3 ms·

Are you sure about that? The dot product gives a measure of the orientation between two vectors, the cross product gives perpendicularity and is vector valued (

by Dn_Ab 14y ago

Are you sure about that? The dot product gives a measure of the orientation between two vectors, the cross product gives perpendicularity and is vector valued (kind of, the distinction matters more for physics than graphics). It also only makes sense in 3 and 7 dimensions without generalizing the concept of vector.

moron4hire 14y ago
Balls, you are correct, dot product is the correct operation, not cross product. Been a while since graphics ops were on my table. EDIT: Well, cross product would work, too, if you extended the 2D vectors into 3D vectors, coplanar in the screen. The orientation of the cross product into or out of the screen would also tell you the relative orientation of the two vectors. So then, you'd be looking for the two cross products with differing signs on the Z component to tell you if it's "in the triangle".
- Dn_Ab 14y ago
  Right and excellent explanation. I was more disagreeing with your definition than saying it could not be done with "cross products" (although it's still bad terminology, biased to people that already know the concepts). That's why I couldn't say you were wrong. Intuitively, a measure of how parallel two vectors of arbitrary dimensions are, is a better default when talking about orientation of vectors.
  - gbhn 14y ago
    You need cross product if you want to know sign. Dot product won't tell you which side of the line you're on, and that sounded like a big part of the algorithm.
- moron4hire 14y ago
  After some playing around, dot product won't do it. You can get close with dot product, but it really only gives you the general direction. Cross product (and really, only calculating the Z component of the cross product of coplanar 3D vectors) will get it exactly right every time. So basically, for a moving between two mouse points (Ix, Iy) and (Ax, Ay), and a top left menu corner (Bx, By) and bottom left (Cx, Cy), then: DAx = Ax - Ix DAy = Ay - Iy DBx = Bx - Ix DBy = By - Iy DCx = Cx - Ix DCy = Cy - Iy Z1 = DBx * DAy - DAx * DBy Z2 = DCx * DAy - DAx * DCy Then, you know you're headed towards that menu if Z1 < 0 and Z2 > 0. Could be simplified a little if you assume the X component of the top left corner of the menu will be the same as the X component of the bottom left corner, but the math was clearer to keep them separate.
  - moron4hire 14y ago
    And if you have any convex polygon as an ordered, circular list of points proceeding clockwise, then you can easily loop around the points checking your cross product values making sure they are all greater than 0 to determine if the point is in the polygon. You'll have to tessalate concave polygons into a list of convex polygons, but that's eaaaaaasy ;)
munchbunny 14y ago
No, the dot product won't work in this case. Take the two edges of the triangle as vectors. You want the mouse vector to fall in between. The dot product can't tell you which side of the two edge vectors your mouse vector falls on, but the cross product can if you look at the component of the cross product that points in/out of the screen. Actually, you can make the dot product give you that answer, but you would need use the dot product of the mouse vector and two vectors normal to the edge vectors, which is really just a roundabout way of using the cross product.

by Dn_Ab 14y ago

moron4hire 14y ago

Balls, you are correct, dot product is the correct operation, not cross product. Been a while since graphics ops were on my table. EDIT: Well, cross product would work, too, if you extended the 2D vectors into 3D vectors, coplanar in the screen. The orientation of the cross product into or out of the screen would also tell you the relative orientation of the two vectors. So then, you'd be looking for the two cross products with differing signs on the Z component to tell you if it's "in the triangle".

Dn_Ab 14y ago
Right and excellent explanation. I was more disagreeing with your definition than saying it could not be done with "cross products" (although it's still bad terminology, biased to people that already know the concepts). That's why I couldn't say you were wrong. Intuitively, a measure of how parallel two vectors of arbitrary dimensions are, is a better default when talking about orientation of vectors.
- gbhn 14y ago
  You need cross product if you want to know sign. Dot product won't tell you which side of the line you're on, and that sounded like a big part of the algorithm.
moron4hire 14y ago
After some playing around, dot product won't do it. You can get close with dot product, but it really only gives you the general direction. Cross product (and really, only calculating the Z component of the cross product of coplanar 3D vectors) will get it exactly right every time. So basically, for a moving between two mouse points (Ix, Iy) and (Ax, Ay), and a top left menu corner (Bx, By) and bottom left (Cx, Cy), then: DAx = Ax - Ix DAy = Ay - Iy DBx = Bx - Ix DBy = By - Iy DCx = Cx - Ix DCy = Cy - Iy Z1 = DBx * DAy - DAx * DBy Z2 = DCx * DAy - DAx * DCy Then, you know you're headed towards that menu if Z1 < 0 and Z2 > 0. Could be simplified a little if you assume the X component of the top left corner of the menu will be the same as the X component of the bottom left corner, but the math was clearer to keep them separate.
- moron4hire 14y ago
  And if you have any convex polygon as an ordered, circular list of points proceeding clockwise, then you can easily loop around the points checking your cross product values making sure they are all greater than 0 to determine if the point is in the polygon. You'll have to tessalate concave polygons into a list of convex polygons, but that's eaaaaaasy ;)

munchbunny 14y ago

No, the dot product won't work in this case. Take the two edges of the triangle as vectors. You want the mouse vector to fall in between. The dot product can't tell you which side of the two edge vectors your mouse vector falls on, but the cross product can if you look at the component of the cross product that points in/out of the screen. Actually, you can make the dot product give you that answer, but you would need use the dot product of the mouse vector and two vectors normal to the edge vectors, which is really just a roundabout way of using the cross product.