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I'm attempting to perform an intersection test using ray casting (not sure if correct term so please forgive me if not) and am outputting the intersections as a point cloud, and the point cloud shows curvature (on the Z-axis only, the point cloud is completely flat on the Y axis, and the horizontal axis in this image is the X axis): enter image description here

I borrowed concepts from the Scratchapixel site, specifically http://scratchapixel.com/lessons/3d-basic-rendering/minimal-ray-tracer-rendering-simple-shapes/ray-box-intersection.

Essentially, I am generating 16 rays, all with the same origin vector. The direction vectors start at +15 degrees on the YZ plane, and continue in increments of -2 degrees down to -15. I have an axis aligned bounding box that I am testing intersection with. I use a rotation transform to rotate the 16 rays CCW around the Z axis. I am performing the intersection test for all 16 rays each 0.1 degrees, and if it returns true, I add the point to the point cloud.

Here's my intersection code:

bool test_intersect(Box b, Ray r, Vec3f& intersect_point)
{
    float txmin = 0.0f, txmax = 0.0f, tymin = 0.0f, tymax = 0.0f, tzmin = 0.0f, tzmax = 0.0f;
    float t_min = 0.0f, t_max = 0.0f, t = 0.0f;

    // Determine inverse direction of ray to alleviate 0 = -0 issues
    Vec3f inverse_direction(1 / r.direction.x, 1 / r.direction.y, 1 / r.direction.z);

    // Solving box_min/box_max0 = O + Dt 
    txmin = (b.box_min.x - r.origin.x) * inverse_direction.x;
    txmax = (b.box_max.x - r.origin.x) * inverse_direction.x;
    tymin = (b.box_min.y - r.origin.y) * inverse_direction.y;
    tymax = (b.box_max.y - r.origin.y) * inverse_direction.y;
    tzmin = (b.box_min.z - r.origin.z) * inverse_direction.z;
    tzmax = (b.box_max.z - r.origin.z) * inverse_direction.z;

    // Depending on direction of ray tmin may > tmax, so we may need to swap
    if (txmin > txmax) std::swap(txmin, txmax);
    if (tymin > tymax) std::swap(tymin, tymax);
    if (tzmin > tzmax) std::swap(tzmin, tzmax);

    t_min = txmin;
    t_max = txmax;

    // If t-value of a min is greater than t-value of max,
    // we missed the object in that plane.
    if ((t_min > tymax) || (tymin > t_max))
        return false;

    if (tymin > t_min)
        t_min = tymin;
    if (tymax < t_max)
        t_max = tymax;

    if ((t_min > tzmax) || (tzmin > t_max))
        return false;

    if (tzmin > t_min)
        t_min = tzmin;
    if (tzmax < t_max)
        t_max = tzmax;

    if (t_min > 0)
        t = t_min;
    else
        if (t_max > 0)
            t = t_max;
        else
            return false;

    intersect_point.x = r.origin.x + r.direction.x * t;
    intersect_point.y = r.origin.y + r.direction.y * t;
    intersect_point.z = r.origin.z + r.direction.z * t;

    return true;

}

And my rotation:

// Rotation around z axis, for rotating array and checking beam intersections
void transform_rotate_z(Vec3f& in_vector, float angle)
{
    float radians = angle * (M_PI / 180);

    float result_x = cos(radians) * in_vector.x + -sin(radians) * in_vector.y;
    float result_y = sin(radians) * in_vector.x + cos(radians) * in_vector.y;
    in_vector.x = result_x;
    in_vector.y = result_y;

}

I have racked my brain for quite a while but I can't seem to determine how I can prevent this curvature, I'm sure I'm overlooking something simple. I'd be grateful for any help you can provide.

EDIT 9/18/16: It turned out that I had incorrectly believed that physical LIDAR units had some algorithmic way of "straightening" out the curves, which is not the case! The output is supposed to look like the picture above.

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1 Answer 1

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The direction vectors start at +15 degrees on the YZ plane, and continue in increments of -2 degrees down to -15.

There's your problem—by generating the rays that way, you're doing something more like a spherical projection or fisheye projection, not a linear perspective projection. The nonlinear projection is the origin of the curvature you see.

To get a perspective projection instead, conceptually the way to think about it is that there's an image rectangle floating in front of the camera, divided into grid squares (pixels), and you fire a ray from the camera position toward the center of each pixel. In code, this looks something like:

p = tan(horizontalFOV / 2)
q = tan(verticalFOV / 2)
for y = 0 to imageHeight - 1
    for x = 0 to imageWidth - 1
        rayDirection = Vec3f(1.0,
                             p * ((x + 0.5) / imageWidth * 2.0 - 1.0)
                             q * ((y + 0.5) / imageHeight * 2.0 - 1.0)).normalize()

Note that the directions generated this way are not equally spaced in angle (they're closer together in angle toward the edges of the image than at the center), but they are equally spaced in distance on the image rectangle.

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  • $\begingroup$ Thanks Nathan! I'm attempting to simulate a LIDAR scanner, so that's the reason for the fan beam and the equal angle spacing. The particular scanner I'm modeling my simulation after has its lasers in the fan beam configuration external to the housing, but as the returns pass in they are diverted from a fan beam into parallel beams (by way of a lens). I'm trying to find a way to maintain this model so as to make the simulation as accurate as possible. Any thoughts as to how I could do this? $\endgroup$
    – Irongrave
    Commented Aug 10, 2016 at 15:19
  • $\begingroup$ @Irongrave Sorry, I don't know much about LIDAR scanners. Do you have a diagram or something of the setup with the lens you mention? I'm having a hard time visualizing it. $\endgroup$ Commented Aug 10, 2016 at 20:54
  • $\begingroup$ Here's a picture from the manual: (imgur.com/a/czhpt) $\endgroup$
    – Irongrave
    Commented Aug 11, 2016 at 1:16
  • 1
    $\begingroup$ It turned out that I had incorrectly believed that physical LIDAR units had some algorithmic way of "straightening" out the curves, which is not the case. Your post helped me reach a final conclusion, so thank you, and I selected it as an answer as you were correct. $\endgroup$
    – Irongrave
    Commented Sep 19, 2016 at 1:51

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