I'm sure I'm about to be heavily laughed at, but here goes anyways. I'm trying to understand why a square root is needed to normalize a vector. When I normalize using the square root of the sum of sqaures, I get values that are unequal to 1. They are close to 1, but always off a bit. Is a unit 3-dimensional vector's length not quite the sum of the absolute axes? In other words, is abs(x) + abs(y) + abs(z) the length? If not, what is happening with sqrt(x * x + y * y + z * z) that makes the length correct? I did a test with a brute force negative check: The sqrtf method takes about 5 seconds to execute this, and the caveman (likely incorrect) method takes less than 1 second. Using the abs() functions instead of conditional checks takes around 13 seconds [lol]. The sum of absolute axes is always exactly 1.0 with the negative checks. The compiler may also be doing something that speeds up the negative checks. Hopefully not. Can anyone explain what is incredibly stupid about what I'm doing here? I'm sure there are several things [smile] Thanks! [Edited by - Jiia on October 16, 2004 10:40:11 PM]

I'm sure you are after the Euclidean length, which is indeed

euc = sqrt(x*x+y*y+z*z);


"In other words, is abs(x) + abs(y) + abs(z) the length?"
Not the Euclidean length.

Note:

sqrt(x*x+y*y+z*z) does not equal abs(x)+abs(y)+abs(z)

"When I normalize using the square root of the sum of sqaures, I get values that are unequal to 1..."
When you find the length of the normalised vector it should be equal to one.

Not quite sure what your problem is...?

Quote:Original post by Kuladus
Not quite sure what your problem is...?

You mean I'm an idiot / nutcase and there is little help for me? [lol]

I didn't really have a problem, other than being confused. The idea of a unit vector was throwing me off. I'm thinking unit = 1. But it's one distance point, not that the sum is equal to 1.

So the actual distance from vector(0,0,0) to vector(5,5,5) is not 15, it's around 8.5 or so, and my sum normalization version is way off. Actually, I must be having a really bad day to think this was right.

Thanks for trying to make sense of my half-unbaked idea [smile]

Not completely half-baked :)

What you were describing is the L1 vector norm (whereas the Euclidean distance is the L2 norm.)

Quote:Original post by Kuladus
Not completely half-baked :)

[lol] Perhaps a quarter-baked?

Quote:What you were describing is the L1 vector norm (whereas the Euclidean distance is the L2 norm.)

Only by coincidence. I really had no idea what I was babbling about [smile] Are there actually uses for this L1 norm? Using this as a direction would push objects more slowly in diagonal directions, right?

Thanks again.

Why sqrt(x*x+y*y+z*z) is a length of vector in Euclidean space :
Read somewhere about Pythagorean theorem.(google.)

***********************************
abs(x) + abs(y) + abs(z)
is, IIRC, called Manhattan distance.
In 2D it's

 0 * **   *  *****X,Y

it's number of stars in such ascii path.(note that number of stars doesn't depend to path as long as you don't go back.)
It's not equal to sqrt(x*x+y*y+z*z)

And as about making it be unit-length. If you normalize using /(abs(x) + abs(y) + abs(z)) you make vector be placed on surface of "unit octohedron", surface where (abs(x) + abs(y) + abs(z))=1
Using max(abs(x) , abs(y) , abs(z)) you make it be placed on surface of unit cube, surface where max(abs(x) , abs(y) , abs(z))=1

Using sqrt(x*x+y*y+z*z) , on unit sphere, surface where sqrt(x*x+y*y+z*z)=1

(this =1 will only work for formules that works like normalization - if you normalize once, or normalize twice,result is the same)

**********

if in some space(euclidean or not) you have some norm of vector given by L(x,y,z) , and L(k*x,k*y,k*z)=k*L(x,y,z) , normalization
can be done by nvector=vector*(1/L(vector)) , so we have
L(nvector)=L(vector*(1/L(vector)))=L(vector)*(1/L(vector))=1

and if L(something)=1 , something is said to be unit-length... err, unit-norm, normalized.