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# UV coordinate on a 2D quadrilateral

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#1
Members - Reputation: **122**

Posted 25 February 2011 - 12:02 PM

Given the coordinates a, b, c, d, and p, how would I find the normalized UV coordinates of p? (For example, to sample a texture at that point.)

a, b, c, d, and p are 2D (that is, only X,Y coordinates). p will always be inside abcd.

I have no idea where to start. Any ideas?

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#2
Crossbones+ - Reputation: **3818**

Posted 25 February 2011 - 12:15 PM

*E.g., adc and acb, etc.

Please don't PM me with questions. Post them in the forums for *everyone's* benefit, and I can embarrass myself publicly.

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#4
Members - Reputation: **637**

Posted 25 February 2011 - 12:25 PM

a continous uv mapping could consist of the solutions to:

p = a + ux + vy + uvz

where the four coordinates of your rectangle in order are a,b,c,d, with x = b-a, y = d-a, z = a+c-d-b

i don't however know of a solution to this, and wolframalpha doesn't give you anything pretty for it!

(I come up with that by linearly interpolating accross top and bottom lengths with 'u', then linearly interpolating between those two points with 'v' which gives a unique, continous, but very complex mapping of p to (u,v) coordinates.

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#5
Crossbones+ - Reputation: **3102**

Posted 25 February 2011 - 12:30 PM

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#7
Crossbones+ - Reputation: **3779**

Posted 25 February 2011 - 01:10 PM

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#8
Members - Reputation: **637**

Posted 25 February 2011 - 01:16 PM

ultimately you can reduce the problem to:

( u ) ( 1 a b c )( uv ) = ( 0 ) ( 0 1 d e )( v ) ( 0 ) ( 1 )

which has a very complex closed solution of: http://www.wolframal...,+solve+for+u,v

of which only one of the two top solutions should be the correct one.

a,b,c,d,e being rather complex equatinos involving your coordinate p, and the coordinates of the 4 rectangle points starting with:

( u ) ( bx-ax cx-bx dx-ax ax-px )( uv ) = ( 0 ) ( by-ay cy-by dy-ay ay-py )( v ) ( 0 ) ( 1 )

with rectangle being represented with top-left a, top-right b, bottom-right c, bottom-left d. u running left to right, v running top to bottom

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#9
Members - Reputation: **967**

Posted 25 February 2011 - 01:32 PM

We've had this discussion here in the past; if you search the old forums you can probably find those threads. The solutions always boiled down to one of,

1.) Split into triangles and do barycentric interpolation on each (buckeye's answer)

2.) Bilinear interpolation (luca-deltodesco's answer)

Details were given in those threads for both of these approaches.

Here's a third: [EDIT: Removed; was wrong.]

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#10
Crossbones+ - Reputation: **3779**

Posted 25 February 2011 - 01:45 PM

The approach of the bilinear interpolation

**p**

_{1}:=

**a**+ (

**b**-

**a**) * u

**p**

_{2}:=

**d**+ (

**c**-

**d**) * u

**p**:=

**p**

_{1}+ (

**p**

_{2}-

**p**

_{1}) * v

means that there is a line between

**p**

_{1}and

**p**

_{2}that passes through

**p**. Expressing this line the other way, namely

**p**+

**l**=

**p**

_{1}

**p**+ k *

**l**=

**p**

_{2}

so that

**l**+ k *

**l**(where k is obviously need to be a negative number) is the said line. Considering that this is done in 2D, we have 4 equations with 4 unkowns (k, l

_{x}, l

_{y}, and u) where k * l

_{x}and k * l

_{y}are the problems. Fortunately, from the lower of the both equations, we can isolate

k * l

_{x}= d

_{x}- p

_{x}+ ( c

_{x}- d

_{x}) * u

k * l

_{y}= d

_{y}- p

_{y}+ ( c

_{y}- d

_{y}) * u

and divide both of these, so that

l

_{x}/ l

_{y }= [ d

_{x}- p

_{x}+ ( c

_{x}- d

_{x}) * u ] / [ d

_{y}- p

_{y}+ ( c

_{y}- d

_{y}) * u ]

Now, using the upper of the equations to get

l

_{x}= a

_{x}- p

_{x}+ ( b

_{x}- a

_{x}) * u

l

_{y}= a

_{y}- p

_{y}+ ( b

_{y}- a

_{y}) * u

and setting these into the above l

_{x}/ l

_{y}, we get a quadratic equation solely in u which can be solved using the famous p,q-formula.

This gives, as luca-deltodesco has mentioned, 0 or 1 real solutions in general. But due to the fact that

**p**always lies inside the quad, I expect 1 real solution. With that, v can be determined easily.

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#11
Members - Reputation: **122**

Posted 25 February 2011 - 08:31 PM

Yeah, I though of looking at it the following way:

The approach of the bilinear interpolationp_{1}:=a+ (b-a) * up_{2}:=d+ (c-d) * up:=p_{1}+ (p_{2}-p_{1}) * v

means that there is a line betweenp_{1}andp_{2}that passes throughp. Expressing this line the other way, namelyp+l=p_{1}p+ k *l=p_{2}

so thatl+ k *l(where k is obviously need to be a negative number) is the said line. Considering that this is done in 2D, we have 4 equations with 4 unkowns (k, l_{x}, l_{y}, and u) where k * l_{x}and k * l_{y}are the problems. Fortunately, from the lower of the both equations, we can isolate

k * l_{x}= d_{x}- p_{x}+ ( c_{x}- d_{x}) * u

k * l_{y}= d_{y}- p_{y}+ ( c_{y}- d_{y}) * u

and divide both of these, so that

l_{x}/ l_{y }= [ d_{x}- p_{x}+ ( c_{x}- d_{x}) * u ] / [ d_{y}- p_{y}+ ( c_{y}- d_{y}) * u ]

Now, using the upper of the equations to get

l_{x}= a_{x}- p_{x}+ ( b_{x}- a_{x}) * u

l_{y}= a_{y}- p_{y}+ ( b_{y}- a_{y}) * u

and setting these into the above l_{x}/ l_{y}, we get a quadratic equation solely in u which can be solved using the famous p,q-formula.

This gives, as luca-deltodesco has mentioned, 0 or 1 real solutions in general. But due to the fact thatpalways lies inside the quad, I expect 1 real solution. With that, v can be determined easily.

Thank you very, very much! This came out to be the easiest solution to implement and it works great!

In case anyone is interested in the same problem, here's some code:

double C = (double)(a.Y - p.Y) * (d.X - p.X) - (double)(a.X - p.X) * (d.Y - p.Y); double B = (double)(a.Y - p.Y) * (c.X - d.X) + (double)(b.Y - a.Y) * (d.X - p.X) - (double)(a.X - p.X) * (c.Y - d.Y) - (double)(b.X - a.X) * (d.Y - p.Y); double A = (double)(b.Y - a.Y) * (c.X - d.X) - (double)(b.X - a.X) * (c.Y - d.Y); double D = B * B - 4 * A * C; double u = (-B - Math.Sqrt(D)) / (2 * A); double p1x = a.X + (b.X - a.X) * u; double p2x = d.X + (c.X - d.X) * u; double px = p.X; double v = (px - p1x) / (p2x - p1x);u and v are then [0,1] coordinates you can sample a texture from, etc.

Thanks again!