This is not a question specific to one programming language, more of a mathematically conceptual, though just in case, I'm using C++ on Visual Studio.
Basically, my current code draws a line, that starts at the centre (of the window), and ends at my mouse position at any time, every frame - I end up with a line that follows my mouse, starting from the centre.
My question is, how would I end up with the exact same system, except that no matter how far my mouse goes from the centre, the line will still follow the direction of the vector 'centerToMouse', but its length will only ever be 100 units (once the distance between MousePos and centre exceeds 100), such that I end up with a line that follows (extends/shrinks) my mouse, but once I reach above 100 units away from the centre, the line stays 100 units long as long as my mouse is further than 100 away.
I'm sorry if the question is badly phrased, in my head it makes sense, and I don't know how else to word it.
I don't necessarily need a code answer for C++, just the concept. I've tried a few methods involving normalizing, unit vectors etc. But I'm just stuck.
Thanks a lot for taking the time!
Paraphrasing from my above comment:
radius = 100;
angle = atan2(mouse_position.y-center.y, mouse_position.x-center.x);
if (distance(center, mouse_position) < radius){
line_position = mouse_position;
}
else{
line_position = center + Vector(radius*cos(angle), radius*sin(angle));
}
Related
I am trying to create a physically plausible 2d physics engine. I have read many documents about detection of collisions, contact resolving, interpenetrations, projection, separating axis theorem (SAT) methods, etc.
Projection via SAT appears to be one physically plausible method for dealing with overlapping ("penetrating") objects. This works fine for objects with no rotation, but I can't figure out how to deal with rotations.
Imagine two polygons in rotation that will collide:
I need to understand how to project the point of contact and the time when this happens.
any help will be appreciated!
The problem is that you are assuming both bodies are moving at the same time, which will be a huge pain to calculate the exact collision point.
You can update the position of each body sequentially, and apply the collisions accordingly:
(I know the bodies are not rotated in this example, but I have no means right now to draw rotated forms, but the same idea applies, sorry)
First, the rectangle is moved, and checks for collision.
Then, the pentagon moves, and collides with the rectangle, updating its position.
And last, the triangle position is also checked for collition and updated.
About how to calculate the collision point, you can apply this:
Position calculateValidPosition(Position start, Position end)
Position middlePoint = (start + end) /2
if (middlePoint == start || middlePoint == end)
return start
if( isColliding(middlePont) )
return calculateValidPosition(start, middlePoint)
else
return calculate(middlePoint, end)
Note that
Position middlePoint = (start + end) /2
not only calculates the body's middle position, but also should calculate it's middle rotation:
If rotation is X at the begining of the movement, and Y at the end, the middle point rotatiln is just (X+Y)/2
Note this algorithm leaves a lot of room for optimization (like making it non-recursive)
This solution might seem not really accurate, but it will only work incorrectly when the speed of the bodies is way greater than the body's size, which will only happen if a really small body moves really fast. In the rest of scenarios the result is "good enough" for a game.
I have a complicated problem and it involves an understanding of Maths I'm not confident with.
Some slight context may help. I'm building a 3D train simulator for children and it will run in the browser using WebGL. I'm trying to create a network of points to place the track assets (see image) and provide reference for the train to move along.
To help explain my problem I have created a visual representation as I am a designer who can script and not really a programmer or a mathematician:
Basically, I have 3 shapes (Figs. A, B & C) and although they have width, can be represented as a straight line for A and curves (B & C). Curves B & C are derived (bend modified) from A so are all the same length (l) which is 112. The curves (B & C) each have a radius (r) of 285.5 and the (a) angle they were bent at was 22.5°.
Each shape (A, B & C) has a registration point (start point) illustrated by the centre of the green boxes attached to each of them.
What I am trying to do is create a network of "track" starting at 0, 0 (using standard Cartesian coordinates).
My problem is where to place the next element after a curve. If it were straight track then there is no problem as I can use the length as a constant offset along the y axis but that would be boring so I need to add curves.
Fig. D. demonstrates an example of a possible track layout but please understand that I am not looking for a static answer (based on where everything is positioned in the image), I need a formula that can be applied no matter how I configure the track.
Using Fig. D. I tried to work out where to place the second curved element after the first one. I used the formula for plotting a point of the circumference of a circle given its centre coordinates and radius (Fig. E.).
I had point 1 as that was simply a case of setting the length (y position) of the straight line. I could easily work out the centre of the circle because that's just the offset y position, the offset of the radius (r) (x position) and the angle (a) which is always 22.5° (which, incidentally, was converted to Radians as per formula requirements).
After passing the values through the formula I didn't get the correct result because the formula assumed I was working anti-clockwise starting at 3 o'clock so I had to deduct 180 from (a) and convert that to Radians to get the expected result.
That did work and if I wanted to create a 180° track curve I could use the same centre point and simply deducted 22.5° from the angle each time. Great. But I want a more dynamic track layout like in Figs. D & E.
So, how would I go about working point 5 in Fig. E. because that represents the centre point for that curve segment? I simply have no idea.
Also, as a bonus question, is this the correct way to be doing this or am I over-complicating things?
This problem is the only issue stopping me from building my game and, as you can appreciate, it is a bit of a biggie so I thank anyone for their contribution in advance.
As you build up the track, the position of the next piece of track to be placed needs to be relative to location and direction of the current end of the track.
I would store an (x,y) position and an angle a to indicate the current point (with x,y starting at 0, and a starting at pi/2 radians, which corresponds to straight up in the "anticlockwise from 3-o'clock" system).
Then construct
fx = cos(a);
fy = sin(a);
lx = -sin(a);
ly = cos(a);
which correspond to the x and y components of 'forward' and 'left' vectors relative to the direction we are currently facing. If we wanted to move our position one unit forward, we would increment (x,y) by (fx, fy).
In your case, the rule for placing a straight section of track is then:
x=x+112*fx
y=y+112*fy
The rule for placing a curve is slightly more complex. For a curve turning right, we need to move forward 112*sin(22.5°), then side-step right 112*(1-cos(22.5°), then turn clockwise by 22.5°. In code,
x=x+285.206*sin(22.5*pi/180)*fx // Move forward
y=y+285.206*sin(22.5*pi/180)*fy
x=x+285.206*(1-cos(22.5*pi/180))*(-lx) // Side-step right
y=y+285.206*(1-cos(22.5*pi/180))*(-ly)
a=a-22.5*pi/180 // Turn to face new direction
Turning left is just like turning right, but with a negative angle.
To place the subsequent pieces, just run this procedure again, calculating fx,fy, lx and ly with the now-updated value of a, and then incrementing x and y depending on what type of track piece is next.
There is one other point that you might consider; in my experience, building tracks which form closed loops with these sort of pieces usually works if you stick to making 90° turns or rather symmetric layouts. However, it's quite easy to make tracks which don't quite join up, and it's not obvious to see how they should be modified to allow them to join. Something to bear in mind perhaps if your program allows children to design their own layouts.
Point 5 is equidistant from 3 as 2, but in the opposite direction.
So I have a ship, that has thrusters at the bottom and that can only use these to move forward. It can also rotate itself around its center. Its thrusters gives it acceleration, so it doesn't move at a constant velocity. What I want to do is to tell it "move to point B".
I have come up with a solution but it doesn't work very well and it doesn't rotate smoothly, it moves jerkily and it doesn't end up exactly where it should be, so I have to have a big margin of error.
Is this a normal problem, and if so is there a "standard" way of doing it? Is this an easy problem? I want to make it look like the ship is steering itself to that point, using the constraints (thrusters, rotation) the player has. This excludes just lerping it from point A to B. Or does it?
I'd love some help in solving this problem. Positions are stored in vectors, and it's a 2D problem. Just for reference I'm including my solution, which basically is accelerating the ship until and rotating it to point to the point. I think my implementation of this idea is the problem:
Vector diff = vector_sub(to_point, pos);
float angle = vector_getangle(diff);
float current_angle = vector_getangle(dir);
float angle_diff = rightrange(angle) - rightrange(current_angle);
float len = vector_getlength(diff);
// "Margin of error"
float margin = 15.0;
// Adjust direction, only if we're not stopping the next thing we do (len <= margin)
if ( len > margin && fabs(angle_diff) > 2.0 )
{
dir = vector_setangle(dir, current_angle + (angle_diff)*delta*(MY_PI) - MY_PI/2);
}
else if ( len > margin )
{
dir = vector_normalize(diff);
}
// accelerate ship (if needed)
acc.x = acc.y = speed;
acc = vector_setangle(acc, vector_getangle(dir));
if ( len <= margin )
{
// Player is within margin of error
}
If you are not looking for a very general solution that works online, then there is a simple solution. What I mean by online is continuously re-calculating the actions along the complete trajectory.
Assuming the ship is at rest at start, simply rotate it towards your target point (while still at rest). Now, your ship can reach the target by accelerating for t seconds, rotating back while in motion (for 0.5 seconds as per your constraint), and decelerating for another t seconds. If the distance between current point and destination is d, then the equation you need to solve is:
d = 0.5*a*t^2 + 0.5*a*t + 0.5*a*t^2
The first term is distance traveled while accelerating. The second term is distance traveled while rotating (v*t_rot, v=a*t, t_rot=0.5). The final term is the distance traveled while decelerating. Solve the above for t, and you have your trajectory.
If the ship is moving at start, I would first stop it (just rotate in opposite direction of its speed vector, and decelerate until at rest). Now we know how to reach destination.
The problem with offline trajectory calculation is that it is not very accurate. There is a good chance that you will end up in the vicinity of the target, but not exactly on top of it.
Let's make the problem a little more interesting: the ship cannot rotate without acceleration. Let's call this acceleration vector a_r, a vector that is at a certain angle against the ship's direction (somewhat like having a thruster at an angle at the back). Your task now is to rotate the ship and accelerate in such a direction that the speed component perpendicular to the vector connecting the current position to the target is canceled out. Instead of trying to calculate the vectors offline, I would go with an online approach with this.
The easiest thing to do would be to add the following algorithm calculated at every time interval:
Calculate the vector pointing from ship to destination.
Split your current speed vector into two components: towards the destination, and perpendicular to it.
If perpendicular speed is zero, skip 4
Start rotating towards the negative of the perpendicular vector's direction. If already looking away from it (not exact opposite, but just looking away), also fire main thruster.
This will oscillate a bit, I suspect it will also stabilize after a while. I must admit, I don't know how I would make it stop at destination.
And the final approach is to model the ship's dynamics, and try to linearize it. It will be a non-linear system, so the second step will be necessary. Then convert the model to a discrete time system. And finally apply a control rule to make it reach target point. For this, you can change your state-space from position and speed to error in position and (maybe) error in speed, and finally add a regulation control (a control loop that takes the current state, and generates an input such that the state variables will approach zero).
This last one is fairly difficult in the maths compartment, and you'd probably need to study control engineering a bit to do it. However, you'll get much better results than the above simplistic algorithm - which admittedly might not even work. In addition, you can now apply various optimization rules to it: minimize time to reach target, minimize fuel consumption, minimize distance traveled, etc.
I'm having trouble wrapping my mind around how to calculate the normal for a moving circle in a 2d space. I've gotten as far as that I'm suppose to calculate the Normal of the Velocity(Directional Speed) of the object, but that's where my college algebra mind over-heats, any I'm working with to 2d Circles that I have the centerpoint, radius, velocity, and position.
Ultimately I'm wanting to use the Vector2.Reflect Method to get a bit more realistic physics out of this exercise.
thanks ahead of time.
EDIT: Added some code trying out suggestion(with no avail), probably misunderstanding the suggestion. Here I'm using a basketball and a baseball, hence base and basket. I also have Position, and Velocity which is being added to position to create the movement.
if ((Vector2.Distance(baseMid, basketMid)) < baseRadius + basketRadius)
{
Vector2 baseNorm = basketMid - baseMid;
baseNorm.Normalize();
Vector2 basketNorm = baseMid - basketMid;
basketNorm.Normalize();
baseVelocity = Vector2.Reflect(baseVelocity, baseNorm);
basketVelocity = Vector2.Reflect(basketVelocity, basketNorm);
}
basePos.Y += baseVelocity.Y;
basePos.X += baseVelocity.X;
basketPos.Y += basketVelocity.Y;
basketPos.X += basketVelocity.X;
basketMid = new Vector2((basketballTex.Width / 2 + basketPos.X), (basketballTex.Height / 2 + basketPos.Y));
baseMid = new Vector2((baseballTex.Width / 2 + basePos.X), (baseballTex.Height / 2 + basePos.Y));
First the reflection. If I'm reading your code right, the second argument to Vector2.Reflect is a normal to a surface. A level floor has a normal of (0,1), and a ball with velocity (4,-3) hits it and flies away with velocity (4,3). Is that right? If that's not right then we'll have to change the body of the if statement. (Note that you can save some cycles by setting basketNorm = -baseNorm.)
Now the physics. As written, when the two balls collide, each bounces off as if it had hit a glass wall tangent to both spheres, and that's not realistic. Imagine playing pool: a fast red ball hits a stationary blue ball dead center. Does the red ball rebound and leave the blue ball where it was? No, the blue ball gets knocked away and the red ball loses most of its speed (all, in the perfect case). How about a cannonball and a golf ball, both moving at the same speed but in opposite directions, colliding head-on. Will they both bounce equally? No, the cannonball will continue, barely noticing the impact, but the golf ball will reverse direction and fly away faster than it came.
To understand these collisions you have to understand momentum (and if you want collisions that aren't perfectly elastic, like when beanbags collide, you also have to understand energy). A basic physics textbook will cover this in an early chapter. If you just want to be able to simulate these things, use the center-of-mass frame:
Vector2 CMVelocity = (basket.Mass*basket.Velocity + base.Mass*base.Velocity)/(basket.Mass + base.Mass);
baseVelocity -= CMVelocity;
baseVelocity = Vector2.Reflect(baseVelocity, baseNorm);
baseVelocity += CMVelocity;
basketVelocity -= CMVelocity;
basketVelocity = Vector2.Reflect(basketVelocity, basketNorm);
basketVelocity += CMVelocity;
The normal of a circle at a given point on its edge is going to be the direction from its center to that point. Assuming that you're working with collisions of circles here, then one easy "shorthand" way to work this out would be that at the time of collision (when the circles are touching), the following will hold true:
Let A be the center of one circle and B the center of the other. The normal for circle A will be normalize(B-A) and the normal for circle B will be normalize(A-B). This is true because the point where they touch will always be colinear with the centers of the two circles.
Caveat: I'm not going to assume that this is completely correct. Physics are not my specialty.
Movement has no effect on a normal. Typically, a normal is just a normalized (length 1) vector indicating a direction, typically the direction that a poly faces on a 3d object.
What I think you want to do is find the collision normal between two circles, yes? If so, one of the cool properties of spheres is that if you find the distance between the centers of them, you can normalize that to get the normal of the sphere.
What seems correct for 2d physics is that you take the velocity * mass (energy) of a sphere, and multiply that by the normalized vector to the other sphere. Add the result to the destination sphere's energy, subtract it from the original sphere's energy, and divide each, individually, by mass to get the resulting velocity. If the other sphere is moving, do the same in reverse. You can probably simplify the math down from there, of course, but it's late and I don't feel like doing it :)
If both spheres are moving, repeat the process for the other sphere (though you could probably simplify that equation to get some more efficient math).
This is just back-of-the-napkin math, but it seems to give the correct results. And, hey, I once derived Euler angles on my own, so sometimes my back-of-the-napkin math actually works out.
This also assumes perfectly elastic collisions.
If I'm incorrect, I'd be happy to find out where :)
See also: Why is my image rotation algorithm not working?
This question isn't language specific, and is a math problem. I will however use some C++ code to explain what I need as I'm not experienced with the mathematic equations needed to express the problem (but if you know about this, I’d be interested to learn).
Here's how the image is composed:
ImageMatrix image;
image[0][0][0] = 1;
image[0][1][0] = 2;
image[0][2][0] = 1;
image[1][0][0] = 0;
image[1][1][0] = 0;
image[1][2][0] = 0;
image[2][0][0] = -1;
image[2][1][0] = -2;
image[2][2][0] = -1;
Here's the prototype for the function I'm trying to create:
ImageMatrix rotateImage(ImageMatrix image, double angle);
I'd like to rotate only the first two indices (rows and columns) but not the channel.
The usual way to solve this is by doing it backwards. Instead of calculating where each pixel in the input image ends up in the output image, you calculate where each pixel in the output image is located in the input image (by rotationg the same amount in the other direction. This way you can be sure that all pixels in the output image will have a value.
output = new Image(input.size())
for each pixel in input:
{
p2 = rotate(pixel, -angle);
value = interpolate(input, p2)
output(pixel) = value
}
There are different ways to do interpolation. For the formula of rotation I think you should check https://en.wikipedia.org/wiki/Rotation_matrix#In_two_dimensions
But just to be nice, here it is (rotation of point (x,y) angle degrees/radians):
newX = cos(angle)*x - sin(angle)*y
newY = sin(angle)*x + cos(angle)*y
To rotate an image, you create 3 points:
A----B
|
|
C
and rotate that around A. To get the new rotated image you do this:
rotate ABC around A in 2D, so this is a single euler rotation
traverse in the rotated state from A to B. For every pixel you traverse also from left to right over the horizontal line in the original image. So if the image is an image of width 100, height 50, you'll traverse from A to B in 100 steps and from A to C in 50 steps, drawing 50 lines of 100 pixels in the area formed by ABC in their rotated state.
This might sound complicated but it's not. Please see this C# code I wrote some time ago:
rotoZoomer by me
When drawing, I alter the source pointers a bit to get a rubber-like effect, but if you disable that, you'll see the code rotates the image without problems. Of course, on some angles you'll get an image which looks slightly distorted. The sourcecode contains comments what's going on so you should be able to grab the math/logic behind it easily.
If you like Java better, I also have made a java version once, 14 or so years ago ;) ->
http://www.xs4all.nl/~perseus/zoom/zoom.java
Note there's another solution apart from rotation matrices, that doesn't loose image information through aliasing.
You can separate 2D image rotation into skews and scalings, which preserve the image quality.
Here's a simpler explanation
It seems like the example you've provided is some edge detection kernel. So if what you want to is detect edges of different angles you'd better choose some continuous function (which in your case might be a parametrized gaussian of x1 multiplied by x2) and then rotate it according to formulae provided by kigurai. As a result you would be able to produce a diskrete kernel more efficiently and without aliasing.