I have a game called downhill rush https://www.roblox.com/games/7236421113/Downhill-Rush#!/about
It's working nicely however the map involves twists and turns and I am trying to keep the car stuck to the road.
Anyway, when the player is going down the hill, a raycast is shot downwards in the direction of the gravity and this raycast returns a normal.
I use this normal and compare it to the UpVector of what is called prevface. "prevface" is simply a part that is aligned with the road that tells the players cart which way to face or go.
My intention is to have the cart's ROLL or Z angle be the same as the angle between the ray normal and the upvector of prevface.
I have made following code and put 3 parts in the workspace to test it out. Each part is called a, b, c
prevface = workspace.b
upvec = workspace.a.CFrame.UpVector
local function getRotationBetween(u, v, axis)
local dot, uxv = u:Dot(v), u:Cross(v)
if (dot < -0.99999) then return CFrame.fromAxisAngle(axis, math.pi) end
return CFrame.new(0, 0, 0, uxv.x, uxv.y, uxv.z, 1 + dot)
end
local randomAxis = Vector3.new(1,0,0)
local rotateToFloorCFrame = getRotationBetween(prevface.CFrame.UpVector,upvec,randomAxis)
local x,y,goalangle = (rotateToFloorCFrame * prevface.CFrame):ToEulerAnglesXYZ()
workspace.c.Position = workspace.b.Position
workspace.c.CFrame = prevface.CFrame * CFrame.Angles(0,0,goalangle + math.pi)
workspace.c.Position = workspace.c.Position + Vector3.new(0,5,0)
The red block is named "a" has an UpVector that is pointing upwards, this represents the ray normal to the ground
The green block is named "b" and this is the "prevface" which is directly in alignment with the road
The blue block is named "c" and this is going to rotate in the same direction as the ray normal. NOT the same direction as the part a because we are only using part a's ray normal
Here are difference scenarios:
All the parts are aligned nicely, so in the game the floor would be flat and the cart is perfectly aligned to the road
We pitch the red part a tiny bit forward now, the blue part is perfectly aligned with the road because the upvector is aligned with the upvector of the green part
We tilt the red part a bit to the side, as this represents the player driving along a curve of the road, all good
Now here's the tricky part
The road will turn left and right as well as going up and down for more fun in the game.
This looks good to me, perfectly aligned
Now here's an issue: The green part represents the general direction of the road, the road is going down at 10 degrees and may occasionally rotate to the left or right relative to that grey part you see which is aligned with the world at an orinetation of 0,0,0...
So now why isn't the blue part aligned with the red part in this case?
A scenario where the cart is driving along a curve too in a hill turn
Related
I am currently trying to write a shader in unity that draws a triangular pattern around countries in a risk-styled game if both countries are not owned by the same player (visual aid to see your borders).
Right now, I'm having an issue with making the shader set the countries properly.
It always sets country 0 to the left, and country 1 to the right - country 0 and 1 are set programically.
The line, a border, can be between 0 and 359 degrees.
How I find the countries 0 and 1 is I draw 3 points to the left and right of the midpoint of the line, one .01f, one .1f and one 1f away from the midpoints in each direction, then spin them around the midpoint to the appropriate location.
After that I do an even-odd check to see if the points are inside or outside of each country, and compare the weight results (closest gets 3 points, mid gets 2, furthest gets 1, just in case someone builds a really screwed up country that flanks the other country).
In my test map, a close to equally sliced octagon, the borders showed up correctly (after I reversed the positions of country 0 and 1 in the event the angle was over 90 and less then or equal 180). Worked without a flaw, but in other maps it doesn't work very well.
Everything but the country allocation works well, so I'm curious if anyone knows of a better way to figure out which point is to the left or a spun line, or a better conceptual way to handle this.
That above is basically when I'm doing, red being left right being blue, then I'm just checking 3 different spots then weighing in the lefts and rights found with even/odding it into the appropriate countries (one at +/- .01, the other at +/- .1 and the third 1, in case of even/odd rounding issues with closeness).
I then flip them if I find that country A is to the right, as it is on the left according to the angles I had draw. (my shader renders left first and right second, hence why I do this).
which way is left/right on a line?
From last edit is this not your case. Why not use dot product?
So if the line goes in -x direction the result is negative and if in the +x direction then the result is positive. if the result is zero that means the line goes up or down only or it is juts a point. If you need specific direction instead of left/right then use appropriate a vector instead of x axis.
dot(a,b)=a.x*b.x+a.y*b.y in 2D
dot(a,b)=a.x*b.x+a.y*b.y+a.z*b.z in 3D
Image is relevant for cases where a vector is in unit size in that case the result of dot is perpendicular projection of b into a just like on image
on which side is some point?
I think this is what you need.
As you can see if you handle line (P0,P1) and point P you want to classify as triangle then its polygon winding determines also the side of the line. So for implicit axis directions:
CW(clockwise) polygon winding means right side of the line
CCW(counter-clockwise) polygon winding means left side of the line
How to get winding? ... simply compute normal vector and get its Z coordinate. Its polarity (sign) determines winding (CW/CCW or the other way around depends on the coordinate system).
normal vector is computed as cross product of the two vertices of triangle (P1-P0)x(P-P1)
No need to compute other axises just the z so:
normal.z = ((P1.x-P0.x)*(P.y-P1.y)) - ((P1.y-P0.y)*(P.x-P1.x))
Now just do if (normal.z<0) ... else ... it should never be zero unless you call it for point on the line or the line is a point ... look here at similar question: Determine rotation direction /toward/ variable point on a circle
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.
How do I make a infinite/repeating world that handles rotation, just like in this game:
http://bloodfromastone.co.uk/retaliation.html
I have coded my rotating moving world by having a hierarchy like this:
Scene
- mainLayer (CCLayer)
- rotationLayer(CCNode)
- positionLayer(CCNode)
The rotationLayer and positionLayer have the same size (4000x4000 px right now).
I rotate the whole world by rotating the rotationLayer, and I move the whole world by moving the positionLayer, so that the player always stays centered on the device screen and it is the world that moves and rotates.
Now I would like to make it so that if the player reaches the bounds of the world (the world is moved so that the worlds bounds gets in to contact with the device screen bounds), then the world is "wrapped" to the opposite bounds so that the world is infinite. If the world did not rotate that would be easy, but now that it does I have no idea how to do this. I am a fool at math and in thinking mathematically, so I need some help here.
Now I do not think I need any cocos2d-iphone related help here. What I need is some way to calculate if my player is outside the bounds of the world, and then some way to calculate what new position I must give the world to wrap the world.
I think I have to calculate a radius for a circle that will be my foundry inside the square world, that no matter what angle the square world is in, will ensure that the visible rectangle (the screen) will always be inside the bounds of the world square. And then I need a way to calculate if the visible rectangle bounds are outside the bounds circle, and if so I need a way to calculate the new opposite position in the bounds circle to move the world to. So to illustrate I have added 5 images.
Visible rectangle well inside bounds circle inside a rotated square world:
Top of visible rectangle hitting bounds circle inside a rotated square world:
Rotated square world moved to opposite vertical position so that bottom of visible rectangle now hitting bounds circle inside rotated world:
Another example of top of visible rectangle hitting bounds circle inside a rotated square world to illustrate a different scenario:
And again rotated square world moved to opposite vertical position so that bottom of visible rectangle now hitting bounds circle inside rotated world:
Moving the positionLayer in a non-rotated situation is the math that I did figure out, as I said I can figure this one out as long as the world does not get rotate, but it does. The world/CCNode (positionLayer) that gets moved/positioned is inside a world/CCNode (rotationLayer) that gets rotated. The anchor point for the rotationLayer that rotates is on the center of screen always, but as the positionLayer that gets moved is inside the rotating rotationLayer it gets rotated around the rotationLayer's anchor point. And then I am lost... When I e.g. move the positionLayer down enough so that its top border hits the top of the screen I need to wrap that positionLayer as JohnPS describes but not so simple, I need it to wrap in a vector based on the rotation of the rotationLayer CCNode. This I do not know how to do.
Thank you
Søren
Like John said, the easiest thing to do is to build a torus world. Imagine that your ship is a point on the surface of the donut and it can only move on the surface. Say you are located at the point where the two circles (red and purple in the picture) intersect:
.
If you follow those circles you'll end up where you started. Also, notice that, no matter how you move on the surface, there is no way you're going to reach an "edge". The surface of the torus has no such thing, which is why it's useful to use as an infinite 2D world. The other reason it's useful is because the equations are quite simple. You specify where on the torus you are by two angles: the angle you travel from the "origin" on the purple circle to find the red circle and the angle you travel on the red circle to find the point you are interested in. Both those angles wrap at 360 degrees. Let's call the two angles theta and phi. They are your ship's coordinates in the world, and what you change when you change velocities, etc. You basically use them as your x and y, except you have to make sure to always use the modulus when you change them (your world will only be 360 degrees in each direction, it will then wrap around).
Suppose now that your ship is at coordinates (theta_ship,phi_ship) and has orientation gamma_ship. You want to draw a square window with the ship at its center and length/width equal to some percentage n of the whole world (say you only want to see a quarter of the world at a time, then you'd set n = sqrt(1/4) = 1/2 and have the length and width of the window set to n*2*pi = pi). To do this you need a function that takes a point represented in the screen coordinates (x and y) and spits out a point in the world coordinates (theta and phi). For example, if you asked it what part of the world corresponds to (0,0) it should return back the coordinates of the ship (theta_ship,phi_ship). If the orientation of the ship is zero (x and y will be aligned with theta and phi) then some coordinate (x_0,y_0) will correspond to (theta_ship+k*x_0, phi_ship+k*y_0), where k is some scaling factor related to how much of the world one can see in a screen and the boundaries on x and y. The rotation by gamma_ship introduces a little bit of trig, detailed in the function below. See the picture for exact definitions of the quantities.
!Blue is the screen coordinate system, red is the world coordinate system and the configuration variables (the things that describe where in the world the ship is). The object
represented in world coordinates is green.
The coordinate transformation function might look something like this:
# takes a screen coordinate and returns a world coordinate
function screen2world(x,y)
# this is the angle between the (x,y) vector and the center of the screen
alpha = atan2(x,y);
radius = sqrt(x^2 + y^2); # and the distance to the center of the screen
# this takes into account the rotation of the ship with respect to the torus coords
beta = alpha - pi/2 + gamma_ship;
# find the coordinates
theta = theta_ship + n*radius*cos(beta)/(2*pi);
phi = phi_ship + n*radius*sin(beta)/(2*pi));
# return the answer, making sure it is between 0 and 2pi
return (theta%(2*pi),phi%(2*pi))
and that's pretty much it, I think. The math is just some relatively easy trig, you should make a little drawing to convince yourself that it's right. Alternatively you can get the same answer in a somewhat more automated fashion by using rotations matrices and their bigger brother, rigid body transformations (the special Euclidian group SE(2)). For the latter, I suggest reading the first few chapters of Murray, Li, Sastry, which is free online.
If you want to do the opposite (go from world coordinates to screen coordinates) you'd have to do more or less the same thing, but in reverse:
beta = atan2(phi-phi_ship, theta-theta_ship);
radius = 2*pi*(theta-theta_ship)/(n*cos(beta));
alpha = beta + pi/2 - gamma_ship;
x = radius*cos(alpha);
y = radius*sin(alpha);
You need to define what you want "opposite bounds" to mean. For 2-dimensional examples see Fundamental polygon. There are 4 ways that you can map the sides of a square to the other sides, and you get a sphere, real projective plane, Klein bottle, or torus. The classic arcade game Asteroids actually has a torus playing surface.
The idea is you need glue each of your boundary points to some other boundary point that will make sense and be consistent.
If your world is truly 3-dimensional (not just 3-D on a 2-D surface map), then I think your task becomes considerably more difficult to determine how you want to glue your edges together--your edges are now surfaces embedded in the 3-D world.
Edit:
Say you have a 2-D map and want to wrap around like in Asteroids.
If the map is 1000x1000 units, x=0 is the left border of the map, x=999 the right border, and you are looking to the right and see 20 units ahead. Then at x=995 you want to see up to 1015, but this is off the right side of the map, so 1015 should become 15.
If you are at x=5 and look to the left 20 units, then you see x=-15 which you really want to be 985.
To get these numbers (always between 0 and 999) when you are looking past the border of your map you need to use the modulo operator.
new_x = x % 1000; // in many programming languages
When x is negative each programming language handles the result of x % 1000 differently. It can even be implementation defined. i.e. it will not always be positive (between 0 and 999), so using this would be safer:
new_x = (x + 1000) % 1000; // result 0 to 999, when x >= -1000
So every time you move or change view you need to recompute the coordinates of your position and coordinates of anything in your view. You apply this operation to get back a coordinate on the map for both x and y coordinates.
I'm new to Cocos2d, but I think I can give it a try on helping you with the geometry calculation issue, since, as you said, it's not a framework question.
I'd start off by setting the anchor point of every layer you're using in the visual center of them all.
Then let's agree on the assumption that the first part to touch the edge will always be a corner.
In case you just want to check IF it's inside the circle, just check if all the four edges are inside the circle.
In case you want to know which edge is touching the circumference of the circle, just check for the one that is the furthest from point x=0 y=0, since the anchor will be at the center.
If you have a reason for not putting the anchor in the middle, you can use the same logic, just as long as you include half of the width of each object on everything.
Im making a game in xna where a tank has to move over a landscape.
I need to be able find the bottom of the tank when it is rotated so I can make it move up and down as the player goes over the landscape.
for example if i have a sprite at with its top left corner at 400,300 and i rotate it around its center by 45 degrees around its center, how do i find the new locations of the bottom track.
Thanks
Thanks for the reply Langaurd.
I have looked at the article link before but didnt understand how it works.
Im making a 2d side scrolling game. As the player moves left and right, the tank has to also tilt to follow the contour of the terrain.
I have two vectors that store the back bottom of the track and one that stores the front bottom of the track.
I have tried
Vector2 backBottom = new Vector2(5, 25);
Vector2 frontBottom = new Vector2(5, 32);
backBottom = Vector2.Transform(backBottom+position, Matrix.CreateRotationZ(angle));
frontBottom = Vector2.Transform(frontBottom+position, Matrix.CreateRotationZ(angle));
but that gave me some strange values
Not 100% clear on exactly what it is you are trying to do. You mention a sprite, which is 2D, but your description is in 3D terms. If you are doing a 2D side view, then you can't tell the tank is rotated 45 degrees. If you are doing a 2D top-down view, then you shouldn't really care where the bottom of the tred is.
In any case, two suggestions. If you are die-hard on tracking rotated pixels, then read this article: 2D collision with Transformed Pixels from the creators.xna.com site. However I would recommend tracking vectors. Use two vectors to represent the track locations, and then use Vector2.Transform to rotate them with the tank. You could then use the vectors to check to see if the tracks have hit something, what angle they are at, ect.
You need to define a clearer orientation for you sprite. I would use a Front and Up Vector for the tank. Now you rotate both of them together with the angle your tank drives depending on the terrain. Lets say these vectors are at the center of your sprite. and your sprite is rotated, exactly like your up and front vectors. Now just multiply your Halfheight with -Up vector and you should have your local bottom center, add your tank position and you have your world bottom track position.
Important: Don't mix up a point, which can be expressed by a vector, or a real vector which has no position and only shows the direction. For directions its important to normalize the vector.
Sorry for the vague answer but you question is a little bit vague too.
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 :)