I'm creating simple game and reached the point where I feel helpless. I was good in geometry but it was long time back in school, now trying to refresh my mind.
Let's say i have iPad screen. Object's xy position at one given point of time and xy position at another point of time stored in 2 variables .
Question:
how to find the third position of the object at the end of the screen being given previous 2 position, considering the object moves in the same direction (line) from point 1 to point 2.
Thanks in advance.
Let us have that v1 and v2 are the vectors representing the two points. Let t0 be the time between the two points. Let t be the current time.
Then our location vector v3 is given by v3 = v1 + (v2 - v1)t/t0
If the object is moving in the same direction and you have an horizontal line, the next position given x and y would be
x+1, y
If the object is moving in the same direction in a vertical line it would be
x, y+1
If the object is moving in a diagonal up-right
x+1,y+1
diagonal down-right
x+1, y+1
diagonal down-left
x-1, y-1
diagonal up-left
x-1, y+1
So something general would be :
newPosition = (x+1,y) //if you wish to move forward to the right, try to handle all
cases
All the cases above work if the object is moving forward, if it is moving backwards just change the + by - . Basically think of the object as moving in a cartesian coordinate system, where x is horizontal and y is vertical.
I think you can get the idea out of this three cases ;)
Related
I have a lot of tutorials & books, but I'm unable to understand how my viewport, my near & far distance etc are used to calc perspective / frustum matrix.
I have the learningwebgl lessons, but.... I dont understand what viewport & 3D space adjustments are made.... What is my initial window projection size ? Why I see the triangle & square placed at z = -7.
Another thing I dont understand . A near plane of 0.001 creates the window projection just in front of my nose ? So what is my projection window dimension ?
I need a very deeper and basic help....
Can anybody help me ? Some really usefull links? I need graphical examples showing & teaching how frustum is calculated.
Thanks
There's this
http://games.greggman.com/game/webgl-3d-perspective/
Imagine you're in 2D. You have a canvas that's 200x100 pixels. If you draw at x = 201 it will be off the canvas. Similarly at x = -1 it will be off the canvas.
In WebGL it works in a 3D space that goes from -1 to +1 in x, y and z. The perspective / frustum matrix is the matrix that takes your 3d scene and converts it to this -1 / +1 space. The near and far values define what range in world space get converted to the -1 / +1 "clipspace". Anything outside that range will be clipped just like the 2D example. If you set near to 10 and far to 100 then something at Z = 9 will be clipped because it's too near and something at 101 will also be clipped as something that's too far. More specifically the near and far settings will form a matrix such that when a point is at Z = near it will become -1 when multiplied by the matrix and when it's at Z = far it will become +1 when multiplied by the matrix.
The viewport setting tells WebGL how to convert from the -1 to +1 space back into pixels.
I'm making a program in which many weird shapes are drawn onto a canvas. Right now i'm trying to implement the last, and possebly hardest, one.
In this particular shape i need a way to find the location (on a 2d canvas) where the line hits the shape. The following image is an example of what i have right now.
The black dots are the points that a known to me (i also have the location of the center of the three open circles and the radius of these circles). Each of the three outer lines needs a line towards the center dot, ending at the point that it hits the circle. This shape can be turned 90, 180 or 270 degrees.
The shape should look something like the following:
If you need any other information, please ask me in the comments. I'm not very good at math so please be gentle, thanks!
If A and B are points forming a line, then you can describe any point on that line using coordinates:
x = t·Ax + (1−t)·Bx
y = t·Ay + (1−t)·By
0 ≤ t ≤ 1
You can also describe the circle with center M and radius r as
(x − Mx)2 + (y − My)2 = r2
So take the x and y from the equations of the line, and plug them into the equation of the circle. You obtain a quadratic equation in t. Its two solutions describe the two points of intersection between the line and circle. In your example, only one of them lies on the line segment, i.e. satisfies 0 ≤ t ≤ 1. The other describes a point on the extension of the segment past its endpoint. Take the correct value for t back to the equations of the line, and you obtain the x and y coordinates of the point of intersection.
If you don't know up front which circle you want to intersect with a given line, then intersect all three and choose the most appropriate point afterwards. Probably that is the point closest to the outside starting point of the line segment. The same goes in cases where both points of intersection lie on the segment.
does anyone know the method or code to add a second x axis to a TGraph in CERN's ROOT program? Ive been searching the root website and its documentation almost always confuses me. What i need is just one plot of data, but a second X axis on top whose values are a function of the bottom x axis' values. Its basically so lazy people dont have to convert from the numbers of the bottom x axis to the top x axis.
For a simple example (if i wasnt clear)
Say you have a sine curve which is some function of theta. On the top x axis we could have degrees whereas on the bottom we could have radians with 360deg corresponding to 2pi rad...
Any help would be appreciated!
TGaxis is the class you are looking for to draw extra axes wherever you desire. Grabbing the world coordinate for your pad you can then superimpose like so. Replace low and high with the appropriate limits.
// your graph code here...
TGraph->Draw("AP");
TGaxis *axis = new TGaxis(gPad->GetUxmin(),gPad->GetUymax(),gPad->GetUxmax(),gPad->GetUymax(),low,high,510,"+L");
axis->Draw();
Check out TGaxis documentation for more examples.
(A previous answer I had was deleted as it was just a link to the site listed as a reference below. I hope this is more in line with the community guidelines.)
I think this might do what you want.
void axis2() {
TH1F *h = new TH1F("h","test",30,-3,3);
h->FillRandom("gaus",10000);
h->Draw();
TText t;
t.SetTextSize(0.02);
t.SetTextAlign(22);
Double_t yt = - h->GetMaximum()/15.;
for (Int_t i=1;i<=30;i++) t.DrawText(h->GetBinCenter(i),yt,Form("%d",i%10));
}
It doesn't create another taxis but shows you how to draw text at the same location of the axis. The answer comes from Rene Brun himself (one of the main authors of root) so I don't think you can have two x axes.
Source:
http://root.cern.ch/phpBB3/viewtopic.php?f=3&t=7110
Here is an example showing how to proceed.
https://root.cern/doc/master/twoscales_8C.html
Let's say I have circle bouncing around inside a rectangular area. At some point this circle will collide with one of the surfaces of the rectangle and reflect back. The usual way I'd do this would be to let the circle overlap that boundary and then reflect the velocity vector. The fact that the circle actually overlaps the boundary isn't usually a problem, nor really noticeable at low velocity. At high velocity it becomes quite clear that the circle is doing something it shouldn't.
What I'd like to do is to programmatically take reflection into account and place the circle at it's proper position before displaying it on the screen. This means that I have to calculate the point where it hits the boundary between it's current position and it's future position -- rather than calculating it's new position and then checking if it has hit the boundary.
This is a little bit more complicated than the usual circle/rectangle collision problem. I have a vague idea of how I should do it -- basically create a bounding rectangle between the current position and the new position, which brings up a slew of problems of it's own (Since the rectangle is rotated according to the direction of the circle's velocity). However, I'm thinking that this is a common problem, and that a common solution already exists.
Is there a common solution to this kind of problem? Perhaps some basic theories which I should look into?
Since you just have a circle and a rectangle, it's actually pretty simple. A circle of radius r bouncing around inside a rectangle of dimensions w, h can be treated the same as a point p at the circle's center, inside a rectangle (w-r), (h-r).
Now position update becomes simple. Given your point at position x, y and a per-frame velocity of dx, dy, the updated position is x+dx, y+dy - except when you cross a boundary. If, say, you end up with x+dx > W (letting W = w-r), then you do the following:
crossover = (x+dx) - W // this is how far "past" the edge your ball went
x = W - crossover // so you bring it back the same amount on the correct side
dx = -dx // and flip the velocity to the opposite direction
And similarly for y. You'll have to set up a similar (reflected) check for the opposite boundaries in each dimension.
At each step, you can calculate the projected/expected position of the circle for the next frame.
If this lies outside the rectangle, then you can then use the distance from the old circle position to the rectangle's edge and the amount "past" the rectangle's edge that the next position lies at (the interpenetration) to linearly interpolate and determine the precise time when the circle "hits" the rectangle edge.
For example, if the circle is 10 pixels away from the rectangle's edge, then is predicted to move to 5 pixels beyond it, you know that for 2/3rds of the timestep (10/15ths) it moves on its orginal path, then is reflected and continues on its new path for the remaining 1/3rd of the timestep (5/15ths). By calculating these two parts of the motion and "adding" the translations together, you can find the correct new position.
(Of course, it gets more complicated if you hit near a corner, as there may be several collisions during the timestep, off different edges. And if you have more than one circle moving, things get a lot more complex. But that's where you can start for the case you've asked about)
Reflection across a rectangular boundary is incredibly simple. Just take the amount that the object passed the boundary and subtract it from the boundary position. If the position without reflecting would be (-0.8,-0.2) for example and the upper left corner is at (0,0), the reflected position would be (0.8,0.2).
I was asking myself if there are examples online which covers how you can for instance detect shapes in touch gestures.
for example a rectangle or a circle (or more complex a heart .. )
or determine the speed of swiping (over time ( like i'm swiping my iphone against 50mph ))
For very simple gestures (horizontal vs. vertical swipe), calculate the difference in x and y between two touches.
dy = abs(y2 - y1)
dx = abs(x2 - x1)
f = dy/dx
An f close to zero is a horizontal swipe. An f close to 1 is a diagonal swipe. And a very large f is a vertical swipe (keep in mind that dx could be zero, so the above won't yield valid results for all x and y).
If you're interested in speed, pythagoras can help. The length of the distance travelled between two touches is:
l = sqrt(dx*dx + dy*dy)
If the touches happened at times t1 and t2, the speed is:
tdiff = abs(t2 - t1)
s = l/tdiff
It's up to you to determine which value of s you interpret as fast or slow.
You can extend this approach for more complex figures, e.g. your square shape could be a horizontal/vertical/horizontal/vertical swipe with start/end points where the previous swipe stopped.
For more complex figures, it's probably better to work with an idealized shape. One could consider a polygon shape as the ideal, and check if a range of touches
don't have too high a distance to their closest point on the pologyon's outline, and
all touches follow the same direction along the polygon's outline.
You can refine things further from there.
There does exist other methods for detecting non-simple touches on a touchscreen. Check out the $1 unistroke gesture recognizer at the University of Washington. http://depts.washington.edu/aimgroup/proj/dollar/
It basically works like this:
Resample the recorded path into a fixed number of points that are evenly spaced along the path
Rotating the path so that the first point is directly to the right of the path’s center of mass
Scaling the path (non-uniformly) to a fixed height and width
For each reference path, calculating the average distance for the corresponding points in the input path. The path with the lowest average point distance is the match.
What’s great is that the output of steps 1-3 is a reference path that can be added to the array of known gestures. This makes it extremely easy to give your application gesture support and create your own set of custom gestures, as you see fit.
This has been ported to iOS by Adam Preble, repo on github:
http://github.com/preble/GLGestureRecognizer