I don't know of any sources, but here's some background.
Let's use smaller numbers to demonstrate. Let's say we have 4 bit numb
+ers.
+++++
    
+++++
As a little refresher, let's forget decimal numbers for a second.
+++++
 0  1  1  1  7 = 0*2^3 + 1*2^2 + 1*2^1 + 0*2^0
+++++
I'm going to call that B4, since 4 bits are used by the integer portio
+n of the number.
+++++
 0  1  1  1  7 B4
+++++
So how would you store a decimal number? Let's take 1.75, for example.
+++++ + +
 0  0  0  1  1 1 1.75 = 1*2^0 + 1*2^(1) + 1*2^(2)
+++++ + +
Unfortunately, I can't just add bits to my register.
What if we shifted the bits?
+++++
 0  1  1  1  1.75 * 2^2 = [ 1*2^0 + 1*2^(1) + 1*2^(2) ] * 2
+^2
+++++
Another way of phrasing that is in terms of number of integer bits lef
+t.
+++++
 0  1  1  1  1.75 B2 = [ 1*2^0 + 1*2^(1) + 1*2^(2) ] * 2^(4
+2)
+++++
In other words, using smaller Bscalings increases precision:
The following show the precision for a given scaling.
+++++
 0  0  0  1  [ 1*2^0 ] * 2^(44) = 1 B4
+++++
+++++
 0  0  0  1  [ 1*2^(1) ] * 2^(43) = 0.5 B3
+++++
+++++
 0  0  0  1  [ 1*2^(2) ] * 2^(42) = 0.25 B2
+++++
+++++
 0  0  0  1  [ 1*2^(3) ] * 2^(41) = 0.125 B1
+++++
+++++
 0  0  0  1  [ 1*2^(4) ] * 2^(40) = 0.0625 B0
+++++
Nothing says we have to stop at 0.
+++++
 0  0  0  1  [ 1*2^(5) ] * 2^(41) = 0.03125 B1
+++++
The downside is that biggest number we can represent goes down
as the precision increases.
+++++
 1  1  1  1  [ 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0 ] *
+2^(44) = 15 B4
+++++
+++++
 1  1  1  1  [ 1*2^2 + 1*2^1 + 1*2^0 + 1*2^(1) ] *
+2^(43) = 7.5 B3
+++++
+++++
 1  1  1  1  [ 1*2^1 + 1*2^0 + 1*2^(1) + 1*2^(2) ] *
+2^(42) = 3.75 B2
+++++
+++++
 1  1  1  1  [ 1*2^0 + 1*2^(1) + 1*2^(2) + 1*2^(3) ] *
+2^(41) = 1.875 B1
+++++
+++++
 1  1  1  1  [ 1*2^(1) + 1*2^(2) + 1*2^(3) + 1*2^(4) ] *
+2^(40) = 0.9375 B0
+++++
+++++
 1  1  1  1  [ 1*2^(2) + 1*2^(3) + 1*2^(4) + 1*2^(5) ] *
+2^(41) = 0.46875 B1
+++++
We can go the other way, if need be.
+++++
 1  1  1  1  [ 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 ] *
+2^(45) = 30 B5
+++++
+++++
 0  0  0  1  [ 1*2^1 ] * 2^(45) = 2 B5
+++++
<blockquote><i>How does multiplying by 4294967296 not include 1 but mu
+ltiplying by 2147483648 does?</i></blockquote>
Multiplying by 4294967296 (2^32) gives 32 bits for the decimal portion
+,
leaving no bits (3232) for the integer portion. In other words, B0.
1 does not fit into no bits.
+ ++
1  0  ... 1 B0
+ ++
Multiplying by 2147483648 (2^31) gives 31 bits for the decimal portion
+,
leaving 1 bit (3231) for the integer portion. In other words, B1.
1 fits in one bit.
+++
 1  0  ... 1 B1
+++
So what's the advantage over floatingpoint numbers?
Floatingpoint numbers does this the same way, and it does it for you.
However, in order to do that, it must save the scaling (called "expone
+nt")
along with the number. That means floats require more memory to store
+than these.
So what good is any of this anyway?
You can use integer arithmetic on these numbers!
X Bi + Y Bi = (X+Y) Bi
X Bi  Y Bi = (XY) Bi
X Bi * Y Bj = (X*Y) B(i+j)
X Bi / Y Bj = (X/Y) B(ij)
X Bi >> j = X B(i+j)
X Bi << j = X B(ij)
You can also compare numbers of the same scaling.
X Bi < Y Bi
X Bi > Y Bi
X Bi == Y Bi
etc.
Do be careful about overflow!
A = ... # B1
B = ... # B1
C = A + B # POTENTIAL OVERFLOW!
# 1 B1 + 1 B1 will overflow, for example
There are two ways of avoiding overflow.
You can make sure in advance that the numbers won't cause an overflow,
+ or
you can switch to a different Bscaling (at the cost of a bit of preci
+sion).
A = ... # B1
A = A >> 1 # B2
B = ... # B1
B = B >> 1 # B2
C = A + B # B2
We do lots of works with values in [0.0..1.0]. When we need to calcula
+te by
how much a valve should be open, we calculate it using values in [0.0.
+.1.0].
Later on, we convert them to the right value to send to the Analog Out
+put.
It sounds like you want to deal with numbers in this same range, so I
+thought
the following would be very useful to you:
A = ... # B1 Must be between 0.0 and 1.0.
B = ... # B1 Must be between 0.0 and 1.0.
C = A * B # B2
C = C << 1 # B1 Safe, since 1.0 * 1.0 = 1.0.
Although we've only dealt unsigned numbers, everything
I've said so far applies to signed 2s complement numbers.
