My Math Forum Integral of sin(distance of reflected light) in 3D space?

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 July 30th, 2007, 09:02 PM #1 Newbie   Joined: Jul 2007 Posts: 19 Thanks: 0 Integral of sin(distance of reflected light) in 3D space? Hello everyone... I'm new to the forum, but it looks very substantial. I'm impressed. My problem: (All objects are hypothetical, allowing for absolute calculations) There is point-source of light in 3D space which emits coherent light in every direction (point A). There is a triangle in 3D space. There is another point in the 3D space that recieves the light that is diffuse-reflected from the triangle (point B). Points A and B are not coplanar with the triangle. There is no time, so the phase of the light is measured instantaneously (as though a standing wave) using the formula f(d)=sin(2*π*f*d). Where f(d) is the distance from the light source to a point d units away. f is the frequency of the light source (in cycles per unit). In my examples, I'll use f=1; The question is how to calculate the net value of the light recieved by point B at an instant . At first, I only took into account the light reflected from the points (V) of the triangle using the formula: sin(dist(A, V1)+dist(B, V1))+sin(dist(A, V2)+dist(B, V2))+sin(dist(A, V3)+dist(B, V3)). Where dist(v1, v2) is a function representing the distance between two points, v1 and v2. The approach is adequate, but if I wanted to do the same calculation using points inside the triangle in order to get more precise results, I would have to do enormous amounts of calculation for high densities of intra-triangular points. While I currently have a means of calculating the value of a beam of light that is reflected from a single point, what I can't figure out is how to make some sort of integral to represent the total value of light, reflected from an edge of the triangle, that hits B. That is the next step, but I might as well think about the final goal, which is to make an integral that represents the total light from the triangle. As far as calculating light reflected by an edge: ∫(V1->V2) sin(dist(B, V)), which is equal to -cos(dist(B, V2))+cos(dist(B, V1)). I can use the integral to calculate the sum of the sine of the distance between the infinitely many points along the edge connecting V1 and V2, and point B, but I don't think that this helps me any, because I still don't know how to "add-in" the initial values at that edge. Plus, this approach does not take into account the face that, while the distances between A and B, and the edge can be taken into account, I still do not know how to factor in the fact that the angle of the face, relative to point A, determines the relative frequency of the light that is reflected from the edge which will, in turn, affect the value of the sum of the sine of the distances between different points on the edge and point B in an unknown way. Anywho... Please help me clarify my question, if possible, and if you have any insight regarding my problem, that would be nice as well! Thank you very much. Andy
 July 31st, 2007, 11:39 AM #2 Newbie   Joined: Jul 2007 Posts: 19 Thanks: 0 The distance that determines a value recieved by point B is the sum of the distance between the light-source and a given point on the triangle, and the distance between that given point and point B: dist(A,Vn) + dist(Vn,B) where Vn is some point on the triangle and dist(x,y) is an abbreviation of the distance between two points, x and y, in 3D space. value(Vn) = sin(2*pi*f*(dist(A,Vn) + dist(Vn,B))) is a single value for a point on the triangle. What I need is to find the value of: ∑(n=1, ∞)sin(2*pi*f*(dist(A,Vn) + dist(Vn,B))) I guess I need to evaluate the this sum. Is this right? Andy

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