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this question is in theoretical math perspective.
In the space S = [-1, 1]^d, given a volume V, can I always define object in S with exactly this volume? also - can I surround any x in S with an object of volume V?
I would answer yes on these 2 questions because I only need to find multiplication of d positive real numbers that would be lengths of lines, and I assume I can construct it around any x in S, but I just want to be sure, and get some nice explanation.
thanks
Basically yes, let's say n is ceil(log2(V)), so V is c 2^n with 0.5 < c <= 1,
then one possible solution is the c th part of an n dimensional cube,
so for 7 you take 7/8ths of a cube, for 9 you have 9/16ths of a hypercube, etc...
The only assumption here is that "Volume" should be interpreted as the nth-dimensional measure in n-dimensional space.
If I get it right your space S has coordinates in range <-1,+1> which limits its volumes to:
V = 2^3
HV = 2^d
where V is standard 3D volume and HV is d-dimensional hyper-volume that can fit into your space S. So you can construct objects with volumes and hypervolumes up to this limit.
So if you want to construct object with volumes V or HV then you can create axis aligned cube of size a:
a*a*a = V
a = V^(1/3)
a^d = HV
a = HV^(1/d)
if a<=2 otherwise your S is too small ...
I want to plot a function in scilab in order to find the maximum over a range of numbers:
function y=pr(a,b)
m=1/(1/270000+1/a);
n=1/(1/150000+1/a);
y=5*(b/(n+b)-b/(m+b))
endfunction
x=linspace(10,80000,50)
y=linspace(10,200000,50)
z=feval(x,y,pr)
surf(x,y,z);
disp( max(z))
For these values this is the plot:
It's obvious that increasing the X axis will not increase the maximum but Y axis will.
However from my tests it seems the two axis are mixed up. Increasing the X axis will actually double the max Z value.
For example, this is what happens when I increase the Y axis by a factor of ten (which intuitively should increase the function value):
It seems to increase the other axis (in the sense that z vector is calculated for y,x pair of numbers instead of x,y)!
What am I doing wrong here?
With Scilab's surf you have to use transposed z if comming from feval. It is easy so realize if you use a different number of points in X and Y directions, as surf will complain about the size of the third argument. So in your case, use:
surf(x,y,z')
For more information see the help page of surf.
Stephane's answer is correct, but I thought I'd try to explain better why / what is happening.
From the help surf page (emphasis mine):
X,Y:
two vectors of real numbers, of lengths nx and ny ; or two real matrices of sizes ny x nx: They define the data grid (horizontal coordinates of the grid nodes). All grid cells are quadrangular but not necessarily rectangular. By default, X = 1:size(Z,2) and Y = 1:size(Z,1) are used.
Z:
a real matrix explicitly defining the heights of nodes, of sizes ny x nx.
In other words, think of surf as surf( Col, Row, Z )
From the help feval page (changed notation for convenience):
z=feval(u,v,f):
returns the matrix z such as z(i,j)=f(u(i),v(j))
In other words, in your z output, the i become rows (and therefore u should represent your rows), and j becomes your columns (and therefore v should represent your columns).
Therefore, you can see that you've called feval with the x, y arguments the other way round. In a sense, you should have designed pr so that it should have expected to be called as pr(y,x) instead, so that when passed to feval as feval(y,x,pr), you would end up with an output whose rows increase with y, and columns increase with x.
Then you could have called surf(x, y, z) normally, knowing that x corresponds to columns, and y corresponds to rows.
However, if you don't want to change your whole function just for this, which presumably you don't want to, then you simply have to transpose z in the call to surf, to ensure that you match x to the columns or z' (i.e, the rows of z), and y to the rows of z' (i.e. the columns of z).
Having said all that, it would probably be much better to make your function vectorized, and just use the surf(x, y, pr) syntax directly.
I am trying to workout the interception of two objects and try to code it into an application
A cannon is in Position A and a plane is in position B
The plane is moving with vector b (without affected by gravity)
A cannonball is shot (affected by gravity) with unit vector a and magnitude of m
They intercept at position C at T seconds
Knowns: A, m, B, b
Unknowns: a, C, T
The only thing I can think of to solve it in terms of code is split the equations into X, Y and Z component and substitute T as a value and increment it.
It will be nice if someone can kindly tell how to find one of the unknowns
Thanks
The answer is in here:
https://blog.forrestthewoods.com/solving-ballistic-trajectories-b0165523348c
from meowgoesthedog
How can I get the width and height of a character in usr coordinates? I found something on R-help, but that appears not to make it completely clear. I assumed that
plot(NULL, xlim=c(-1,1), ylim=c(-1,1))
h <- par()$cxy
rect(-h[1]/2, -h[2]/2, h[1]/2, h[2]/2)
text(0,0,"M")
would be the answer but the rectangle is slightly too big. Additionally I want the size also to respect different cex values. Thanks for your time!
I finally discovered an answer in the par docu:
cxy
R.O.; size of default character (width, height) in user coordinate
units. par("cxy") is par("cin")/par("pin") scaled to user coordinates.
Note that c(strwidth(ch), strheight(ch)) for a given string ch is
usually much more precise.
Using strwidth and strheight instead of par()$cxy gives much better results.
plot(NULL, xlim=c(-1,1), ylim=c(-1,1))
h <- c(strwidth("M"), strheight("M"))
rect(-h[1]/2, -h[2]/2, h[1]/2, h[2]/2)
text(0,0,"M")
Ok, I know this sounds really daft to be asking here, but it is programming related.
I'm working on a game, and I'm thinking of implementing a system that allows users to triangulate their 3D coordinates to locate something (eg for a task).
I also want to be able to let the user make the coordinates of the points they are using for triangulation have user-determined coordinates (so the location's coordinate is relative, probably by setting up a beacon or something).
I have a method in place for calculating the distance between the points, so essentially I can calculate the lengths of the sides of the triangle/pyramid as well as all but the coordinate I am after.
It has been a long time since I have done any trigonometry and I am rusty with the sin, cos and tan functions, I have a feeling they are required but have no clue how to implement them.
Can anyone give me a demonstration as to how I would go about doing this in a mathematical/programatical way?
extra info:
My function returns the exact distance between the two points, so say you set two points to 0,0,0 and 4,4,0 respectively, and those points are set to scale(the game world is divided into a very large 3d grid, with each 'block' area being represented by a 3d coordinate) then it would give back a value at around 5.6.
The key point about it varying is that the user can set the points, so say they set a point to read 0,0,0, the actual location could be something like 52, 85, 93. However, providing they then count the blocks and set their other points correctly (eg, set a point 4,4,0 at the real point 56, 89, 93) then the final result will return the relative position (eg the object they are trying to locate is at real point 152, 185, 93, it will return the relative value 100,100,0). I need to be able to calculate it knowing every point but the one it's trying to locate, as well as the distances between all points.
Also, please don't ask why I can't just calculate it by using the real coordinates, I'm hoping to show the equation up on screen as it calculates the result.7
Example:
Here is a diagram
Imagine these are points in my game on a flat plain.
I want to know the point f.
I know the values of points d and e, and the sides A,B and C.
Using only the data I know, I need to find out how to do this.
Answered Edit:
After many days of working on this, Sean Kenny has provided me with his time, patience and intellect, and thus I have now got a working implementation of a triangulation method.
I hope to place the different language equivalents of the code as I test them so that future coders may use this code and not have the same problem I have had.
I spent a bit of time working on a solution but I think the implementer, i.e you, should know what it's doing, so any errors encountered can be tackled later on. As such, I'll give my answer in the form of strong hints.
First off, we have a vector from d to e which we can work out: if we consider the coordinates as position vectors rather than absolute coordinates, how can we determine what the vector pointing from d to e is? Think about how you would determine the displacement you had moved if you only knew where you started and where you ended up? Displacement is a straight line, point A to B, no deviation, not: I had to walk around that house so I walked further. A straight line. If you started at the point (0,0) it would be easy.
Secondly, the cosine rule. Do you know what it is? If not, read up on it. How can we rearrange the form given in the link to find the angle d between vectors DE and DF? Remember you need the angle, not a function of the angle (cos is a function remember).
Next we can use a vector 'trick' called the scalar product. Notice there is a cos function in there. Now, you may be thinking, we've just found the angle, why are we doing it again?
Define DQ = [1,0]. DQ is a vector of length 1, a unit vector, along the x-axis. Which other vector do we know? Do we know of two position vectors?
Once we have two vectors (I hope you worked out the other one) we can use the scalar product to find the angle; again, just the angle, not a function of it.
Now, hopefully, we have 2 angles. Could we take one from the other to get yet another angle to our desired coordinate DF? The choice of using a unit vector earlier was not arbitrary.
The scalar product, after some cancelling, gives us this : cos(theta) = x / r
Where x is the x ordinate for F and r is the length of side A.
The end result being:
theta = arccos( xe / B ) - arccos( ( (A^2) + (B^2) - (C^2) ) / ( 2*A*B ) )
Where theta is the angle formed between a unit vector along the line y = 0 where the origin is at point d.
With this information we can find the x and y coordinates of point f relative to d. How?
Again, with the scalar product. The rest is fairly easy, so I'll give it to you.
x = r.cos(theta)
y = r.sin(theta)
From basic trigonometry.
I wouldn't advise trying to code this into one value.
Instead, try this:
//pseudo code
dx = 0
dy = 0 //initialise coordinates somehow
ex = ex
ey = ey
A = A
B = B
C = C
cosd = ex / B
cosfi = ((A^2) + (B^2) - (C^2)) / ( 2*A*B)
d = acos(cosd) //acos is a method in java.math
fi = acos(cosfi) //you will have to find an equivalent in your chosen language
//look for a method of inverse cos
theta = fi - d
x = A cos(theta)
y = A sin(theta)
Initialise all variables as those which can take decimals. e.g float or double in Java.
The green along the x-axis represents the x ordinate of f, and the purple the y ordinate.
The blue angle is the one we are trying to find because, hopefully you can see, we can then use simple trig to work out x and y, given that we know the length of the hypotenuse.
This yellow line up to 1 is the unit vector for which scalar products are taken, this runs along the x-axis.
We need to find the black and red angles so we can deduce the blue angle by simple subtraction.
Hope this helps. Extensions can be made to 3D, all the vector functions work basically the same for 3D.
If you have the displacements from an origin, regardless of whether this is another user defined coordinate or not, the coordinate for that 3D point are simply (x, y, z).
If you are defining these lengths from a point, which also has a coordinate to take into account, you can simply write (x, y, z) + (x1, y1, z1) = (x2, y2, z2) where x2, y2 and z2 are the displacements from the (0, 0, 0) origin.
If you wish to find the length of this vector, i.e if you defined the line from A to B to be the x axis, what would the x displacement be, you can use Pythagoras for 3D vectors, it works just the same as with 2D:
Length l = sqrt((x^2) + (y^2) + (z^2))
EDIT:
Say you have a user defined point A (x1, y1, z1) and you want to define this as the origin (0,0,0). You have another user chosen point B (x2, y2, z2) and you know the distance from A to B in the x, y and z plane. If you want to work out what this point is, in relation to the new origin, you can simply do
B relative to A = (x2, y2, z2) - (x1, y1, z1) = (x2-x1, y2-y1, z2-z1) = C
C is the vector A>B, a vector is a quantity which has a magnitude (the length of the lines) and a direction (the angle from A which points to B).
If you want to work out the position of B relative to the origin O, you can do the opposite:
B relative to O = (x2, y2, z2) + (x1, y1, z1) = (x1+x2, y1+y2, z1+z2) = D
D is the vector O>B.
Edit 2:
//pseudo code
userx = x;
usery = y;
userz = z;
//move origin
for (every block i){
xi = xi-x;
yi = yi - y;
zi = zi -z;
}