I'm not so good in English, so maybe you do not understand my note. I
like koeficient 1e7, because it is human readable form of coordinate
13.5 -> 135000000. I do not want to use (2^32/360).
I just react to difference between float and int. float has only 24 bits
of significant which will result in at least same error as int. Many
times not so good as int.
Example:
double a;
a = 89.00001;
printf("%.10f %.10f %.10f\n", (float)a, a, a - ((float)a));
a = 89.00002;
printf("%.10f %.10f %.10f\n", (float)a, a, a - ((float)a));
a = 89.0001;
printf("%.10f %.10f %.10f\n", (float)a, a, a - ((float)a));
89.0000076294 89.0000100000 0.0000023706
89.0000228882 89.0000200000 -0.0000028882
89.0000991821 89.0001000000 0.0000008179
So you can see that float has error on six place of number. But if you
use int, worst case is on eight.
So once more. It was just note between float and int....
Tom
Stefan de Konink napsal(a):
On Wed, 12 Nov 2008, Tomas Kolda wrote:
Integer has better precision if you are far from zero, because floats
use only 24 bits for significant, other bits is for sign and exponent.
So 32 bit integer is better if you know extends (-180, .... 180). Min
and max of int is -2.1e9 to 2.1e9. So if you multiply longitude and
latitude by 1e7 you get almost maximum from 32bit int in all range (from
-1.8e9 to 1.8e9).
It is also faster to process. And if you are doing some difference
compression, it is easier to do with ints.
Integers are only a better precision if all the bits are used. Thus only
improve the situation if a multiplication with 2^32 / 360 is used.
Now the minor detail is that this operation has an error of 10^-7, but the
more preciese the input the better the storage resolution.
89 * (2^32 / 360)
1061811359.28889 ---> 1061811359
1061811359 / (2^32 / 360) ---> 1061811359 * 360 / 2^32 (bit shift)
88.9999999757856
So what you suggest is also a naive method; but that method just killed
the most significant 10^-8 digit.
Basically it says;
360.0000000 =/ 3600000000
Now what was mentioned before; is the hard cut of a problem? Not if the
input is typically only significant till 10^-6. Should we overmodel it
now? Probably not until we have data that is that significant.
Stefan
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