Calculate a 3D trajectory by start point, end point and height

Question

I've already figured out how to make a 3D trajectory using a start point and an angle. However, I am trying to make a trajectory from a start point, an end point, and a height.

I tried taking the approach of a parabola on a 2D plane in a 3D space. I calculated the Prabola's A, B, and C values as well as the plane it's on given 3 points on the Parabola. However, I've had a few complications with this sort of calculation, I assume it has to do with the inability to properly calculate a Z-axis without a plane but I cannot tell.

Other than a 2D parabola on a plane google did not provide another possible answer and a 3D trajectory yields a formula using a start point, an angle, and a power multiplier.

• Is there any way to calculate a 3D trajectory given the start point, end point, and height?

Appreciating your help

Edit :

My code to calculate a parabola using 3 points (in case someone would like to know how I've done that and perhaps fix what I've done wrong)

public Parabola(Vector3 pa, Vector3 pb, Vector3 pc)
        {
            this.pa = pa;
            this.pc = pc;
            float a1 = -pa.x * pa.x + pb.x * pb.x, b1 = -pa.x + pb.x, c1 = -pa.y + pb.y;
            float a2 = -pb.x * pb.x + pc.x * pc.x, b2 = -pb.x + pc.x, c2 = -pb.y + pc.y;
            float bm = -(b2 / b1), a3 = bm * a1 + a2, c3 = bm * c1 + c2;
            float a = c3 / a3, b = (c1 - a1 * a) / b1, c = pa.y - a * pa.x * pa.x - b * pa.x;
            this.a = a; this.b = b; this.c = c;
            plane = Vector3.Cross(pb - pa, pc - pa);
        }

public Vector3 GetPoint(float x) 
        {
            float angle = Mathf.Atan2(pc.z - pa.z, pc.x - pa.x) * Mathf.Rad2Deg;
            float xs = Mathf.Cos(angle * Mathf.Deg2Rad) * x, zs = Mathf.Sin(angle * Mathf.Deg2Rad) * x;
            return new Vector3(xs, a * x * x + b * x + c, zs); 
        }

public Vector3 ProjectOn(float x) => Vector3.ProjectOnPlane(GetPoint(x), plane);

The result looks ok when it's only on 2 Axis, but not 3. here are 2 images for demonstration:

Answer 1

Looking at the second image, the parabola seems to be correct aside from being scaled incorrectly. Let's take a look at your code.

public Vector3 GetPoint(float x) 
    {
        float angle = Mathf.Atan2(pc.z - pa.z, pc.x - pa.x) * Mathf.Rad2Deg;
        float xs = Mathf.Cos(angle * Mathf.Deg2Rad) * x, zs = Mathf.Sin(angle * Mathf.Deg2Rad) * x;
        return new Vector3(xs, a * x * x + b * x + c, zs); 
    }

I'm making a lot of assumptions here, but it seems like x is meant as a value that goes from 0.0 to 1.0, representing the start and end of the parabola, respectively. If so, you are determining the X and Z coordinates of this point based exclusively on the sine/cosine of the angle and x . This means that the values xs and zs should only ever be able to be between -1 and 1, limiting yourself to the confines of the unit circle.

The values xs and zs look like they need to be scaled by a factor s calculated by measuring the 2D distance of the start and end points when projected onto the XZ plane. This should stretch the parabola just enough to reach the end point.

Answer 2

I found an answer, but it's kinda a workaround.

Before messing around with Parabolas in 3D, I messed around with linear equations in 3D. Unlike parabolas, lines have a defined equation even in 3D( Pn = P0 + tx V )(Pn vector containing XYZ, P0 initial point containing XYZ, t float, V Vector3)

In addition, there's only ONE line that goes through 2 points, even in 3D.

I used that to make a trajectory that's made out of 2 points and a height. I make a new point in the center of those two points and add the height value to the highest Y value of the points, thus creating an Apex.

then I use the same calculations as before to calculate the A, B, and C values that a parabola with those 3 points would have had.

I made a method that takes in an X value and returns a Vector3 containing the point this X is on a linear equation, but instead, changing the vector's Y value based on the parabola's equation.

Practically creating an elevated line, I made something that looks and behaves perfectly like a 3D parabola.

If you're in C#, here is the code(images):

FIX!!

in the Linear's GetX(float x) method. it should be:

public Vector3 GetX(float x) => => r0 + (x - r0.x)/v.x * v;

I made a slight mistake in the calculations which I noticed immediately and changed.