Perspective-Correct Barycentric Interpolation in Triangles

The image illustrates a 2D side view of a perspective projection. It shows a line segment in "view space" with endpoints $(X_0, Z_0)$ and $(X_1, Z_1)$, with an intermediate point $(X, Z)$ defined by a weight $t$. These are projected onto an image plane at $Z=1$, resulting in screen coordinates $S_0, S, S_1$. The diagram demonstrates that the interpolation weight $q$ in screen space is not equal to the weight $t$ in world/view space due to the non-linear nature of the perspective divide ($1/Z$).

Answer

To perform perspective-correct interpolation in a triangle, you must interpolate the values divided by depth ($I_i/Z_i$) and the reciprocal of depth ($1/Z_i$) using screen-space barycentric coordinates $(\lambda_0, \lambda_1, \lambda_2)$. The final attribute is recovered by dividing the interpolated product by the interpolated reciprocal depth.

Explanation

1. Analyze the non-linearity of projection In perspective projection, a 3D point $(X, Y, Z)$ projects to screen coordinates $s_x = X/Z$ and $s_y = Y/Z$. Because $Z$ varies linearly across a triangle in 3D space, $1/Z$ varies linearly in screen space. Linear interpolation of an attribute $A$ directly in screen space leads to "texture warping" because screen horizontal/vertical steps do not correspond to equal steps in 3D space.

2. Define the perspective-correct formulation Let a triangle have vertices $V_0, V_1, V_2$ with depths $Z_0, Z_1, Z_2$ and attribute values $I_0, I_1, I_2$. In screen space, we calculate barycentric coordinates $(\lambda_0, \lambda_1, \lambda_2)$ such that $\sum \lambda_i = 1$. The attribute $I$ at a pixel is found using: $\frac{I(s)}{Z(s)} = \lambda_0 \frac{I_0}{Z_0} + \lambda_1 \frac{I_1}{Z_1} + \lambda_2 \frac{I_2}{Z_2}$ The interpolated attribute divided by depth is a linear combination of the vertices' attributes divided by their respective depths.

3. Interpolate the reciprocal depth To isolate $I(s)$, we need the depth $Z(s)$ at that specific pixel. We use the property that $1/Z$ is linear in screen space: $\frac{1}{Z(s)} = \lambda_0 \frac{1}{Z_0} + \lambda_1 \frac{1}{Z_1} + \lambda_2 \frac{1}{Z_2}$ The reciprocal of the projected depth is the linearly interpolated sum of the reciprocal depths of the vertices.

4. Recover the final attribute value Finally, we divide the result of Step 2 by the result of Step 3. This cancels out the $1/Z$ term, leaving the perspective-correct attribute. $I(s) = \frac{\sum_{i=0}^2 \lambda_i \frac{I_i}{Z_i}}{\sum_{i=0}^2 \lambda_i \frac{1}{Z_i}}$ The final attribute is the weighted sum of depth-divided attributes normalized by the weighted sum of reciprocal depths.

Final Answer

The perspective-correct barycentric interpolation for an attribute $I$ across a triangle is given by: $\boxed{I_{pixel} = \frac{\frac{\lambda_0 I_0}{Z_0} + \frac{\lambda_1 I_1}{Z_1} + \frac{\lambda_2 I_2}{Z_2}}{\frac{\lambda_0}{Z_0} + \frac{\lambda_1}{Z_1} + \frac{\lambda_2}{Z_2}}}$

Common Mistakes

• Linear Screen Interpolation: Assuming that $\lambda_0 I_0 + \lambda_1 I_1 + \lambda_2 I_2$ will yield the correct result. This causes "affine mapping" artifacts, which are highly visible in low-poly models with textures (textures appear to "swim" or bend as the camera moves).
• Using $Z$ instead of $W$: In modern graphics APIs (OpenGL/DirectX), the value used for the division is usually the clip-space $W$ component. While $W$ is proportional to view-space $Z$, using the wrong depth buffer value (like a non-linear $Z_{buffer}$ value) will break the math.

Related Topics: Rasterization, Hyperbolic Interpolation, Texture Mapping, Homogeneous Coordinates.