coordinate

I am trying to find the new value of a coordinate if I rotate around the origin. For example, say I have the point (1,1). If I rotate the coordinate axis 45 degrees around the origin, the transformed coordinate would be (0,1.414) Is there a way to do this efficiently in cocos2d, or in objective-c ? Even answers explaining the math to do this would be helpful. See this page: http://www.siggraph.org/education/materials/HyperGraph/modeling/mod_tran/2drota.htm This is the formula: x' = x cos f - y sin f y' = y cos f + x sin f Remember that sin and cos takes radians, so you have to do like this:

I have position (x,y,z) and velocity (Vx,Vy,Vz) vectors in Earth Centered Inertial Coordinates (ECI) for a satellite orbit, and ultimately want to end up with geodetic coordinates (Latitude, Longitude, & Altitude). According to this other Stack Overflow question it seems that I need to convert to Earth Centered Earth Fixed (ECEF) coordinates as an intermediate step (so ECI --> ECEF --> Lat/Lon/Alt). I know ECI and ECEF share the same origin point (the center of mass of Earth) and the same z-axis that points to the North Pole. However, I am not sure what actual equations or adjustments I need

my previous question ( Integral of Intensity function in python ) you can see the diffraction model in image below: I want to calculate integral of intensity in each pixel (square), so I can't use R and Theta as variable. How can I do this in X-Y coordinate. Our function: instead of sin(theta) we can use: sintheta= (np.sqrt((x)**2 + (y)**2)/(np.sqrt((x)**2 + (y)**2 + d**2))) Other constants: lamb=550*10**(-9) k=2.0*np.pi/lamb a=5.5*2.54*10**(-2) d=2.8 when you plot function, the result is something like this:(The image above is a view from top) the method in previous topic: calculate integrate

This is just a extension for a old question ggplot2 polar plot arrows You will find the x axis is out of the most_out circle. In ggplot2, I use "panel.grid.major = theme_line(colour = "black", size = 0.2, linetype=2)" to get the dashed circle, just as below: So my question is how to make the axis label (180, 135, 90, .....) outside of the circle, because the text are merge with the circular lines. I try to use "hjust" or "vjust" to adjust the distance between text and axis. But it does not work. So do you have some ideas about this problem? Thanks first!!!! You have not provided code to

My team needs a DBMS like DynamoDB to store large amount of data, principally places and coordinates. I've considered to use some GIS-based DBMS (like PostGIS) with an index on the POINT, but DynamoDB seems great for our use. What is the best method to store the coordinate and quickly retrieve all objects in a particular radius? In PostGIS it's easy, something like this: SELECT * FROM places WHERE ST_DWithin(coordinate, ST_GeomFromText('POINT(45.07085 7.68434)', 4326), 100.0); How can I do something like that in a NoSQL DBMS? We had the same issue, we're using AWS and DynamoDB in particular.

How can I get the bottom left coordinate, i.e. the maximum y coordinate, of a view in obj-c and Swift? Gabriele Petronella Supposing that you're talking about UIView , you may want to look at the CGGeometry reference. Talking about 'left' and y coordinate doesn't make much sense, but for instance CGRectGetMaxY(view.frame) // will return the bottommost y coordinate of the view CGRectGetMinX(view.frame) // will return the leftmost x coordinate of the view and so on. You get the idea. In Swift, this is even more convenient, as you can do view.frame.maxY // will return the bottommost y coordinate

I need to find out Y coordinate of specific point of bezier curve in canvas. Do you know, how to find it out? Thank you Using de Casteljau's algorithm you can find the coordinates x and y of a bezier curve for any t, the percentage or interpolation step. So a t of .1 would give you the x and y at 10% of the curve from the beginning. A t of .9 would be 90% from the beginning, and so on. In our cubic bezier we have p0 (point 0), cp0 (control point 0), cp1 (control point 1), and p1 (point 1). In the first step of the algorithm we draw a line connecting p0 and cp0, another line connecting cp0 and