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Polar coordinate point generation function upper bound is not 2Pi for theta?

So I wrote the following function to take a frame, and polar coordinate function and to graph it out by generating the cartesian coordinates within that frame. Here's the code.

func cartesianCoordsForPolarFunc(frame: CGRect, thetaCoefficient:Double, cosScalar:Double, iPrecision:Double, largestScalar:Double) -> Array<CGPoint> {

    // Frame: The frame in which to fit this curve.
    // thetaCoefficient: The number to scale theta by in the cos.
    // cosScalar: The number to multiply the cos by.
    // largestScalar: Largest cosScalar used in this frame so that scaling is relative.
    // iPrecision: The step for continuity. 0 < iPrecision <= 2.pi. Defaults to 0.1

    // Clean inputs
    var precision:Double = 0.1 // Default precision
    if iPrecision != 0 {// Can't be 0.
        precision = iPrecision
    }

    // This is ther polar function
    // var theta: Double = 0 //  0 <= theta <= 2pi
    // let r = cosScalar * cos(thetaCoefficient * theta)

    var points:Array<CGPoint> = [] // We store the points here
    for theta in stride(from: 0, to: Double.pi * 2 , by: precision) { //TODO: Try to recreate continuity. WHY IS IT NOT 2PI
        let x = cosScalar * cos(thetaCoefficient * theta) * cos(theta) // Convert to cartesian
        let y = cosScalar * cos(thetaCoefficient * theta) * sin(theta) // Convert to cartesian

        // newvalue = (max'-min')/(max-min)*(value-max)+max'
        let scaled_x = (Double(frame.width) - 0)/(largestScalar*2)*(x-largestScalar)+Double(frame.width) // Scale to the frame
        let scaled_y = (Double(frame.height) - 0)/(largestScalar*2)*(y-largestScalar)+Double(frame.height) // Scale to the frame

        points.append(CGPoint(x: scaled_x, y:scaled_y)) // Add the result

    }
    print("Done points")

    return points
}

The polar function I'm passing is r = 100*cos(9/4*theta) which looks like this.Graph of the polar function

I'm wondering why my function returns the following when theta goes from 0 to 2. (Please note I'm in this image I'm drawing different sizes flowers hence the repetition of the pattern) Partial graph of the function

As you can see it's wrong. Weird thing is that when theta goes from 0 to 2Pi*100 (Also works for other random values such as 2Pi*4, 2Pi*20 but not 2Pi*2 or 2Pi*10)it works and I get this.Correct graph

Why is this? Is the domain not 0 to 2Pi? I noticed that when going to 2Pi*100 it redraws some petals so there is a limit, but what is it?

PS: Precision here is 0.01 (enough to act like it's continuous). In my images I'm drawing the function in different sizes and overlapping (last image has 2 inner flowers).

Solution

  • No, the domain is not going to be 2π. Set up your code to draw slowly, taking 2 seconds for each 2π, and watch. It makes a whole series of full circles, and each time the local maxima and minima land at different points. That's what your petals are. It looks like your formula repeats after 8π.

    It looks like the period is the denominator of the theta coefficient * 2π. Your theta coefficient is 9/4, the denominator is 4, so the coefficient is 4*2π, or 8π.

    (That is based on playing in Wolfram Alpha and observing the results. I may be wrong.)