So i have this vector in for a rating system in PaintCode
I would like to be able to fill the colors partially. Like this:
Can this be accomplished in PaintCode?
Yes, it’s possible.
How did I do that:
Made the stars as a clipping shape for a simple rectangle in the background.
Created a numeric variable Stars that is supposed to be from 0 to 5. This will be a parameter of the canvas:
- (void)drawRatingWithStars: (CGFloat)stars;
Created an expression variable Width like 100 / 5 * stars, where 100 is the width of all 5 stars.
Attached this Width variable to the width of the rectangle.
Edit: The GIF animation was exported from PaintCode using a helper fraction variable Animation, which I animated from 0 to 1 (in GIF export settings). I then changed Stars to be an expression with this content 5 - 10 * abs(animation - 0.5). That maps numbers in 0..1 to 0..5..0.
Related
I've noticed that (by changing column width) that the column width measured in points is not proportional to the pixel size. For example, at 21.44 points the pixel width of a column is 200. But at 20 pixels the width becomes 1.44 points, not the expected 2.14 points.
This is very confusing as I'm trying to write a code in VBA which will divide a particular size in 'n' different columns of equal size. Can anyone explain this abnormality? How can I write a code to divide the width (since the parameters for the column width are in points)?
Thanks
So I just was trying things and stumbled across this.
Maybe the numbers are off from a "true" width. If the theory is correct, then there must exist such an offset.
(21.44 + x) = 10 (1.44 + x)
x = 0.7822
Now let's see if this offset works for other some other lengths. For 80 pixels, the length mentioned by MS Excel is 8.11 points. Thus the true length is 8.89. The true length of a column with a width 20 pixels is 1.44 + 0.7822 = 2.222. Note that 2.222 * 4 = 8.89 approx. And this works for some other numbers as well, so I guess the theory should be correct.
Thus to answer the question, add the offset 0.7822 to the observed column width that you need to divide. Then divide it by 'n'. Subtract the offset to obtain the length 'x'. Then use the command Columns(var).ColumnWidth = x for each of the n columns
This is a geometrical question based on a programming problem I have. Basically, I have a MySQL database full of latitude and longitude points, spaced out to be 1km from each other, corresponding to a population of people who live within the square kilometer around each point. I then want to know the relative fraction of each of those grids taken up by a circle of arbitrary size that overlaps them, so I can figure out how many people roughly live within a given circle.
Here is a practical example of one form of the problem (distances not to scale):
I am interested in knowing the population of people who live within a radius of point X. My database figures out that its entries for points A and B are close enough to point X to be relevant. Point A in this example is something like 40.7458, -74.0375, and point B is something like 40.7458, -74.0292. Each of those green lines from A and B to its grid edge represents 0.5 km, so that the gray circle around A and B each represent 1 km^2 respectively.
Point X is at around 40.744, -74.032, and has a radius (in purple) of 0.05 km.
Now I can easily calculate the red lines shown using geographic trig functions. So I know that the line AX is about .504 km, and the distance line BX is about .309 km, for whatever that gets me.
So my question is thus: what's a solid way for calculating the fraction of grid A and the fraction of grid B taken up by the purple circle inscribed around X?
Ultimately I will be taking the population totals and multiplying them by this fraction. So in this case, the 1 km^2 grid around corresponds to 9561 people, and the grid around B is 10763 people. So if I knew (just hypothetically) that the radius around X covered 1% of the area of A and 3% of the area of B, I could make a reasonable back-of-the-envelope estimate of the total population covered by that circle by multiplying the A and B populations by their respective fractions and just summing them.
I've only done it with two squares above, but depending on the size of the radius (which can be arbitrary), there may be a whole host of possible squares, like so, making it a more general problem:
In some cases, where it is easy to figure out that the square grid in question is 100% encompassed by the radius, it is in principle pretty easy (e.g. if the distance between AX was smaller than the radius around X, I know I don't have to do any further math).
Now, it's easy enough to figure out which points are within the range of the circle. But I'm a little stuck on figuring out what fractions of their corresponding areas are.
Thank you for your help.
I ended up coming up with what worked out to be a pretty good approximate solution, I think. Here is how it looks in PHP:
//$p is an array of latitude, longitude, value, and distance from the centerpoint
//$cx,$cy are the lat/lon of the center point, $cr is the radius of the circle
//$pdist is the distance from each node to its edge (in this case, .5 km, since it is a 1km x 1km grid)
function sum_circle($p, $cx, $cy, $cr, $pdist) {
$total = 0; //initialize the total
$hyp = sqrt(($pdist*$pdist)+($pdist*$pdist)); //hypotenuse of distance
for($i=0; $i<count($p); $i++) { //cycle over all points
$px = $p[$i][0]; //x value of point
$py = $p[$i][1]; //y value of point
$pv = $p[$i][2]; //associated value of point (e.g. population)
$dist = $p[$i][3]; //calculated distance of point coordinate to centerpoint
//first, the easy case — items that are well outside the maximum distance
if($dist>$cr+$hyp) { //if the distance is greater than circle radius plus the hypoteneuse
$per = 0; //then use 0% of its associated value
} else if($dist+$hyp<=$cr) { //other easy case - completely inside circle (distance + hypotenuse <= radius)
$per = 1; //then use 100% of its associated value
} else { //the edge cases
$mx = ($cx-$px); $my = ($cy-$py); //calculate the angle of the difference
$theta = abs(rad2deg(atan2($my,$mx)));
$theta = abs((($theta + 89) % 90 + 90) % 90 - 89); //reduce it to a positive degree between 0 and 90
$tf = abs(1-($theta/45)); //this basically makes it so that if the angle is close to 45, it returns 0,
//if it is close to 0 or 90, it returns 1
$hyp_adjust = ($hyp*(1-$tf)+($pdist*$tf)); //now we create a mixed value that is weighted by whether the
//hypotenuse or the distance between cells should be used
$per = ($cr-$dist+$hyp_adjust)/100; //lastly, we use the above numbers to estimate what percentage of
//the square associated with the centerpoint is covered
if($per>1) $per = 1; //normalize for over 100% or under 0%
if($per<0) $per = 0;
}
$total+=$per*$pv; //add the value multiplied by the percentage to the total
}
return $total;
}
This seems to work and is pretty fast (even though it does use some trig on the edge cases). The basic logic is that when calculating edge cases, the two extreme possibilities is that the circle radius is either exactly perpendicular to the grid, or exactly at 45 degree angles from it. So it figures out roughly where between those extremes it falls and then uses that to figure out roughly what percentage of the grid square is covered. It gives plausible results in my testing.
For the size of the squares and circles I am using, this seems to be adequate?
I wrote a little application in Processing.js to try and help me work this out. Without explaining all of it, you can see how the algorithm is "thinking" by looking at this screenshot:
Basically, if the circle is yellow it means it has already figured out it is 100% in, and if it is red it is already quickly screened as 100% out. The other cases are the edge cases. The number (ranging from 0 to 1) under the dot is the (rounded) percentage of coverage calculated using the above method, while the number under that is the calculated theta value used in the above code.
For my purposes I think this approximation is workable.
With enough classification (sketched below) all computations can be reduced to a primitive calculation, the one that provides the angular area of the orange region depicted in the image
When y0 > 0, as illustrated above, and regardless of whether x0 is positive or negative, the orange area can be calculated accurately as the integral from x0 to x1 of sqrt(r^2 - y^2) minus the rectangular area (x1 - x0) * (y1 - y0). The integral has a well known closed expression and therefore there is no need to use any numerical algorithm for calculating it.
Other intersections between a circle and a square can be reduced to a combination of rectangles and right-angular shapes as the one painted in orange above. For instance, an intersection delimited by the horizontal and vertical orange rays in the following picture can be expressed by summing the area of the red rectangle plus two angular shapes: the blue and the green.
The blue area results from a direct application of the primitive case identified above (where the inferior rectangle collapses to nothing.) The green one can also be measured in the same way, once the negative y coordinate is replaced by its absolute value (the other y being 0).
Applying these ideas one could enumerate all cases. Basically, one should consider the case where just one, two, three or four corners of the square lie inside the circle, while the remaining (if any) fall outside. The enumeration is a problem in itself, but it can be solved, at least in theory, by considering a relatively small number of "typical" configurations.
For each of the cases enumerated as described a decomposition on some few rectangles and angular areas has to be calculated and the parts added up (or subtracted) as shown in the three-color example above. The area of every part would reduce to rectangular or primitive angular areas.
A considerably amount of work has to be done to turn this line of attack into a working algorithm. A deeper analysis could shed some light on how to minimize the number of "typical" configurations to consider. If not, I think that the amount of combinations to consider, however large, should be manageable.
In case your problem admits an approximate answer there is another technique you could use which is much simpler to program. The whole idea of this problem reduces to calculate the area of the intersection of a square and a circle. I didn't explain this in my other answer, but finding the squares that are likely to intercept the circle shouldn't be a problem, otherwise, let us know.
The idea of calculating the approximate area of the intersection is very simple. Generate enough points in the square at random and check how many of them belong in the circle. The ratio between the number of points in the circle and the total number of random points in the square will give you the proportion of the intersection with respect to the square's area.
Now, given that you have to repeat the same routine for all squares surrounding the circle (i.e., squares which center has a distance to the circle's center not very different from the circle's radius) you could re-use the random points by translating them from one square to the other.
I don't want to go into details if this method is not appropriate for your problem, so let me just indicate that generating random points uniformly distributed in the square is fairly easy. You only need to generate random numbers for the x coordinate and, independently, random numbers for y. Then just consider all pairs (x, y). Then, for every (x, y) verify whether (x - a)^2 + (y - b)^2 <= r^2 or not, where (a, b) stands for the circle's center and r for the radius.
I insert a shape of type rounded rectangular callout at run time and initially its pointer's direction is downward.If i check its value using
shp.Adjustments.Item(1)
code then it shows its value -0.20833
I want to change its direction to upward.I recorded a macro to adjust that and got this code
Selection.ShapeRange.Adjustments.Item(1) = -0.20223
But still its direction is downward.
Please help to set its direction to upward.
I would suggest just rotating the shape 180 degrees on the y axis. This rotates the shape, but as you note it also rotates the text which may not be desirable.
shp.ShapeRange.ThreeD.RotationY = -180
If you simply want to adjust the pointer from the callout, this is Item(2). When I create a shape, it has a value like:
shp.Adjustments.Item(2) = 0.625
To reverse its location, so that it is on the top of the rectangle, change it to:
shp.Adjustments.Item(2) = -0.625
I was wondering if someone could help with explaining in simple terms what interpolation is and how its used in 3d computer graphics
Simply put: given two points A and B, find a point between them.
For example, if I want to move something along a line from a position x=1 to x=4 in one step:
1-----------------------4
The first step is at location 1, the second step is at location 4, so the object moves instantly from one location to the other. However, if I want the object to take a certain amount of time or number of frames to make the transition, I'll need to refine that by finding intermediate points that are evenly spaced.
If I want the object to take two steps (or frames) to move from 1 to 4,
1-----------X-----------4
I need to calculate what the new point (X) is so I can draw the object there at the appropriate time. In this case, the point X will be
(max-min)
location = min + (current_step) * --------
steps
location is what we're trying to find. min=1, max=4, and in this example steps=2 since we want to divide the span into two steps:
step: location:
0 1
1 2.5
2 4
1------------(2.5)-----------4
If we want to take 4 steps:
step: location:
0 1
1 1.75
2 2.5
3 3.25
4 4
1---(1.75)---(2.5)---(3.25)---4
And so forth. For four steps, the object moves 25% of the total distance per frame. For 10 steps, 10%, etc ad nauseum.
For multiple dimensions (when an object has a 2- or 3-dimensional trajectory), just apply this to each X,Y,Z axis independently.
This is linear interpolation. There are other kinds. As always, Google can help you out.
Other applications include texture mapping, anti-aliasing, image smoothing and scaling, etc., and of course many other uses outside of games and graphics.
Note: a lot of frameworks already provide this. In XNA, for instance, it's Matrix.Lerp.
Interpolation is the smooth adjustment from one thing to another. It is used in animation.
For example, if an object is at location 1, and we want to move it to location 2 over the course of six seconds, we need to slowly interpolate its location between the two endpoints. Interpolation also refers to any search for a location on that path.
Interpolation is the 'guessing' of points based on other points.
for example when you have the points (0,0) and (2,2) you might 'guess' that the point (1,1) also belongs to the set.
The simples application is to deduce a line from two points.
The same thing works in 3 or actually n-dimension.
In 3D graphics it will be used
for animations, to calculate the position of things based on start and end coordinations
calculating lines
gradients
scaling of graphics
and probably many more
General Definition
Interpolation (in mathematics) can be regarded as a transition from one value to another. Interpolation usually uses a value in the 0 to 1 range like a percentage. 0 is the starting value and 1 is the end value. The main purpose of interpolation is to find values in between given values.
Types of Interpolation
There are many types of interpolation used in various programs, the most common being linear interpolation. This type of interpolation is the most simple and straight-forward; It is used to find values in a line segment between two points or numbers. There are also: cubic interpolation, quadratic interpolation, bilinear, trilinear, etc. For more information go here: https://en.wikipedia.org/wiki/Interpolation.
Application in 3D Graphics
Interpolation, especially linear, bilinear and trilinear, is important for computing fragments in geometry (the textures and visuals of the geometry), blending volumetric textures, mip-mapping (a depth of field effect on texture), and lighting (like unreal engine's volumetric lightmaps). The results of the interpolation may vary, but it could potentially yield very realistic results. It is a rather large computation, especially when the interpolation is in 3-dimensions or above (hyperspace).
Example of Interpolation
In 1 Dimension:
n1 = 1
n2 = 2
i = 0.5
n3 = (n1 - n1 * i) + n2 * i
///////////////////////////////////////
n3
├────────┼────────┼────────┼────────┤
1 1.25 1.5 1.75 2
///////////////////////////////////////
In 2 Dimensions:
v1 = {1, 1}
v2 = {1.5, 2}
i = 0.5
d = √((v1.x - v2.x)^2 + (v1.y - v2.y)^2)
v3 = {v1.x + -d * i * ((v1.x - v2.x) / d),v1.y + -d * i * ((v1.y - v2.y) / d)}
///////////////////////////////
2 ┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼
┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼
┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼
┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼
┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼ v2
1.5 ┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─●
┼─┼─┼─┼─┼─┼─┼v3─┼─┼─┼─┼─┼
┼─┼─┼─┼─┼─┼─●─┼─┼─┼─┼─┼─┼
┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼
┼v1─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼
●─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼─┼
1 1.5 2
///////////////////////////////
Basically, I have a context where I can't programatically tint an image, though I can change it's alpha value. With some experimentation, I've found that I can layer a red, blue, and green version of the image, with specific alpha values, to produce a wide range of colors. However, I am wondering if it's possible to achieve a true RGB representation through this method? If so, what is the formula for converting an RGB value into different alpha values for the red, blue, and green layers.
The basic "equation" of alpha combination is:
alpha * (R1,G1,B1) + (1-alpha) * (R2,G2,B2)
When you have three layers with alpha you are actually combining 4 layers (the 4th one is black) so the final color is:
alpha1 * (R1,G1,B1) + (1-alpha1) * (
alpha2 * (R2,G2,B2) + (1-alpha2) * (
alpha3 * (R3,G3,B3) + (1-alpha2) * (0,0,0) ) )
Provided you have the same image on each layer and layer1 is the red channel (G1=B1=0) and layer2 is green and layer3 is blue you get:
(alpha1 * R, (1-alpha1)*alpha2 * G, (1-alpha1)*(1-alpha2)*alpha3 * B)
For a white pixel you can do any possible color. For a black pixel you cannot do anything but black. For any other pixel, you are restricted by the values of R, G and B.
Say you wanted to achieve (Rd, Gd, Bd) at a pixel where the current color is (R, G, B) then you would have to choose:
alpha1 = Rd/R
alpha2 = Gd/(G*(1-alpha1))
alpha3 = Bd/(B*(1-alpha1)*(1-alpha2))
The problem is that alpha can typically only be between 0 and 1. So, for example, if Rd > R there is nothing you can do.
You can do better if you can control the blending function (for example, in Photoshop).
I don't think that's possible, if I understand you correctly.
If, for a certain pixel, your image's red value is, say, 0.5, you can combine that with an alpha in the (typical) range [0,1] to form any value up to and including 0.5, but you can't go above, to get e.g. 0.6 or so as the output value.
If you're looking to create 3 layers that blended together add up to the original image, it's quite possible. Here's a link to a Python script that calls functions in Paint Shop Pro to do the hard parts.
http://pixelnook.home.comcast.net/~pixelnook/SplitToRGB.htm
If you need more detail than that, leave a comment and I'll try to fill it in later.