CODEWITHSUNDEEP
c#mathcurve
Garlik
Garlik

Reputation: 13

Calculate a specific curve with specific rotation, c#

I am trying to code for a game I am working on a specific curve with a specific rotation. I am not a great mathematician... At all... Tried searching for solutions for a few hours, but I'm affraid I do not find any solution.

So, a small picture to illustrate first:

enter image description here

This is an eighth of a circle, radius of 9, beggining is (0,0)

The end is now at about 6.364, -2.636. But I need this same curve, with a 45° direction at the end, but ending at aexactly 6.0,-3.0.

Could any of you show me how to do this? I need to be able to calculate precisly any point on this curve & its exact length. I would suppose using some kind of eliptical math could be a solution? I admit my math class are reaaaly far now and have now good clue for now...

Thank for any possible help

Upvotes: 0

Views: 985

Answers (2)

AlexDev
AlexDev

Reputation: 4717

I think I found a quadratic curve which sastisfies your requirement:

f(x) = -1/12 x^2 + 9

Copy the following into https://www.desmos.com/calculator to see it:

-\frac{1}{12}x^2+9

f'(x) would be -1/6x, so when x=6, the derivative would be -1, which corresponds to a -45° inclination. There are probably infinite curves that satisfy your requirement but if my calculus isn't too rusty this is one of them.

I tried to fit an ellipse with foci starting at y=6 here and starting at y=9 here to your points but the slope doesn't look like 45°. Also starting at any height k, here doesn't seem to work.

Upvotes: 1

meowgoesthedog
meowgoesthedog

Reputation: 15035

I don't think you've fully understood the question I asked in the comments about the "inclination" angle. So I will give a general case solution, where you have an explicit tangent vector for the end of the curve. (You can calculate this using the inclination angle; if we clarify what you mean by it then I will be happy to edit with a formula to calculate the tangent vector if necessary)

Let's draw a diagram of how the setup can look:

enter image description here

(Not 100% accurate)

A and B are your fixed points. T is the unit tangent vector. r and C are the radius and center of the arc we need to calculate.

The angle θ is given by the angle between BA and T minus π/2 radians (90 degrees). We can calculate it using the dot product:

enter image description here

The (signed) distance from the center of AB to C is given by:

enter image description here

Note that this is negative for the case on the right, and positive for the left. The radius is given by:

enter image description here

(You can simplify by substituting and using a cosine addition rule, but I prefer to keep things in terms of variables in the diagram). To obtain the point C, we need the perpendicular vector to AB (call it n):

enter image description here

Now that we have the radius and center of the circular arc, we still need to determine which direction we are moving in, i.e. whether we are moving clockwise or anti-clockwise when going from A to B. This is a simple test, using the cross-product:

enter image description here

If this is negative, then T is as in the diagram, and we need to move clockwise, and vice versa. The length of the arc l, and the angular displacement γ when we move by a distance x along the arc:

enter image description here

Nearly there! Just one more step - we need to work out how to rotate the point A by angle γ around point C, to get the point we want (call it D):

enter image description here

(Adapted from this Wikipedia page)

Now for some code, in case the above was confusing (it probably was!):

public Vector2 getPointOnArc(Vector2 A, Vector2 B, Vector2 T, double x)
{
    // calculate preliminaries
    Vector2 BA = B - A;
    double d = BA.Length();
    double theta = Math.Acos(Vector2.DotProduct(BA, T) / d) - Math.PI * 0.5;

    // calculate radius
    double r = d / (2.0 * Math.Cos(theta));

    // calculate center 
    Vector2 n = new Vector2(BA.y, -BA.x);
    Vector2 C = 0.5 * (A + B + n * Math.Tan(theta));

    // calculate displacement angle from point A
    double l = (Math.PI - 2.0 * theta) * r;
    double gamma = (2.0 * Math.PI * x) / l;

    // sign change as discussed
    double cross = T.x * BA.y - T.y * BA.x;
    if (cross < 0.0) gamma = -gamma;

    // finally return the point we want
    Vector2 disp = A - C;
    double c_g = Math.Cos(gamma), s_g = Math.Sin(gamma);
    return new Vector2(disp.X * c_g + disp.Y * s_g + C.X,
                       disp.Y * c_g - disp.X * s_g + C.Y);
}

Upvotes: 0

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