Average Radial Distance of Points within a CircleDate: 03/26/2003 at 03:35:51 From: Dashiel Subject: The Average Radial Distance of Points Within a Circle I'm trying to determine the average value of a circular/radial gradient that is at full value (white, call it 100% brightness) in the center, and drops in a linear fashion to zero (black, call it 0% brightness) at the radius. Frankly, I'm just not sure where to start, or what kind of math I'm getting into here. My math education leaves off at college level statistics, so if this is getting into calculus I’m probably out of my league, but I have a suspicion that it can be reduced at least to something within my level of understanding, and hopefully that’s where you good folks come in. Mathematically speaking, I'm trying to determine the average radial distance (distance from the center) of all points within a circle with a specified radius. Essentially what I need is a simple equation for this that can be evaluated for various radii. Date: 03/26/2003 at 12:24:58 From: Doctor Douglas Subject: Re: The Average Radial Distance of Points Within a Circle Hi, Dashiel, Thanks for submitting your question to the Math Forum. Yes, the most straightforward way of determining the answer does in fact use calculus. I will try to explain where this comes from, and identify it with concepts that you're probably familiar with from your statistics course. The average value of something (say a radius r) is given by r1 w1 + r2 w2 + ... <r> = ---------------------- w1 + w2 + ... where the r's are the radii of various pieces, and the w's are weights that are used in the averaging. This should be very familiar to you from your college statistics - it looks like an expectation of r: E(r). In the case of your problem, the weights are the *areas* of pieces with a given radius. So you identify these areas with a "distribution function" (of r): f(r) = (2*pi*r) 0 < r < R This distribution function weights each radius according to the circumference of a (very thin) annulus with that radius. Now, the integral of a distribution function is 1, because the total probability over all possible outcomes is 1. For this problem, the integral is Integral(r=0,r=R) f(r) dr = Integral(r=0,r=R) 2*pi*r dr = pi*r^2 |(r=0,r=R) = pi*R^2. This should be expected - the sum of all of the individual areas had better be the total area of the circle, i.e., pi*R^2. Now, let's calculate the quantity you want - i.e., the average radius. To do this, we simply compute its expectation: Integral(r=0,r=R) f(r) r dr = Integral(r=0,r=R) 2*pi*r^2 dr = pi*(2/3)r^3 |(r=0,r=R) = 2*pi*R^3/3 This is not quite the expectation, because this is only the numerator of the sum - we still have to divide by the sum of the weights (which we've already calculated above as pi*R^2). So, in the end we obtain Integral(r=0,r=R) f(r) r dr 2*pi*R^3/3 2R <R> = --------------------------- = ---------- = ---- Integral(r=0,r=R) f(r) dr pi*R^2 3 There's your formula for the "average" radius among all the points in a circle of radius R. It's certainly reassuring that this number is between zero and R. In fact we also know that the outer radii of the circle possess more area than the innermost radii - so that the final answer should be somewhere between R/2 (which simply splits the difference between the minimum and maximum radii), and R. I hope this helps! - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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