tham_kathy Posted November 9, 2018 Share Posted November 9, 2018 Question is regarding example https://www.babylonjs-playground.com/#4GBWI5. I don't understand how outer walls and footprint is created. I'm confused with the below code and can anyone please explain the below code step by step? Thanks for(var w = 0; w <= nbWalls; w++) { angle = Math.acos(BABYLON.Vector3.Dot(line, nextLine)/(line.length() * nextLine.length())); direction = BABYLON.Vector3.Cross(nextLine, line).normalize().y; lineNormal = new BABYLON.Vector3(line.z, 0, -1 * line.x).normalize(); line.normalize(); outerData[(w + 1) % nbWalls] = walls[(w + 1) % nbWalls].corner.add(lineNormal.scale(ply)).add(line.scale(direction * ply/Math.tan(angle/2))); line = nextLine.clone(); walls[(w + 3) % nbWalls].corner.subtractToRef(walls[(w + 2) % nbWalls].corner, nextLine); } Quote Link to comment Share on other sites More sharing options...

JohnK Posted November 9, 2018 Share Posted November 9, 2018 In this image the blue walls are the inner walls and the red ones the outer walls, the position vectors ai, bi, ci are known and hence so will line L = bi - ai and next line N = ci - bi. We need to find the position vector of bo. Knowing the distance between the inner and outer wall we can use the right angled triangles (as shown) to determine bo. Using the vector dot product angle = Math.acos(BABYLON.Vector3.Dot(line, nextLine)/(line.length() * nextLine.length())); we can determine the angle between L and N. However that angle can be acute (left hand image), obtuse or reflex (right hand image) whereas the Math.acos returns an angle between 0 and PI. Using the cross product direction = BABYLON.Vector3.Cross(nextLine, line).normalize().y; will give a vector perpendicular to L and N ie of the form (0, y, 0) and normalizing will give (0, 1, 0) or (0, -1, 0) depending whether the angle is reflex or not. We need to find out the unit vector perpendicular to L (forms the right angle in the images). Since taking the dot product of (x1, y1, z1) with (x2, y2, z2) gives x1x2 + y1y2 + z1z2 which will be zero when the two vectors are perpendicular then for L = (x, 0, z) the perpendicular (or normal line) to L in the plane LN will have the form (-z, 0, x) as these give a dot product of zero. Hence lineNormal = new BABYLON.Vector3(line.z, 0, -1 * line.x).normalize(); The we need the unit vector in the direction of L line.normalize(); Current ai tells us we are at wall w and bo is for wall w + 1. Since the walls are closed at the end wall w + 1 = 0. Hence the need to work modulo the number of walls (w + 1) % nbWalls Now we can use the right angled triangle to determine the position vector of bo. Starting from bi (corner of wall w + 1 mod nbWalls) travel along base and along height to get to bo (vector addition is commutative so can add in opposite order) as below Walls] = walls[(w + 1) % nbWalls].corner.add(lineNormal.scale(ply)).add(line.scale(direction * ply/Math.tan(angle/2))); remember ply is distance between inner and outer wall so vector along height is scaled by ply. Knowing ply and the angle we can use tan to get the base length. Multiplying by direction deals with when angle is reflex or not. we can then move on to the next line along line = nextLine.clone(); and find the next line. Note we are using subtractToRef V.subtractToRef(W, R) which takes W from V and puts the result into R walls[(w + 3) % nbWalls].corner.subtractToRef(walls[(w + 2) % nbWalls].corner, nextLine); note that as ai is wall w, bi is wall w + 1 and ci is wall w + 2 Hope that helps Arte, NasimiAsl, Wingnut and 2 others 4 1 Quote Link to comment Share on other sites More sharing options...

tham_kathy Posted November 9, 2018 Author Share Posted November 9, 2018 One last thing; What is the purpose of using indices array? for(var w = 0; w <nbWalls; w++) { indices.push(w, (w + 1) % nbWalls, nbWalls + (w + 1) % nbWalls, w, nbWalls + (w + 1) % nbWalls, w + nbWalls); // base indices } currentLength = indices.length; for(var i = 0; i <currentLength/3; i++) { indices.push(indices[3*i + 2] + 2*nbWalls, indices[3*i + 1] + 2*nbWalls, indices[3*i] + 2*nbWalls ); // top indices } for(var w = 0; w <nbWalls; w++) { indices.push(w, w + 2 *nbWalls, (w + 1) % nbWalls + 2*nbWalls, w, (w + 1) % nbWalls + 2*nbWalls, (w + 1) % nbWalls); // inner wall indices indices.push((w + 1) % nbWalls + 3*nbWalls, w + 3 *nbWalls, w + nbWalls, (w + 1) % nbWalls + nbWalls, (w + 1) % nbWalls + 3*nbWalls, w + nbWalls); // outer wall indices } And Quote To form the mesh, the base, top, inner wall and outer wall have to be split into triangular facets by grouping sets of three corners for each and pushing these into the indices array. what does this mean? Thanks Quote Link to comment Share on other sites More sharing options...

JohnK Posted November 10, 2018 Share Posted November 10, 2018 Indices explained https://doc.babylonjs.com/how_to/custom#positions-and-indices tham_kathy 1 Quote Link to comment Share on other sites More sharing options...

tham_kathy Posted November 11, 2018 Author Share Posted November 11, 2018 On 11/9/2018 at 6:13 PM, JohnK said: Current ai tells us we are at wall w and bo is for wall w + 1. Since the walls are closed at the end wall w + 1 = 0. Hence the need to work modulo the number of walls (w + 1) % nbWalls Now we can use the right angled triangle to determine the position vector of bo. Starting from bi (corner of wall w + 1 mod nbWalls) travel along base and along height to get to bo (vector addition is commutative so can add in opposite order) as below Walls] = walls[(w + 1) % nbWalls].corner.add(lineNormal.scale(ply)).add(line.scale(direction * ply/Math.tan(angle/2))); remember ply is distance between inner and outer wall so vector along height is scaled by ply. Knowing ply and the angle we can use tan to get the base length. Multiplying by direction deals with when angle is reflex or not. @JohnK Can you explain this bit more if you don't mind ? Quote Link to comment Share on other sites More sharing options...

JohnK Posted November 11, 2018 Share Posted November 11, 2018 How much trigonometry and vectors do you know? 6 walls 0,1,2,3,4,5 x % 6 means x modulo 6 ie remainder when x is divded by 6 Walls join 0-1-2-3-4-5-0 get by adding 1 each time 5+1=6 Divide 6 by 6 remainder is 0 and (5+1)%6=0 tham_kathy 1 Quote Link to comment Share on other sites More sharing options...

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