I'm trying to draw a vertical scrollbar for my G15 applet, but am having difficulty positioning it properly (if you haven't done anything for the G15 LCD screen, think of it as drawing on a 160x43 pixel image).
This is my current code for positioning:
perc = (float)Math.Round( range / Items.Count+1 );
y = ( perc * SelectedID+1 );
The upper edge of the scrollbar is at 5px from the top, the bottom edge is at 32px.
Y in this case would be the upper end of the scrollbar, and I'm using a length of 2 pixels; I did try to impliment a variable length bar, it went about as well as the code above.
SelectionID is 0 based.
All I need is the math to figure out the position, no need for code for drawing it.
Thanks.
So you're just after straightforward linear interpolation, right?
As in you have a value c in the range a..b and you need a resulting value in the range x..y based on its linear position between a and b?
The equation for this situation is (assuming the numbers in question are floats or doubles):
// c is between a and b
pos = (c-a)/(b-a) // pos is between 0 and 1
result = pos * (y-x) + x // result is between x and y
Now if everything is 0-indexed, you can omit a and x to get
pos = c/b
result = pos * y
If any of the numbers you're working with are integer types, you'll need to cast them to doubles or floats (or any real number type) before the divisions.
You can combine all of the equations together if you can't cast anything to doubles, though:
result = (c * y) / b
This will guarantee that c and y are multiplied together before the integer division takes place, which will reduce the error associated with integer division.
Related
Being not very good at math, I have a problem with my project.
The objective is boundary correction on 3D files.
In my application, the user moves a 3D point on X-axis in order to correct or modify the boundary of the object.
I want to move the nearest boundary points in the same direction but decreasingly. I mean no point should move more than the main point. The nearest points move most and, the farthest points should move less.
On the image, the red dots represent the initial status of points. And the user pulls the P0 in the x-direction. And the other points follow it. The last status of the points is represented by violet dots.
Here is what I tried.
//On point moved event
//Get nearest boundary Points (Uses Geometry3D to get boundary points).
(var clothDMesh, _) = Utilities3D.BuildDMesh(baseMesh);
CreateClothModel(clothDMesh);
var boundryVertices = nodes.Where(ro => ro.Value.isBorder).Select(ro => ro.Value.vertex).ToList();
var refPoint = CustomPoint.FromPoint3D(movedPoint);
//Gets the delta X.
var deltaX = p.X - initialX;
//Gets nearest country points, so 15 points above and 15 points below to move only a given number of points (I know maybe this should be calculated according to delta).
var nearestPoints = refPoint.GetNearesPoints(boundryVertices, 30);
foreach (var item in nearestPoints)
{
//This is only one of what I tried as a function. None of them worked correctly.
item.X += deltaX - (deltaX * 1/ Math.Pow(item.Distance, 2));
}
Any help will be appreciated.
Thanks in advance.
Here's the math part:
I call "a" your "deltaX".
We also need a second parameter: "b", the maximum height of the red dots. I assume it is symetrical and "-b" would be the minimum height of the red dots.
So, if you look for the value X, horizontal move, in fonction of the coordinate Y of the dot:
X = a - a * Y * Y / (b * b);
You can verify that for Y = 0, you obtain X = a and for Y = b (or -b) you get X = 0.
You have your parabola (X is function of Y^2).
Well, I'm continuing this question without answer (Smoothing random noises with different amplitudes) and I have another question.
I have opted to use the contour/shadow of a shape (Translating/transforming? list of points from its center with an offset/distance).
This contour/shadow is bigger than the current path. I used this repository (https://github.com/n-yoda/unity-vertex-effects) to recreate the shadow. And this works pretty well, except for one fact.
To know the height of all points (obtained by this shadow algorithm (Line 13 of ModifiedShadow.cs & Line 69 of CircleOutline.cs)) I get the distance of the current point to the center and I divide between the maximum distance to the center:
float dist = orig.Max(v => (v - Center).magnitude);
foreach Point in poly --> float d = 1f - (Center - p).magnitude / dist;
Where orig is the entire list of points obtained by the shadow algorithm.
D is the height of the shadow.
But the problem is obvious I get a perfect circle:
In red and black to see the contrast:
And this is not what I want:
As you can see this not a perfect gradient. Let's explain what's happening.
I use this library to generate noises: https://github.com/Auburns/FastNoise_CSharp
Note: If you want to know what I use to get noises with different amplitude: Smoothing random noises with different amplitudes (see first block of code), to see this in action, see this repo
Green background color represent noises with a mean height of -0.25 and an amplitude of 0.3
White background color represent noises with a mean height of 0 and an amplitude of 0.1
Red means 1 (total interpolation for noises corresponding to white pixels)
Black means 0 (total interpolation for noises corresponding to green pixels)
That's why we have this output:
Actually, I have tried comparing distances of each individual point to the center, but this output a weird and unexpected result.
Actually, I don't know what to try...
The problem is that the lerp percentage (e.g., from high/low or "red" to "black" in your visualization) is only a function of the point's distance from the center, which is divided by a constant (which happens to be the maximum distance of any point from the center). That's why it appears circular.
For instance, the centermost point on the left side of the polygon might be 300 pixels away from the center, while the centermost point on the right might be 5 pixels. Both need to be red, but basing it off of 0 distance from center = red won't have either be red, and basing it off the min distance from center = red will only have red on the right side.
The relevant minimum and maximum distances will change depending on where the point is
One alternative method is for each point: find the closest white pixel, and find the closest green pixel, (or, the closest shadow pixel that is adjacent to green/white, such as here). Then, choose your redness depending on how the distances compare between those two points and the current point.
Therefore, you could do this (pseudo-C#):
foreach pixel p in shadow_region {
// technically, closest shadow pixel which is adjacent to x Pixel:
float closestGreen_distance = +inf;
float closestWhite_distance = +inf;
// Possibly: find all shadow-adjacent pixels prior to the outer loop
// and cache them. Then, you only have to loop through those pixels.
foreach pixel p2 in shadow {
float p2Dist = (p-p2).magnitude;
if (p2 is adjacent to green) {
if (p2Dist < closestGreen_distance) {
closestGreen_distance = p2Dist;
}
}
if (p2 is adjacent to white) {
if (p2Dist < closestWhite_distance) {
closestWhite_distance = p2Dist;
}
}
}
float d = 1f - closestWhite_distance / (closestWhite_distance + closestGreen_distance)
}
Using the code you've posted in the comments, this might look like:
foreach (Point p in value)
{
float minOuterDistance = outerPoints.Min(p2 => (p - p2).magnitude);
float minInnerDistance = innerPoints.Min(p2 => (p - p2).magnitude);
float d = 1f - minInnerDistance / (minInnerDistance + minOuterDistance);
Color32? colorValue = func?.Invoke(p.x, p.y, d);
if (colorValue.HasValue)
target[F.P(p.x, p.y, width, height)] = colorValue.Value;
}
The above part was chosen for the solution. The below part, mentioned as another option, turned out to be unnecessary.
If you can't determine if a shadow pixel is adjacent to white/green, here's an alternative that only requires the calculation of the normals of each vertex in your pink (original) outline.
Create outer "yellow" vertices by going to each pink vertex and following its normal outward. Create inner "blue" vertices by going to each pink vertex and following its normal inward.
Then, when looping through each pixel in the shadow, loop through the yellow vertices to get your "closest to green" and through the blue to get "closest to white".
The problem is that since your shapes aren't fully convex, these projected blue and yellow outlines might be inside-out in some places, so you would need to deal with that somehow. I'm having trouble determining an exact method of dealing with that but here's what I have so far:
One step is to ignore any blues/yellows that have outward-normals that point towards the current shadow pixel.
However, if the current pixel is inside of a point where the yellow/blue shape is inside out, I'm not sure how to proceed. There might be something to ignoring blue/yellow vertexes that are closer to the closest pink vertex than they should be.
extremely rough pseudocode:
list yellow_vertex_list = new list
list blue_vertex_list = new list
foreach pink vertex p:
given float dist;
vertex yellowvertex = new vertex(p+normal*dist)
vertex bluevertex = new vertex(p-normal*dist)
yellow_vertex_list.add(yellowvertex)
blue_vertex_list.add(bluevertex)
create shadow
for each pixel p in shadow:
foreach vertex v in blue_vertex_list
if v.normal points towards v: break;
if v is the wrong side of inside-out region: break;
if v is closest so far:
closest_blue = v
closest_blue_dist = (v-p).magnitude
foreach vertex v in yellow_vertex_list
if v.normal points towards v break;
if v is the wrong side of inside-out region: break;
if v is closest so far:
closest_yellow = v
closest_yellow_dist = (v-p).magnitude
float d = 1f - closest_blue_dist / (closest_blue_dist + closest_yellow_dist)
How can i calulate a valid range (RED) for my object's (BLACK) traveling direction (GREEN). The green is a Vector2 where x and y range is -1 to 1.
What I'm trying to do here is to create rocket fuel burn effekt. So what i got is
rocket speed (float)
rocket direction (Vector2 x = [-1, 1], y = [-1, 1])
I may think that rocket speed does not matter as fuel burn effect (particle) is created on position with its own speed.
A cheap and cheerful trick with 2D vectors is to transpose the x and y, then flip the sign on one of them to get the perpendicular vector (pseudo code):
Vector2 perpendicular ( -original.y, original.x ) // Or original.y, -original.x
Then you could do something like:
direction + perpendicular * rand(-0.3 , 0.3)
Update: having realised the question asks for the opposite vector (too busy looking at the picture!) I figure I had better answer that too. Multiply 'direction' by -1 to get the opposite vector. So this:
perpendicular * rand(-0.3 , 0.3) - direction
should give you a random direction vector somewhere in your range (not normalised, but close enough for these purposes). Then you can multiply that result by a random number depending on how long you want the tail.
If to expend upon OlduwanSteve's answer, you can make is such that it's somewhat physically accurate.
You want to create several vectors that will represent the expulsion (the red lines).
First define the number of vectors you want to represent the expulsion with - lets mark it n.
You want to get a set of n numbers which sum up to Vx. These numbers will be the x components of the expulsion vectors. You can do this like so (semi-pseudo code):
SumX = Vx;
for (i = 0; i < n; i++)
{
Ax[i] = -rand(0..SumX); // Ax is the array of all expulsion vectors x components
SumX -= Ax[i];
}
Now you'll want to calculate Ay (the y components of the expulsion vectors). This is quite similar to calculating the, except that SumY = 0.
Here instead of splitting up SumY among n elements, you need to decide a maximal y component. Best way I can think of to select this is to define a maximal allowed angle for the expulsion vectors and define the maximal Vy using: maxVy = minVx*tan(maxAlpha).
Now you can get Ay using this (semi-pseudo code):
SumY = maxVy*2; // The actual range is (-maxVy, maxVy), but using (0, 2*maxVy) is simpler IMO
for (i = 0; i < n; i++)
{
Ay[i] = rand(0..SumY);
SumY -= Ay[i];
}
for (i = 0; i < n; i++)
{
Ay[i] -= maxVy; // Translate the range back to (-maxVy, maxVy) from (0, 2*maxVy)
}
Now you have arrays of both the x and y components of the expulsion vectors. Iterate over both arrays and pair up elements to create the vectors (you don't have to iterate both arrays in the same order).
Notes:
• I align the axes in my calculations such that X is parallel to the objects speed vector (the green line).
• The calculation for maxVy does NOT guarantee that a vector of angle maxAlpha will be produced, it only guarantees that no vector of larger angle will be.
• The lines Ay[i] = rand(0..SumY) and Ax[i] = -rand(0..SumX) may lead to vectors with components of size 0. This may lead to annoying scenarios, I'd recommend to handle away such cases (for instance "while rand returns zero, call it again").
I have a rectangle.
Its height (RH) is 400.
Its width (RW) is 500.
I have circle.
Its height (CH) is 10.
Its width (CW) is 10.
Its starting location (CX1, CY1) is 20, 20.
The circle has moved.
Its new location (CX2, CY2) is 30, 35.
Assuming my circle continues to move in a straight line.
What is the circle's location when its edge reaches the boundary?
Hopefully you can provide a reusable formula.
Perhaps some C# method with a signature like this?
point GetDest(size itemSize, point itemPos1, point itemPos2, size boundarySize)
I need to calculate what that location WILL be once it arrives - knowing that it is not there yet.
Thank you.
PS: I need this because my application is watching the accelerometer on my Windows Phone. I am calculating the target necessary to animate the motion of the circle inside the rectangle as the user is tilting their device.
It is 1 radius away from the boundar(y/ies).
The answer is X=270 Y=395
first define the slope V as dy/dx =(y2-y1)/(x2-x1). In your example: (35-20)/(30-20)=1.5
the line equation is
y = V * (x-x1) + y1. You are interested in the horizontal locations x at:
y= CH/2 OR y= H-CH/2
so (not code, just math)
if (y2-y1)<0:
x=(CH/2 -y1)/V +x1 10 for your example. OR
if (y2-y1)>0:
x=(H-CH/2 -y1)/V +x1 270 for your example
else (that is: y2==y1)
the upper or lower lines were not hit.
if CH/2 <= x <= W-CH/2 the circle did hit the that upper or lower side: since V>0, we use x=270 and that is within CH/2 and W-CH/2.
So the answer to your question is y=H-CH/2 = 395 , X=270
For the side lines it's similar:
(if (x2-x1)<0)
y=(CH/2 -x1)*V +y1
(if (x2-x1)>0)
y=(W-CH/2 -x1)*V +y1
else (that is: x2==x1)
the side lines were not hit.
if CH/2 <= y <= H-CH/2 the circle did hit that side at that y.
be careful with the trivial cases of completely horizontal or vertical movement so that you don't divide by zero. when calculating V or 1/V. Also deal with the case where the circle did not move at all.
Since you now asked, here's metacode which you should easily be able to convert to a real method. It deals with the special cases too. The input is all the variables you listed in your example. I here use just one symbol for the circle size, since it's a circle not an ellipse.
method returning a pair of doubles getzy(x1,y1,W,H,CH){
if (y2!=y1){ // test for hitting upper or lower edges
Vinverse=(x2-x1)/(y2-y1)
if ((y2-y1)<0){
xout=(CH/2 -y1)*Vinverse +x1
if (CH/2 <= y <= H-CH/2) {
yout=CH/2
return xout,yout
}
}
if ((y2-y1)>0){
xout=(H-CH/2 -y1)*Vinverse +x1
if (CH/2 <= y <= H-CH/2) {
yout=H-CH/2
return xout,yout
}
}
}
// reaching here means upper or lower lines were not hit.
if (x2!=x1){ // test for hitting upper or lower edges
V=(y2-y1)/(x2-x1)
if ((x2-x1)<0){
yout=(CH/2 -x1)*V +y1
if (CH/2 <= x <= W-CH/2) {
xout=CH/2
return xout,yout
}
}
if ((x2-x1)>0){
yout=(H-CH/2 -x1)*V +y1
if (CH/2 <= x <= W-CH/2) {
xout=H-CH/2
return xout,yout
}
}
}
// if you reach here that means the circle does not move...
deal with using exceptions or some other way.
}
It's easy; no calculus required.
Your circle has radius R = CW/2 = CH/2, since the diameter of the circle D = CW = CH.
In order to have the circle touch the vertical edge of the rectangle at a tangent point, you have to move the circle to the right by a distance (W - (CX1 + CW/2))
Likewise, the circle will touch the bottom edge of the rectangle at a tangent point when you move it down by a distance (H - (CY1 + CH/2)).
If you do this in two separate translations (e.g., first to the right by the amount given, then down by the amount given or visa versa), you'll see that the circle will touch both the right hand vertical and the bottom horizontal edges at tangent points.
When the moving circle arrives at a wall (boundary) then it will be tangent at one of four points on the circle, call them N, S, E, and W. You know their initial coordinates.
The points travel in a line with a slope known to you: m=(y2-y1)/(x2-x1); where in your example (x1, y1) - (20,20) and (x2, y2)= (30, 35).
Your problem is to find the trajectory of the first point N, S, E, or W which reaches any wall. The trajectory will be a line with slope m.
You can do this by adding (or subtracting) the direction vector for the line to the point N, S, E, or W, scaled by some t.
For example, N is (20, 15). The direction vector is (x2-x1,y2-y1) = (10, 15). Then (20, 15) + t * (10, 15) will hit the boundary lines at different t's. You can solve for these; for example 20 + t*10 = 0, and 20 + t*10 = 400, etc.
The t that is smallest in magnitude, over all four trajectories, gives you your tangent point.
Not sure its calculus..wouldn't it just be the following:
if y >= 390 then it reached the top edge of the rectangle
if x >= 490 then it reached the right edge of the rectangle
if y <= 0 then it reached the bottom edge of the rectangle
if x <= 0 then it reached the left edge of the rectangle
I'm looking to do some complex part analysis within Seimens NX. I'm looking to implement the double caliper method of measuring a model in order to find the minimum possible box that it could possibly fit into(for machining purposes). I've got all of my measurement code in place, but I am completely baffled by the idea of a construct that can randomly output normalized 3x3 vectors for use as coordinate systems. The part is measured with respect to this coordinate system, so each coordinate system gives a unique "minimum part envelope". Once analyzed, the smallest envelope is selected and displayed.
this is the type of vector I am talking about:
1 0 0
0 1 0
0 0 1
numbers can be any value between -1 and 1, with decimals not only being accepted but pretty much required.
and no, this isn't my homework. More of an individual pursuit in my free time at work.
If you apply a rotation matrix to an already orthogonal matrix, then the result should also be orthogonal.
So you can redefine your problem as applying a random rotation matrix to the identity matrix.
Perhaps do one random rotation matrix for each axis (x,y,z) and then apply the matrices themselves in a random order?
If you don't mind to consider only a special subset of the orthogonal matrices, there is an easier way to achieve this, which is to take advantage of the Rodrigues' rotation formula to generate rotation matrices (which has an additional constraint that its determinant is equal to 1).
With this, you only need to generate a random 3x1 unit vector (as the rotation axis) and specify a rotation angle. This formula will transform them into a valid rotation matrix.
MATLAB example:
function R = rot(w, theta)
bw = [0, -w(3), w(2); w(3), 0, -w(1); -w(2), w(1), 0];
R = eye(3) + sin(theta)*bw + (1-cos(theta))*bw*bw;
end
w = rand(3,1)
w = w/norm(w)
R = rot(w, 3.14)
C++ example:
// w: the unit vector indicating the rotation axis
// theta: the rotation angle in radian
Eigen::Matrix3d MatrixExp3 (Eigen::Vector3d w, float theta){
Eigen::Matrix3d bw, R;
bw << 0, -w(2), w(1), w(2), 0, -w(0), -w(1), w(0), 0;
R << Eigen::Matrix3d::Identity() + std::sin(theta)*bw + (1-std::cos(theta))*bw*bw;
return R;
}
int main() {
std::srand((unsigned int) time(0));
Eigen::Vector3d w = Eigen::Vector3d::Random();
Eigen::Matrix3d R = MatrixExp3(w.normalized(), 3.14f);
std::cout << R << std::endl;
}