I have a grass texture:
I use it in my 2d-game. I want to animate it by code, without any predefined animations.
The grass should interact with wind. So when the wind is stronger, the grass should stoop into need side more.
First version of animation I made using sinusoid function, but such animation is a bit ugly, because the base of the grass moves left/right like all another part of picture. And with sinusoid I'm not able to regulate stoop of the image.
Any advices?
This is what you may get easily by shifting pixels:
Although probably not very useful for you, here is a Mathematica program:
f[l_, sh_] := Module[{c = l, k = (Dimensions#l)[[1]]},
For[i = 1, i <= k, i++,
c[[i]] = RotateRight[l[[i]], IntegerPart[(k + 1 - i)/sh]]];
Return[c];]
b = ArrayPad[ImageData#a, {{40}, {40}, {0}}, {1, 1, 1}];
Export["c:\\anim.gif",
{Image#b, Image#f[b, 7],
Image#f[b, 5], Image#f[b, 3],
Image#f[b, 5], Image#f[b, 7], Image#b}, "DisplayDurations" -> .3]
Edit
But you can get a better effect by applying a circular transform:
b = Image#ArrayPad[ImageData#a, {{40, 0}, {40}, {0}}, {1, 1, 1}];
f[image_, angleMult_] := ImageForwardTransformation[image, (
fi = ArcTan[Abs[#[[2]]/(#[[1]] - .5)]];
fi1 = angleMult fi (#[[1]]^2 + #[[2]]^2)/2;
{(1/2 - Sin[fi1] #[[2]] - Cos[fi1]/2 +
Cos[fi1] #[[1]]), -Sin[fi1]/2 + Sin[fi1] #[[1]] +
Cos[fi1] #[[2]]}) &]
t = Table[f[b, x], {x, 0, .2, .02}];
t1 = Reverse#t;
Export["c:\\anim.gif", Join[t, t1], "DisplayDurations" -> .15];
Import["c:\\anim.gif", "Animation"]
You could just shift the rows, so that e.g. every 3rd row is shifted 1px to the right, beginning with the bottom.
How are you displaying the texture? When using a billboard you could manipulate the vertices of the billboard and even triangulate the billboard for more control.
Related
I am trying to visualize the result of numerical integration in Wolfram Mathematica using DensityPlot. But dark artifacts appear inside the circle on the graph, this is incorrect. How to fix this?
f[u_, v_]:=(1/1.2)*(1/(3.14159*0.02^2)*E^(-((x-u)^2+(y-v)^2)/0.02^2)+0.2/(3.14159*2^2)*E^(-((x-u)^2+(y-v)^2)/2^2));
i[x_, y_]:=NIntegrate[f[u, v],{u, v} \[Element] Disk[{0,0},2.5], AccuracyGoal -> 30];
DensityPlot[i[x, y], {x, -3, 3}, {y, -3, 3},ColorFunction->"SunsetColors",PlotPoints -> 20, PlotLegends -> Automatic]
Link to the image: https://postimg.cc/47MJTYj7
Well, that is not an artifact, or is it. First I'd like to point out that f is a function of u, v, x and y. So the way it is programmed is not very clean. Now to the main problem. There are two Gaussian peaks. One is very broad, the other very sharp. In many cases the numerical integral just does not get points near the sharp peak, therefore not detecting it. As the rest is smooth and easy, no subset is generated and the integral is just missing this contribution. Luckily, there is an option specifically for this. MinRecursion: NIntegrate may miss sharp peaks of integrands:.... So this works
g[x_, y_, s_] := Exp[ -( x^2 + y^2 ) / s^2] / (s^2 Pi)
f[x_, y_, u_, v_] := g[ x - u, y - v, 0.02 ] / 1.2 + 0.2 g[ x - u, y - v, 2 ]
i[x_, y_] := NIntegrate[ f[x, y, u, v], {u, v} \[Element] Disk[{0, 0}, 2.5], MinRecursion -> 5]
nn = 30;
t = Table[i[x, y], {x, -3, 3, 6/nn}, {y, -3, 3, 6/nn}];
ListDensityPlot[t, ColorFunction -> "SunsetColors", PlotLegends -> Automatic]
Takes quite some time, though. Probably, it would be better to split this up in two integrals and invest some thought on the integration boundaries.
Assuming I have a vector of say four dimensions in which every variable lays in a special interval. Thus we got:
Vector k = (x1,x2,x3,x4) with x1 = (-2,2), x2 = (0,2), x3 = (-4,1), x4 = (-1,1)
I am only interested in the points constraint by the intervals.
So to say v1 = (0,1,2,0) is important where v2 = (-5,-5,5,5) is not.
In additon to that the point i+1 should be relatively close to point i among my journey. Therefore I dont want to jump around in space.
Is there a proper way of walking through those interesting points?
For example in 2D space with x1,x2 = (-2,2) like so:
Note: The frequenz of the red line could be higher
There are many ways to create a space-filling curve while preserving closeness. See the Wikipedia article for a few examples (some have associated algorithms for generating them): https://en.wikipedia.org/wiki/Space-filling_curve
Regardless, let's work with your zig-zag pattern for 2D and work on extending it to 3D and 4D. To extend it into 3D, we just add another zig to the zig-zag. Take a look at the (rough) diagram below:
Essentially, we repeat the pattern that we had in 2D but we now have multiple layers that represent the third dimension. The extra zig that we need to add is the switch between bottom-to-top and top-to-bottom every layer. This is pretty simple to abstract:
In 2D, we have x and y axes.
We move across the x domain switching between positive and negative
directions most frequently.
We move across the y domain once.
In 3D, we have x, y, and z axes.
We move across the x domain switching between positive and negative directions most frequently.
We move across the y domain switching between positive and negative directions second most frequently.
We move across the z domain once.
It should be clear how this generalizes to higher dimensions. Now, I'll present some (Python 3) code that implements the zig-zag pattern for 4D. Let's represent the position in 4D space as (x, y, z, w) and the ranges in each dimension as (x0, x1), (y0, y1), (z0, z1), (w0, w1). These are our inputs. Then, we also define xdir, ydir, and zdir to keep track of the direction of the zig-zag.
x, y, z, w = x0, y0, z0, w0
xdir, ydir, zdir = +1, +1, +1
for iw in range(w1 - w0):
for iz in range(z1 - z0):
for iy in range(y1 - y0):
for ix in range(x1 - x0):
print(x, y, z, w)
x = x + xdir
xdir = -xdir
print(x, y, z, w)
y = y + ydir
ydir = -ydir
print(x, y, z, w)
z = z + zdir
zdir = -zdir
print(x, y, z, w)
w = w + 1
This algorithm has the guarantee that no two points printed out after each other have a distance greater than 1.
Using recursion, you can clean this up to make a very nice generalizable method. I hope this helps; let me know if you have any questions.
With the work of #Matthew Miller I implemented this generalization for any given multidimenisonal space:
'''assuming that we take three points out of our intervals [0,2] for a,b,c
which every one of them is corresponding to one dimension i.e. a 3D-space'''
a = [0,1,2]
b = [0,1,2]
c = [0,1,2]
vec_in = []
vec_in.append(a)
vec_in.append(b)
vec_in.append(c)
result = []
hold = []
dir = [False] * len(vec_in)
def create_points(vec , index, temp, desc):
if (desc):
loop_x = len(vec[index])-1
loop_y = -1
loop_z = -1
else:
loop_x = 0
loop_y = len(vec[index])
loop_z = 1
for i in range(loop_x,loop_y,loop_z):
temp.append(vec[index][i])
if (index < (len(vec) - 1)):
create_points(vec, index + 1, temp, dir[index])
else:
u = []
for k in temp:
u.append(k)
result.append(u)
temp.pop()
if (dir[index] == False):
dir[index] = True
else:
dir[index] = False
if len(temp) != 0:
temp.pop()
#render
create_points(vec_in, 0, hold, dir[0])
for x in (result):
print(x)
The result is a journey which covers every possible postion in a continous way:
[0, 0, 0]
[0, 0, 1]
[0, 0, 2]
[0, 1, 2]
[0, 1, 1]
[0, 1, 0]
[0, 2, 0]
[0, 2, 1]
[0, 2, 2]
[1, 2, 2]
[1, 2, 1]
[1, 2, 0]
[1, 1, 0]
[1, 1, 1]
[1, 1, 2]
[1, 0, 2]
[1, 0, 1]
[1, 0, 0]
[2, 0, 0]
[2, 0, 1]
[2, 0, 2]
[2, 1, 2]
[2, 1, 1]
[2, 1, 0]
[2, 2, 0]
[2, 2, 1]
[2, 2, 2]
Mathematica: Filling an infinitely deep potential well
Comment: The proper page for Mathematica questions is this one
I would like to visualize a potential well for a particle in a box in Mathematica similar to the second picture from Wikipedia here.
I have defined my function piecewise
(*Length of the box*)
L = 4;
(*Infinitly deep potential well between 0 and L*)
V[x_] := Piecewise[{
{\[Infinity], x <= 0},
{0, 0 < x < L},
{\[Infinity], L <= x}}]
and would like to obtain a plot function which gives a filled area where the potential goes to infinity.
Unfortunately my tries end up in shaded areas between the "zero region" of the potential, while I would like to have the shading in the infinity region.
Table[Plot[V[x], {x, -5, 10},
Filling -> f], {f, {Top, Bottom, Axis, 0.3}}]
The problem is that Infinity is too much for plot. So let's just give it some other big number. But to prevent it from rescaling the y axis we need to be specific with the upper plot range
Block[{\[Infinity] = 1*^1},
Plot[V[x], {x, -5, 10}, Filling -> Bottom,
PlotRange -> {Automatic, 1}]
]
Alternatively you could plot V[x]/.\[Infinity]->1*^1 instead of Block but I like Block's way better
Just give it values instead of infinity:
(*Length of the box*)L = 4;
(*Infinitly deep potential well between 0 and L*)
V[x_] := Piecewise[{{1, x <= 0}, {0, 0 < x < L}, {1, L <= x}}]
Plot[V[x], {x, -5, 10}, Filling -> Bottom]
Another way using graphic primitives:
wellLeft = 0;
leftBorder = wellLeft - 1;
rightBorder = L + 1;
wellRight = L;
top = 5;
Graphics[{
Hue[0.67, 0.6, 0.6],
Opacity[0.2],
Rectangle[{leftBorder, 0}, {wellLeft, top}],
Rectangle[{wellRight, 0}, {rightBorder, top}]
}, Axes -> True]
I've gotten stuck getting my euler angles out my rotation matrix.
My conventions are:
Left-handed (x right, z back, y up)
YZX
Left handed angle rotation
My rotation matrix is built up from Euler angles like (from my code):
var xRotationMatrix = $M([
[1, 0, 0, 0],
[0, cx, -sx, 0],
[0, sx, cx, 0],
[0, 0, 0, 1]
]);
var yRotationMatrix = $M([
[ cy, 0, sy, 0],
[ 0, 1, 0, 0],
[-sy, 0, cy, 0],
[ 0, 0, 0, 1]
]);
var zRotationMatrix = $M([
[cz, -sz, 0, 0],
[sz, cz, 0, 0],
[ 0, 0, 1, 0],
[ 0, 0, 0, 1]
]);
Which results in a final rotation matrix as:
R(YZX) = | cy.cz, -cy.sz.cx + sy.sx, cy.sz.sx + sy.cx, 0|
| sz, cz.cx, -cz.sx, 0|
|-sy.cz, sy.sz.cx + cy.sx, -sy.sz.sx + cy.cx, 0|
| 0, 0, 0, 1|
I'm calculating my euler angles back from this matrix using this code:
this.anglesFromMatrix = function(m) {
var y = 0, x = 0, z = 0;
if (m.e(2, 1) > 0.999) {
y = Math.atan2(m.e(1, 3), m.e(3, 3));
z = Math.PI / 2;
x = 0;
} else if (m.e(2, 1) < -0.999) {
y = Math.atan2(m.e(1, 3), m.e(3, 3));
z = -Math.PI / 2;
x = 0;
} else {
y = Math.atan2(-m.e(3, 1), -m.e(1, 1));
x = Math.atan2(-m.e(2, 3), m.e(2, 2));
z = Math.asin(m.e(2, 1));
}
return {theta: this.deg(x), phi: this.deg(y), psi: this.deg(z)};
};
I've done the maths backwards and forwards a few times, but I can't see what's wrong. Any help would hugely appreciated.
Your matrix and euler angles aren't consistent. It looks like you should be using
y = Math.atan2(-m.e(3, 1), m.e(1, 1));
instead of
y = Math.atan2(-m.e(3, 1), -m.e(1, 1));
for the general case (the else branch).
I said "looks like" because -- what language is this? I'm assuming you have the indexing correct for this language. Are you sure about atan2? There is no single convention for atan2. In some programming languages the sine term is the first argument, in others, the cosine term is the first argument.
The last and most important branch of the anglesFromMatrix function has a small sign error but otherwise works correctly. Use
y = Math.atan2(-m.e(3, 1), m.e(1, 1))
since only m.e(3, 1) of m.e(1, 1) = cy.cz and m.e(3, 1) = -sy.cz should be inverted. I haven't checked the other branches for errors.
Beware that since sz = m.e(2, 1) has two solutions, the angles (x, y, z) used to construct the matrix m might not be the same as the angles (rx, ry, rz) returned by anglesFromMatrix(m). Instead we can test that the matrix rm constructed from (rx, ry, rz) does indeed equal m.
I worked on this problem extensively to come up with the correct angles for a given matrix. The problem in the math comes from the inability to determine a precise value for the SIN since -SIN(x) = SIN(-x) and this will affect the other values of the matrix. The solution I came up with comes up with two equally valid solutions out of eight possible solutions. I used a standard Z . Y . X matrix form but it should be adaptable to any matrix. Start by findng the three angles from: X = atan(m32,m33): Y = -asin(m31) : Z = atan(m21,m11) : Then create angles X' = -sign(X)*PI+X : Y'= sign(Y)*PI-Y : Z = -sign(Z)*pi+Z . Using these angles create eight set of angle groups : XYZ : X'YZ : XYZ' : X'YZ' : X'Y'Z' : XY'Z' : X'Y'Z : XY'Z
Use these set to create the eight corresponding matrixes. Then do a sum of the difference between the unknown matrix and each matrix. This is a sum of each element of the unknown minus the same element of the test matrix. After doing this, two of the sums will be zero and those matrixes will represent the solution angles to the original matrix. This works for all possible angle combinations including 0's. As 0's are introduced, more of the eight test matrixes become valid. At 0,0,0 they all become idenity matrixes!
Hope this helps, it worked very well for my application.
Bruce
update
After finding problems with Y = -90 or 90 degrees in the solution above. I came up with this solution that seems to reproduce the matrix at all values!
X = if(or(m31=1,m31=-1),0,atan(m33+1e-24,m32))
Y = -asin(m31)
Z = if(or(m31=1,m31=-1),-atan2(m22,m12),atan2(m11+1e-24,m21))
I went the long way around to find this solution, but it wa very enlightening :o)
Hope this helps!
Bruce
How can I write the code in mathematica to see the result like this:
As you see we have the complex function w=f(z), where z=x+iy and w=u+iv.
In this example w=sin z, and we see the image of vertical line x=c is hyperbola. (Left)
and the image of horizontal line y=c is an elliptic. (Right)
This picture took from the book "Complex Variables and Applications, by James Ward Brown, Ruel Vance Churchill", 8th edition: pages 331 and 333 or third edition pages 96-97
Something like this?
ClearAll[u, v, f];
f[z_] := Sin[z]
u[x_, y_] := Re#f[x + I*y];
v[x_, y_] := Im#f[x + I*y];
EDIT: This just produces the whole thing. If you want to just see what happens for a single path parallel to the imaginary axis, try
ParametricPlot[{u[5, y], v[5, y]}, {y, -3, 3}]
or for the same parallel to the real axis try
ParametricPlot[{u[x, 1], v[x, 1]}, {x, -3, 3}]
EDIT2: Interactive:
ClearAll[u, v, f];
f[z_] := Sin[z]
u[x_, y_] := Re#f[x + I*y];
v[x_, y_] := Im#f[x + I*y];
Manipulate[
Show[
Graphics[{Line[{p1, p2}]}, PlotRange \[Rule] 3, Axes \[Rule] True],
ParametricPlot[
{u[p1[[1]] + t (p2[[1]] - p1[[1]]),
p1[[2]] + t (p2[[2]] - p1[[2]])],
v[p1[[1]] + t (p2[[1]] - p1[[1]]),
p1[[2]] + t (p2[[2]] - p1[[2]])]},
{t, 0, 1},
PlotRange \[Rule] 3]],
{{p1, {0, 1}}, Locator},
{{p2, {1, 2}}, Locator}]
(ugly, yes, but no time to fix it now). Typical output:
or
The idea is that you can vary the line on the left hand side of the figures you give (by clicking around the plot, which amounts to clicking on the Argand diagram...) and see the corresponding images.
Depending on what you want to do with the representations, it might sometimes be helpful to visualize the Riemann surface in 3D. Here's the surface for w=sin(z) in 3D, neatly showing the branch cuts and the different branches (same as acl's first plot, but in 3D).
ParametricPlot3D[
Evaluate[{Re#Sin[z], Im#Sin[z], y} /. z -> x + I y], {x, -2,
2}, {y, -2, 2}]