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Calculate Position as 0 to 1 between two planes.


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#1 AverageJoeSSU   Members   -  Reputation: 516

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Posted 12 February 2014 - 06:44 PM

I'm struggling to understand the equation on page 55 of the following paper.

 

http://web.cs.wpi.edu/~rich/courses/imgd4000-d10/lectures/camera.pdf

 

the previous page 54 explains the premise behind the equation.

 

Here is my code in Unity3d to try and reproduce it:

zForward = new Vector4(gameObject.transform.forward.x,gameObject.transform.forward.y,gameObject.transform.forward.z,0);
        plane = new Vector4(gameObject.transform.up.x, gameObject.transform.up.y, gameObject.transform.up.z, 0);

Vector4 playerPos = new Vector4(player.transform.position.x,player.transform.position.y,player.transform.position.z,1);

Debug.Log(Vector4.Dot(plane,playerPos)/Vector4.Dot(plane,zForward));

gameObject has a collider that is triggered when the player walks in it. I then try to calculate the field from that paper...

 

The author suggests that the math in the equation is wrong, but i'm not sure what he is trying to do WITH that exact math there. seems to me like it is missing some sort of position or something of the plane??? just a guess.


Edited by AverageJoeSSU, 13 February 2014 - 01:36 PM.

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#2 Buckeye   Crossbones+   -  Reputation: 5680

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Posted 12 February 2014 - 08:29 PM

First, please note that the topic title isn't what you want. Per the article you reference, it's about how to "Calculate position as 0 to 1 between planes."

 

The equation ( which is poorly written and even more poorly described IMHO ) appears to represent the player's position "value" as a ratio of the distance from the player to a plane along a predefined direction to the full distance between the planes along that same direction, passing through the player's position. E.g., if a string is stretched(*) from one plane to the other through the player, and the player is two-thirds of the way along that string, the "value" is 0.333 for camera1 and 0.667 for camera 2.

 

(*) always in the same direction.

 

[In the following discussion, if you want to have a ratio = 1 when the player reaches the far plane, switch 1's and 2's around. It's moot however, because you'll need the ratio and 1-ratio in either case to weight the two "cameras"]

 

The algorithm he mentions is to calculate the intersection point in both plane 1 (back) and plane 2 (forward), calling them point1 and point2, defined by a vector from the player's position along the direction vector to each plane. The length (absolute value) of a vector defined with those two points ( p2 minus p1 ) is the "full" distance between the planes. Define a vector from the player's position to point 2 ( p2 minus playerPos ). Now calculate the dot product of the player vector with the normal to plane 2 (player-distance). Calculate the dot product of the p2-p1 vector with the normal to plane 2 (full-distance). The ratio player-distance / full-distance is what (I believe) the author is trying to represent as the "value" for weighting. That ratio will vary from 1 to 0 as the player hits plane 1 and progresses to plane 2. So, as the player progresses from plane 1 to plane 2, camera 1 is weighted by ratio; camera 2 is weighted by 1 - ratio.

 

He shows two parallel planes in his example, but that algorithm will work for any two planes. He (apparently) also assumes there's a predefined direction vector that the user will have to come up with. You can pick whatever you'd like, provided that vector is not parallel with either plane. It's tough to calculate the intersection of a vector with a plane if it never intersects that plane.

 

The part of the math he (conveniently) skips over is how to determine the points in the planes. For that you'll have to have the predefined vector (best to use a unit vector) and solve two equations using the two equations for the planes, one for the distance from the player to plane1; the other for the distance to plane 2. N.B., those are not dot product distances, but distances along a direction!


Edited by Buckeye, 12 February 2014 - 08:52 PM.

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


#3 AverageJoeSSU   Members   -  Reputation: 516

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Posted 13 February 2014 - 01:47 PM

Thanks for the replay,

 

A friend helped me understand it last night and we were able to come up with some working Unity code. After reading your post, I am pretty sure we are doing the exact same thing.

 

I'll be sure to post it here when I get the chance.

 

We actuallly just used the bounds of the box collider in Unity to derive points for the two planes, and then since we want the colliders to be point to the "front plane", we just take the forward vector of the unity gameobject that has the box collider. Ends up pretty clean.

 

Although i really appriciate that guy doing that paper, and am grateful for it as a resource, the lack of clarity on some of the parts has made it extremely painful.

Luckily we are almost there!!!!!!!!!! And the results are soooo worth it.


Edited by AverageJoeSSU, 13 February 2014 - 03:15 PM.

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