This may be ridiculously obvious, but math wasn't my strong suit in school. I've been banging my head against the wall long enough that I finally figured I'd ask.
I'm trying to animate a sprite moving along a 2D parabolic path, from point A to point B. Both points are at the same y-coordinate. The desired height of the parabola from the starting/ending y-coordinate is also given (or, if you prefer, a desired velocity). Currently in my code I have a timer firing at a high frequency. I would like to calculate the new location of the ball based on the amount of time that has passed. So a parametric parabola equation should work nicely.
I found this answer from GameDev adequate, until my requirements grew (although I'm not sure its really a parabolic path... I can't follow the derivation of the final equations there provided). Now I would like to squish/stretch the sprite at different points along the parabolic path. But to get the effect to work right, I'll need to rotate the sprite so that it's primary axis is tangential to the path. So I need to be able to derive the angle of the tangent at any given location/time.
I can find all sorts of equations for each of these requirements (parametric parabola, tangent at a point, etc.), but I just can't figure out how to unify them all. Could someone with more math skills help a fellow coder out and provide a set of equations that will work? Thanks ever so much in advance.
What you are missing is this:
Slope = TAN(angle) // in radians
What is the slope? It is how much up/down you move per how much across you move ( dy/dx on some other answers ). For you it is actually (dy/dt)/(dx/dt) since both x and y are functions of time.
So for a trajectory x(t)=Vx*t and y(t)=Vy*t-1/2*g*t^2 the slope is Slope=(Vy-g*t)/Vx where Vx is the initial horizontal velocity, and Vy the initial vertical velocity. g is the gravity (vertical acceleration down). So your rotation in degrees shall be
angle = ATAN( (Vy-g*t)/Vx ) * 180/PI
Basically the slope is equal to the ratio of vertical velocity to horizontal velocity.
Let X be the distance from A to B, Y the desired height of the parabola, V the horizontal speed.
x = Vt
y = Y - (4Y/X^2) (X/2-Vt)^2
tangent dy/dx = (8Y/X^2) (X/2-Vt)
Related
I recently had to convert euler rotation rates to vectorial angular velocity.
From what I understand, in a local referential, we can express the vectorial angular velocity by:
R = [rollRate, pitchRate, yawRate] (which is the correct order relative to the referential I want to use).
I also know that we can convert angular velocities to rotations (quaternion) for a given time-step via:
alpha = |R| * ts
nR = R / |R| * sin(alpha) <-- normalize and multiply each element by sin(alpha)
Q = [nRx i, nRy j, nRz k, cos(alpha)]
When I test this for each axis individually, I find results that I totally expect (i.e. 90°pitch/time-unit for 1 time unit => 90° pitch angle).
When I use two axes for my rotation rates however, I don't fully understand the results:
For example, if I use rollRate = 0, pitchRate = 90, yawRate = 90, apply the rotation for a given time-step and convert the resulting quaternion back to euler, I obtain the following results:
(ts = 0.1) Roll: 0.712676, Pitch: 8.96267, Yaw: 9.07438
(ts = 0.5) Roll: 21.058, Pitch: 39.3148, Yaw: 54.9771
(ts = 1.0) Roll: 76.2033, Pitch: 34.2386, Yaw: 137.111
I Understand that a "smooth" continuous rotation might change the roll component mid way.
What I don't understand however is after a full unit of time with a 90°/time-unit pitchRate combined with a 90°/time-unit yawRate I end up with these pitch and yaw angles and why I still have roll (I would have expected them to end up at [0°, 90°, 90°].
I am pretty confident on both my axis + angle to quaternion and on my quaternion to euler formulas as I've tested these extensively (both via unit-testing and via field testing), I'm not sure however about the euler rotation rate to angular-velocity "conversion".
My first bet would be that I do not understand how euler rotation-rates axes interacts on themselves, my second would be that this "conversion" between euler rotation-rates and angular velocity vector is incorrect.
Euler angles are not good way of representing arbitrary angular movement. Its just a simplification used for graphics,games and robotics. They got some pretty hard restrictions like your rotations consist of only N perpendicular axises in ND space. That is not how rotation works in real world. On top of this spherical representation of reper endpoint it creates a lot of singularities (you know when you cross poles ...).
The rotation movement is analogy for translation:
position speed acceleration
pos = Integral(vel) = Integral(Integral(acc))
ang = Integral(omg) = Integral(Integral(eps))
That in some update timer can be rewritten to this:
vel+=acc*dt; pos+=vel*dt;
omg+=eps*dt; ang+=omg*dt;
where dt is elapsed time (Timer interval).
The problem with rotation is that you can not superimpose it like translation. As each rotation has its own axis (and it does not need to be axis aligned, nor centered) and each rotation affect the axis orientation of all others too so the order of them matters a lot. On top of all this there is also gyroscopic moment creating 3th rotation from any two that has not parallel axis. Put all of this together and suddenly you see Euler angles does not match the real geometrics/physics of rotation. They can describe orientation and fake its rotation up to a degree but do not expect to make real sense once used for physic simulation.
The real simulation would require list of rotations described by the axis (not just direction but also origin), angular speed (and its change) and in each simulation step the recomputing of the axis as it will change (unless only single rotation is present).
This can be done by using coumulative homogenous transform matrices along with incremental rotations.
Sadly the majority of programmers prefers Euler angles and Quaternions simply by not knowing that there are better and simpler options and once they do they stick to Euler angles anyway as matrix math seem to be more complicated to them... That is why most nowadays games have gimbal locks, major rotation errors and glitches, unrealistic physics.
Do not get me wrong they still have their use (liek for example restrict free look for camera etc ... but they missused for stuff they are the worse option to use for.
I am stuck on a particular problem. I am learning on how to create a very basic game, where a ball will travel diagonally from either top left corner of a square or a rectangular down to the bottom right corner in a straight line (As shown in Fig 1 & 2). Now I know that the ball x and y position will both need to be changed frame by frame but I am unsure on how to go about this.
enter image description here
Math is not my strong point and I am unsure how do I calculate the exact route, especially since both the square and rectangle will have a different angles. Are there any math formulas I can use to calculate the diagonal line and by how much each of the x and y coordinates of the ball will need to be adjusted frame by frame.
From the research that I have done I think that I will most likely need to calculate the angle using the sin or cos functions but I am not sure how everything fits together. Have been using https://www.mathsisfun.com/sine-cosine-tangent.html to try and learn more.
I am planning on starting to code this but would really appreciate answers to these basic questions. I am trying to learn both the programming and the mathematical aspect at the same time and I feel that this approach would be the best fit.
Many Thanks for any suggestions/help, I would really appreciate it.
Since it's rectangular, just calculate the slope: rise (Y) / run (X). That will give you how much to increase the object's location in each direction per frame. Depending on how fast or slow you want the object to move, you'll need to apply some modifier to that (e.g., if you want the object to move twice as fast, you'll need to multiple 2 by the change in a particular direction before you actually change the object's location.
For square :
If you are using Frame or JFrame, you have coordinate with you.
You can move ball from left top to right down as follow ->
Suppose ur top left corner is at (0,0), add 1in both coordinate until you reach right bottom corner.
U can do this using for loop
You don't technically need the angle for this mapping. You know that the formula for a line is "y = m * x + b." I presume that you can calculate m,b. If not, let me know.
Given that - you can simply increment x based on anything you like (timer, event, etc. ). You can place your incremented x into the equation above to get your respective y.
Now, that won't be quite enough as you are dealing with pixels instead of actual numbers. For example, lest assume that in your game x/y are in feet. You will need to know how many pixels represent a foot. Then when you draw to the screen you adjust your coordinates by dividing by pixels per foot.
So...
1. Calculate your m and b for your path.
2. Use a timer. At each tick, adjust your x value
3. Use your x value to calculate your y value
4. Divide x and y by a scaling number
5. Use the new scaled x and y to plot your object
Now...There are all kinds of tricks you can play with the math, but that should get you started.
Let's left bottom corner of rectangle (or square) has coordinates (x0, y0), and right top corner (x1, y1). Then diagonal has equation
X(t) = x0 + t * (x1 - x0)
Y(t) = y0 + t * (y1 - y0)
where t is parameter in range 0..1. Point at t=0 corresponds to (x0, y0), at t=1 to (x1, y1), and at t=0.5 - to the center of rectangle. So all you need is vary parameter t and calculate position.
When your object will move with constant absolute speed in arbitrary direction, use horizontal and vertical components of velocity vx and vy. At every moment get coordinates as x_new = x_old + delta_time * vx. Note that reflection from vertical edge just changes horizontal component of velocity 'vx = - vx' and so on.
Yeah, yeah, I checked out the suggested questions/answers that were given to me but most involved quaternions, or had symbols in them that I don't even HAVE on my keyboard.
I failed at high school trig, and while I understand the basic concepts of sin and cos in 2D space, I'm at a loss when throwing in a third plane to deal with.
Basically, I have these things: centerpoint, distance, and angles for each of the three axes. Given that information, I want to calculate the point that is -distance- away from the center point, at the specified angles.
I'm not sure I'm explaining this correctly. My intent is to get what amounts to electrons orbiting around a nucleus, if anyone happens to know how to do that. I am working with Java, JRE 6, if there are any utility classes in there that can help.
I don't want just an answer, but also the how and why of the answer. If I'm going to learn something, i want to learn ABOUT it as well. I am not afraid to take a lesson in trigonometry, or how quaternions work, etc. I'm not looking for an entire course on the answer, but at least some basic understanding would be cool.
If you did this in 2D, you would have a point on a plane with certain x and y coordinates. The distance from the origin would be sqrt(x^2+y^2), and the angle atan(y/2).
If you were given angle phi and distance r you would compute x= r*cos(phi); y=r*sin(phi);
To do this in three dimensions you need two angles - angle in XY plane and angle relative to Z axis. Calling these phi and theta, you compute coordinates as
X = r*cos(phi)*sin(theta);
Y = r*sin(phi)*sin(theta);
Z = r*cos(theta);
When I have a chance I will make a sketch to show how that works.
I know there are plenty of questions about 3d rotation that have been answered here but all of them seem to deal with rotational matrices and quaternions in OpenGL (and I don't really care if I get gimbal lock). I need to get 3d coordinates EX:(x,y,z) of a point that always must be the same distance, I'll call it "d" for now, from the origin. The only information I have as input is the deltax and deltay of the mouse across the screen. So far here is what I have tried:
First:
thetaxz+=(omousex-mouseX)/( width );
thetaxy+=(omousey-mouseY)/( height);
(thetaxy is the angle in radians on the x,y axis and thetaxz on the x,z axis)
(I limit both angles so that if they are less than or equal to 0 they equal 2*PI)
Second:
pointX=cos(thetaxz)*d;
pointY=sin(thetaxy)*d;
(pointX is the point's x coordinate and pointY is the y)
Third:
if(thetaxz)<PI){
pointZ=sqrt(sq(d)-sq(eyeX/d)-sq(eyeY/d));
}else{
pointZ=-sqrt(abs(sq(d)-sq(eyeX/d)-sq(eyeY/d)));
}
(sq() is a function that squares and abs() is an absolute value function)
(pointZ should be the point's z coordinate and it is except at crossing between the positive z hemisphere and negative z hemisphere. As it approaches the edge the point gets stretched further than the distance that it is always supposed to be at in the x and y and seemingly randomly around 0.1-0.2 radians of thetaxz the z coordinate becomes NAN or undefined)
I have thought about this for awhile, and truthfully I'm having difficulty warping my head around the concept of quaternions and rotational matrices however if you can show me how to use them to generate actual coordinates I would be glad to learn. I would still prefer it if I could just use some trigonometry in a few axis. Thank you in advance for any help and if you need more information please just ask.
Hint/last minute idea: I think it may have something to do with the z position affecting the x and y positions back but I am not sure.
EDIT: I drew a diagram:
If you truly want any success in this, you're going to have to bite the bullet and learn about rotation matrices and / or quaternion rotations. There may be other ways to do what you want, but rotation matrices and quaternion rotations are used simply because they are widely understood and among the simplest means of expressing and applying rotations to vectors. Any other representation somebody can come up with will probably be a more complex reformulation of one or both of these. In fact it can be shown rotation is a linear transformation and so can be expressed as a matrix. Quaternion rotations are just a simplified means of rotating vectors in 3D, and therefore have equivalent matrix representations.
That said, it sounds like you're interested in grabbing an object in your scene with a mouse click and rotating in a natural sort of way. If that's the case, you should look at the ArcBall method (there are numerous examples you may want to look over). This still requires you know something of quaternions. You will also find that an at least minimal comprehension of the basic aspects of linear algebra will be helpful.
Update: Based on your diagram and the comments it contains, it looks like all you are really trying to do is to convert Spherical Coordinates to Cartesian Coordinates. As long as we agree on the the notation, that's easy. Let θ be the angle you're calling XY, that is, the angle between the X axis rotated about the Z axis; this is called the azimuth angle and will be in the range [0, 2π) radians or [0°, 360°). Let Φ be an angle between the XY plane and your vector; this is called the elevation angle and will be in the range [-π/2, +π/2] or [-90°, +90°] and it corresponds to the angle you're calling the XZ angle (rotation in the XZ plane about the Y axis). There are other conventions, so make sure you're consistent. Anyway, the conversion is simply:
x = d∙cos(Φ)∙cos(θ)
y = d∙cos(Φ)∙sin(θ)
z = d∙sin(Φ)
Okay I'm working on a Space sim and as most space sims I need to work out where the opponents ship will be (the 3d position) when my bullet reaches it. How do I calculate this from the velocity that bullets travel at and the velocity of the opponents ship?
Calculate the relative velocity vector between him and yourself: this could be considered his movement if you were standing still. Calculate his relative distance vector. Now you know that he is already D away and is moving V each time unit. You have V' to calculate, and you know it's length but not it's direction.
Now you are constructing a triangle with these two constraints, his V and your bullet's V'. In two dimensions it'd look like:
Dx+Vx*t = V'x*t
Dy+Vy*t = V'y*t
V'x^2 + V'y^2 = C^2
Which simplifies to:
(Dx/t+Vx)^2 + (Dy/t+Vx)^2 = C^2
And you can use the quadratic formula to solve that. You can apply this technique in three dimensions similarly. There are other ways to solve this, but this is just simple algebra instead of vector calculus.
Collision Detection by Kurt Miller
http://www.gamespp.com/algorithms/collisionDetection.html
Add the negative velocity of the ship to the bullet, so that only the bullet moves. Then calculate the intersection of the ship's shape and the line along which the bullet travels (*pos --> pos + vel * dt*).
The question probably shouldn't be "where the ship will be when the bullet hits it," but IF the bullet hits it. Assuming linear trajectory and constant velocity, calculate the intersection of the two vectors, one representing the projectile path and another representing that of the ship. You can then determine the time that each object (ship and bullet) reach that point by dividing the distance from the original position to the intersection position by the velocity of each. If the times match, you have a collision and the location at which it occurs.
If you need more precise collision detection, you can use something like a simple BSP tree which will give you not only a fast way to determine collisions, but what surface the collision occurred on and, if handled correctly, the exact 3d location of the collision. However, it can be challenging to maintain such a tree in a dynamic environment.