I am writing a simulation code in MATLAB. I am using a set of formulas
called 'Pacejka 94', they are formulas used to simulate a vehicle
tire.

The code I am copying from is 'I believe' C++, but it may be something
else similar. The text I am copying the code from does not
distinquish.

The script in question is:

E=((B6*FZ^2+B7)*FZ+B8)*(1.D0-(B13*SIGN(1.D0,X1))))

The real issue I have is with the variable 1.D0 or 1.d0. What does
this stand for? I have a variable named D already, and it may be the
initial value for D, but I still don't know what the 1 and period are
for. Is this a single variable, a dot product, or?

Also, I do not understand the function 'SIGN()'. MATLAB has a SIGN
function but it only operates for a single variable, not a pair (as it
seems to be). Can you explain the SIGN function.

I understand all the other variables in the equation (X1, B6, FZ,
etc), they are just integers.

The D is used instead of E  to express double precision numbers, e.g.
in FORTRAN, as shown here
REAL X
DATA X /.01/
IF ( X * 100.d0 .NE. 1.0 ) THEN
   PRINT *, 'Many systems print this surprising result. '
ELSE
   PRINT *, 'And some may print this.'
ENDIF
http://www.lahey.com/float.htm

FORTRAN also has sign with 2 arguments, explained e.g. here
Sign(A, B)

 Sign: INTEGER or REAL function, the exact type being the result of
cross-promoting the types of all the arguments.
Returns `ABS(A)*s', where s is +1 if `B.GE.0', -1 otherwise. 
http://cclib.nsu.ru/projects/gnudocs/gnudocs/g77/g77_298.html

or here 
The function sign in Fortran is called the sign transfer function. It
is a function of two variables, sign(x,y)
If we substitute x = 1 in the sign transfer function, we get the sign of y; 
http://www.math.hawaii.edu/lab/197/fortran/fort4.htm

So, it looks like Hans was writing in Fortran
http://www.amazon.co.uk/exec/obidos/ASIN/0750651415/wwwlink-software-21/026-9120172-4002819

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Request for Answer Clarification by cping3-ga on 08 Jul 2004 22:00 PDT

Ok, so the 1.d0 is some kind of precision, and sign() is some kind of
an assignable negative.

So if I understand this correctly:

(1.D0-(B13*SIGN(1.D0,X1)))

is roughly equivalent to

(1-(B13*-1)) if X1 is negative
(1-(B13*1)) if X1 is positive

Is this what you see as well? I am very rusty on my Fortran. The
double precision call-out is simply so you can do the math in Fortran
correct? (It won't subtract a double precision from an integer).

Thanks

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Clarification of Answer by hedgie-ga on 09 Jul 2004 08:26 PDT

Basically yes.
You can do calculation with single precision, but it is
recommended and customary to do numerical, nontrivial ones in
double, to get reasonable accuracy.
http://www.webopedia.com/TERM/D/double_precision.html
 "...It won't subtract a double precision from an integer ."
and neither it will subtract single precision (both are floats)
(so double in FORTAN is long float in c) .You need to cast your integer to
suitable float before add/subtract from floats.
 So what your wrote  is basically correct, except for the
type (you got integer) and he got that same number as a double float.

Very simply,  you wrote 1  , which, when cast as single precision number 
is                      1.00000000000
and as double           1.000000000000000000000000000