I’ve been working these last few months on a few personal projects, both are mini games developed using Unreal Engine 5: CyberPong, Toy Plane and Team Puzzle.
Another Unity game
I just uploaded my version of a 2D shooter game here: Galaxy Shooter Game.
Galaxy Shooter Splash Screen
Unity game
I’ve been working on a game using the Unity game engine and I finally have a very (very) simple playable level! The game features a dog whose goal is to collect all the treats in the level as fast as possible. Give it a try here.
Perrito 0.1 Splash Screen
Lego RCX C++ programming on Ubuntu 16.04
These are the steps I followed in order to program the RCX brick under Ubuntu 16.04 in C++. The documentation from http://brickos.sourceforge.net/docs/install-Debian.html is mostly still accurate with a few changes:
- The package “legos” is no longer available, but you can find it as “brickos” so you install it with
$ sudo apt-get install brickos
- The makefile /usr/lib/brickos/Makefile.common used by the building process did not work for me out of the box. I had to comment out a couple of lines to make it work:
#ensureConfigured:
# @if [ ! -f $(BRICKOS_ROOT)/.configured.flg ]; then \
- The c++ examples from /usr/share/doc/brickos/examples/demo/c++/ did not build, but removing the temperatureSensor.C file fixed the problem.
Hope this helps!
Triangle and plane intersection
How to determine if a triangle intersects a plane? How to form a triangle (or two triangles) from one side of the intersection?
Let’s start by defining a triangle as a set of three points \(a,b,c\) and a plane as a point \(q\) and a normal vector \(\hat{n}\). To answer the first question – how to determine if a triangle intersects a plane – we can simplify by figuring out how to determine if a segment intersects a plane.
With a segment defined by two points \(s_1, s_2\) and a plane defined as before (a point \(q\) and a normal vector \(\hat{n}\)), test if both points lie in different sides of the plane. Take each segment point and subtract the plane point $$\vec{v_1} = s_1 – q, \vec{v_2} = s_2 – q$$ and then compute the dot product with the plane normal $$d_1 = \hat{n} \cdot \vec{v_1}, d_2 = \hat{n} \cdot \vec{v_2}$$. If the signs of both dot products are different, then the segment intersects the plane.
Run the test with the three segments that build the triangle. Some edge cases to consider: what if one segment lies in the plane? what if all three segments lie in the plane?
How to form a triangle (or two triangles) from one side of the intersection?
Once you know that a segment intersects a plane, computing the intersection point is easy. Take the two dot products computed earlier \(d_1,d_2\) and the intersection point \(i\) can be computed as follows: $$ t=d_1/(d_2 – d_1) $$ and $$ i = s_1 + t(s_2 – s_1) $$
Now, to form a triangle (or two triangles) from one side of the intersection (usually the side \(\hat{n}\) points to, the “positive” side) you need to test for intersection in a clockwise or counter clockwise (again, depending on \(\hat{n}\) direction and the coordinate system handedness) and keep track of which point of the segment under test is on the “positive” side (the dot product is greater than zero) and add it to the list of “visible” points and add the intersection point to that list as well. You should end up with three or four points in the list of “visible” points. From there, you can build one or two triangles.