There are (apparently) several types of gradients that can be interesting to support, but are not part of the SVG standard. This page is meant to document which gradient types Inkscape may want to support and how these could be simulated.

Separating shape/pattern from color

Any gradient can be defined by a mapping $\displaystyle{ f:\mathbb{R}^2\rightarrow[0,1] }$ from any point in the plane to a value in the range $\displaystyle{ [0,1] }$, combined with a mapping from the range $\displaystyle{ [0,1] }$ to colors. The former is defined by the type of gradient (and its parameters) and the latter by the gradient stops.

For each new gradient type it suffices to find a good method of generating $\displaystyle{ f }$ (over all four color channels). The desired colors can then be determined by using an feComponentTransfer filter with the table transfer function type. This allows the gradient to be approximated to any required accuracy. An example for a gradient from red to green to blue to transparent black (assumes we are in a filter definition and an image with $\displaystyle{ f }$ on all four channels is in "f"):

<feComponentTransfer in="f">
<feFuncR type="table" tableValues="1 0 0 0" />
<feFuncG type="table" tableValues="0 1 0 0" />
<feFuncB type="table" tableValues="0 0 1 0" />
<feFuncA type="table" tableValues="1 1 1 0" />
</feComponentTransfer>


Alternatively the mapping to colors of the gradient could be done using feDisplacementMap, if $\displaystyle{ f }$ is pre-processed properly. Specifically, the values of $\displaystyle{ f }$ are meant to be interpreted in an absolute sense, while feDisplacementMap uses them as an offset relative to the current position. So if $\displaystyle{ (x1,y1),(x2,y2) }$ defines the bounding box of the object to which the gradient is applied, and a (horizontal) linear gradient is applied to a rectangle using these same coordinates, then the scale parameter should be set in such a way that $\displaystyle{ x1+scale*0.5=x2 }$ and $\displaystyle{ x2-scale*0.5=x1 }$, so $\displaystyle{ scale=2(x2-x1) }$. The x displacement channel should then be set to a channel corresponding to the pre-processed $\displaystyle{ f }$ and the y displacement channel should then be set to some channel that is 0.5 everywhere. An example for the same gradient as above (we are in a defs element, the red channel is assumed to correspond to $\displaystyle{ f }$ and the green channel to one, the bounding box is assumed to be $\displaystyle{ (0,0),(1,1) }$):

<g id="symbolstops">
<stop offset="0" stop-color="#f00" />
<stop offset="0.33" stop-color="#0f0" />
<stop offset="0.67" stop-color="#00f" />
<stop offset="1" stop-color="#000" stop-opacity="0" />
<rect width="150" height="150" fill="url(#gradientstops)" />
</g>
<g id="symbolxoffset">
<stop offset="0" stop-color="#F00" />
<stop offset="1" stop-color="#000" />
<rect width="1" height="1" fill="url(#gradientxoffset)" />
</g>
<filter ...>
...
<feImage xlink:href="#symbolxoffset" result="xoffset" />
<feComposite in="f" in2="xoffset" result="fp" operator="arithmetic" k2="0.5" k3="0.5" />
<feImage xlink:href="#symbolstops" result="stops" />
<feDisplacementMap in="stops" in2="fp" scale="2" xChannelSelector="R" yChannelSelector="G" />
</filter>


This last method has the advantage that there is a more direct correspondence with the gradient stops defined by the user (and does not use an approximation to the gradient as the feComponentTransfer method), but it is obviously also more complex. Also, if the mapping $\displaystyle{ f }$ needs a (non-linear) correction this can directly be taken into account in the feComponentTransfer method. However, the feComponentTransfer method can give a lower quality result if the alpha channel is also part of the gradient (and the viewer uses premultiplied alpha).

"A gradient which goes along the circular arc around a center." In other words (ignoring rotations and so on), given a center $\displaystyle{ (cx,cy) }$, the mapping $\displaystyle{ f }$ is defined as $\displaystyle{ f(x,y)=\frac{1}{2\pi}\arctan(y-cy,x-cx)+\frac{1}{2} }$.

One reasonably obvious way to mimic a conical gradient is to use a circular gradient to define a conical surface and then use feDiffuseLighting. This doesn't directly result in the right mapping, but it at least shares the property that the pixels on radial lines from the center have the same value. In addition, the generated gradient will be symmetrical, so if this is not wanted the region should be split in two along the symmetry axis.

To derive the right parameters, first define the (ideal) surface of the cone created by the gradient as $\displaystyle{ Z(x,y)=s-\frac{s}{r}\sqrt{x^2+y^2} }$ (setting the center to $\displaystyle{ (0,0) }$ without loss of generality). Per the SVG 1.1 specification the surface normal is defined by using a sobel filter, however, as this is a reasonable approximation of a directional derivative we use the derivative instead. Specifically, taking the right scale factors into account, the surface normal is defined as:

\displaystyle{ \begin{align} N'_x(x,y)&=-2 \frac{\partial}{\partial x}Z(x,y) \\&=\frac{2sx}{\sqrt{x^2+y^2}} \\N'_y(x,y)&=\frac{2sy}{\sqrt{x^2+y^2}} \\N'_z(x,y)&=1 \\N(x,y)&=\frac{N'}{\|N\|} \end{align} }

Assuming a distant (white) light source with a zero azimuth and a diffuse lighting constant of one, the intensity of the resulting image can now be computed as:

\displaystyle{ \begin{align} D(x,y)&=N(x,y)\cdot L(e) \\&=\frac{1}{\|N'(x,y)\|}\left(\frac{2sx}{r\sqrt{x^2+y^2}}\cos(e)+\sin(e)\right) \\&=\frac{1}{\sqrt{4s^2+r^2}\sqrt{x^2+y^2}}\left(2sx\cos(e)+r\sqrt{x^2+y^2}\sin(e)\right) \end{align} }

To get a nice gradient we can now demand that $\displaystyle{ D(x,0) }$ is zero for negative $\displaystyle{ x }$ and one for positive $\displaystyle{ x }$ and solve for $\displaystyle{ s }$ and $\displaystyle{ e }$:

\displaystyle{ \begin{align} D(x,0)&=\frac{1}{\sqrt{4s^2+r^2}|x|}\left(2sx\cos(e)+r|x|\sin(e)\right) \\&=\frac{1}{\sqrt{4s^2+r^2}}\left(2s\mbox{sign}(x)\cos(e)+r\sin(e)\right) \\D(x_-,0)&=0 \\0&=\frac{1}{\sqrt{4s^2+r^2}}\left(-2s\cos(e)+r\sin(e)\right) \\2s\cos(e)&=r\sin(e) \\\frac{\sin(e)}{\cos{e}}&=\frac{2s}{r} \\e&=\arctan(\frac{2s}{r}) \\D(x_+,0)&=1 \\1&=\frac{1}{\sqrt{4s^2+r^2}}\left(2s\cos(e)+r\sin(e)\right) \\\sqrt{4s^2+r^2}&=2r\sin(e) \\\sqrt{4s^2+r^2}&=\frac{4rs}{r\sqrt{\frac{4s^2}{r^2}+1}} \\\sqrt{4s^2+r^2}&=\frac{4rs}{\sqrt{\frac{4s^2}+r^2}} \\4s^2+r^2&=4rs \\4s^2-4rs+r^2&=0 \\(2s-r)^2&=0 \\s&=\frac{1}{2}r \\e&=\arctan(1) \\&=\frac{1}{4}\pi \end{align} }

So surfaceScale should be half of the radius of the radial gradient and the elevation of the light source should be 45 degrees. By filling in this result in $\displaystyle{ D }$ it is apparent that the intensity now corresponds exactly to the cosine of the angle we are really interested in with a conical gradient (scaled and translated to fit in the range $\displaystyle{ [0,1] }$):

\displaystyle{ \begin{align} D(x,y)&=\frac{1}{\sqrt{4(\frac{1}{2}r)^2+r^2}\sqrt{x^2+y^2}}\left(2\frac{1}{2}rx\cos(\frac{1}{4}\pi)+r\sqrt{x^2+y^2}\sin(\frac{1}{4}\pi)\right) \\&=\frac{1}{\sqrt{2}\sqrt{x^2+y^2}}\left(x\frac{1}{\sqrt{2}}+\sqrt{x^2+y^2}\frac{1}{\sqrt{2}}\right) \\&=\frac{x}{2\sqrt{x^2+y^2}}+\frac{1}{2} \end{align} }

For a perfect result it should now be taken into account that we have a cosine and not an angle, but the difference is not very big and it may not always be an issue.

Here defined as similar to a conical gradient, but with the mapping $\displaystyle{ f }$ also depending on the distance to the center. Specifically $\displaystyle{ f(x,y)=\frac{1}{2\pi}\arctan(y-cy,x-cx)+\sqrt{(x-cx)^2+(y-cy)^2}\mod 1 }$.