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Question by W1k3 · May 28, 2015 at 04:15 AM · c#rotationraycastraycastingraycasthit

Help finding out which side of a cube a raycast hits

I'm making a voxel based game, and I'm trying to make placing blocks act similar to minecraft in the sense that the voxel is created adjacent to the side of a preexisting voxel that the player clicks. I successfully did this part by comparing the origin of the 1 meter cube and checking which direction it was 0.5 meters away from. The catch is that I need to get this to work if the cube is at any rotation. If I use RaycastHit.point, those number are less valuable without taking the rotation into account and I'm really unsure of how to go about doing that. I'm using c#

Any advice would be great.

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Answer by crohr · May 28, 2015 at 05:33 AM

I believe this is what you are looking for. Or something like that.

  var newBlockPostion = hit.transform.position + hit.normal;
  var newBlockRotation = hit.transform.rotation;
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avatar image W1k3 · May 28, 2015 at 04:43 PM 0
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Thanks for the reply, but would you $$anonymous$$d explaining a little bit more how hit.normal would work? I understand that it's a Vector3 that points away from the collider, but how far and such?

avatar image crohr · May 28, 2015 at 05:43 PM 0
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Yes, so hit.normal is a unit vector, so it has a length of 1 unit. Thus, (hit.transform.position + hit.normal) will give you a vector position that is 1 unit distance away from the center of hit object in the direction of hit.normal. I did not test this for your specific needs, but I believe this should work.

avatar image W1k3 · May 28, 2015 at 06:22 PM 0
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It worked out great! Thanks a lot. Just curious though, is there any way to change the unit to 2 or more units away?

avatar image crohr · May 28, 2015 at 06:30 PM 0
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Yes there is. hit.normal * (whatever x number of units you want). Since, a vector multiplied by a scalar simply modifies the length of the vector, you are given a vector that is x units in length.

avatar image W1k3 · May 28, 2015 at 11:03 PM 0
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Awesome, thanks for the help.

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