, any intersection curve of a cone or a cylinder (they are generated by lines) with a quadric consists of intersection points of lines and the quadric (see pictures).
The intersection points are: (−0.8587, 0.7374, −0.6332), (0.8587, 0.7374, 0.6332). A line–sphere intersection is a simple special case. Like the case of a line and a plane, the intersection of a curve and a surface in general position consists of discrete points, but a curve may be partly or totally contained in a surface.
If the initial point is at the North or South pole, then the first equation is indeterminate. If the initial azimuth is due East or West, then the second equation is indeterminate. If the standard 2-argument arctangent atan2 function is used, then these values are usually handled correctly. [clarification needed]
Linear interpolation on a data set (red points) consists of pieces of linear interpolants (blue lines). Linear interpolation on a set of data points (x 0, y 0), (x 1, y 1), ..., (x n, y n) is defined as piecewise linear, resulting from the concatenation of linear segment interpolants between each pair of data points.
One solution using the even-odd rule is to transform (complex) polygons into simpler ones that are even-odd-equivalent before the intersection check. [11] This, however, is computationally expensive. It is less expensive to use the fast non-zero winding number algorithm, which gives the correct result even when the polygon overlaps itself.
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