Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[lemon10]

Derp, I messed up the parentheses on my previous equations.
I put it as:

x2=cos(180+deflection angle+tan(y1/x1)*sqrt(x1^2+y1^2))
y2=sin(180+deflection angle+tan(y1/x1)*sqrt(x1^2+y1^2))

But it should be:
x2=cos(180+deflection angle+tan(y1/x1))*sqrt(x1^2+y1^2)
y2=sin(180+deflection angle+tan(y1/x1))*sqrt(x1^2+y1^2)


Also, keep in mind that you get small rounding errors using the basic sqrt distance equations, so if accuracy is particularly important it might be worth it to import a function that makes sure that there isn't any rounding errors.

Pretty sure it should be atan rather than tan.
This formula looks a great deal like what I'm currently using for calculating my reflections.
I remember reading somewhere that the answer can be calculated with less trigonometry somehow by using the dot product, but I can't quite recall how to do so

It should indeed be arctan, it just didn't copy properly when I took it from my last post.

EDIT: Not sure how you missed both Ruler's post (detailing dot products) and my first post (which didn't have any errors despite being slightly longer then post with just the equations) on though.

[ejseto]

But thats pretty much what I just gave you in my previous post. 
Well then, here it is in two easy equations:
x2=cos(180+deflection angle+tan(y1/x1)*sqrt(x1^2+y1^2))
y2=sin(180+deflection angle+tan(y1/x1)*sqrt(x1^2+y1^2))

This isn't right. You can easily show it doesn't work for a vertical surface, in which case (180+deflection angle+tan(y1/x1)) = 270. Then x2 = 0 which makes no sense. In general, once you have the angle of incidence (angle between incoming ray and the normal), you just run the coordinates through a rotation matrix with an angle twice that. From the form of the rotation matrix it should be clear your solution cannot work, since in general x2 depends on both x1 AND y1. You also completely ignored the orientation of the surface.

Anyway, Ruler's vector solution is correct and also avoids all sorts of divide-by-zero silliness. I'd avoid the rotation matrix altogether, but just in case:

[[x2],      [[ cos(a), -sin(a)],     [[x1],
 [y2]]  =   [ sin(a), cos(a)]]   *  [y1]]

for angle of rotation 'a.'

Or equivalently, since idk if you can read that matrix equation typed out like that,
x2 = x1cos(a) - y1sin(a)
y2 = x1sin(a) + y1cos(a)

You can also just "unrotate" the system by its orientation angle (thus aligning the normal with the x-axis, assuming 0 radians = x-axis), set y2=-y1, and rotate it back, as someone else suggested.

[lemon10]

But thats pretty much what I just gave you in my previous post. 
Well then, here it is in two easy equations:
x2=cos(180+deflection angle+tan(y1/x1)*sqrt(x1^2+y1^2))
y2=sin(180+deflection angle+tan(y1/x1)*sqrt(x1^2+y1^2))

This isn't right. You can easily show it doesn't work for a vertical surface, in which case (180+deflection angle+tan(y1/x1)) = 270. Then x2 = 0 which makes no sense. In general, once you have the angle of incidence (angle between incoming ray and the normal), you just run the coordinates through a rotation matrix with an angle twice that. From the form of the rotation matrix it should be clear your solution cannot work, since in general x2 depends on both x1 AND y1. You also completely ignored the orientation of the surface.

Anyway, Ruler's vector solution is correct and also avoids all sorts of divide-by-zero silliness. I'd avoid the rotation matrix altogether, but just in case:

[[x2],      [[ cos(a), -sin(a)],     [[x1],
 [y2]]  =   [ sin(a), cos(a)]]   *  [y1]]

for angle of rotation 'a.'

Or equivalently, since idk if you can read that matrix equation typed out like that,
x2 = x1cos(a) - y1sin(a)
y2 = x1sin(a) + y1cos(a)

You can also just "unrotate" the system by its orientation angle (thus aligning the normal with the x-axis, assuming 0 radians = x-axis), set y2=-y1, and rotate it back, as someone else suggested.

You can also easily show that you didn't read my post just before yours.

It should indeed be arctan, it just didn't copy properly when I took it from my last post.

So yeah, tan doesn't work. arctan does.

EDIT: I will also note that its typically better to use the arctan2(y,x) function instead of arctan(y/x) in order to avoid the problems of dividing by 0. But yeah, doing it the dot product way is probably better.

[ejseto]

You can also easily show that you didn't read my post just before yours.

Actually I did. I'm not taking issue with whether it's arctan or not. I assumed that's what you meant, which actually is implied by what I posted. It's still not right. Just from my example you can see your equations don't work. If a ray is incident on a vertical surface (e.g. y-axis), say, at positive angle theta with respect to the normal (x-axis in this case), it should reflect off at angle -theta, meaning x2=x1 and y2=-y1.

You defined "deflection angle" as the complement of theta, i.e. pi/2 - theta
arctan(y1/x1) = theta (in this example)
pi + deflection angle + arctan(y1/x1) = 3pi/2
cos(3pi/2) = 0

Therefore, x2 = 0 with your method.

[LoSboccacc]

can't we just use an engine? 

aaanyway
for a ball moving on velocity v bouncing to a plane that has a normal of n

u = (v · n / n · n) n  (velocity parallel to the normal)
w = v − u              (remainder)

· is a dot product, all other are the usual vector operations

v′ = w − u (velocity parallel to the normal is reversed by the collision, the other component remains the same)

[noodle0117]

But using an engine means you miss out half the fun...
Anyways I'll try implementing the dot product variation and see if I can get stuff to work.

[alway]

can't we just use an engine? 

aaanyway
for a ball moving on velocity v bouncing to a plane that has a normal of n

u = (v · n / n · n) n  (velocity parallel to the normal)
w = v − u              (remainder)

· is a dot product, all other are the usual vector operations

v′ = w − u (velocity parallel to the normal is reversed by the collision, the other component remains the same)

Or to finish the equations:
v' = v - 2 ((v · n) / (n · n)) * n
Which would probably take about as much time to compute as a single trig function.
And in general, if you're using trig functions for a game engine physics system outside of an initial conversion, you're probably doing it wrong. http://www.iquilezles.org/www/articles/noacos/noacos.htm

[LoSboccacc]

If you are fixing trigonometry moving to vector you better go to quaternion directly, or werid surprises will await you on the animation road

[alway]

If you are fixing trigonometry moving to vector you better go to quaternion directly, or werid surprises will await you on the animation road

And if you're feeling extra special, go for Dual Quaternions; the fun numbers where positions meet rotations at the corner of dual numbers (n^2 = 0, n != 0) and imaginary numbers (n^2 = -1) to make everything as simple as middleschool algebra.