I am using a special Map SDK for iOS and I am adding a custom shape to the map. The shape is always a different size and it could be a circle, square, star etc. the point being it is always dynamic whenever the app is run.
After adding this shape to the map, I can access it's property called overlayBounds which is described as: This property contains the smallest rectangle that completely encompasses the overlay.
The overlay is my shape that I'm adding to the map.
Whenever a location update is generated by CLLocationManager, I want to check and see if the most recent coordinate is inside of that overlayBounds property of the shape.
When accessing overlayBounds, it has an ne property and a sw property. Both of these are just CLLocationCoordinate2D's
So, if the overlayBounds is made up of two CLLocationCoordinate2D's and the CLLocationManager is always updating the user's location and giving me the most recent coordinate(CLLocationCoordinate2D), how can I check if that most recent coordinate is within the overlayBounds?
After doing a lot of research I have only found one potential solution to go off of which is this: https://stackoverflow.com/a/30434618/3344977
But that answer assumes that my overlayBounds property has 4 coordinates(CLLocationCoordinate2D's), when I only have 2.
Your description seems much harder then the actual question. So if I am getting this correctly your question is only to check if the point is inside the rectangle described in overlayBounds.
You have only 2 points as it is enough to define a rectangle. So NE and SW are the two points where the other two are received as (NE.x, SE.y) and (SE.x, NE.y). With this you may use the answer you linked or you may simply construct a MKMapRect where origin is NE and size is SE-NE. So in this case you may simply use MKMapRectMake and then use MKMapRectContainsPoint. BUT watch out when computing size as SE-NE might produce negative results in which cases you need to add degrees to the size. That is 180 to x (latitude) and 360 to y (longitude)...
MKMapRect rect = MKMapRectMake(NE.latitude, NE.longitude, SE.latitude-NE.latitude, SE.longitude-NE.longitude);
if(rect.width < .0) rect.width += 180.0;
if(rect.height < .0) rect.height += 360.0;
BOOL pointInside = MKMapRectContainsPoint(rect, pointOnMap);
Something like this should do the trick.
Now if you are trying to check if the point is inside the shape itself it really depends on how your shape is defined. If this is some form of analytic representation you might find some method already made for you to return the value but if not then your best shot would most likely be drawing the shape to some canvas and checking the color of canvas at the location you need to check. In any case the bigger problem here is converting the point and the rect to a Cartesian coordinate system. If that is the case then just add a comment and I will try to help you on that...
Related
I'm looking to create a Q-Q plot within Rascal using the Vis library. I have been told there is no positional system. Is this true? If true, how would I go about plotting this or any scatterplot? Does anyone have an example of this in use?
That's an excellent question. Certainly Rascal's Vis library is "point free" in the sense that its layout mechanism has no absolute coordinate system. However, there are certain Figure kinds which have a relative coordinate system wrt their own "origin". When you combine several of those using horizontal, vertical or overlay boxes (and align them properly), you can create the effect of bar charts, scatterplots and whatever you desire.
In particular the overlay Figure composition is interesting: http://tutor.rascal-mpl.org/Rascal/Libraries/Vis/Figure/Figure.html#/Rascal/Libraries/Vis/Figure/Figures/overlay/overlay.html
Figure point(num x, num y){ return ellipse(shrink(0.05),fillColor("red"),align(x,y));}
coords = [<0.0,0.0>,<0.5,0.5>,<0.8,0.5>,<1.0,0.0>];
ovl = overlay([point(x,y) | <x,y> <- coords]);
render(ovl);
Produces this (both code and image taken from the documentation linked above):
Each point is an ellipse which is aligned at the (x, y) position relative to the origin of the enclosing overlay box.
The origin by default of this overlay seems to be the upper-left corner, when no other FProperty's are given to the overlay. It's possible other alignment options for the overlay Figure also change the position of its origin.
With the help of Jurgen Vinju I wrote this code, hope it helps someone: https://gist.github.com/rlmhermans/c9e82a6a623b65f0c6957ab3ff2742cf
So I am working on a project that includes many uses placing annotations all around a map. The annotation, (which is a custom image with a much larger circular range) appears on the screen and, ideally, I would like for a user to be:
Notified if they are within the range of a annotation
and
Not be allowed to place another annotation within the range of another one if the circular pins overlap by, say, more than 25%
I think this is a pretty unique question and should be fun for somebody to help out with, so have fun! Thanks everybody!
You can check the distance from each annotation using
- (CLLocationDistance)distanceFromLocation:(const CLLocation *)location
This method measures the distance between the two locations by tracing
a line between them that follows the curvature of the Earth. The
resulting arc is a smooth curve and does not take into account
specific altitude changes between the two locations.
For more details refer Link
Try this:
let location = CLLocation(latitude: 1, longitude: 1)//Or user's location
let distance = location.distance(from: anotherLocation)
Edit:
As mentioned in the comments, you wanted to create an equidistant point. I suggest manually doing that:
Subtract the annotation's location from he user's location. Then add your distance back to the original one. For example:
The user's location = (1, 1)
The annotation's location = (3, 2)
Vertical difference would be 2
Horizontal difference would be 1
Then:
(3 + 2, 2 + 1)
Your result: (5, 3)
Now you would have two points (the one you just created and the user's location) at each end with a center point (original annotation)
I would like draw a curved line and attach an object to it. Is it possible to create fraction (from 0.0 to 1.0) which makes move my object on the path? When fraction is 0 then object is on the beginning, when 0.5 is on half way and finally when is on 1.0 it is at the end. Of course i want a curved path, not a straight line :) Is it possible to do in PaintCode?
If you need it only as a progress bar, it is possible in PaintCode. The trick is to use dashed stroke with very large Gap and then just change the Dash.
Then just attach a Variable and you are done.
Edit: Regarding the discussion under the original post, this solution uses points as the unit, so it will be distributed equally along the curve, no matter how curved the bezier is.
Based on the fact that you're going to walk along the curve using linear distance, a thing Bezier curves are terrible for, you need to build the linear mapping yourself. That's fairly simple though:
When you draw the curve, also build a look-up table that samples the curve once, at say 100 points (t=0, t=0.01, t=0.02, etc). In pseudocode:
lut = [];
lut[0] = 0;
tlen = curve.length();
for(v=0; v<=100; v++) {
t = v/100;
clen = curve.split(0,t).length();
percent = 100*clen/tlen;
lut[percent] = t;
}
This may leave gaps in your LUT - you can either fix those as a secondary step, or just leave them in and do a binary scan on your array to find the nearest "does have a value" percentage.
Then, when you need to show your progress as some percentage value, you just look up the corresponding t value: say you need to show 83%, you look up lut[83] and draw your object at the value that gives you.
Depending on where the anchorPoint is with UIAttachmentBehavior, the view can be quite rotated, so it's in more of a diamond shape than a square. In these situations, where it's rotated say 90°, how do I find what the lowest or highest point of this UIView is (in relation to the window)?
It's easy enough when the UIView is a (non-rotated) square, as I can just use CGRectGetMaxY (or min) and the x value doesn't matter, and then use convertPoint, but with the rotation the x value seems to have a real importance, as if I choose maxX, it will tell me the bottom right's point, while minX will give me the bottom left's.
I just want the lowest point that exists in the UIView. How do I get this?
EDIT: Here's some code showing how I'm attempting it currently:
CGPoint lowestPoint = CGPointMake(CGRectGetMinX(weakSelf.imageView.bounds), CGRectGetMaxY(weakSelf.imageView.bounds));
CGPoint convertedPoint = [weakSelf.imageView convertPoint:lowestPoint toView:nil];
The tracking of convertedPoint's y value completely changes depending on what I supply for the x value in lowestPoint's CGPointMake.
The view's frame is its bounding box. Normally you should be careful about using the frame of a transformed view, which is precisely what a rotated-by-UIKit-Dynamics view is. But it does give the info you are after.
How to get Random Geo-points[ lat/long in decimal], placed anywhere inside a 100 meter radius circle? The center of the circle is another reference GeoPoint.
Is there any Function/Formulae that implements this?
Basically I am reading the GPS input of my android device and need to generate random Geo-Points around the device [In a circle of radius 100 meters centered at my device].
Please note : There are no Geo-Points pre-stored in Database. I need to create all the Geo-points on the fly as mentioned above.
I just wrote a a Ruby method which extends Random to provide this.
Caveat: The points all lay within a box, not a circle.
class Random
def location(lat, lng, max_dist_meters)
This is called with Random.new.location(mid_lat, mid_lng, dist). It will return a point which is probably within max_dist_meters of a the mid point.
max_radius = Math.sqrt((max_dist_meters ** 2) / 2.0)
Without adjusting to max_radius we'd get points inside a square outside the circle (i.e. in the corners), where the distance would be greater than max_dist_meters. This constrains us to a square inside the circle which is probably more what you want.
lat_offset = rand(10 ** (Math.log10(max_radius / 1.11)-5))
lng_offset = rand(10 ** (Math.log10(max_radius / 1.11)-5))
The 1.11 and 5 come from here.
lat += [1,-1].sample * lat_offset
lng += [1,-1].sample * lng_offset
lat = [[-90, lat].max, 90].min
lng = [[-180, lng].max, 180].min
We should probably wrap around here instead of just clamping the value, fixes welcome.
[lat, lng]
end
end
Comments / clean up welcome!
Sample output here which you can see nicely if you paste the lat/lngs here.
Pick random points on a square (i.e. pairs of uniform random numbers), then discard any that don't lie within a circle inscribed in that square. Given (x,y) pairs, a point is within your circle if:
(x - c_x)^2 + (y - c_y)^2 < r,
where (c_x, c_y) is the centre of your circle and r is its radius.
Start here: Generate a random point within a circle (uniformly). Then figure out how to center that circle at the reference lat/long. Then figure out how to map the randomly generated points to lat/long values. Hint: you want to add the individual components (say, x and y on your circle) to the circle's center.