i'm struggling with this geometry problem right now.
Let's say we have a line defined by point A(x1,y1) and point B(x2,y2)
We also have a point C(x3,y3).
What function written in SWIFT could give me the coordinates (X,Y) of the point that has the smallest distance from the line ? In other words, the point on the line which is the intersection between a perpendicular segment and the other point.
func getCoordsOfPointsWithSmallestDistanceBetweenLineAndPoint(lineX1: Double, lineY1: Double, lineX2: Double, lineY2: Double, pointX3: Double, pointY3: Double) -> [Double] {
// ???
return [x,y]
}
In a mathematical point of view you can :
first find the equation of the line :
y1 = a1x1+b1
a1 = (y2-y1) / (x2-x1)
b1 = y1-a1*x1
Then calculate the gradient of the second line knowing :
a1 * a2 = -1 <->
a2 = -1/a1
with a2 you can find the value of b for the second equation :
y3 = a2*x3 + b2 <->
b2 = y3 - a2*x3
Finally calculate the intersection of the 2 lines :
xi = (b2-b1) / (a1-a2)
y = a1*xi + b1
Then it's quite straightforward to bring that to swift :
typealias Line = (gradient:CGFloat, intercept:CGFloat)
func getLineEquation(point1:CGPoint, point2:CGPoint) -> Line {
guard point1.x != point2.x else {
if(point1.y != point2.y)
{
print("Vertical line : x = \(point1.x)")
}
return (gradient: .nan, intercept: .nan)
}
let gradient = (point2.y - point1.y)/(point2.x-point1.x)
let intercept = point1.y - gradient*point1.x
return (gradient: gradient, intercept: intercept)
}
func getPerpendicularGradient(gradient:CGFloat) -> CGFloat
{
guard gradient != 0 else {
print("horizontal line, the perpendicilar line is vertical")
return .nan
}
return -1/gradient
}
func getIntercept(forPoint point:CGPoint, withGradient gradient:CGFloat) -> CGFloat
{
return point.y - gradient * point.x
}
func getIntersectionPoint(line1:Line, line2:Line)-> CGPoint
{
guard line1.gradient != line2.gradient else {return CGPoint(x: CGFloat.nan, y: CGFloat.nan)}
let x = (line2.intercept - line1.intercept)/(line1.gradient-line2.gradient)
return CGPoint(x:x, y: line1.gradient*x + line1.intercept)
}
func getClosestIntersectionPoint(forLine line:Line, point:CGPoint) -> CGPoint
{
let line2Gradient = getPerpendicularGradient(gradient:line.gradient)
let line2 = (
gradient: line2Gradient,
intercept: getIntercept(forPoint: point, withGradient: line2Gradient))
return getIntersectionPoint(line1:line, line2:line2)
}
func getClosestIntersectionPoint(forLinePoint1 linePoint1:CGPoint, linePoint2:CGPoint, point:CGPoint) -> CGPoint
{
return getClosestIntersectionPoint(
forLine:getLineEquation(point1: linePoint1, point2: linePoint2),
point:point)
}
You can minimize the squared distance of C to a point on the straight line AB:
(CA + t.AB)² = t²AB² + 2t AB.CA + CA²
The minimum is achieved by
t = - AB.CA / AB²
and
CP = CA + t.AB
To elaborate on Yves Daoust answer which if converted to a function has the form
func closestPnt(x: Double, y: Double, x1: Double, y1: Double, px: Double, py: Double)->[Double]{
let vx = x1 - x // vector of line
let vy = y1 - y
let ax = px - x // vector from line start to point
let ay = py - y
let u = (ax * vx + ay * vy) / (vx * vx + vy * vy) // unit distance on line
if u >= 0 && u <= 1 { // is on line segment
return [x + vx * u, y + vy * u] // return closest point on line
}
if u < 0 {
return [x, y] // point is before start of line segment so return start point
}
return [x1, y1] // point is past end of line so return end
}
Note that the function is for line segments, if the closest points unit distance is behind the start or past the end then an end point is the closest.
If you want the point on a line (finitely long) then the following will do that.
func closestPnt(x: Double, y: Double, x1: Double, y1: Double, px: Double, py: Double)->[Double]{
let vx = x1 - x // vector of line
let vy = y1 - y
let ax = px - x // vector from line start to point
let ay = py - y
let u = (ax * vx + ay * vy) / (vx * vx + vy * vy) // unit distance on line
return [x + vx * u, y + vy * u] // return closest point on line
}
Note That both functions assume that !(x1 == x && y1 == y) is be true. IE the line segment MUST have a length > 0.
I'm trying to write a function in swift, which returns a CGPoint where the extension of a vector (which is within a screen) will intersect the screen. Let's assume that the screen is 800 x 600. It's like the scheme:
The function should have the following parameters:
func calcPoint(start: CGPoint, end: CGPoint) -> CGPoint
start: CGPoint(x: x1, y: y1) - this is the beginning of the vector.
end: CGPoint(x: x1, y: y1) - this is the end point of the vector.
the return point is the one at which the vector intersects the screen (CGPoint(x: x3, y: y3) as shown at the scheme).
The values for the vector start and end are aways points within the screen (the rectangle 0, 0, 800, 600).
EDIT (for Alexander):
Is there a formula, which in the given situation will make it easy to write the function, in not the obvious way using if ... else ... and triangle vertices ratio?
To compute point E you can look at the triangles given by your setting. You have the Triangle ABC and DBE. Note that they are similar, such that we can set up following relation AB : AC = DB : DE using the intercept theorem (AB etc. stands for the line segment between A and B). In the given setting you know all points but E.
Using start and end Points from given setting:
In case start and end have the same x or y-coordinate it is only the top bottom or left right border with the same coordinate.
Using the absolute values it should work for all four corners of your rectangle. Then of course you have to consider E being out of your rectangle, again the same relation can be used AB : AC = D'B : D'E'
A pure swift solution for everyone interested in such (thanks to Ivo Ivanoff):
// Example for iOS
/// The height of the screen
let screenHeight = UIScreen.main.bounds.height
/// The width of the screen
let screenWidth = UIScreen.main.bounds.width
func calculateExitPoint(from anchor : CGPoint, to point: CGPoint) -> CGPoint {
var exitPoint : CGPoint = CGPoint()
let directionV: CGFloat = anchor.y < point.y ? 1 : -1
let directionH: CGFloat = anchor.x < point.x ? 1 : -1
let a = directionV > 0 ? screenHeight - anchor.y : anchor.y
let a1 = directionV > 0 ? point.y - anchor.y : anchor.y - point.y
let b1 = directionH > 0 ? point.x - anchor.x : anchor.x - point.x
let b = a / (a1 / b1)
let tgAlpha = b / a
let b2 = directionH > 0 ? screenWidth - point.x : point.x
let a2 = b2 / tgAlpha
exitPoint.x = anchor.x + b * directionH
exitPoint.y = point.y + a2 * directionV
if (exitPoint.x > screenWidth) {
exitPoint.x = screenWidth
} else if (exitPoint.x < 0) {
exitPoint.x = 0;
} else {
exitPoint.y = directionV > 0 ? screenHeight : 0
}
return exitPoint
}
Any kind of optimizations are welcomed ;-)
There is no single formula, because intersection depends on starting point position, line slope and rectangle size, and it may occur at any rectangle edge.
Here is approach based on parametric representation of line. Works for any slope (including horizontal and vertical). Finds what border is intersected first, calculates intersection point.
dx = end.x - start.x
dy = end.y - start.y
//parametric equations for reference:
//x = start.x + dx * t
//y = start.y + dy * t
//prerequisites: potential border positions
if dx > 0 then
bx = width
else
bx = 0
if dy > 0 then
by = height
else
by = 0
//first check for horizontal/vertical lines
if dx = 0 then
return ix = start.x, iy = by
if dy = 0 then
return iy = start.y, ix = bx
//in general case find parameters of intersection with horizontal and vertical edge
tx = (bx - start.x) / dx
ty = (by - start.y) / dy
//and get intersection for smaller parameter value
if tx <= ty then
ix = bx
iy = start.y + tx * dy
else
iy = by
ix = start.x + ty * dx
return ix, iy
How does one create a path of a sine wave between two points?
I am able to create a path of a sine wave from an origin, but am not sure how the direction can be transformed so that the sine wave ends at a target CGPoint.
I would like to animate a SKNode along the path using SKAction.followPath
The simplest way to think about this is to transform the coordinate system, rotating by the angle between the two points, scaling by the distance between them and translating by the first point (assuming the sine starts at 0,0).
The OP has specified that he doesn't just want to draw the curve (in which case all one needs to do is apply the transform to the graphics context), but rather to use the curve in a SpriteKit SKAction.followPath call, so the transform has to be applied to the coordinates in the path, not to the context.
Here's a solution using CGPath rather than UIBezierPath, but they are equivalent, and you can get the UI version simply by let uip = UIBezierPath(cgPath: path). (I prefer CoreGraphics as they are cross-platform).
Playground code...
class MyView: UIView {
override func draw(_ rect: CGRect) {
guard let context = UIGraphicsGetCurrentContext() else { return }
context.setFillColor(UIColor.red.cgColor)
context.fill(self.bounds)
// Calculate the transform
let p1 = CGPoint(x: 100, y: 100)
let p2 = CGPoint(x: 400, y: 400)
let dx = p2.x - p1.x
let dy = p2.y - p1.y
let d = sqrt(dx * dx + dy * dy)
let a = atan2(dy, dx)
let cosa = cos(a) // Calculate only once...
let sina = sin(a) // Ditto
// Initialise our path
let path = CGMutablePath()
path.move(to: p1)
// Plot a parametric function with 100 points
let nPoints = 100
for t in 0 ... nPoints {
// Calculate the un-transformed x,y
let tx = CGFloat(t) / CGFloat(nPoints) // 0 ... 1
let ty = 0.1 * sin(tx * 2 * CGFloat.pi ) // 0 ... 2π, arbitrary amplitude
// Apply the transform
let x = p1.x + d * (tx * cosa - ty * sina)
let y = p1.y + d * (tx * sina + ty * cosa)
// Add the transformed point to the path
path.addLine(to: CGPoint(x: x, y: y))
}
// Draw the path
context.setStrokeColor(UIColor.blue.cgColor)
context.addPath(path)
context.strokePath()
}
}
let v = MyView(frame: CGRect(origin: CGPoint(x: 0, y:0), size: CGSize(width: 500, height: 500)))
Not crystal clear what you want but here's one possibility assuming you want a tilted sin curve:
Assume that the start point is (0, 0) and the end point is (x, y).
Let L be the distance between the origin and your point: L = sqrt(x^2 + y^2)
Write a loop that starts at 0 and ends at L, with increment dL and running sum l (which ends up running between 0 and L). This loop will allow us to create the points on your Bezier.
Then the x coordinate of your sin graph will be:
x_P = l * cos(theta), ranging from 0 to L * cos(theta) = x
To get the y coordinate, we add a sin function with the correct period to the sloping line between the origin and your point:
y_P = l * sin(theta) + sin(2 * PI * l / L)
note that, at l = L, (x_P, y_P) = (x, y) which is as it should be.
Was this what you wanted? It is not a rotation and so will not behave when the angle theta is large.
Having made some progress in the geometry side of things I'm moving on to putting together an entire scene. That scene has a couple dozen objects, each defined by a bounding cube whose corners are specified by two SCNVector3s (originally two sets of x,y,z).
Here's an example of what I have so far - it's an 11-element log-periodic antenna, like the old school TV antennas from the 70s. Each of the grey lines is an "element", typically made of aluminum rod. I used SCNCylinders from +ve to -ve Y and the entire thing is less than 100 lines (SK is pretty amazing).
The problem is what happens if the elements are not symmetrical across X and thus the SCNCylinder has to be rotated. I found this example, but I can't understand the specifics... it appears to take advantage of the fact that a sphere is symmetric so angles kind of "go away".
Does anyone have a general function that will take two 3D points and return the SCNVector3 suitable for setting the node's eulerAngle, or a similar solution?
Both solutions mentioned above work very well and I can contribute third solution to this question.
//extension code starts
func normalizeVector(_ iv: SCNVector3) -> SCNVector3 {
let length = sqrt(iv.x * iv.x + iv.y * iv.y + iv.z * iv.z)
if length == 0 {
return SCNVector3(0.0, 0.0, 0.0)
}
return SCNVector3( iv.x / length, iv.y / length, iv.z / length)
}
extension SCNNode {
func buildLineInTwoPointsWithRotation(from startPoint: SCNVector3,
to endPoint: SCNVector3,
radius: CGFloat,
color: UIColor) -> SCNNode {
let w = SCNVector3(x: endPoint.x-startPoint.x,
y: endPoint.y-startPoint.y,
z: endPoint.z-startPoint.z)
let l = CGFloat(sqrt(w.x * w.x + w.y * w.y + w.z * w.z))
if l == 0.0 {
// two points together.
let sphere = SCNSphere(radius: radius)
sphere.firstMaterial?.diffuse.contents = color
self.geometry = sphere
self.position = startPoint
return self
}
let cyl = SCNCylinder(radius: radius, height: l)
cyl.firstMaterial?.diffuse.contents = color
self.geometry = cyl
//original vector of cylinder above 0,0,0
let ov = SCNVector3(0, l/2.0,0)
//target vector, in new coordination
let nv = SCNVector3((endPoint.x - startPoint.x)/2.0, (endPoint.y - startPoint.y)/2.0,
(endPoint.z-startPoint.z)/2.0)
// axis between two vector
let av = SCNVector3( (ov.x + nv.x)/2.0, (ov.y+nv.y)/2.0, (ov.z+nv.z)/2.0)
//normalized axis vector
let av_normalized = normalizeVector(av)
let q0 = Float(0.0) //cos(angel/2), angle is always 180 or M_PI
let q1 = Float(av_normalized.x) // x' * sin(angle/2)
let q2 = Float(av_normalized.y) // y' * sin(angle/2)
let q3 = Float(av_normalized.z) // z' * sin(angle/2)
let r_m11 = q0 * q0 + q1 * q1 - q2 * q2 - q3 * q3
let r_m12 = 2 * q1 * q2 + 2 * q0 * q3
let r_m13 = 2 * q1 * q3 - 2 * q0 * q2
let r_m21 = 2 * q1 * q2 - 2 * q0 * q3
let r_m22 = q0 * q0 - q1 * q1 + q2 * q2 - q3 * q3
let r_m23 = 2 * q2 * q3 + 2 * q0 * q1
let r_m31 = 2 * q1 * q3 + 2 * q0 * q2
let r_m32 = 2 * q2 * q3 - 2 * q0 * q1
let r_m33 = q0 * q0 - q1 * q1 - q2 * q2 + q3 * q3
self.transform.m11 = r_m11
self.transform.m12 = r_m12
self.transform.m13 = r_m13
self.transform.m14 = 0.0
self.transform.m21 = r_m21
self.transform.m22 = r_m22
self.transform.m23 = r_m23
self.transform.m24 = 0.0
self.transform.m31 = r_m31
self.transform.m32 = r_m32
self.transform.m33 = r_m33
self.transform.m34 = 0.0
self.transform.m41 = (startPoint.x + endPoint.x) / 2.0
self.transform.m42 = (startPoint.y + endPoint.y) / 2.0
self.transform.m43 = (startPoint.z + endPoint.z) / 2.0
self.transform.m44 = 1.0
return self
}
}
//extension ended.
//in your code, you can like this.
let twoPointsNode1 = SCNNode()
scene.rootNode.addChildNode(twoPointsNode1.buildLineInTwoPointsWithRotation(
from: SCNVector3(1,-1,3), to: SCNVector3( 7,11,7), radius: 0.2, color: .cyan))
//end
you can reference http://danceswithcode.net/engineeringnotes/quaternions/quaternions.html
BTW, you will get same result when you use a cylinder to make a line between two points from above 3 methods. But indeed, they will have different normal lines. In another words, if you use box between two points, sides of box, except top and bottom, will face different direction from above 3 methods.
let me know pls if you need further explanation.
EDIT: For under or equal to IOS 11
I've good news for you ! You can link two points and put a SCNNode on this Vector !
Take this and enjoy drawing line between two point !
class CylinderLine: SCNNode
{
init( parent: SCNNode,//Needed to add destination point of your line
v1: SCNVector3,//source
v2: SCNVector3,//destination
radius: CGFloat,//somes option for the cylinder
radSegmentCount: Int, //other option
color: UIColor )// color of your node object
{
super.init()
//Calcul the height of our line
let height = v1.distance(v2)
//set position to v1 coordonate
position = v1
//Create the second node to draw direction vector
let nodeV2 = SCNNode()
//define his position
nodeV2.position = v2
//add it to parent
parent.addChildNode(nodeV2)
//Align Z axis
let zAlign = SCNNode()
zAlign.eulerAngles.x = Float(M_PI_2)
//create our cylinder
let cyl = SCNCylinder(radius: radius, height: CGFloat(height))
cyl.radialSegmentCount = radSegmentCount
cyl.firstMaterial?.diffuse.contents = color
//Create node with cylinder
let nodeCyl = SCNNode(geometry: cyl )
nodeCyl.position.y = -height/2
zAlign.addChildNode(nodeCyl)
//Add it to child
addChildNode(zAlign)
//set contrainte direction to our vector
constraints = [SCNLookAtConstraint(target: nodeV2)]
}
override init() {
super.init()
}
required init?(coder aDecoder: NSCoder) {
super.init(coder: aDecoder)
}
}
private extension SCNVector3{
func distance(receiver:SCNVector3) -> Float{
let xd = receiver.x - self.x
let yd = receiver.y - self.y
let zd = receiver.z - self.z
let distance = Float(sqrt(xd * xd + yd * yd + zd * zd))
if (distance < 0){
return (distance * -1)
} else {
return (distance)
}
}
}
#maury-markowitz's answer worked for me, here is the latest (Swift4) version of it.
To anyone working with SCNVector3 in Swift I can only recommend to add the +-*/ operator overloads somewhere in your code (e.g. from here).
extension SCNNode {
static func lineNode(from: SCNVector3, to: SCNVector3, radius: CGFloat = 0.25) -> SCNNode {
let vector = to - from
let height = vector.length()
let cylinder = SCNCylinder(radius: radius, height: CGFloat(height))
cylinder.radialSegmentCount = 4
let node = SCNNode(geometry: cylinder)
node.position = (to + from) / 2
node.eulerAngles = SCNVector3.lineEulerAngles(vector: vector)
return node
}
}
extension SCNVector3 {
static func lineEulerAngles(vector: SCNVector3) -> SCNVector3 {
let height = vector.length()
let lxz = sqrtf(vector.x * vector.x + vector.z * vector.z)
let pitchB = vector.y < 0 ? Float.pi - asinf(lxz/height) : asinf(lxz/height)
let pitch = vector.z == 0 ? pitchB : sign(vector.z) * pitchB
var yaw: Float = 0
if vector.x != 0 || vector.z != 0 {
let inner = vector.x / (height * sinf(pitch))
if inner > 1 || inner < -1 {
yaw = Float.pi / 2
} else {
yaw = asinf(inner)
}
}
return SCNVector3(CGFloat(pitch), CGFloat(yaw), 0)
}
}
For the sake of another method, I achieved this through trigonometry. This made the code very minimal. Here is the end result:
In my case the nodes are always placed on a fixed plane that slices the Y-Axis.
// Create Cylinder Geometry
let line = SCNCylinder(radius: 0.002, height: node1.distance(to: node2))
// Create Material
let material = SCNMaterial()
material.diffuse.contents = UIColor.red
material.lightingModel = .phong
line.materials = [material]
// Create Cylinder(line) Node
let newLine = SCNNode()
newLine.geometry = line
newLine.position = posBetween(first: node1, second: node2)
// This is the change in x,y and z between node1 and node2
let dirVector = SCNVector3Make(node2.x - node1.x, node2.y - node1.y, node2.z - node1.z)
// Get Y rotation in radians
let yAngle = atan(dirVector.x / dirVector.z)
// Rotate cylinder node about X axis so cylinder is laying down
currentLine.eulerAngles.x = .pi / 2
// Rotate cylinder node about Y axis so cylinder is pointing to each node
currentLine.eulerAngles.y = yAngle
This is the function to get the position between two nodes, place it within your class:
func posBetween(first: SCNVector3, second: SCNVector3) -> SCNVector3 {
return SCNVector3Make((first.x + second.x) / 2, (first.y + second.y) / 2, (first.z + second.z) / 2)
}
This is the extension to get the distance between nodes for the cylinder height, place it somewhere outside of your class:
extension SCNVector3 {
func distance(to destination: SCNVector3) -> CGFloat {
let dx = destination.x - x
let dy = destination.y - y
let dz = destination.z - z
return CGFloat(sqrt(dx*dx + dy*dy + dz*dz))
}
}
If you don't have one fixed axis like myself then you could do the extra trig to use this method.
Here's a solution using simd and quaternions for the rotation. I based the extension off of the answer by #Bersaelor.
I used this derivation (https://stackoverflow.com/a/1171995/6693924) to create the quaternion from two vectors. Hope this helps.
extension SCNNode {
static func lineNode(from: simd_float3, to: simd_float3, radius : CGFloat = 0.25) -> SCNNode
{
let vector = to - from
let height = simd_length(vector)
//cylinder
let cylinder = SCNCylinder(radius: radius, height: CGFloat(height))
cylinder.firstMaterial?.diffuse.contents = UIColor.white
//line node
let lineNode = SCNNode(geometry: cylinder)
//adjust line position
let line_axis = simd_float3(0, height/2, 0)
lineNode.simdPosition = from + line_axis
let vector_cross = simd_cross(line_axis, vector)
let qw = simd_length(line_axis) * simd_length(vector) + simd_dot(line_axis, vector)
let q = simd_quatf(ix: vector_cross.x, iy: vector_cross.y, iz: vector_cross.z, r: qw).normalized
lineNode.simdRotate(by: q, aroundTarget: from)
return lineNode
}
}
Sprout's (wow, the autocorrect will not allow me to actually type in his name!) post is indeed a solution, but I have implemented a very different solution in my code.
What I do is calculate the length of the line and the two endpoints, based on the X, Y and Z locations from the two ends:
let w = SCNVector3(x: CGFloat(x2m-x1m), y: CGFloat(y2m-y1m), z: CGFloat(z2m-z1m))
let l = w.length()
The length is simply pythag. Now I make an SCNNode that will hold the SCNCylinder, and position it in the middle of the line:
let node = SCNNode(geometry: cyl)
node.position = SCNVector3(x: CGFloat((x1m+x2m)/2.0), y: CGFloat((y1m+y2m)/2.0), z: CGFloat((z1m+z2m)/2.0))
And now the nasty part, where we calculate the Euler angles and rotate the node:
let lxz = (Double(w.x)**2 + Double(w.z)**2)**0.5
var pitch, pitchB: Double
if w.y < 0 {
pitchB = M_PI - asin(Double(lxz)/Double(l))
} else {
pitchB = asin(Double(lxz)/Double(l))
}
if w.z == 0 {
pitch = pitchB
} else {
pitch = sign(Double(w.z)) * pitchB
}
var yaw: Double
if w.x == 0 && w.z == 0 {
yaw = 0
} else {
let inner = Double(w.x) / (Double(l) * sin (pitch))
if inner > 1 {
yaw = M_PI_2
} else if inner < -1 {
yaw = M_PI_2
} else {
yaw = asin(inner)
}
}
node.eulerAngles = SCNVector3(CGFloat(pitch), CGFloat(yaw), 0)
I suspect there is a much simpler way to do this using one of the other rotation inputs, but this works and working is a feature!
Draw the line between two nodes:
func generateLine( startPoint: SCNVector3, endPoint: SCNVector3) -> SCNGeometry {
let vertices: [SCNVector3] = [startPoint, endPoint]
let data = NSData(bytes: vertices, length: MemoryLayout<SCNVector3>.size * vertices.count) as Data
let vertexSource = SCNGeometrySource(data: data,
semantic: .vertex,
vectorCount: vertices.count,
usesFloatComponents: true,
componentsPerVector: 3,
bytesPerComponent: MemoryLayout<Float>.size,
dataOffset: 0,
dataStride: MemoryLayout<SCNVector3>.stride)
let indices: [Int32] = [ 0, 1]
let indexData = NSData(bytes: indices, length: MemoryLayout<Int32>.size * indices.count) as Data
let element = SCNGeometryElement(data: indexData,
primitiveType: .line,
primitiveCount: indices.count/2,
bytesPerIndex: MemoryLayout<Int32>.size)
return SCNGeometry(sources: [vertexSource], elements: [element])
}
How To Use
let line = generateLine(startPoint: SCNVector3Make(1, 1, 1), endPoint: SCNVector3Make(8, 8, 8))
let lineNode = SCNNode(geometry: line)
lineNode.position = SCNVector3Make(15, 15, 10)
scene.rootNode.addChildNode(lineNode)
The thickness of the line requires implementing the SCNSceneRendererDelegate, in particular:
func renderer(_ renderer: SCNSceneRenderer, willRenderScene scene: SCNScene, atTime time: TimeInterval){
glLineWidth(10)
}
Objective-C version of Winchill's answer:
-(void)lineNodeFrom:(SCNVector3)to to:(SCNVector3)from radius:(float)radius{
SCNVector3 w = SCNVector3Make(to.x - from.x, to.y - from.y, from.z - to.z);
float l = sqrtf(powf(w.x, 2) + powf(w.y, 2) + powf(w.z, 2.0f));
SCNCylinder * cylinder = [SCNCylinder cylinderWithRadius:radius height:l];
SCNMaterial * material = [SCNMaterial material];
material.diffuse.contents = [[UIColor darkGrayColor] colorWithAlphaComponent:0.75f];
cylinder.materials = #[material];
[self setGeometry:cylinder];
//original vector of cylinder above 0,0,0
SCNVector3 ov = SCNVector3Make(0, l/2.0,0);
//target vector, in new coordination
SCNVector3 nv = SCNVector3Make((from.x - to.x)/2.0, (from.y - to.y)/2.0, (from.z-to.z)/2.0);
// axis between two vector
SCNVector3 av = SCNVector3Make((ov.x + nv.x)/2.0, (ov.y+nv.y)/2.0, (ov.z+nv.z)/2.0);
//normalized axis vector
SCNVector3 av_normalized = [self normaliseVector:av];
float q0 = 0.0f; //cos(angel/2), angle is always 180 or M_PI
float q1 = av_normalized.x; // x' * sin(angle/2)
float q2 = av_normalized.y; // y' * sin(angle/2)
float q3 = av_normalized.z; // z' * sin(angle/2)
float r_m11 = q0 * q0 + q1 * q1 - q2 * q2 - q3 * q3;
float r_m12 = 2 * q1 * q2 + 2 * q0 * q3;
float r_m13 = 2 * q1 * q3 - 2 * q0 * q2;
float r_m21 = 2 * q1 * q2 - 2 * q0 * q3;
float r_m22 = q0 * q0 - q1 * q1 + q2 * q2 - q3 * q3;
float r_m23 = 2 * q2 * q3 + 2 * q0 * q1;
float r_m31 = 2 * q1 * q3 + 2 * q0 * q2;
float r_m32 = 2 * q2 * q3 - 2 * q0 * q1;
float r_m33 = q0 * q0 - q1 * q1 - q2 * q2 + q3 * q3;
SCNMatrix4 transform;
transform.m11 = r_m11;
transform.m12 = r_m12;
transform.m13 = r_m13;
transform.m14 = 0.0;
transform.m21 = r_m21;
transform.m22 = r_m22;
transform.m23 = r_m23;
transform.m24 = 0.0;
transform.m31 = r_m31;
transform.m32 = r_m32;
transform.m33 = r_m33;
transform.m34 = 0.0;
transform.m41 = (to.x + from.x) / 2.0;
transform.m42 = (to.y + from.y) / 2.0;
transform.m43 = (to.z + from.z) / 2.0;
transform.m44 = 1.0;
self.transform = transform;
}
-(SCNVector3)normaliseVector:(SCNVector3)iv{
float length = sqrt(iv.x * iv.x + iv.y * iv.y + iv.z * iv.z);
if (length == 0){
return SCNVector3Make(0.0, 0.0, 0.0);
}
return SCNVector3Make(iv.x / length, iv.y / length, iv.z / length);
}
Refer to
Fastest Inverse Square Root on iPhone
I need do a "Fastest Inverse Square Root" on iPhone iOS Swift, which is supposed to be faster than 1/sqrt(float).
How do I do it?
In embedded C programming, it is:
// Fast inverse square-root
// See: http://en.wikipedia.org/wiki/Fast_inverse_square_root
func invSqrt(x: Float) -> Float {
var halfx : Float = 0.5 * x
var y : Float = x
long i = *(long*)&y
i = 0x5f3759df - (i>>1)
y = *(float*)&i
y = y * (1.5 - (halfx * y * y))
return y
}
The only tricky part is how to do the forced conversions between floating
point numbers and integer types, and the easiest way is to use
memcpy():
// Fast inverse square-root
// See: http://en.wikipedia.org/wiki/Fast_inverse_square_root
func invSqrt(x: Float) -> Float {
let halfx = 0.5 * x
var y = x
var i : Int32 = 0
memcpy(&i, &y, 4)
i = 0x5f3759df - (i >> 1)
memcpy(&y, &i, 4)
y = y * (1.5 - (halfx * y * y))
return y
}
I made some performance tests on an iPhone 6s with 1.000.000 random
floating point numbers in the range 0 ... 1000, and it turned out
that invSqrt(x) is about 40% faster than 1.0/sqrt(x).
The maximal relative error was below 0.176%, confirming the bound in
the Wikipedia article.
I also made a test with vvrsqrtf from the
Accelerate framework, but this was actually slower than
calling 1.0/sqrt(x), at least when called with single floating
point numbers.
As of Swift 3, memcpy() can be replaced by the bitPattern:
method of Float and the corresponding constructor from UInt32:
func invSqrt(x: Float) -> Float {
let halfx = 0.5 * x
var i = x.bitPattern
i = 0x5f3759df - (i >> 1)
var y = Float(bitPattern: i)
y = y * (1.5 - (halfx * y * y))
return y
}