3D Rotation – Calculating Rotation Around a Random Axis

3dmathproblem solving

This is actually a solved problem, but I want to understand why my original method didn't work (hoping someone with more knowledge can explain).

(Keep in mind, I've not very experienced in 3d programming, having only played with the very basic for a little bit…nor do I have a lot of mathematical experience in this area).

I wanted to animate a point rotating around another point at a random axis, say a 45 degrees along the y axis (think of an electron around a nucleus). I know how to rotate using the transform matrix along the X, Y and Z axis, but not an arbitrary (45 degree) axis.

Eventually after some research I found a suggestion: Rotate the point by -45 degrees around the Z so that it is aligned. Then rotate by some increment along the Y axis, then rotate it back +45 degrees for every frame tick. While this certainly worked, I felt that it seemed to be more work then needed (too many method calls, math, etc) and would probably be pretty slow at runtime with many points to deal with.

I thought maybe it was possible to combine all the rotation matrixes involve into 1 rotation matrix and use that as a single operation.

Something like:

 [ cos(-45)    -sin(-45)         0]
 [ sin(-45)     cos(-45)         0]      rotate by -45 along Z 
 [    0            0             1]

multiply by

 [ cos(2)          0       -sin(2)]
 [    0            1            0 ]      rotate by 2 degrees (my increment) along Y
 [ sin(2)          0        cos(2)]

then multiply that result by (in that order)

 [ cos(45)      -sin(45)         0]
 [ sin(45)       cos(45)         0]      rotate by 45 along Z 
 [    0            0             1]

I get 1 mess of a matrix of numbers (since I was working with unknowns and 2 angles), but I felt like it should work. It did not and I found a solution on wiki using a different matirx, but that is something else.

I'm not sure if maybe I made an error in multiplying, but my question is: this is actually a viable way to solve the problem, to take all the separate transformations, combine them via multiplying, then use that or not?

Best Answer

You can indeed compound ('add up') the rotations by multiplication.

I'd consider using Quaternions instead though. They're much nicer to work with and they avoid problems with Euler type rotations (e.g. gimbal lock). You can plug in arbitrary axes of rotation, rather than worrying about X, Y, Z rotations. Quaternion compound rotations nicely -- and if you have interactive 3D objects the user can spin etc., you'll want to compound rotations.

Lots on the web about Quaternions, e.g.:

Related Solutions

How to Get a Weighted Random Item in Ruby

The conceptually simplest solution would be to create a list where each element occurs as many times as its weight, so

fruits = [apple, apple, apple, apple, orange, orange, lemon]

Then use whatever functions you have at your disposal to pick a random element from that list (e.g. generate a random index within the proper range). This is of course not very memory efficient and requires integer weights.

Another, slightly more complicated approach would look like this:

Calculate the cumulative sums of weights:
```
intervals = [4, 6, 7]
```

Where an index of below 4 represents an apple, 4 to below 6 an orange and 6 to below 7 a lemon.

Generate a random number n in the range of 0 to sum(weights).
Go through the list of cumulative sums from start to finish to find the first item whose cumulative sum is above n. The corresponding fruit is your result.

This approach requires more complicated code than the first, but less memory and computation and supports floating-point weights.

For either algorithm, the setup-step can be done once for an arbitrary number of random selections.

Algorithms – Using Pi as a Crude Random Number Generator

Digging from http://www.befria.nu/elias/pi/binpi.html to get the binary value of pi (so that it was easier to convert into bytes rather than trying to use decimal digits) and then running it through ent I get the following for an analysis of the random distribution of the bytes:

Entropy = 7.954093 bits per byte.

Optimum compression would reduce the size of this 4096 byte file by 0 percent.

Chi square distribution for 4096 samples is 253.00, and randomly would exceed this value 52.36 percent of the times.

Arithmetic mean value of data bytes is 126.6736 (127.5 = random).

Monte Carlo value for Pi is 3.120234604 (error 0.68 percent).

Serial correlation coefficient is 0.028195 (totally uncorrelated = 0.0).

So yes, using pi for random data would give you fairly random data... realizing that it is well known random data.

From a comment above...

Depending on what you are doing, but I think you can use the decimals of the square root of any prime number as a random number generator. These should at least have evenly distributed digits. – Paxinum

So, I computed the square root of 2 in binary to undetake the same set of problems. Using Wolfram's Iteration I wrote a simple perl script

#!/usr/bin/perl
use strict;
use Math::BigInt;

my $u = Math::BigInt->new("2");
my $v = Math::BigInt->new("0");
my $i = 0;

while(1) {
    my $unew;
    my $vnew;

    if($u->bcmp($v) != 1) { # $u <= $v
        $unew = $u->bmul(4);
        $vnew = $v->bmul(2);
    } else {
        $unew = ($u->bsub($v)->bsub(1))->bmul(4);
        $vnew = ($v->badd(2))->bmul(2);
    }   

    $v = $vnew;
    $u = $unew;

    #print $i,"  ",$v,"\n";
    if($i++ > 10000) { last; }
}

open (BITS,"> bits.txt");
print BITS $v->as_bin();
close(BITS);

Running this for the first 10 matched A095804 so I was confident I had the sequence. The value v_n as when written in binary with the binary point placed after the first digit gives an approximation of the square root of 2.

Using ent against this binary data produces:

Entropy = 7.840501 bits per byte.

Optimum compression would reduce the size
of this 1251 byte file by 1 percent.

Chi square distribution for 1251 samples is 277.84, and randomly
would exceed this value 15.58 percent of the times.

Arithmetic mean value of data bytes is 130.0616 (127.5 = random).
Monte Carlo value for Pi is 3.153846154 (error 0.39 percent).
Serial correlation coefficient is -0.045767 (totally uncorrelated = 0.0).

Best Answer

Related Solutions

How to Get a Weighted Random Item in Ruby

Algorithms – Using Pi as a Crude Random Number Generator

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