How I can define/ calculate reflection of the ray correctly? As my reflection makes mistake in direction

Question

Aknur 2023-3-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1925930-how-i-can-define-calculate-reflection-of-the-ray-correctly-as-my-reflection-makes-mistake-in-dire

评论： Aknur 2023-4-11

Hello, everyone. Kinsly ask about advice or suggestion about how to find reflection of incidence ray and plot it

How I can make reflected ray in 3d indoor. I wrote function step by step but when it reflects it does not work properly. Maybe you can advice me function or documentation/source. I did everything with geometry. And tried then to solve using Rr = Ri - 2 N (Ri . N)

and my R become like 0.7 1.5 -157 which is not coreect.

The main problem that my ray does not reflect from the surface. I found intersection. Maybe something wrong with my directional vector and angles theta

Thank you in advance

19 个评论
显示 17更早的评论隐藏 17更早的评论

William Rose 2023-4-2

@Aknur,

I will look at your code if you add a lot of comment to it - to the main script and to the funciton. For the script, explain in the comments what you are trying to do and what the different variable are. Are you reflecting a ray off or a series of mirrors? Each mirror should be described by a normal vector and by a point that lies in the plane. Please make it clear in the code which variables are the normals to each plane and the points in each plane. Each ray should be described by a direction vector and by a point on the ray. Please make it clear in the code which variables are the directions of each ray and the points on each ray.

For the function, please explain the role of each input variable and each output variable in the comments. That is a good general practice.

You said "I have failed with my attemp to solve Phi angle of reflected ray." Did you mean to say "theta angle of reflected ray"? I ask this because you refer to theta and 90-theta when discussing angle of incidence.

You said " I still have issue with my azimuth angle Phi." Please explain what issue you have with the azimuth angle. Azimuth angle is measured with respect to some coordinate system. What coordinate system do you want to use when measuring azimuth?

The formula which I provided as my original answer gives the reflected ray direction and a point on the ray (the reflection point, where it insrersects the plane) in "absolute" coordinates, i.e. in terms of an absolute sets of x,y,z axes. The angle of incidence and angle of reflection, which are the subjects of the law of reflection, are measured with respect to the normal to the plane. Therefore it is not true that the angle incidence is the angle down from the z axis, unless the reflecting plane is parallel to the x-y plane.

Aknur 2023-4-3

编辑：Aknur 2023-4-4

Dear, @William Rose

Thank you for your response and help, very appreciate. I changed and add comments to my code and attached it to this comments.

Also Please find my answers for your questions

1) "Are you reflecting a ray off or a series of mirrors? Each mirror should be described by a normal vector and by a point that lies in the plane. Please make it clear in the code which variables are the normals to each plane and the points in each plane"

- I am reflecting inside of 3*3*3 dimension box from almost each plane which will be hit by ray. I calculated normal for all six plane in box and comment it in code 'n'

2) "Each ray should be described by a direction vector and by a point on the ray. Please make it clear in the code which variables are the directions of each ray and the points on each ray."

-XBar, YBar, ZBar are direction vector. And points on each ray will be incident ray.

3)"You said "I have failed with my attemp to solve Phi angle of reflected ray." Did you mean to say "theta angle of reflected ray"? I ask this because you refer to theta and 90-theta when discussing angle of incidence."

-I have two angle Theta0 and Phi0, and they are incidnet angle for the first ray. Then I read that Theta reflected = 180 - Theta incident (Theta0 in my case). And Phi reflected = 180 +Phi incident (Phi0 in my case). But when I tried these formulas I got strange result.

Then I tried to apply another formula geomtrically. For odd number of plane Theta reflected = 90- Theta0, and for even number of plane Theta reflected = Theta0

4) "You said " I still have issue with my azimuth angle Phi." Please explain what issue you have with the azimuth angle. Azimuth angle is measured with respect to some coordinate system. What coordinate system do you want to use when measuring azimuth?"

For azimuth Phi I use angle measured in the XY. When I rotate the box Phi will change also, but I dont know how it will change according to Phi0, which formula I have to use.

Difficulties for me that this box rotate and ray reflected from all planes, and I could exactly understand how two angles Theta and Phi will be changes.

I have attach editted code with comments. Please find it below

I used this sample function to find intersection point and then your formulas.

https://www.mathworks.com/matlabcentral/fileexchange/17751-straight-line-and-plane-intersection

Thank you for your time and consideration

clc

clear all

close all

X0 = 1.5; %Start point, point of first ray starts

Y0 = 1.5;

Z0 = 3.0;

Theta0 = 30; %Angle measured from the Z axis

Phi0 = 90; %Angle measured in the X-Y plane

K = 5; %Number of Reflection

X = [X0 zeros(1,K)];

Y = [Y0 zeros(1,K)];

Z = [Z0 zeros(1,K)];

Theta = [Theta0 zeros(1,K)];

Phi = [Phi0 zeros(1,K)];

for ii = 1:K

ii

[XBar, YBar, ZBar, p, Xr, Yr, Zr, ThetaBar, PhiBar, Rr] = planeLocation7(X(ii), Y(ii), Z(ii),Theta(ii), Phi(ii));

X(ii+1) = Xr; %Xr, Yr, Zr point of intersection and point on reflected ray

Y(ii+1) = Yr; %Further this point will be considered as start point for the next ray

Z(ii+1) = Zr;

Theta(ii+1) = ThetaBar; %ThetaBar angle of reflection

Phi(ii+1) = PhiBar; %PhiBar angle of reflection in XY plane

end

fprintf('X(%d)=%f\n', ii, X(ii));

fprintf('Y(%d)=%f\n', ii, Y(ii));

fprintf('Z(%d)=%f\n', ii, Z(ii));

fprintf('Theta(%d)=%f\n', ii, Theta(ii));

fprintf('Phi(%d)=%f\n', ii, Phi(ii));

disp("===");

function [XBar, YBar, ZBar, p, Xr, Yr, Zr, ThetaBar, PhiBar, Rr] = planeLocation7(X0,Y0,Z0,Theta0, Phi0)

% Input

% X0, Y0, Z0 start point of intersection

% Theta0 Angle measured from the Z axis

% Phi0 Angle measured in the X-Y plane

XBar = sind(Theta0) * cosd(Phi0); %direction vector with unit length name them as XBar, YBar, ZBar

YBar = sind(Theta0) * sind(Phi0); %[sin (θ)cos(φ), sin (θ)sin(φ), cos(θ)]

ZBar = cosd(Theta0);

% Each plane of cube 3*3*3 with coordinates of four points of each plane,

% and fifth point is any point lies on the plane

planes(:,:,1) = [0 3 3; 0 0 3; 0 3 0; 0 0 0; 0 0 0];

planes(:,:,2) = [0 0 3; 3 0 3; 0 0 0; 3 0 0; 0 0 0];

planes(:,:,3) = [3 0 3; 3 3 3; 3 0 0; 3 3 0; 3 0 0];

planes(:,:,4) = [3 3 3; 0 3 3; 3 3 0; 0 3 0; 0 3 3];

planes(:,:,5) = [0 3 0; 3 3 0; 0 0 0; 3 0 0; 0 0 0];

planes(:,:,6) = [0 3 3; 3 3 3; 0 0 3; 3 0 3; 0 0 3];

location_plane = 6;

asd=[]; %create array for I-intersection point of each ray with one of the plane

jjj = []; % create array for 1 - 6 planes

% Checking planes 1-6 for intersection with incident ray Ri which

% was computed using two angles

for j=1:6 % j is number of plane

j

plane = planes(:,:,j);

p0 = plane(1,:); %p0 is top left point of plane

p1 = plane(2,:); %p1 is top right point of plane

p2 = plane(3,:); %p2 is bottom left point of plane

p3 = plane(4,:); %p3 is bottom right point of plane

V0 = plane(5,:); %point on the plane

% Pi is initial start point on the ray

Pi = [X0 Y0 Z0]; %initial start point

Ri = [XBar YBar ZBar]; %direction vector with unit length

A = p0-p2; %calculate A and B then

B = p0-p3; %then to calculate Normal of each plane

n=cross(A,B); % Normal for each Plane

n

w = V0-Pi; %V0 is point on plane

u = Ri;

N = dot(n,w);

D = dot(n,u);

sI = N/D; % t or slope of equation

%Equation of incident ray

I = Pi+sI.*u; % Pint intersection of incident ray Ri with plane

% Pint is point of the reflected ray

Pint = Pi + sI.*u;

fprintf('Pint: %f\n',Pint);

Rr = Ri - 2 * (dot(Ri, n)) * n;

Rr

%Equation of reflected ray - Ref_ray

Rray= Pint + sI.*Rr;

if any(isnan(I))== 0 || any(isinf(I))== 0

%any(I > 0 & I < 3, "all")

jjj = [jjj j];

asd = [asd; I];

end

%fprintf('I: %f\n',I);

%X=I(1);

%Y=I(2);

%Z=I(3);

%F = sqrt((X-X0)^2 + (Y-Y0)^2 + (Z-Z0)^2);

p = j;

fprintf('Rr: %f\n',Rr);

%E=I(1);

%F=I(2);

%G=I(3);

end

% PPP represent points of intersection of Ri incident ray with any of 6

% planes. asd=[ ] is array of intersection points, jjj = [] array of 6

% planes

PPP = [jjj' asd] %array without NaN

B = PPP(:,end-2:end) % New array of PPP without column, which disp. number planes

% Nr extarct row one where is X, Y, Z coordinates of intersection presented

% and then set crieria for them. As intersection should be inside of box

% 3*3*3

Nr = size(B,1); %rows in B

C = B(1:Nr,:) >= 0 & B(1:Nr,:) <= 3; %criteria that Intersected point should be in 3*3*3 dimension room

C

C = [ones(length(jjj),1), C] %keep column of N planes

D = PPP(all(C,2),:) % cleaned coordinates from PPP of intersection points which

% are inside of room accoridng to the dimension 3*3*3

% Because of start point X0, Y0, Z0 also inside of room dimension then D

% shows two points wwhich are valid for our condition. One of them is

% X0,Y0,Z0

for i = 1:size(D,1)

if (D(i,2) ~= X0 || D(i,3) ~= Y0 || D(i,4) ~= Z0) % using thiis condition we dont consider X0,Y0,Z0

disp(D(i,:))

A = D(i,:) %Final intersection point as vector

end

% Theta angle calculation consider Theta0. According to the rule Theta

% reflected = 180-Theta0. Also if we rotate the cube then Theta for odd plane =

% 90-Theta0 with 'minus' and in case if plane is even number than Theta =

% Theta0 with 'minus'.

plane = A(:,1)

if mod(plane,2) == 0 %odd

%ThetaBar = -(Theta0);

ThetaBar = -(90 - Theta0); %before

fprintf('ThetaBar odd: %f\n',ThetaBar);

PhiBar = Phi0;

%fprintf('ThetaBar even: %f\n',ThetaBar);

else

ThetaBar = -( Theta0); %even before

fprintf('ThetaBar even: %f\n',ThetaBar);

%ThetaBar = -(90 - Theta0); %odd

PhiBar = Phi0;

%fprintf('ThetaBar odd: %f\n',ThetaBar);

end

% Here code extarct number of plane to see for the next ray

tempPlane = A(:,1)

% And value of Xr, Yr, Zr will be considered as X0, Y0, Z0 for the next

% ray

% This code extract value of Xr, Yr, Zr from the vector A

Xr = A(:,2);

Yr = A(:,3);

Zr = A(:,4);

%PhiBar = Phi0; %edit it and add where ThetaBar calculated

fprintf('plane: %f\n',plane);

fprintf('Xr: %f\n',Xr);

fprintf('Yr: %f\n',Yr);

fprintf('Zr: %f\n',Zr);

plot3([X0 Xr], [Y0 Yr], [Z0 Zr])

% Below I put incident and reflected ray to this box

Room = [

3 3 0;

0 3 0;

0 3 3;

3 3 3;

0 0 3;

3 0 3;

3 0 0;

0 0 0];

sections_per_edge = 1;

weights = ((1:sections_per_edge-1) / sections_per_edge).';

edges = [];

points = [];

n = size(Room, 1);

for i = 1:n-1

pointA = Room(i, :);

for j = i+1:n

pointB = Room(j, :);

if nnz(pointA - pointB) == 1

edges = [edges; i, j];

points = [points; weights * pointA + (1 - weights) * pointB];

end

plot3(Room(:,1), Room(:,2), Room(:,3), '.b')

hold on

plot3(points(:,1), points(:,1), points(:,1), '.r', 'markersize', 10)

hold on

line([Room(edges(:,1), 1), Room(edges(:,2), 1)].', ...

[Room(edges(:,1), 2), Room(edges(:,2), 2)].', ...

[Room(edges(:,1), 3), Room(edges(:,2), 3)].', 'color', 'k')

plot3(1.5, 1.5, 3.0, '*', 'markersize', 15)

plot3(1.5, 0, 0, 'o', 'markersize', 7, 'MarkerFaceColor', 'black')

plot3(0, 1.5, 0, 'o', 'markersize', 7, 'MarkerFaceColor', 'black')

plot3(3, 1.5, 0,'o', 'markersize', 7, 'MarkerFaceColor', 'black')

plot3(1.5, 3, 0, 'o', 'markersize', 7, 'MarkerFaceColor', 'black')

plot3([X0 Xr], [Y0 Yr], [Z0 Zr])

end

William Rose 2023-4-5

rayReflection.m

@Aknur,

Thank you for your updated code with comments added.

I am not able to follow the logic of this code, and I cannot determine why you are computing thisnhgs the way you are. YOur orignal quesiton was how to compute the formula for a ray reflected by a plane. My earlier answer did not inclide azimuth or angle. You are interested in azimuth and angle and I am not sure why. I don;t understand why they are relevant for your problem.

Therefore I wrote a script that computes N rays (i.e. N- reflections) from M planes, to demonstrate that you do not need azimuth or angle to answer your quesiton. My code also illustrates that there are additional issues you must address when reflecting from multiple planes: 1. You must determine which plane a ray will strike first, because a given ray may intersect more than one plane. 2. You must propagate the ray forward, and not backward, so that intersections with planes behind the ray are not mistakenly identified as the first reflecting plane. (A plane behind could be closer, but if it is behind then we do not want to reflect off of it.) 3. If the planes do not form a bounding box, then it is possible that the ray esscapes from the planes before N-1 reflections have occurred. The code must deal with this possibility correctly.

You do not need azimuth or angle to do the calculations above. The inclusion of azimuth or angle leads to confusion, in my opinion.

My code, attached, deals with the above issues successfully. By adding and removing comments, you can demonstrate that is gives the expected answers. In al three examples, the ray begins at point (1,1,1). In all three examples, there are three planes passing through the origin: plane 1=Y-Z plane though x=0, plane 2=X-Z plane through y=0, plane 3=X-Y plane through z=0. In exaple 1 and example 2, there is also plane 4=X-Y plane thorugh z=+3. The normal vector of plane 1 is (0,0,1); the normal vector of plane 4 is (0,0,-1).

Example 1: 4 planes, initial ray direction=(-.8, 0, -.6).

Example 2: 4 planes, initial ray direction = (-.8, -.6, 0).

Example 3: 3 planes, initial ray direction = (-2/7,-3/7, -6/7).

William Rose 2023-4-11

@Aknur,

I do not understand why phi becomes important if the box is rotating. If I were simulating reflections inside a rotating box, I would not use phi to compute reflections. I would continue to compute reflections with code similar to the code I provided. The difference would be that the reflection computation would be inside a loop that corresponds to time steps, and the orientation vectors of the reflecting planes would be functions of time.

Aknur 2023-4-11

Dear @William Rose thank you so much for your question

Because I set the direction vector using two angles theta and phi and reflected angles of theta and phi will set a new direction vector for the next ray. I hope I was able to explain. However, there is also an option to do without any changes in angle phi, but I have to see how it will go with changing angle phi. Because when the plane changes then the angle phi also has to be changed as the plane is different from the incident plane.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

William Rose 2023-3-9

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1925930-how-i-can-define-calculate-reflection-of-the-ray-correctly-as-my-reflection-makes-mistake-in-dire#answer_1189610

I meant the comment above to be an answer; please consider it to be one. t is a scalar, t>=0 so that the reflected ray is only on one side of the plane.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

How I can define/ calculate reflection of the ray correctly? As my reflection makes mistake in direction

19 个评论
显示 17更早的评论隐藏 17更早的评论

采纳的回答

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

How I can define/ calculate reflection of the ray correctly? As my reflection makes mistake in direction

19 个评论 显示 17更早的评论隐藏 17更早的评论

采纳的回答

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

19 个评论
显示 17更早的评论隐藏 17更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论