Linearize an exponential function

[mfersaoui]

Hi,
I have created an user data slider (with percent unit) to controlling subdivisions number of a spline (spline with an Adaptive interpolation). So I have used a MapRange function to convert a percentage to a degree but this dosen't returnning a natural result. because the number of an intermediate subdivision points is increasing exponentially.
I want to know how to fit an exponential number of intermediate points to a linear.

        SplineObject *sp = SplineObject::Alloc(4, SPLINETYPE_BEZIER); // create a spline with 3 verts
        Vector *gp = sp->GetPointW();

        gp[0] = Vector(0, 0, 0);
        gp[1] = Vector(0, 0, 900);
        gp[2] = Vector(0, 250, 1200);
        gp[3] = Vector(0, 900, 1200);

        sp->GetDataInstance()->SetBool(SPLINEOBJECT_CLOSED, false);

        Float value = data->GetFloat(BSK_OBJECT_CURVE_SUB);
	Float min_input = 0;
	Float max_input = 1;
	Float min_output = 90;
	Float max_output = 1;

	Float inrange = max_input - min_input;
	if (CompareFloatTolerant(inrange, 0.0)) value = 0.0;
	else value = (value - min_input) / inrange;
	curve_sub = DegToRad(min_output + (max_output - min_output) * value);

	sp->SetParameter(SPLINEOBJECT_ANGLE, curve_sub, DESCFLAGS_SET_0);

[r_gigante]

Hi mfersaoui, thanks for reaching out us.

With regard to your issue I've a few notes:

• the Adaptive interpolation is by definition strictly dependent on the shape of the spline with regard to the tessellation of the curve: the curve will be tessellated more where the curvature is larger and this generates evidently non-uniform segments distribution - check here for further notes -;
• standing on the code presented the in_range variable is always 1.0 which makes this

if (CompareFloatTolerant(inrange, 0.0)) value = 0.0;
	else value = (value - min_input) / inrange;

pretty useless.

I would recommend, assuming that the adaptive strategy is the one you want to keep to consider ( sorry for the python code, but porting should be trivial)

        # swap the max/min values of the slider
        val = op[c4d.ID_USERDATA, 1]
        # set the interpolation method
        interp = 3
        # allocate the spline
        spline = c4d.SplineObject(4, c4d.SPLINETYPE_BSPLINE)
        # set interpolation method
        spline[c4d.SPLINEOBJECT_INTERPOLATION] = interp
        # switch according to interpolation method
        if interp == 2: # uniform
            # set the max number of segs
            maxSegs = 100
            spline[c4d.SPLINEOBJECT_SUB] = int(maxSegs * val)
        if interp == 3: # adaptive
            # get the rad representation of the slider's value
            radVal = c4d.utils.DegToRad(90 * (1.0 - val))
            # set the adaptive angle value
            spline[c4d.SPLINEOBJECT_ANGLE] = radVal
        # set the points
        point = spline.SetAllPoints((c4d.Vector(0,0,0),c4d.Vector(0,0,900),c4d.Vector(0,900,900), c4d.Vector(0,900,0)))
        # return the spline
        return spline

If instead you'd like to linearly increase the subdivision, I warmly recommend to switch to the uniform method and you'll see that a linear behavior is established

Best, R

[mfersaoui]

@r_gigante Thank you for the detailed reply,

Sorry, I didn't put the last part of my code in my question above. I will mention that I want to set a tangent on the points 2 and 3, here is the last part of my code in python:

curve_offset = op[c4d.BSK_OBJ_CURVE_OFFSET] # correspond to the value "250" in my above code
spline.SetTangent(1, Vector(0, 0, 0), Vector(0, 0, curve_offset * 0.5))
spline.SetTangent(2, Vector(0, -curve_offset * 0.5, 0), Vector(0, 0, 0))

So, I try to add the lines above on your code but this returning this error message:

IndexError: tangent index out of range

Thanks Again.

[mfersaoui]

@r_gigante
Finally I found the solution to linearize the subdivision expansion on Adaptive spline interpolation on the angle deviation.

I just replaced the line :

radVal = c4d.utils.DegToRad(90 * (1.0 - val)

By :

radVal = math.pow(val, -0.85)

I changed also the slider user data unit to REAL
if I want to keep it with a PERCENT unit I just multiply the value by 100

radVal = math.pow(val*100, -0.85)

Here is demo: