does anyone know the method or code to add a second x axis to a TGraph in CERN's ROOT program? Ive been searching the root website and its documentation almost always confuses me. What i need is just one plot of data, but a second X axis on top whose values are a function of the bottom x axis' values. Its basically so lazy people dont have to convert from the numbers of the bottom x axis to the top x axis.
For a simple example (if i wasnt clear)
Say you have a sine curve which is some function of theta. On the top x axis we could have degrees whereas on the bottom we could have radians with 360deg corresponding to 2pi rad...
Any help would be appreciated!
TGaxis is the class you are looking for to draw extra axes wherever you desire. Grabbing the world coordinate for your pad you can then superimpose like so. Replace low and high with the appropriate limits.
// your graph code here...
TGraph->Draw("AP");
TGaxis *axis = new TGaxis(gPad->GetUxmin(),gPad->GetUymax(),gPad->GetUxmax(),gPad->GetUymax(),low,high,510,"+L");
axis->Draw();
Check out TGaxis documentation for more examples.
(A previous answer I had was deleted as it was just a link to the site listed as a reference below. I hope this is more in line with the community guidelines.)
I think this might do what you want.
void axis2() {
TH1F *h = new TH1F("h","test",30,-3,3);
h->FillRandom("gaus",10000);
h->Draw();
TText t;
t.SetTextSize(0.02);
t.SetTextAlign(22);
Double_t yt = - h->GetMaximum()/15.;
for (Int_t i=1;i<=30;i++) t.DrawText(h->GetBinCenter(i),yt,Form("%d",i%10));
}
It doesn't create another taxis but shows you how to draw text at the same location of the axis. The answer comes from Rene Brun himself (one of the main authors of root) so I don't think you can have two x axes.
Source:
http://root.cern.ch/phpBB3/viewtopic.php?f=3&t=7110
Here is an example showing how to proceed.
https://root.cern/doc/master/twoscales_8C.html
Related
First off, this is a homework question. The problem is ex. 2.6 from pg.26 of An Introduction to Applied Multivariate Analysis. It's laid out as:
Construct a bubble plot of the earthquake data using latitude and longitude as the scatterplot and depth as the circles, with greater depths giving smaller circles. In addition, divide the magnitudes into three equal ranges and label the points in your bubble plot with a different symbol depending on the magnitude group into which the point falls.
I have figured out that symbols, which is in base graphics does not work well with lattice. Also, I haven't figured out if lattice has the functionality to change symbol size (i.e. bubble size). I bought the lattice book in a fit of desperation last night, and as I see in some of the examples, it is possible to symbol color and shape for each "cut" or panel. I am then working under the assumption that symbol size could then also be manipulated, but I haven't been able to figure out how.
My code looks like:
plot(xyplot(lat ~ long | cut(mag, 3), data=quakes,
layout=c(3,1), xlab="Longitude", ylab="Latitude",
panel = function(x,y){
grid.circle(x,y,r=sqrt(quakes$depth),draw=TRUE)
}
))
Where I attempt to use the grid package to draw the circles, but when this executes, I just get a blank plot. Could anyone please point me in the right direction? I would be very grateful!
Here is the some code for creating the plot that you need without using the lattice package. I obviously had to generate my own fake data so you can disregard all of that stuff and go straight to the plotting commands if you want.
####################################################################
#Pseudo Data
n = 20
latitude = sample(1:100,n)
longitude = sample(1:100,n)
depth = runif(n,0,.5)
magnitude = sample(1:100,n)
groups = rep(NA,n)
for(i in 1:n){
if(magnitude[i] <= 33){
groups[i] = 1
}else if (magnitude[i] > 33 & magnitude[i] <=66){
groups[i] = 2
}else{
groups[i] = 3
}
}
####################################################################
#The actual code for generating the plot
plot(latitude[groups==1],longitude[groups==1],col="blue",pch=19,ylim=c(0,100),xlim=c(0,100),
xlab="Latitude",ylab="Longitude")
points(latitude[groups==2],longitude[groups==2],col="red",pch=15)
points(latitude[groups==3],longitude[groups==3],col="green",pch=17)
points(latitude[groups==1],longitude[groups==1],col="blue",cex=1/depth[groups==1])
points(latitude[groups==2],longitude[groups==2],col="red",cex=1/depth[groups==2])
points(latitude[groups==3],longitude[groups==3],col="green",cex=1/depth[groups==3])
You just need to add default.units = "native" to grid.circle()
plot(xyplot(lat ~ long | cut(mag, 3), data=quakes,
layout=c(3,1), xlab="Longitude", ylab="Latitude",
panel = function(x,y){
grid.circle(x,y,r=sqrt(quakes$depth),draw=TRUE, default.units = "native")
}
))
Obviously you need to tinker with some of the settings to get what you want.
I have written a package called tactile that adds a function for producing bubbleplots using lattice.
tactile::bubbleplot(depth ~ lat*long | cut(mag, 3), data=quakes,
layout=c(3,1), xlab="Longitude", ylab="Latitude")
I am programming a simple ball projectile in a game.
the update pretty much looks like:
velocity += gravity;
velocity *=0.9;
pos += vel;
Is there a way to set the angle and power of the launch in order to hit a point that is specified with the mouse?
like peggle, http://youtu.be/4KbNiWsgJck?t=45s
I know there is a solution that I have used several years ago, but I can't find it.
I believed it turned my update into a quadratic formula, or derived it or something.
It had two solutions that was solved with the quadratic equation.
ps- hopefully this could be in 3D, but I could use a 2D solution too because my curve would be 2D
any help?
thanks,
Dan
Yes, you can do this. If you can change the angle and speed, you have more variability than you need, so you have to find a reasonable set of parameters that will work, which isn't hard. The basic equations are:
x = x0 + t*v0x
y = y0 + v0yt + (1/2)ayt2
Here, x and y will be the points you want to hit, and t will be the time that you hit them. t won't show up in the final solution, but you'll use it as in intermediary to calculate the values you want.
Basically, then, pick a reasonable value for v0x. Using the x-equation, find what t will be when the target is hit. Then plug this value into the y-equation, and solve for v0y. This then will give you a pair of values of v0x and v0y that will work to hit the target.
I'm creating simple game and reached the point where I feel helpless. I was good in geometry but it was long time back in school, now trying to refresh my mind.
Let's say i have iPad screen. Object's xy position at one given point of time and xy position at another point of time stored in 2 variables .
Question:
how to find the third position of the object at the end of the screen being given previous 2 position, considering the object moves in the same direction (line) from point 1 to point 2.
Thanks in advance.
Let us have that v1 and v2 are the vectors representing the two points. Let t0 be the time between the two points. Let t be the current time.
Then our location vector v3 is given by v3 = v1 + (v2 - v1)t/t0
If the object is moving in the same direction and you have an horizontal line, the next position given x and y would be
x+1, y
If the object is moving in the same direction in a vertical line it would be
x, y+1
If the object is moving in a diagonal up-right
x+1,y+1
diagonal down-right
x+1, y+1
diagonal down-left
x-1, y-1
diagonal up-left
x-1, y+1
So something general would be :
newPosition = (x+1,y) //if you wish to move forward to the right, try to handle all
cases
All the cases above work if the object is moving forward, if it is moving backwards just change the + by - . Basically think of the object as moving in a cartesian coordinate system, where x is horizontal and y is vertical.
I think you can get the idea out of this three cases ;)
I've been having a bit of a problem with making the Y axis my up axis when exporting mesh and scenes from Blender. Both Blender and my export target use right handed transformation matrices. Z is the up axis in Blender while Y is the up axis in my target. The problem exists in 2 places though. The scene's transformations can't just be shifted on the X axis to fix this, because I also need to do the Y/Z switch for the vertices in the mesh (export as vertex.x, vertex.z, vertex.y). I need to have the actual Y and Z rotations switched, so that if the Y and Z rotations are the same, no change will occur (ie. identity). Thanks for your help in advance. Feel free to ask questions if I was not thorough enough.
Blender does two things different than the rest of the known world!
1. It uses Z axis for vertical (should be Y); Y axis for horizontal (should b X); and X axis for in and out (should b Z).
Very weird! Every high school graph since the beginning of time uses X for horizontal and Y for vertical.
It uses the right mouse button for selections.
U can change the selection btn in Preferences, but not the crazy axis arrangement!
no,
you do this
y=z
z=-y
no rotation of 90 degrees can make you go from left to right hand.
I ran into a similar issue when working with cinema4d and blender. In cinema4d Y is the up axis and rotations are heading,pitch and bank.
Blender's system looks like a right handed system, but rotated by 90 degrees on x axis.
I did the same thing for coordinates(exported as vertex.x,vertex.z,vertex.y). For rotations,
I think you should add 90 degrees(math.pi * 0.5) for rotations on X axis and the rest should be fine.
HTH
Have you tried just using Select All (the 'a' key) and then r x 90 to rotate everything 90 degrees around the X axis and the pivot point? (your pivot point is choosable in the menu bar of the 3D view if you need to control that).
You could do that, export, and then undo.
Just Download Wings3D. Export from Blender as .3ds and then Import this file in Wings3D.
Now you can just export it from Wings3D, again to .3ds. But instead of clicking directly on .3ds, click on the small icon in the right of the ".3ds" menu. now you can unchecked the Box Swap y und z axis and import the .3ds in another program.
There is no way that would be possible. Coordinate system was innately selected as hard coded from the blender source and there are no explicit option has been made in blender to switch it. It would also affected many of the hard coded functionality of any function blender was used or has been made by assume that coordinate
However, in theory, it would be possible to access blender source code and rebuild the blender to have it use another coordinate we would like. Albeit we need to carefully swap everything related to coordinate system
I too wish that left handed coordinate system (as of Unity3D) would be industrial standard and blender should at least have another version that work in left handed coordinate. People should just graduated from table coordinate to screen coordinate already
In blender, you could add empty plain axes, that will correct your orientation when exporting to unity, or try exporting as fbx file and change orientation in export options
how can i calculate angle to reach particular height?
suppose i want height 320.time is increasing as 0.1.
i am using h = (u sin(angle))^2 / 2g;
where can i put the time?
The inverse of the sin() function is called the arcsine, or sin-1 in mathematical notation. In many programming languages, it's available as asin().
From my answer to your previous question:
Where y is the height you want to reach (320 in this case), and assuming you're starting at y=0:
angle = arctan( 2*y / x )
where x is the distance on the X-AXIS between your starting point and the point where you want to reach that height, which is necessary in order to specify an angle.
If you really want me to, I can derive this one for you, but it follows directly from my answer to your previous question.
Also (since I can't comment answers yet I'm saying this here), you may be having issues getting an angle "less than 1" because you're trying to use degrees instead of radians. Many math libraries work in radians, so convert your angles.