From: ram@tiac.net (robert a. moeser)
Newsgroups: sci.math
Subject: Re: Equation for a  Baseball's Seam
Date: Sat, 20 Jul 1996 13:07:18 -0500

In article <31EFCDFE.4416@transport.com>, balou@transport.com wrote:

:What would be the equation for the seam of a baseball?  I've looked at 
:the knot pages off of Yahoo but didn't see what I was looking for.  Any 
:idea what this curve might be called?

surf

http://www.mathsoft.com/asolve/baseball/baseball.html

for all (!) you ever wanted to know about making a baseball.


-- rob

(tried email to balou@transport.com, no go!)

==============================================================================

From: jasonp@wam.umd.edu (Jason Stratos Papadopoulos)
Newsgroups: sci.math
Subject: Re: baseball stitching pattern function?
Date: 20 Aug 1996 18:25:25 GMT

Try the following: I used this approach about two years ago, and it made
convincing pictures. The idea is to figure out parametric equations for
the angles of latitude and longitude (in terms of a parameter "t") and
then convert to spherical coordinates. Apologies for the lack of ascii
art, but drawing circles with a keyboard is too awkward for me. (Tracing
things out with a real baseball may help).

The latitude function is easy: measuring the angle as going DOWN from
the vertical, your baseball stitching will start at a positive latitude of
"a" radians at t=0 (the "crest" of one of the "hills") and zoom down to an
angle of pi-a radians at t=pi/2 (again as measured from the north pole;
this corresponds to hitting a "valley" at 1/4 turn around the baseball).
A suitable function to do this is  latitude(t) = pi/2 + (a - pi/2)cos(2t);
it's not too complicated, but choosing this function means that to get a
complete circuit of the baseball stitching t must vary between 0 and 2*pi.
Note also that the weird way of choosing the angle for latitude is to 
match conventions with spherical coordinates.

Now for the longitude: if you look "down" on a baseball from the "north
pole" and start at t=0 as above, the longitude starts at 0 radians and
moves to pi/2 radians (1/4 turn) but wiggles back and forth once, with the
wiggle centered at pi/4 radians. It then does the same from pi/2 to pi 
radians, and so forth around the baseball. Draw the longitude out and you
will find a good match in longitude(t) = t + b sin(2t), where b is a 
parameter that controls how much the original poster's "semicircles"
flare out. Fiddling with b (and with a above) produces a whole family of
baseball-like curves.

Finally, converting to cartesian from spherical coordinates and assuming 
the baseball has radius r,

lat(t) = pi/2 + (a-pi/2)cos(2t)
long(t) = t + b sin(2t)

and so ...
    x(t) = r sin( lat(t) )  sin( long(t) )
    y(t) = r sin( lat(t) )  cos( long(t) )
    z(t) = r cos( lat(t) )

And there you have it. My only regret is needing 2 weeks to see all this
the first time! If I remember correctly, you can get a semi-realistic
baseball for a~=.2 or .3 and b~=.4 or .6

Hope this helps.
jasonp