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Add documentation for barycentric coordinates

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Doyle Thai 2017-05-12 14:45:13 +10:00
parent 0738326384
commit afcd158d7d
1 changed files with 68 additions and 23 deletions
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91
src/DRenderer.cpp
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@ -143,21 +143,22 @@ FILE_SCOPE void DrawTriangle(PlatformRenderBuffer *const renderBuffer,
} }
/* /*
NOTE(doyle): Given two points that form a line and an extra point /////////////////////////////////////////////////////////////////////////
to test, we can determine whether a point lies on the line, or is // Rearranging the Determinant
to the left or right of a the line. /////////////////////////////////////////////////////////////////////////
Given two points that form a line and an extra point to test, we can
determine whether a point lies on the line, or is to the left or right of
a the line.
First forming a 3x3 matrix of our terms and deriving a 2x2 matrix First forming a 3x3 matrix of our terms and deriving a 2x2 matrix by
by subtracting the 1st column from the 2nd and 1st column from subtracting the 1st column from the 2nd and 1st column from the third.
the third.
| ax bx cx | | (bx - ax) (cx - ax) | | ax bx cx | | (bx - ax) (cx - ax) |
m = | ay by cy | ==> | (by - ay) (cy - ay) | m = | ay by cy | ==> | (by - ay) (cy - ay) |
| 1 1 1 | | 1 1 1 |
From our 2x2 representation we can calculate the determinant From our 2x2 representation we can calculate the determinant which gives
which gives us the signed area of the triangle extended into us the signed area of the triangle extended into a parallelogram.
a parallelogram.
det(m) = (bx - ax)(cy - ay) - (by - ay)(cx - ax) det(m) = (bx - ax)(cy - ay) - (by - ay)(cx - ax)
@ -167,38 +168,77 @@ FILE_SCOPE void DrawTriangle(PlatformRenderBuffer *const renderBuffer,
- CW and c(x,y) is outside the triangle, the signed area is positive - CW and c(x,y) is outside the triangle, the signed area is positive
- CW and c(x,y) is inside the triangle, the signed area is negative - CW and c(x,y) is inside the triangle, the signed area is negative
NOTE(doyle): The det(m) can be rearranged if expanded to be /////////////////////////////////////////////////////////////////////////
SignedArea(cx, cy) = (ay - by)cx + (bx - ay)cy + (ax*by + ay*bx) // Optimising the Determinant Calculation
/////////////////////////////////////////////////////////////////////////
The det(m) can be rearranged if expanded to be
SignedArea(cx, cy) = (ay - by)cx + (bx - ay)cy + (ax*by - ay*bx)
When we scan to fill our triangle we go pixel by pixel, left to right, When we scan to fill our triangle we go pixel by pixel, left to right,
bottom to top, notice that this translates to +1 for x and +1 for y, i.e. bottom to top, notice that this translates to +1 for x and +1 for y, i.e.
The first pixel's signed area is cx, then cx+1, cx+2 .. etc The first pixel's signed area is cx, then cx+1, cx+2 .. etc
SignedArea(cx, cy) = (ay - by)cx + (bx - ax)cy + (ax*by + ay*bx) SignedArea(cx, cy) = (ay - by)cx + (bx - ax)cy + (ax*by - ay*bx)
SignedArea(cx+1, cy) = (ay - by)cx+1 + (bx - ax)cy + (ax*by + ay*bx) SignedArea(cx+1, cy) = (ay - by)cx+1 + (bx - ax)cy + (ax*by - ay*bx)
Then Then
SignedArea(cx+1, cy) - SignedArea(cx, cy) = SignedArea(cx+1, cy) - SignedArea(cx, cy) =
(ay - by)cx+1 + (bx - ax)cy + (ax*by + ay*bx) (ay - by)cx+1 + (bx - ax)cy + (ax*by - ay*bx)
- (ay - by)cx + (bx - ax)cy + (ax*by + ay*bx) - (ay - by)cx + (bx - ax)cy + (ax*by - ay*bx)
= (ay - by)cx+1 - (ay - by)cx = (ay - by)cx+1 - (ay - by)cx
= (ay - by)(cx+1 - cx) = (ay - by)(cx+1 - cx)
= (ay - by)(1) = (ay - by) = (ay - by)(1) = (ay - by)
Similarly when progressing in y Similarly when progressing in y
SignedArea(cx, cy) = (ay - by)cx + (bx - ay)cy + (ax*by + ay*bx) SignedArea(cx, cy) = (ay - by)cx + (bx - ay)cy + (ax*by - ay*bx)
SignedArea(cx, cy+1) = (ay - by)cx + (bx - ay)cy+1 + (ax*by + ay*bx) SignedArea(cx, cy+1) = (ay - by)cx + (bx - ay)cy+1 + (ax*by - ay*bx)
Then Then
SignedArea(cx, cy+1) - SignedArea(cx, cy) = SignedArea(cx, cy+1) - SignedArea(cx, cy) =
(ay - by)cx + (bx - ax)cy+1 + (ax*by + ay*bx) (ay - by)cx + (bx - ax)cy+1 + (ax*by - ay*bx)
- (ay - by)cx + (bx - ax)cy + (ax*by + ay*bx) - (ay - by)cx + (bx - ax)cy + (ax*by - ay*bx)
= (bx - ax)cy+1 - (bx - ax)cy = (bx - ax)cy+1 - (bx - ax)cy
= (bx - ax)(cy+1 - cy) = (bx - ax)(cy+1 - cy)
= (bx - ax)(1) = (bx - ax) = (bx - ax)(1) = (bx - ax)
Then we can see that when we progress along x, we only need to change by Then we can see that when we progress along x, we only need to change by
the value of SignedArea by (ay - by) and similarly for y, (bx - ay) the value of SignedArea by (ay - by) and similarly for y, (bx - ay)
/////////////////////////////////////////////////////////////////////////
// Barycentric Coordinates
/////////////////////////////////////////////////////////////////////////
At this point we have an equation that can be used to calculate the
2x the signed area of a triangle, or the signed area of a parallelogram,
the two of which are equivalent.
det(m) = (bx - ax)(cy - ay) - (by - ay)(cx - ax)
SignedArea(cx, cy) = (ay - by)cx + (bx - ay)cy + (ax*by - ay*bx)
A barycentric coordinate is some coefficient on A, B, C that allows us to
specify an arbitrary point in the triangle as a linear combination of the
three usually with some coefficient [0, 1].
The SignedArea turns out to be actually the barycentric coord for c(x, y)
normalised to the sum of the parallelogram area. For example a triangle
with points, A, B, C and an arbitrary point P inside the triangle. Then
SignedArea(P) with vertex A and B = Barycentric Coordinate for C
SignedArea(P) with vertex B and C = Barycentric Coordinate for A
SignedArea(P) with vertex C and A = Barycentric Coordinate for B
B
/ \
/ \
/ P \
/_______\
A C
This is normalised to the area's sum, but we can trivially turn this into
a normalised version by dividing the area of the parallelogram, i.e.
BaryCentricA(P) = (SignedArea(P) w. vertex C and B)/SignedArea(of the orig triangle)
BaryCentricB(P) = (SignedArea(P) w. vertex A and C)/SignedArea(of the orig triangle)
BaryCentricC(P) = (SignedArea(P) w. vertex A and B)/SignedArea(of the orig triangle)
*/ */
DqnV2i scanP = DqnV2i_2i(min.x, min.y); DqnV2i scanP = DqnV2i_2i(min.x, min.y);
@ -263,7 +303,6 @@ FILE_SCOPE void DrawText(PlatformRenderBuffer *const renderBuffer,
stbtt_aligned_quad alignedQuad = {}; stbtt_aligned_quad alignedQuad = {};
stbtt_GetPackedQuad(font.atlas, font.bitmapDim.w, font.bitmapDim.h, stbtt_GetPackedQuad(font.atlas, font.bitmapDim.w, font.bitmapDim.h,
charIndex, &pos.x, &pos.y, &alignedQuad, true); charIndex, &pos.x, &pos.y, &alignedQuad, true);
stbtt_packedchar *charData = font.atlas + charIndex;
DqnRect fontRect = {}; DqnRect fontRect = {};
fontRect.min = DqnV2_2f(alignedQuad.s0 * font.bitmapDim.w, alignedQuad.t1 * font.bitmapDim.h); fontRect.min = DqnV2_2f(alignedQuad.s0 * font.bitmapDim.w, alignedQuad.t1 * font.bitmapDim.h);
@ -279,7 +318,13 @@ FILE_SCOPE void DrawText(PlatformRenderBuffer *const renderBuffer,
u8 *fontPtr = font.bitmap + fontOffset; u8 *fontPtr = font.bitmap + fontOffset;
DQN_ASSERT(sizeof(u32) == renderBuffer->bytesPerPixel); DQN_ASSERT(sizeof(u32) == renderBuffer->bytesPerPixel);
f32 fontHeightOffset = charData->yoff2 + charData->yoff;
// NOTE(doyle): This offset, yOffset and flipping t1, t0 is necessary
// for reversing the order of the font since its convention is 0,0 top
// left and -ve Y.
stbtt_packedchar *const charData = font.atlas + charIndex;
f32 fontHeightOffset = charData->yoff2 + charData->yoff;
u32 screenOffset = (u32)(screenRect.min.x + (screenRect.min.y - fontHeightOffset) * renderBuffer->width); u32 screenOffset = (u32)(screenRect.min.x + (screenRect.min.y - fontHeightOffset) * renderBuffer->width);
u32 *screenPtr = ((u32 *)renderBuffer->memory) + screenOffset; u32 *screenPtr = ((u32 *)renderBuffer->memory) + screenOffset;