I have a set of 3D normal vectors for points on a 3D mesh, and I need to calculate the slope of the area below each of them. I have no idea how to do this. I don't need X or Y slope, I just need the total incline of the point in question (although to be fair, I don't know how to derive total slope from X and Y slope individually, which is part of my problem). I did see This article, but I couldn't really make heads or tails of it... The vectors are outward-facing. If anyone can explain this one to me, I'd be really grateful.
If you already have the normal vector, you're almost there. What you now need is the angle (look for dot product) between the normal and a vertical line (what exactly vertical means depends on your application).
If your normal vectors are actually normalized (have length 1) and the vertical is (0 0 1), the cosine of the slope angle is simply the z coordinate of the normal vector.
To demonstrate this: Take a pen and let it stand on your table. This is your table's normal vector. The angle between this vector and a vertical line is zero, as your table has no slope at all. If you tilt your table by a certain amount, the angle between the normal and a vertical line will increase by the same amount.
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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.
If there is a point that is given in x,y co-ordinate, we can pass infinite number of straight lines from it by varying slope a and intercept b.
But when we fix our x and y and vary our a and b in a,b co-ordinate we get a single unique line. Why?
I found this annoying while studying Hough transform.
I don't fully understand what you want to ask....but yes this may clear things a bit. When you fix a point you can make infinite lines pass through it.This is obvious. Now you fix the a,b as co-ordinates that is you have fixed for one particular case the slope and the y-intercept both.
Now note that the slope can vary between a definite range. If you consider the argument, it will vary from 0 degrees to 180 degrees. now out of the infinite lines having different slopes that pass through a point (in this case the point on the y-axis i.e. the y-intercept of your line) you are essentially selecting one line in particular having a specific slope. now only one such line is possible.
I hope this answers your question...
Okay first I wasn't sure if this was better suited to the MathSO so apologies if it needs migrating.
I have a 3D grid of points (representing the centers of voxels) with pitch varying in each dimension, but regular. For example resolution may be 100 by 50 by 40 for a cube shaped object.
Giving me nVox = 200,000.
For each voxel - I would like to cast (nVox - 1) rays, ending at the center, and originating from each of the other voxels.
Now there is obviously a lot of overlap here but I am having trouble finding how to calculate the minimum set of rays required. This sounds like a problem that has an elegant solution, I am however struggling to find it.
As a start, it is obvious that you only need to compute
[nVox * (nVox - 1)] / 2
of the rays, as the other half will simply be in the opposite directions. It is also easy in the 2D case to combine all of those parallel to one of the grid axes (and the two diagonals).
So how do I find the minimum set of rays I need, to pass from all voxel centers, to all others?
If someone could point me in the right direction that'd be great. Any and all help will be much appreciated.
Your problem really isn't about three dimensions in any specific way. All the conceptual complexity is present in the two dimensional case.
Instead of connecting points individually, think about the set of lines that pass through at least two points on your grid. Thus instead of thinking about points initially, think about directions. For 2-D these directions are slopes of lines. These slopes have to be rational numbers, since they intersect points on an integer lattice. Since you have a finite lattice, the numerator and the denominator of the slope can be bounded by the size of the figure. So your underlying problem is enumerating possible slopes for rational numbers of bounded "height" (math jargon).
There's an algorithm for that. It's the one used to generate the Farey sequence of reduced fractions. If your figure is N pixels wide, there will (in general) be a slope with denominator N in the somewhere, but there can't be a slope in reduced form with denominator >N; it wouldn't fit.
It's easier to deal with slopes between 0 and 1 directly. You get the other directions by two operations: negating the slope and by interchanging axes. For three dimensions, you need two slopes to define a direction.
Given an arbitrary direction (no necessarily a rational one as above), there's a perpendicular linear space of dimension k-1; for 3-D that's a plane. Projecting a 3-D parallelpiped onto this plane yields a hexagon in general; two vertices project onto the interior, six project to the vertices of the hexagon.
For a given discrete direction, there's a minimal bounding box on the integer lattice such that two opposite vertices lie along that direction. As long as that bounding box fits within your original grid, each of the interior points of the projection each correspond to a line that intersects your grid in at least two points.
In summary, enumerate directions, then for each direction enumerate where that direction intersects your grid in at least two points.
From this threa Determine the centroid of multiple points I came to know that area of polygon can also be negative if we start in clockwise direction. Why it can be negative?
It is a product of the maths. You can use the sign if you wish to, or use an absolute value for the area.
You often get a similar effect with dot products and cross products. This can be effective, for example determining the orientation of a polygon in 3d (does the 'outside' side of the polygon face towards me or away from me?)
The sign tells you some useful information, that you can either use or discard. For example, what is the area below the curve sin(x) and above the x axis, for x over the interval [0,pi]. Yes, this is simply a definite integral. In MATLAB, I'd do it as:
>> quad(#sin,0,pi)
ans =
2
But suppose I computed that same definite integral, with limits of integration [pi,0]? Clearly, we would get -2.
>> quad(#sin,pi,0)
ans =
-2
And of course this makes sense. In either case, we can assure that we get the positive area by ignoring the sign. But the sign tells us something in that integral.
The sign computed for the area of a polygon is indeed useful in some problems. In the case of a triangle, a cross product will yield a vector that points in the direction orthogonal to the plane of the triangle containing the vectors. The magnitude of the vector will be twice the area of that triangle. Note that this vector can point in one of two directions orthogonal to a given plane, which one is indicated by the right hand rule. You can think of that sign as indicating which direction the vector pointed.
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)