I have a component that changes its location based on other elements. I'm trying to find its x and y position at different intervals, so I tried compname.x and compname.y.
The x position seems to be working, but the y position is always 0. I'm guessing I need to play with localToGlobal or contentToGlobal or one of those conversions. Is that the problem?
A component's x and y values are relative to it's parent. Flex calls this the content coordinate system; and the contentToGlobal should give you the answers you need.
Read up on positioning components, which explains the content, local, and viewable coordinates:
http://livedocs.adobe.com/flex/3/html/help.html?content=size_position_2.html
And read up on the Flex coordinate system:
http://livedocs.adobe.com/flex/3/html/help.html?content=containers_intro_5.html#254752
If you had a working example it may be easier to give a specific answer.
...also, have a look at getBounds() - this method returns you the position relative to other display objects.
simon
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How to get the sill height (height above the floor) of a Ifcwindow in ifc file
There is no solution to get the height above the floor directly. This is because the height above for depends on several factors, like how the wall is created in which the window resides, etc.
It could be that the sillHeight is exported by the original modelling software to a custom IFC property. You could check for that, but since there is no common standard for it, it's risky.
Your best bet is to look into the ObjectPlacement property which IfcWindow inherits from IfcProduct. The ObjectPlacement defines how a product is placed either in world space or relative to its host. See https://standards.buildingsmart.org/IFC/RELEASE/IFC4/ADD2/HTML/schema/templates/product-local-placement.htm for details.
You need to read the ObjectPlacement property, and check if there is a RelativeTo property, if so, you need to fill into that property as well, and check if it's the placement of a floor. If so, you can stop the looping, and perform a matrix calculation on all the placements you harvested to calculate the placement of window relative to floor.
(Maybe even more simple: calculate world placement of window and floor separately, than subtract the two vector z values to get the height of window from floor)
I am not sure how to put this problem in a single sentence, sorry if the title is misleading.
I am currently developing a simple terrain editor with a circle-shaped brush size. The image below shows a few cases that represent my problem.
additional info: the square size is fixed and uniform and in the current version, my concern is only to find which one is hit and which one is not (the amount of region covered is important for weighting the hit, but probably not right now)
My current solution (which is not even correct for a certain condition) is: given a hit in a position (x, y) with radius r, loop through all square from (x-radius, y-radius) to (x+radius, y+radius) and apply 2-D box to circle collision detection. But I don't think this is optimal (or even correct IMO).
Can anyone help me with this one? Thank you
Since i can't add a simple comment due to bureaucracy on this website i have to type it out here.
Anyway you're in luck since i was trying to do this recently as well! The way i did it is i iterated through the vertex array and check if the current vertex falls inside the radius of the circle. But perhaps what you want is to check it against each quad center and if that center falls inside the radius then add the whole quad as it's being collided.
Of course depending on the size of your grid the performance will vary so it's good to try to iterate through as few quads as needed. Though accessing these quads from the array is something you have to figure out yourself.
A simple question as i am developing a java application based on dcm4che ...
I want to calculate/find the "position" of a dicom image into its sequence (series). By position i mean to find if this image is first, second etc. in its series. More specifically i would like to calculate/find:
Number of slices into a Sequence
Position of each slice (dicom image) into the Sequence
For the first question i know i can use tag 0020,1002 (however it is not always populated) ... For the second one?
If you are dealing with volumetric image series, best way to order your series is to use the Image Position (Patient) (0020, 0032). This is a required Type 1 tag (should always have value) and it is part of the image plane module. It will contain the X, Y and Z values coordinates representing the upper left corner of the image in mm. If the slices are parallel to each other, only one value should change between the slices.
Please note that the Slice Location (0020, 1041) is an optional (Type 3) element and it may not exist in the DICOM file.
We use the InstanceNumber tag (0x0020, 0x0013) as our first choice for the slice position. If there is no InstanceNumber, or if they are all the same, then we use the SliceLocation tag (0x0020, 0x1041). If neither tag is available, then we give up.
We check the InstanceNumber tag such that the Max(InstanceNumber) - Min(InstanceNumber) + 1 is equal to the number of slices we have in the sequence (just in case some manufacturers start counting at 0 or 1, or even some other number). We check the SliceLocation the same way.
This max - min + 1 is then the number of slices in the sequence (substitute for tag ImagesInAcquisition 0x0020, 0x1002).
Without the ImagesInAcquisition tag, we have no way of knowing in advance how many slices to expect...
I would argue that if the slice location is available, use that. It will be more consistent with the image acquisition. If it is not available, then you'll have to use or compute from the image position (patient) attribute. Part 3 section C.7.6.2.1 has details on these attributes.
The main issue comes when you have a series that is oblique. If you just use the z-value of the image position (patient), it may not change by the slice thickenss/spacing between slices attributes, while the slice location typically will. That can cause confusion to end users.
Two.js seems to only support a single value for a scale, not x,y components. Is it possible to just stretch something horizontally?
Yes, under the hood every Two.Group and Two.Polygon has a _matrix object. You can set _matrix.manual and then do a number of other transformations that you can't do with the typical two.js API. Check out this example for a demonstration.
I'm doing some image processing, and I need to find some information on line growing algorithms - not sure if I'm using the right terminology here, so please call me out on this is needs be.
Imagine my input image is simply a circle on a black background. I'd basically like extract the coordinates, so that I may draw this circle elsewhere based on the coordinates.
Note: I am already using edge detection image filters, but I thought it best to explain with a simple example.
Basically what I'm looking to do is detect lines in an image, and store the result in a data type where by I have say a class called Line, and various different Point objects (containing X/Y coordinates).
class Line
{
Point points[];
}
class Point
{
int X, Y;
}
And this is how I'd like to use it...
Line line;
for each pixel in image
{
if pixel should be added to line
{
add pixel coordinates to line;
}
}
I have no idea how to approach this as you can probably establish, so pointers to any subject matter would be greatly appreciated.
I'm not sure if I'm interpreting you right, but the standard way is to use a Hough transform. It's a two step process:
From the given image, determine whether each pixel is an edge pixel (this process creates a new "binary" image). A standard way to do this is Canny edge-detection.
Using the binary image of edge pixels, apply the Hough transform. The basic idea is: for each edge pixel, compute all lines through it, and then take the lines that went through the most edge pixels.
Edit: apparently you're looking for the boundary. Here's how you do that.
Recall that the Canny edge detector actually gives you a gradient also (not just the magnitude). So if you pick an edge pixel and follow along (or against) that vector, you'll find the next edge pixel. Keep going until you don't hit an edge pixel anymore, and there's your boundary.
What you are talking about is not an easy problem! I have found that this website is very helpful in image processing: http://homepages.inf.ed.ac.uk/rbf/HIPR2/wksheets.htm
One thing to try is the Hough Transform, which detects shapes in an image. Mind you, it's not easy to figure out.
For edge detection, the best is Canny edge detection, also a non-trivial task to implement.
Assuming the following is true:
Your image contains a single shape on a background
You can determine which pixels are background and which pixels are the shape
You only want to grab the boundary of the outside of the shape (this excludes donut-like shapes where you want to trace the inside circle)
You can use a contour tracing algorithm such as the Moore-neighbour algorithm.
Steps:
Find an initial boundary pixel. To do this, start from the bottom-left corner of the image, travel all the way up and if you reach the top, start over at the bottom moving right one pixel and repeat, until you find a shape pixel. Make sure you keep track of the location of the pixel that you were at before you found the shape pixel.
Find the next boundary pixel. Travel clockwise around the last visited boundary pixel, starting from the background pixel you last visited before finding the current boundary pixel.
Repeat step 2 until you revisit first boundary pixel. Once you visit the first boundary pixel a second time, you've traced the entire boundary of the shape and can stop.
You could take a look at http://processing.org/ the project was created to teach the fundamentals of computer programming within a visual context. There is the language, based on java, and an IDE to make 'sketches' in. It is a very good package to quickly work with visual objects and has good examples of things like edge detection that would be useful to you.
Just to echo the answers above you want to do edge detection and Hough transform.
Note that a Hough transform for a circle is slightly tricky (you are solving for 3 parameters, x,y,radius) you might want to just use a library like openCV