I'm new for the Gnuplot.
I have some trouble.
For the 2D graph, There is the coordinate (x,y) at the left-bottom but i want to know.
How can i know the coordinate (x,y,z) when i move the mouse pointer on the 3D graph.
I waste my time for search the answer but I'm still cannot find.
thank in advance for your help.
That is not possible, because gnuplot doesn't hold all data points to be accessed by the mouse.
In 2D, the displayed coordinates are extracted from the graph boundaries, but there is no such things to snap the mouse pointer to the nearest plotted data point (like requested in How can I make gnuplot show coordinates of a plotted function which have same x value as the mouse pointer?).
Because of the missing "data snapping", it is not possible to extract the 3D coordinates from the 2D mouse position.
Related
I saw this figure in Leland Wilkinson's book the Grammar of Graphics and was wondering how I could go about creating something similar in R.
I am suspicious this could be done using rgl, persp3d, but there's a couple aspects that are unclear to me like how to create the conformal mapping shown in the coordinates of the XY plane, as well as how to create the 2D color map in a 3D context.
Any advice would be much appreciated. Thanks!
That should be possible with rgl, but there might be some snags in the details. Here's the outline:
The green surface does not appear to have a rectangular base,
so you'll pass matrices for all of x, y and z coordinates to surface3d() to draw it.
I can't tell if the map is on a flat surface with curved edges, or if it's a curved surface. In either case, you plot the surface with a 2D texture showing the map and the contours.
a. To produce that 2D texture, use whatever mapping software you've got, and output the image to a PNG file.
b. To put it on the surface, use surface3d() with arguments texture = <filename>, texture_s = ..., texture_t = ...) where texture_s and texture_t are set to coordinates in the image (bottom left = (0,0), top right = (1,1)) corresponding to each x and y location. The z value is
either constant or varying depending on whether you want it flat
or curved.
The axes will be drawn with axis3d.
I have an input of three [1:360,2] tables that include measurements of the three planes X,Y,Z with each plane having 1:360 degree polar coordinates. A good example of one of those tables being the image here.
image example
I would like to print those three "circles" as different planes circles (x,y,z) as is shown in the link below.
click here
I wrote last night a small example that works on the rgl library (but perhaps
I should "move" this code to the ggplot2) that plots those three circles by converting the polar coordinates to cartesian ones, for simplicity assume that all three circles have radius of 1. You can copy paste the below and see what I mean
require("rgls")
degreeToRadian<-function(degree){
return (0.01745329252*degree)
}
turnPolarToX<-function(Amplitude,Coordinate){
return (Amplitude*cos(degreeToRadian(Coordinate)))
}
turnPolarToY<-function(Amplitude,Coordinate){
return (Amplitude*sin(degreeToRadian(Coordinate)))
}
# first circle
X1<-turnPolarToX(1,1:360)
Y1<-turnPolarToY(1,1:360)
Z1<-rep(0,360)
# second circle
X2<-turnPolarToX(1,1:360)
Y2<-rep(0,360)
Z2<-turnPolarToY(1,1:360)
# third circle
X3<-rep(0,360)
Y3<-turnPolarToX(1,1:360)
Z3<-turnPolarToY(1,1:360)
Min<-min(X1,Y1,Z1,X2,Y2,Z2,X3,Y3,Z3)
Max<-max(X1,Y1,Z1,X2,Y2,Z2,X3,Y3,Z3)
plot3d(X1,Y1,Z1,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=FALSE,add=TRUE,col="red",type="l")
plot3d(X2,Y2,Z2,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=FALSE,add=TRUE,col="green",type="l")
plot3d(X3,Y3,Z3,xlim=c(Min,Max),ylim=c(Min,Max),zlim=c(Min,Max),box=TRUE,axe=TRUE,add=FALSE,col="blue",type="l")
the problem I have now is that the axes contain cartesian coordinate values that are not understandable by the user. I am thinking of how I can remove the cartesian coordinates and use colors that considers the amplitude value (of the initial information at polar corrdinates) of each vector instead of the x,y values)
I would like to thank you in advance for your reply
Regards
Alex
i haven’t been entirely sure what to google or search for to help solve my problem, really hoping someone here can help a little…
currently i have a 3d scene, it has a massive sphere with a texture mapped to it and the camera at the center of the sphere, so it’s much like a qtvr viewer.
i’d like a way to click on the polygons within the sphere and update the texture at that position with something and dot etc..
the only part of the process where i need help is converting the 2d mouse position to a point on the inside of the sphere.
hope this makes sense…
fyi, im only looking for a pure math solution..
The first thing you need to do is convert the screen coordinate into a line in 3d space. This will pass through the point you click and your eyepoint.
Once you have this line you can then intersect this line with your sphere to find the intersection point on the sphere.
You may get 2d coordinates of the polygons (triangles?) that are making up the sphere and then find the one that contains the mouse pointer point.
It's been a while since my math in university, and now I've come to need it like I never thought i would.
So, this is what I want to achieve:
Having a set of 3D points (geographical points, latitude and longitude, altitude doesn't matter), I want to display them on a screen, considering the direction I want to take into account.
This is going to be used along with a camera and a compass , so when I point the camera to the North, I want to display on my computer the points that the camera should "see". It's a kind of Augmented Reality.
Basically what (i think) i need is a way of transforming the 3D points viewed from above (like viewing the points on google maps) into a set of 3d Points viewed from a side.
The conversion of Latitude and longitude to 3-D cartesian (x,y,z) coordinates can be accomplished with the following (Java) code snippet. Hopefully it's easily converted to your language of choice. lat and lng are initially the latitude and longitude in degrees:
lat*=Math.PI/180.0;
lng*=Math.PI/180.0;
z=Math.sin(-lat);
x=Math.cos(lat)*Math.sin(-lng);
y=Math.cos(lat)*Math.cos(-lng);
The vector (x,y,z) will always lie on a sphere of radius 1 (i.e. the Earth's radius has been scaled to 1).
From there, a 3D perspective projection is required to convert the (x,y,z) into (X,Y) screen coordinates, given a camera position and angle. See, for example, http://en.wikipedia.org/wiki/3D_projection
It really depends on the degree of precision you require. If you're working on a high-precision, close-in view of points anywhere on the globe you will need to take the ellipsoidal shape of the earth into account. This is usually done using an algorithm similar to the one descibed here, on page 38 under 'Conversion between Geographical and Cartesian Coordinates':
http://www.icsm.gov.au/gda/gdatm/gdav2.3.pdf
If you don't need high precision the techniques mentioned above work just fine.
could anyone explain me exactly what these params mean ?
I've tried and the results where very weird so i guess i am missunderstanding some of the params for the perspective projection
* {a}_{x,y,z} - the point in 3D space that is to be projected.
* {c}_{x,y,z} - the location of the camera.
* {\theta}_{x,y,z} - The rotation of the camera. When {c}_{x,y,z}=<0,0,0>, and {\theta}_{x,y,z}=<0,0,0>, the 3D vector <1,2,0> is projected to the 2D vector <1,2>.
* {e}_{x,y,z} - the viewer's position relative to the display surface. [1]
Well, you'll want some 3D vector arithmetic to move your origin, and probably some quaternion-based rotation functions to rotate the vectors to match your direction. There are any number of good tutorials on using quaternions to rotate 3D vectors (since they're used a lot for rendering and such), and the 3D vector stuff is pretty simple if you can remember how vectors are represented.
well, just a pice ov advice, you can plot this points into a 3d space (you can do easily this using openGL).
You have to transforrm the lat/long into another system for example polar or cartesian.
So starting from lat/longyou put the origin of your space into the center of the heart, than you have to transform your data in cartesian coord:
z= R * sin(long)
x= R * cos(long) * sin(lat)
y= R * cos(long) * cos(lat)
R is the radius of the world, you can put it at 1 if you need only to cath the direction between yoour point of view anthe points you need "to see"
than put the Virtual camera in a point of the space you've created, and link data from your real camera (simply a vector) to the data of the virtual one.
The next stemp to gain what you want to do is to try to plot timages for your camera overlapped with your "virtual space", definitevly you should have a real camera that is a control to move the virtual one in a virtual space.
Given a list of points that form a simple 2d polygon oriented in 3d space and a normal for that polygon, what is a good way to determine which points are specific 'corner' points?
For example, which point is at the lower left, or the lower right, or the top most point? The polygon may be oriented in any 3d orientation, so I'm pretty sure I need to do something with the normal, but I'm having trouble getting the math right.
Thanks!
You would need more information in order to make that decision. A set of (co-planar) points and a normal is not enough to give you a concept of "lower left" or "top right" or any such relative identification.
Viewing the polygon from the direction of the normal (so that it appears as a simple 2D shape) is a good start, but that shape could be rotated to any arbitrary angle.
Is there some other information in the 3D world that you can use to obtain a coordinate-system reference?
What are you trying to accomplish by knowing the extreme corners of the shape?
Are you looking for a bounding box?
I'm not sure the normal has anything to do with what you are asking.
To get a Bounding box, keep 4 variables: MinX, MaxX, MinY, MaxY
Then loop through all of your points, checking the X values against MaxX and MinX, and your Y values against MaxY and MinY, updating them as needed.
When looping is complete, your box is defined as MinX,MinY as the upper left, MinX, MaxY as upper right, and so on...
Response to your comment:
If you want your box after a projection, what you need is to get the "transformed" points. Then apply bounding box loop as stated above.
Transformed usually implies 2D screen coordinates after a projection(scene render) but it could also mean the 2D points on any plane that you projected on to.
A possible algorithm would be
Find the normal, which you can do by using the cross product of vectors connecting two pairs of different corners
Create a transformation matrix to rotate the polygon so that it is planer in XY space (i.e. normal alligned along the Z axis)
Calculate the coordinates of the bounding box or whatever other definition of corners you are using (as the polygon is now aligned in 2D space this is a considerably simpler problem)
Apply the inverse of the transformation matrix used in step 2 to transform these coordinates back to 3D space.
I believe that your question requires some additional information - namely the coordinate system with respect to which any point could be considered "topmost", or "leftmost".
Don't forget that whilst the normal tells you which way the polygon is facing, it doesn't on its own tell you which way is "up". It's possible to rotate (or "roll") around the normal vector and still be facing in the same direction.
This is why most 3D rendering systems have a camera which contains not only a "view" vector, but also "up" and "right" vectors. Changes to the latter two achieve the effect of the camera "rolling" around the view vector.
Project it onto a plane and get the bounding box.
I have a silly idea, but at the risk of gaining a negative a point, I'll give it a try:
Get the minimum/maximum value from
each three-dimensional axis of each
point on your 2d polygon. A single pass with a loop/iterator over the list of values for every point will suffice, simply replacing the minimum and maximum values as you go. The end result is a list that has the "lowest" X, Y, Z coordinates and "highest" X, Y, Z coordinates.
Iterate through this list of min/max
values to create each point
("corner") of a "bounding box"
around the object. The result
should be a box that always contains
the object regardless of axis
examined or orientation (no point on
the polygon will ever exceed the
maximum or minimums you collect).
Then get the distance of each "2d
polygon" point to each corner
location on the "bounding box"; the
shorter the distance between points,
the "closer" it is to that "corner".
Far from optimal, certainly crummy, but certainly quick. You could probably post-capture this during the object's rotation, by simply looking for the min/max of each rotated x/y/z value, and retaining a list of those values ahead of time.
If you can assume that there is some constraints regarding the shapes, then you might be able to get away with knowing less information. For example, if your shape was the composition of a small square with a long thin triangle on one side (i.e. a simple symmetrical geometry), then you could compare the distance from each list point to the "center of mass." The largest distance would identify the tip of the cone, the second largest would be the two points farthest from the tip of the cone, etc... If there was some order to the list, like points are entered in counter clockwise order (about the normal), you could identify all the points. This sounds like a bit of computation, so it might be reasonable to try to include some extra info with your shapes, like the "center of mass" and a reference point that is located "up" above the COM (but not along the normal). This will give you an "up" vector that you can cross with the normal to define some body coordinates, for example. Also, the normal can be defined by an ordering of the point list. If you can't assume anything about the shapes (or even if the shapes were symmetrical, for example), then you will need more data. It depends on your constraints.
If you know that the polygon in 3D is "flat" you can use the normal to transform all 3D-points of the vertices to a 2D-representation (of the points with respect to the plan in which the polygon is located) - but this still leaves you with defining the origin of this coordinate-system (but this don't really matter for your problem) and with the orientation of at least one of the axes (if you want orthogonal axes you can still rotate them around your choosen origin) - and this is where the trouble starts.
I would recommend using the Y-axis of your 3D-coordinate system, project this on your plane and use the resulting direction as "up" - but then you are in trouble in case your plan is orthogonal to the Y-axis (now you might want to use the projected Z-Axis as "up").
The math is rather simple (you can use the inner product (a.k.a. scalar product) for projection to your plane and some matrix stuff to convert to the 2D-coordinate system - you can get all of it by googling for raytracer algorithms for polygons.