Dakwamine

Catégorie : Unity3D

Woot Unity!

  • Unity3D.tips[2]: Intersection point between two lines in 2D

    The code

    The maths behind

    To get the point where two lines intersect, we will do some maths. To the mathematicians who will come across this post, I’m truly sorry for the heart attacks or strokes you may experience by reading this post.

    Ok, let’s take a look on the figure:

    Study case. We can see two lines crossing in an 2D Euclidian space.
    Figure 1

    First, sorry, I was too lazy to make a proper figure using computer tools so I just put a scan. XD

    Next, here are the definitions:

    • AA & BB: the two lines,
    • A1A_1, B1B_1: the arbitrary starting points of the two lines,
    • A2A_2, B2B_2: the arbitrary points which tells the direction of the two lines,
    • XX: the intersection point,
    • OO: the origin point.

    Kewl. Now, what we want is the intersection point XX between those two lines. In other words, we want to know the position which starts from either lines’ arbitrary starting point, added by the direction of the line multiplied by a scalar value. So in our figure 1, the XX position is:

    • the A1A_1 position added by the components of A1A2→\overrightarrow{A_1A_2} multiplied by an unknown which I named λ\lambda
    • the B1B_1 position added by the components of B1B2→\overrightarrow{B_1B_2} multiplied by an unknown which I named μ\mu

    In this case, it is clear that λ=μ=0.5\lambda = \mu = 0.5 so it will be easy to check if our final formula is correct. 🙂

    At the point of intersection, we know that:

    OX→=OA1→+λ×A1A2→andOX→=OB1→+μ×B1B2→\begin{array}{rcl} \overrightarrow{OX} = \overrightarrow{OA_1} + \lambda\times\overrightarrow{A_1A_2} & and & \overrightarrow{OX} = \overrightarrow{OB_1} + \mu\times\overrightarrow{B_1B_2} \end{array}

    … which gives us:

    OA1→+λ×A1A2→=OB1→+μ×B1B2→OA1→−OB1→=μ×B1B2→−λ×A1A2→\begin{array}{rcl} \overrightarrow{OA_1} + \lambda\times\overrightarrow{A_1A_2} & = & \overrightarrow{OB_1} + \mu\times\overrightarrow{B_1B_2} \\ \overrightarrow{OA_1} – \overrightarrow{OB_1} & = & \mu\times\overrightarrow{B_1B_2} – \lambda\times\overrightarrow{A_1A_2} \end{array}

    But we can’t use this statement exactly like this to solve our equation as we cannot multiply and divide vectors in this raw format. Also we need to reduce the unkowns count to 1. So we will separate our equation into two equations with the magic of matrices (which I don’t understand well at the moment), one for the x component and one for the y:

    {OA1→x−OB1→x=μ×B1B2→x−λ×A1A2→xOA1→y−OB1→y=μ×B1B2→y−λ×A1A2→y\left\{\begin{array}{c}\overrightarrow{OA_1}_x – \overrightarrow{OB_1}_x = \mu\times\overrightarrow{B_1B_2}_x – \lambda\times\overrightarrow{A_1A_2}_x \\ \overrightarrow{OA_1}_y – \overrightarrow{OB_1}_y = \mu\times\overrightarrow{B_1B_2}_y – \lambda\times\overrightarrow{A_1A_2}_y \end{array}\right.

    To make our equation more readable, we will use some shorthands:

    Aα=A1A2→,Bα=B1B2→,C=OA1→−OB1→\begin{array}{r}A_\alpha = \overrightarrow{A_1A_2}, & B_\alpha = \overrightarrow{B_1B_2}, & C = \overrightarrow{OA_1} – \overrightarrow{OB_1}\end{array}

    … so:

    {Cx=μBαx−λAαxCy=μBαy−λAαy\left\{\begin{array}{c} C_x = \mu B_{\alpha x} – \lambda A_{\alpha x} \\ C_y = \mu B_{\alpha y} – \lambda A_{\alpha y} \end{array}\right.

    From this point, we can reduce the unknown count. In my case, I have chosen to keep μ\mu instead of λ\lambda:

    {CxAαy=μBαxAαy−λAαxAαyCyAαx=μBαyAαx−λAαyAαx\left\{\begin{array}{c} C_x A_{\alpha y} = \mu B_{\alpha x} A_{\alpha y} – \lambda A_{\alpha x} A_{\alpha y} \\ C_y A_{\alpha x} = \mu B_{\alpha y} A_{\alpha x} – \lambda A_{\alpha y} A_{\alpha x} \end{array}\right.
    CxAαy−CyAαx=μBαxAαy−λAαxAαy−(μBαyAαx−λAαyAαx)CxAαy−CyAαx=μBαxAαy−λAαxAαy−μBαyAαx+λAαyAαxCxAαy−CyAαx=μBαxAαy−μBαyAαxCxAαy−CyAαx=μ(BαxAαy−BαyAαx)μ=CxAαy−CyAαxBαxAαy−BαyAαx\begin{array}{rcl} C_x A_{\alpha y} – C_y A_{\alpha x} & = & \mu B_{\alpha x} A_{\alpha y} – \lambda A_{\alpha x} A_{\alpha y} – (\mu B_{\alpha y} A_{\alpha x} – \lambda A_{\alpha y} A_{\alpha x}) \\ C_x A_{\alpha y} – C_y A_{\alpha x} & = & \mu B_{\alpha x} A_{\alpha y} – \lambda A_{\alpha x} A_{\alpha y} – \mu B_{\alpha y} A_{\alpha x} + \lambda A_{\alpha y} A_{\alpha x} \\ C_x A_{\alpha y} – C_y A_{\alpha x} & = & \mu B_{\alpha x} A_{\alpha y} – \mu B_{\alpha y} A_{\alpha x} \\ C_x A_{\alpha y} – C_y A_{\alpha x} & = & \mu (B_{\alpha x} A_{\alpha y} – B_{\alpha y} A_{\alpha x}) \\ \mu & = & \frac{C_x A_{\alpha y} – C_y A_{\alpha x}}{B_{\alpha x} A_{\alpha y} – B_{\alpha y} A_{\alpha x}} \end{array}

    And finally, you have to check if BαxAαy−BαyAαx=0B_{\alpha x} A_{\alpha y} – B_{\alpha y} A_{\alpha x} = 0. This will happen if your two lines are parallel or if there is one line defined as a point such as A1=A2A_1 = A_2 or B1=B2B_1 = B_2; no solution exists in those cases.

    Test 1

    Kewl! Now let’s try it with the Figure 1:

    μ=(1−1)×1−(2−4)×22×1−(−3)×2μ=48=0.5\begin{array}{rcl} \mu & = & \frac{(1 – 1) \times 1 – (2 – 4) \times 2}{2 \times 1 – (-3) \times 2} \\ \mu & = & \frac{4}{8} = 0.5 \end{array}

    We can get the XX position:

    {OX→x=OB1→x+Bαx×μ=1+2×0.5=2OX→y=OB1→y+Bαy×μ=4+(−3)×0.5=2.5\left\{\begin{array}{l} \overrightarrow{OX}_x = \overrightarrow{OB_1}_x + B_{\alpha x} \times \mu = 1 + 2 \times 0.5 = 2 \\ \overrightarrow{OX}_y = \overrightarrow{OB_1}_y + B_{\alpha y} \times \mu = 4 + (-3) \times 0.5 = 2.5 \end{array}\right.

    Test 2

    Ok done! Now let’s try with another figure:

    Study case 2. We can see two lines crossing in an 2D Euclidian space.
    Figure 2
    μ=((−2)−2)×(−1)−((−2)−(−3))×0(−1.5)×(−1)−1.5×0μ=41.5≈2.666\begin{array}{rcl} \mu & = & \frac{((-2) – 2) \times (-1) – ((-2) – (-3)) \times 0}{(-1.5) \times (-1) – 1.5 \times 0} \\ \mu & = & \frac{4}{1.5} \approx 2.666 \end{array}

    We can get the XX position:

    {OX→x=OB1→x+Bαx×μ=2+(−1.5)×2.666=−2OX→y=OB1→y+Bαy×μ=(−3)+1.5×2.666=1\left\{\begin{array}{l} \overrightarrow{OX}_x = \overrightarrow{OB_1}_x + B_{\alpha x} \times \mu = 2 + (-1.5) \times 2.666 = -2 \\ \overrightarrow{OX}_y = \overrightarrow{OB_1}_y + B_{\alpha y} \times \mu = (-3) + 1.5 \times 2.666 = 1 \end{array}\right.

  • Unity3D.tips[1]: Strategies to prevent the collision when a ball rolls from a planar surface to another

    Method 1: mega box collider

    Use a single BoxCollider for all your planes. For example, if you have multiple aligned planes, remove their Collider component (Box, Mesh, whichever they have) and on only one of the planes, add a BoxCollider and adjust its center and size values in the component’s inspector to encapsulate all the planes.

    Pros

    • Simple.

    Cons

    • Supports only rectangular surfaces.

    Method 2: generated colliders

    If you are rolling your ball on a non rectangular surface, you will need to find a way to generate a mesh collider for all your planes. For this, choose one of the following:

    Pros

    • Just works.

    Cons

    • Uses noticeable CPU power when updating (baking) the colliding mesh in real time. Update sparingly.

    Method 3: ramps as bridges

    Create a set of meshes from a 3D software such as Blender to make invisible ramps which will be placed partly underground between your planes. They have to be large enough to make the transitions as smooth as possible. The height between the top of the bridge and the ground (the planes) should be approximately the value defined by the Default Contact Offset value in the Physics Manager.

    Pros

    • No real improvement.

    Cons

    • The ball may jump a little, especially at high speed.
    • Doesn’t work with small tiling (or when your successive planes are small): the longer the tiles, the longer bridges can be.
    • Requires a lot of testing.
    • Requires 3D software usage.
    • The result is not consistent.
    • Requires good workflow / toolchain.

    Method 4: make custom extended planes’ colliders

    This one should not work very well.

    Create in a 3D software a planar mesh. Then, on its edges, extrude towards outside and lower down slightly the new edges. To put the words into image:

    Photo of speed cushions.
    Credits: Richard Drdul — cc-by-sa-2.0 (original)

    Make them intersect, and hopefully the bouncing will be invisible.

    Pros

    • Quite simple.

    Cons

    • The ball may jump a little, especially at high speed.
    • Requires a lot of testing.
    • Requires 3D software usage.
    • The result is not consistent.
    • Requires good workflow / toolchain.

    Method 5: turn slightly your planes’ colliders

    Make your planes’ colliders slightly rotated so that, in the direction of movement, the end of a collider is slightly above the next plane’s collider. The height should be approximately the value defined by the Default Contact Offset value in the Physics Manager and hopefully no bouncing will occur.

    Pros

    • Works flawlessly.

    Cons

    • You will see the ball flying and falling a few units from one plane to another.

    Method 6: do everything by script

    Maybe the Physics is not what is the best for your gameplay. Write your own system with accelerations, collision detection. This is the old-fashioned way and is the choice of a lot of developers! ^^

    Pros

    • Works flawlessly.
    • Permits game design driven gameplay and not by physics.

    Cons

    • Not using the PhysX forces and gravity.
    • Requires to work (omg! 😉 ) to simulate fake physics.

    Method 7: Make your ball much larger

    By using a greater ball (x10, x100), this could lead to less edge detection.

    Pros

    • May work.

    Cons

    • The scale of the simulation would change: if you make everything at the same scale as the ball, objects falling by gravity will look like they fall in slow motion.
    • The result is not consistent.

    Method 8: collision layers and raycast

    This one is pretty cool.

    1. Keep single BoxColliders for each plane. Set the layer of the plane prefab to SinglePlanes (create the layer).
    2. Make a big Unity cube with a collider which encapsulates all the planes. Set the layer of the cube to ContinuousPlane.
    3. Create two layers: OnAPlane and NotOnAPlane.
    4. Set the ball prefab’s layer to OnAPlane.
    5. In Physics Manager, set the collision matrix so that:
      • NotOnAPlane and ContinuousPlane are not collidable (unchecked),
      • OnAPlane and SinglePlanes are not collidable (unchecked).
    6. On each Update(), do a raycast from the center of the ball towards Vector3.down on the layer SinglePlanes:
    7. If the raycast returns false, the center of the ball is not on a plane. So now, we change the layer of the ball to NotOnAPlane. The ball will start to fall through the hole.
    8. If the raycast returns true, set the layer of the ball to OnAPlane.

    For how to define a layer, please check the manual.

    Pros

    • Works efficiently with dynamic floor layout.
    • You will learn how to deal with layers and Physics.

    Cons

    • No real flaws.

    Additional Information

    • Methods above may be combined.
    • You can adjust the physics project settings and / or set the physics materials to reduce the bounciness of your objects.
    • Make sure the thickness of the planes collision is not 0f. Make them 1f large or more.
    • There is a possibility to reduce the bounces, but I am not recommending it (may cause jittery and physics artifacts): set the Default Contact Offset to 0.0001. Again, I am not recommending it!
    • All of this post could be useless when the default PhysX embedded in Unity will be upgraded.
    • I originally posted this post as an answer on the Unity Answers website but none of the above methods have been considered as valid by the OP. Hence the removal from the answers website.

  • Unity3D.tips[0]: Avoid naming your MonoBehaviour methods Main()

    Try to avoid naming your MonoBehaviour methods Main(). These would be triggered between Awake() and Start(). This Main() method supports IEnumerator flavors as much as Awake() and alike! But as useful as it could be, I would not recommend to use it as there is no official documentation entry in the manual. And usually, you should not need this method.

    Évitez de nommer Main() les méthodes se situant dans vos MonoBehaviour. Celles-ci seraient déclenchées entre Awake() et Start(). Cette méthode Main() supporte la version en IEnumerator autant que Awake() et les autres ! Mais aussi utile que pourrait être cette méthode, je ne la recommanderais pas à l’usage puisqu’il n’existe aucune documentation officielle dans le manuel. Et normalement, vous ne devriez pas avoir recours à cette méthode.

    Screenshot of Unity console
    Awake() is fired, then Main() and finally Start(). But don’t use Main()!