CODEWITHSUNDEEP
pythonrandomtrigonometryhypotenuse
eric p
eric p

Reputation: 235

2d random walk in python - drawing hypotenuse from distribution

I'm writing a simple 2d brownian motion simulator in Python. It's obviously easy to draw values for x displacement and y displacement from a distribution, but I have to set it up so that the 2d displacement (ie hypotenuse) is drawn from a distribution, and then translate this to new x and y coordinates. This is probably trivial and I'm just too far removed from trigonometry to remember how to do it correctly. Am I going to need to generate a value for the hypotenuse and then translate it into x and y displacements with sin and cos? (How do you do this correctly?)

Upvotes: 1

Views: 1098

Answers (2)

tom10
tom10

Reputation: 69192

This is best done by using polar coordinates (r, theta) for your distributions (where r is your "hypotenuse")), and then converting the result to (x, y), using x = r cos(theta) and y = r sin(theta). That is, select r from whatever distribution you like, and then select a theta, usually from a flat, 0 to 360 deg, distribution, and then convert these values to x and y.

Going the other way around (i.e., constructing correlated (x, y) distributions that gave a direction independent hypotenuse) would be very difficult.

Upvotes: 1

nmichaels
nmichaels

Reputation: 50971

If you have a hypotenuse in the form of a line segment, then you have two points. From two points in the form P0 = (x0, y0) P1 = (x1, y1) you can get the x and y displacements by subtracting x0 from x1 and y0 from y1.

If your hypotenuse is actually a vector in a polar coordinate plane, then yes, you'll have to take the sin of the angle and multiply it by the magnitude of the vector to get the y displacement and likewise with cos for the x displacement.

Upvotes: 0

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