I've been using SDL for my game but rendering order as always been a problem. When you draw something in SDL, it gets drawn over what was drawn before it. Which means, if I draw a character at 10 in the Y coordinate, and I draw a character at 5 in the Y coordinate, the one at 5 appears above the one at 10, even though it should be the other way around.
To demonstrate the problem.
This is what it should be like. The one with lower Y axis value should appear behind no matter which one I draw first.
In 3D, things don't need to be rendered in a specific order, things are drawn behind other things according to the Z buffer. I want SDL to behave the same way, but on the Y axis. How?
Related
I am making a videogame in Blender Game Engline, and want to be able to adjust the direction gravity pulls objects. I can change scene Z gravity in a script, but that's one-dimensional along the Z axis.
I would be fine with the ability to set X, Y, and Z gravity, or an easy way to make everything rotate at the same time around the origin (or an arbitrary point).
I could also build a system if I could have a plane exert gravity or a force field and have it rotate around the center at a set distance (the entire gameworld is encased in a sphere).
Basically, I want to be able to, from a python script, cause a force on all dynamic objects, automatically and without . How can I do this?
I think you're looking for bge.constraints.setGravity(x, y, z)
To set the gravity.
and this one: bge.logic.getCurrentScene().gravity to get the actual gravity. According to this you should calculate the ideal rotation for your objects
Let's say I have a rectangle-shaped object. I want to move it along a path. Is it possible to position this object according to not only one point, but two points on it? For example, the point A on the object is at 125,220, in this case I want point B to be at 140,235.This way I want to set the direction of the object.
In Objective-c (and I assume in other languages too) when we say "Position of a graphical object" we think of only one point, which is usually the bottom-left corner. But positioning
an object according to only that point will just redraw the object with the lower left corner in another point, and the rest part will be determined according to the height and the width of the object, which does not do what I want.
EDIT:
As you can see (and probably it's what you naturally expect) the object will move as a box from one point to another, because there's only one point determining it's position. You ask why I need a different thing. Because I have a timer and a curved path. Each time the timer ticks I need my object to be at a different location(the next position in an array of dumped points). So, instead of adding to X and Y coordinates, I explicitly tell the object to be at certain place. This way I want to achieve normal movement of my object along the curved path. When the front part of the object moves to some point, I need the rear part to move to a certain point as well.
I finally found a way to do do it. I have to rotate the object according the prior and the next points. So, assume there are points A,B,C,D,E,F,G,H on the path that the object will travel along. If the car is at point D, to calculate the rotation angle I do the following:
myObject.rotation=-atan((D.y-C.y)/(D.x-C.x))/3.141592*180;
As you can see it's just maths. A fine tuning can be applied to get a better, smoother rotation. Here for instance I subtract the Y of the previous position from the Y of the current position, then I do the same thing for X and then I get the minus arctangance of their ratio. But you can do
-atan((E.y-C.y)/(E.x-C.x))/3.141592*180;
Choosing the right positions to subtract their x and y coordinates will result in the right and smooth rotation.
I think you can guess that 3.141592 is M_PI;
I'm trying to make a little archer game, and the problem I'm having has to do with 2 pixels in particular, I'll call them _arm and _arrow. When a real archer is pulling back an arrow, he doesn't immediately pull the arrow back as far as his strength allows him, the arrow takes a little bit of time to be pulled back.
The _arm's angle is equal to the vector from a point to where the user touched on the screen. The rotation is perfect, so the _arm is good. The _arrow needs to be on the same line as _arrow, they are 1 pixel wide each so it looks as though the _arrow is exactly on top of the _arm.
I tried to decrement from the x/y coordinates based on a variable that changes with time, and I set the _arrow's location equal to the _arm's location, and tried to make it look like the _arrow was being pulled back. however, if you rotated, the x/y would mess up because it is not proportional on the x and y axis, so basically _arrow will either be slightly above the arm or slightly below it depending on the angle of the vector, based on touch.
How could I used the x/y position of _arm and the vector of touch to make the arrow appear as though it was being pulled back by a small amount, yet keep the arrow on top of the _arm sprite so that it's position would be similar to the arm, but slightly off yet still on top of the _arm pixel at all times. If you need anymore info, just leave a comment.
I'm not sure I've fully understood, but I'll have a go at answering anyway:
To make the arrow move and rotate to the same place as the consider adding the arrow as a child of the arm. You can still render it behind if you like by making its z is less than one: [arm addChild:arrow z:-1]
To then make the arrow move away from the arm as the bow is drawn, you then just set the position of the arrow with respect to the arm.
The problem I do see with this solution however is that this grouping of the sprites may be a little unusual after the arrow leaves the bow. Here you probably don't want the arrow to be a child of the arm as the coordinate systems are no longer related.
Even though they're sure what I "suggested would have solved [the] problem" here is the
Poster's solution
I had to get the x and y coords of the arm based of angle, then I got the sin/cos of a number that was based of the same angle as the arm and subtraced from that.
I want to use the same effect Apple uses in iOS5 for showing new notifications (this flip-in effect). But I don't know how to start or what to google for. Could somebody please give me a hint?
Thanks!
Create two CALayers. One aligned normally along the x and y axis, the other rotated around the x axis so that it "lies flat", i.e. its surface points down along the y axis.
Then rotate both layers by 90 degrees along the x axis. At the end of the rotation, the first layer will now "lie flat", pointing upward along the y axis and the second layer will be aligned normally along the x and y axes.
I know you asked more than a year ago, but if it's useful to you, I would checkout the new and awesome Animate.css css3 transitions! I would take the FlipInX effect, which looks exactly like in iOS5!
I'm using a 3d engine and need to translate between 3d world space and 2d screen space using perspective projection, so I can place 2d text labels on items in 3d space.
I've seen a few posts of various answers to this problem but they seem to use components I don't have.
I have a Camera object, and can only set it's current position and lookat position, it cannot roll. The camera is moving along a path and certain target object may appear in it's view then disappear.
I have only the following values
lookat position
position
vertical FOV
Z far
Z near
and obviously the position of the target object.
Can anyone please give me an algorithm that will do this using just these components?
Many thanks.
all graphics engines use matrices to transform between different coordinats systems. Indeed OpenGL and DirectX uses them, because they are the standard way.
Cameras usually construct the matrices using the parameters you have:
view matrix (transform the world to position in a way you look at it from the camera position), it uses lookat position and camera position (also the up vector which usually is 0,1,0)
projection matrix (transforms from 3D coordinates to 2D Coordinates), it uses the fov, near, far and aspect.
You could find information of how to construct the matrices in internet searching for the opengl functions that create them:
gluLookat creates a viewmatrix
gluPerspective: creates the projection matrix
But I cant imagine an engine that doesnt allow you to get these matrices, because I can ensure you they are somewhere, the engine is using it.
Once you have those matrices, you multiply them, to get the viewprojeciton matrix. This matrix transform from World coordinates to Screen Coordinates. So just multiply the matrix with the position you want to know (in vector 4 format, being the 4ยบ component 1.0).
But wait, the result will be in homogeneous coordinates, you need to divide X,Y,Z of the resulting vector by W, and then you have the position in Normalized screen coordinates (0 means the center, 1 means right, -1 means left, etc).
From here it is easy to transform multiplying by width and height.
I have some slides explaining all this here: https://docs.google.com/presentation/d/13crrSCPonJcxAjGaS5HJOat3MpE0lmEtqxeVr4tVLDs/present?slide=id.i0
Good luck :)
P.S: when you work with 3D it is really important to understand the three matrices (model, view and projection), otherwise you will stumble every time.
so I can place 2d text labels on items
in 3d space
Have you looked up "billboard" techniques? Sometimes just knowing the right term to search under is all you need. This refers to polygons (typically rectangles) that always face the camera, regardless of camera position or orientation.