In a game I have a specific object and two positions the object will move from and to.
I already have the function for calculating current position in specific time.
It works like this:
Inputting 0 will move the object to Position 1.
Inputting 1 will move the object to Position 2.
Inputting 0.5 will move the object in the middle of the two positions.
etc...
(In the examples below, time is varying from 0 to 1)
When I want to start the object slowly and stop it when it is moving fast, I use:
MoveObject(sin(time * 90))
When I want to start the object fast and stop it as it is getting slower, I use:
MoveObject(1 - cos(time * 90))
Without the effects, it's:
MoveObject(time)
How do I make the object start moving slowly, move fast in the center of two positions and then get slower while reaching the second position?
It would be:
MoveObject((time) * (time) * (3 - 2 * (time)))
sol.gfxile.net/interpolation
Related
I have graphs of sets of points like:-
There are up to 1 million points on each graph. You can see that the points are scattered over a grid of cells, each sized 200 x 100 units. So there are 35 cells shown.
Is there an efficient way to count how many points there are in each cell? The brute force approach seems to be to parse the data 35 times with a whole load of combined is less or greater than statements.
Some of the steps below could be optimized in the sense that you could perform some of these as you build up the data set. However I'll assume you are just given a series of points and you have to find which cells they fit into. If you can inject your own code into the step that builds up the graph, you could do the stuff I wrote below along side of building the graph instead of after the fact.
You're stuck with brute force in the case of just being given the data, there's no way you can know otherwise since you have to visit each point at least once to figure out what cell it is in. Therefore we are stuck with O(n). If you have some other knowledge you could exploit, that would be up to you to utilize - but since it wasn't mentioned in the OP I will assume we're stuck with brute force.
The high level strategy would be as follows:
// 1) Set rectangle bounds to have minX/Y at +inf, and maxX/Y to be -inf
// or initialize it with the first point
// 2) For each point:
// Set the set the min with min(point.x, bounds.min.x)
// Same for the max as well
// 3) Now you have your bounds, you divide it by how many cells fit onto each
// axis while taking into account that you might need to round up with division
// truncating the results, unless you cast to float and ceil()
int cols = ceil(float(bounds.max.x - bounds.min.x) / CELL_WIDTH);
int rows = ceil(float(bounds.max.y - bounds.min.y) / CELL_HEIGHT);
// 4) You have the # of cells for the width and height, so make a 2D array of
// some sort that is w * h cells (each cell contains 32-bit int at least) and
// initialize to zero if this is C or C++
// 5) Figure out the cell number by subtracting the bottom left corner of our
// bounds (which should be the min point on the x/y axis that we found from (1))
for (Point p in points):
int col = (p.x - minX) / cellWidth;
int row = (p.y - minY) / cellHeight;
data[row][col]++;
Optimizations:
There are some ways we might be able to speed this up off the top of my head:
If you have powers of two with the cell width/height, you could do some bit shifting. If it's a multiple of ten, this might possibly speed things up if you aren't using C or C++, but I haven't profiled this so maybe hotspot in Java and the like would do this for you anyways (and no idea about Python). Then again 1 million points should be pretty fast.
We don't need to go over the whole range at the beginning, we could just keep resizing our table and adding new rows and columns if we find a bigger value. This way we'd only do one iteration over all the points instead of two.
If you don't care about the extra space usage and your numbers are positive only, you could avoid the "translate to origin" subtraction step by just assuming everything is already relative to the origin and not subtract at all. You could get away with this by modifying step (1) of the code to have the min start at 0 instead of inf (or the first point if you chose that). This might be bad however if your points are really far out on the axis and you end up creating a ton of empty slots. You'd know your data and whether this is possible or not.
There's probably a few more things that can be done but this would get you on the right track to being efficient with it. You'd be able to work back to which cell it is as well.
EDIT: This assumes you won't have some really small cell width compared to the grid size (like your width being 100 units, but your graph could span by 2 million units). If so then you'd need to look into possibly sparse matrices.
I am new to using PsychoPy and I have programmed a few simple tasks. I am currently really struggling to program a word dot probe. I do not want to use coder, simply because the rest of my research team need to be able to easily edit the program, and work and use it.
In case anyone is wondering what my specific problem is, I cannot seem to get the pictures to load at the same time correctly and do not know how to get a probe to appear behind one of the pictures once the pictures have disappeared.
Timing
The timing issue can be solved by inserting an ISI period in the beginning of the trial, e.g. during a fixation cross. This allows psychopy to load the images in the background so that they are ready for presentation.
Truly random dot position
In your case, you want the dot position to be random, independently of image. This is one of the cases that TrialHandler does not handle and I suspect you need to insert a code component to make this work. For true randomness but only 50% probability in the limit of infinite trials, simply put this in a code component under "begin routine":
x = (np.random.binomial(1, prob) - 0.5) * xdist
y = 0
dot.pos = [x, y]
and change dot to the name of your dot stimulus, y is the vertical offset, x is the horizontal offset (here varying between trials), xdist is the distance between the dot positions, and prob is the chance of the dot appearing to the right. You probably want to set this to 0.5, i.e. 50 %.
Balanced dot position
If you want the dot to appear at each side exactly the same number of times, you can do the following in the code component:
Under "begin experiment", make a list with the exact length of the number of trials:
dotPos = [0, 1] * int(round(numberOfTrials/2)) # create the correct number of left/right (coded as 0 and 1). [0,1] yields 50%. [0,0,0,1] and /4 would yield 25 % etc.
np.random.shuffle(dotPos) # randomize order
Then under "begin routine" do something akin to what we did above:
x = (dotPos.pop() - 0.5) * xdist # dotPos.pop() takes returns the last element while removing it from the list.
y = 0
dot.pos = [x, y]
Naturally, if the number of trials is uneven, one position will be occupied one more time than the other.
Two dot positions for each condition
For the record, if the dot position is to be shown at each position for each image-combination, simply count each of these situations as conditions, i.e. give them a separate rows in the conditions file.
So I am making a program, where you can have two objects (circles). I want them to orbit like planets around each other, but only in 2D.
I know that using Newtons Universal Law of Gravitation I can get the force between the two objects. I also know A = F / M. My question is how would I take the A from the previous equation and change it into a vector?
You need to use vector equations:
// init values (per object)
double ax=0.0,ay=0.0,az=0.0; // acceleration [m/s^2]
double vx=0.0,vy=0.0,vz=0.0; // velocity [m/s]
double x=0.0, y=0.0, z=0.0; // position [m]
double m=1.0; // mass [kg]
// iteration inside some timer (dt [seconds] period) ...
int i; double a,dx,dy,dz; // first compute acceleration
for (ax=0.0,ay=0.0,az=0.0,i=0;i<obj.num;i++)
if (obj[i]!=this) // ignore gravity from itself
{
dx=obj[i].x-x;
dy=obj[i].y-y;
dz=obj[i].z-z;
a=sqrt((dx*dx)+(dy*dy)+(dz*dz)); // a=distance to obj[i]
a=6.67384e-11*(obj[i].m*m)/(a*a*a); // a=acceleration/distance to make dx,dy,dz unit vector
ax+=a*dx; // ax,ay,az = actual acceleration vector (integration)
ay+=a*dy;
az+=a*dz;
}
vx+=ax*dt; // update speed via integration of acceleration
vy+=ay*dt;
vz+=az*dt;
x+=vx*dt; // update position via integration of velocity
y+=vy*dt;
z+=vz*dt;
Code is taken from here
obj[] is list of all your objects
obj.num is the count of them
I recommend to create object class with all the variables inside (ax,ay,az,...m), init them once and then continuously update (iteration) in some timer. If you want more accuracy then you should compute ax,ay,az for all objects first and only after update speed and position (to avoid change of position of objects during gravity computation). If you want to drive an object (like with truster) then just add its acceleration to ax,ay,az vector)
Now to setup an orbit just:
place planet object
must be massive enough and also set its position / velocity to what you want
place satellite
Initial position should be somewhere near planet. It should not be too massive. Init also speed vector with tangent direction to orbiting trajectory. If speed is too low it will collapse into planet and if speed is too high it will escape from planet otherwise will be orbiting (circle or ellipse)
timer
lower the interval better the simulation usually 10ms is OK but for massive and far objects is also 100ms and more OK. If you want particles or something then use 1ms (very dynamic sceene).
I strongly recommend to read this related QA:
Is it possible to make realistic n-body solar system simulation in matter of size and mass?
especially [edit3] about the integration precision and creating orbital data.
With two objects you are probably best using an ellipse which is the path the objects will follow about their common center of mass. Read Kepler's laws of planetary motion which gives the background.
If one object is a much greater mass than the other, i.e. a sun and a planet you can have one stationary and the other taking an elliptical path. The equation of the ellipse is given by
r = K e / ( 1 + e cos(theta))
K is a constant giving the size and e is the eccentricity. If you want an elliptical orbit have 0 < e < 1 the smaller it is the more circular the orbit. To get x, y coordinates from this use, x = r cos(theta), y = r sin(theta). The missing bit is time and how the angle is dependant on time. This is where the second and third laws come in. If a and b are the semi-major and semi-minor lengths of the ellipse, and P is the period then
0.5 * P * r^2 theta'= pi a b
theta' is the rate of change of angle with respect to time (d theta/d t). You can use this to get how much theta will change given a increase in time. First work out the current radius r0 given the current angle th0 if the time increment is δt then the angle increment δtheta is
δtheta = 2 pi * a * b / r^2 * δt
and the next angle is th0 + δtheta.
If the masses are of similar magnitude then see two body problem. Both objects will have elliptical orbits, there are two patterns which you can see in animations on that page. The ellipses will follow the same formula as above with the focus at the common center of mass.
If you have three object things get considerably harder and there are not generally neat solutions. See three body problem for this.
I am working on a yet another 3d game engine and have following Problem. In order to improve Performance I want to check if two Objects are coming closer or moving away from each other. Only if they are coming closer to each other collision detection should happen.
At the moment the code calculates the current distance using the positions of both Objects. Then moves both positions about the velocity vector and calculates the expected distance. etc.
public class Model{
public boolean isApproaching(Model model) {
float [] otherPosition = model.getPosition();
float [] otherVelocity = model.getVelocity();
double currentDistance = Math.pow(this.position[0] - otherPosition[0], 2)
+ Math.pow(this.position[1] - otherPosition[1], 2)
+ Math.pow(this.position[2] - otherPosition[2], 2);
double expectedDistance = Math.pow(this.position[0] + this.velocityVector[0]/1000 - otherPosition[0] - otherVelocity[0]/1000, 2)
+ Math.pow(this.position[1] + this.velocityVector[1]/1000 - otherPosition[1] - otherVelocity[1]/1000, 2)
+ Math.pow(this.position[2] + this.velocityVector[2]/1000 - otherPosition[2] - otherVelocity[2]/1000, 2);
if(currentDistance < expectedDistance)
return false;
return true;
}
}
As you can see in the Picture it does not work for fast moving objects. Is there any good way to show if two points are moving towards each other?
Given the two straight line trajectories, there's a unique time when they're going to be closest, which may have been in the past. If it's in the past then it means that the objects are moving away from each other, but if its in the future it means that they're moving closer together. But even it they're moving closer together, I would guess that knowing how soon is the information that you need, so that you can schedule when to worry about it. This methodology should cater for both slow moving and fast moving objects.
It appears that you are using the /1000 versus any other value for some unstated reason that may negate the following.
By changing this scale factor you get better detection with a fast object. Perform your same isApproaching() calculation except with a /(1000*N) instead of /1000. You will then has N times better sensitivity to know if they are approaching or not.
I would like to create a "Gravity Grid" such as in this image:
The closest I have been in creating this grid is shown in this image:
I'm only warping the lines parallel to the Y-Axis until I can solve this problem. But as you can see in the picture, the lines seem to be warping past my "planet".
Here is the code of interest I have for now:
for (each point on a line parallel to the y-axis) {
if (planetPosition.x > currrentPoint.x) {
warpedXPos = currrentPoint.x + (1 / (distance*1000));
}
else {
warpedXPos = currrentPoint.x - (1 / (distance*1000));
}
}
The idea is to pull every point towards the planet by an amount proportional to 1/R, where R is the distance from the planet to the point.
In your code you currentPoint.x, etc, is the absolute position, but you need to warp the position relative to the planet. Therefore your equation should look like:
warped.x = currrentPoint.x + (planetPosition.x-currrentPoint.x)/(1000*distance)
The part after the + is the scaling of the relative distance basically your warp, and then you add this deflection to the original absolute value. You probably also won't want to use the 1000 factor, but this is just keeping with your example. Also, note that there's no need to break this into cases using the >, the sign of the subtraction should make the adjustment in the appropriate direction.
(By the way, this type of operation is super common... first you subtract an absolute factor, then you scale, then you add the original factor back in. It's a trick worth memorizing for use in lots of applications.)