Simple way of making Golf Ball with Solidworks!

Please help me!
Anyone!?
As I know Golf ball has 77 dimples around the surface...
Is there anyway making golfball simply?
I've read lot's of articles with lots of calculations...but it brings
me a headache.

Please..help me out.~
Thank you.

 

Search the newsgroup first. Then check 3D Content Central.

 

Just curious - why is this so popular it comes up every month?
It seems like a largely boring problem that can be solved by any
beginner with just a ltitle patience. I suppose it would be mildly
interesting if there was a battle to see who could pull it off in the
fewest features, but even then it still just a ball with a bunch of
revolved cuts in it.

I just think I must be missing something, because it is endlessly
fascinating to folks on this newsgroup. Or is it just that Spring is
on its way in the Northern Hemisphere
Sign me 'curious'
-Ed

 

ALL of them?
Is it a Scottish standard?

 

Have you actually tried it?

 

Usually 300-500. 336 is very common.
That's the N.Z. standard

Greg

 

http://www.answers.com/topic/golf-ball

 

abab,
77 dimples? Not even close. Of course all manufacturers have a different
design and number of dimples, but mostly they average around 330. There are
approximately 33 at the equator of the ball, then naturally the number
diminishes as the smaller perimeters migrate towards either of the two the
poles.There are approximately 10 perimeters, growing progressively smaller,
and running paralell to the equator, running north and south. It shouldn't
be a very difficult mathematical indexing for you to do, but I wouldn't go
so far as to say that it is "simple".
Good Luck
G. De Angelis

 

A stacking or nesting problem. They are not uniformly spaced ..... are
they even all the same size?
The largest regular (Platonic) 3D solid is a regular icosahedron
which only has 20 faces IIRC. You'd need ....
[
Most balls on sale today have about 300 to 450 dimples. There were a few balls
having over 500 dimples before. The record holder was a ball with 1,070 dimples
-- 414 larger ones (in four different sizes) and 656 pinhead-sized ones. All
brands of balls, except one, have even-numbered dimples. The only odd-numbered
ball on market is a ball with 333 dimples.
] from TOP's link : http://www.answers.com/topic/golf-ball <g>.

 

(On 2 Mar 2006 16:22:39 -0800, wrote:

Have you actually tried it?

Cliff )

---------------------------

If that is the Ed I think it is, believe me, it would be a boring
problem to him!

I think its one of those "fun" kinda issues in that the dimples on a
golf ball has many myths about why they even exist. I like Ed's idea
about creating one in the fewest number of features. Sounds like a
possible upcomming contest for our usergroup!

 

A stacking or nesting problem. They are not uniformly spaced ..... are

I was able to model the golf ball depicted in
http://content.answers.com/main/content/wp/en/f/f0/Two_similar_icosahedron_golf_ball_designs.jpg

It is pretty straightforward if you realize that the dimples are based on
the icosahedron.You actually need to carve only four (4) dimples! The
toughest part was to find out how to pattern the geometry with minimum
amount of features.

Pretty interesting modeling problem, though!

-h-

 

Yeah, I noticed that too - four dimples, then figure out the
mirroring/rotating/patterning required to fill the sphere. Thanks Paul
for posting the link. I have some time to kill on an airplane tomorrow
so I might see how efficiently I can do it. Of course some guy (not
me) will come through with a two feature version (I still remember the
dodecahedron)...

But still, the question remains in my mind - what's with the
fascination with modeling a golf ball? I have faith that anyone could
do it, but it would be tedious. It is a mildly entertaining
time-killer to try to do it in very few features, but I can think of
much more compelling things out there to dink around with. Why a golf
ball? And why is it every month it comes back up?
-Ed

 

Ed,

What I was thinking it to model a circumscribed icosahedron and sketch
the dimple points on the face like on the link. Then project to the
surface of a sphere.

Strange that a gentleman's relaxation has had so much effort put into
it. I guess in the end they don't know how to relax.

The fascination is with the scientific principles involved.

 

I don't think you can project a sketch onto the sphere - it needs to
wrap to be accurate - ther will be distortions, even if they are only
slight. If its worth doing, its worth doing to ten decimal places.

 

If you look at the link it seems to be what they did on some golf
balls. You cannot project a sketch, but you can get points from a
sketch to project onto a surface. Very tedious, but doable.

 

Some time ago I read an article in a magazine talking about how many patents
involve golf. (It was probably in a plastics magazine and involved patents
related to plastics and plastic parts.) I can't remember the numbers now,
but it was absolutely amazing what the percentage was.

Jerry Steiger
Tripod Data Systems
"take the garbage out, dear"

 

The easiest way to do the dimples is to use the hole wizard. Hole wizard is
basically a revolve, and you can edit the hole profile to represent a
dimple. You can model a 1/6 piece of one icosahedron triangle, cut it to
sphere shape, use hole wizard to create four dimples and finally use body
mirrors and patterns to create the ball shape.

-h-

 

While a regular (platonic) icosahedron has 20 faces it has only 12 vertices.

[
Most balls on sale today have about 300 to 450 dimples. There were a few balls
having over 500 dimples before. The record holder was a ball with 1,070 dimples
-- 414 larger ones (in four different sizes) and 656 pinhead-sized ones. All
brands of balls, except one, have even-numbered dimples. The only odd-numbered
ball on market is a ball with 333 dimples.
] from TOP's link : http://www.answers.com/topic/golf-ball <g>.

G. De Angelis chose 330.

12=2*3*2^2
20=5*2^2

330=2*3*5*11
300=(2^2)*3*(5^2)
333=3*111
414= 2*207
450= 2*(3^2)*(5^2)
656 =(2^4)*41
1,070= 2*5*107

You may be able to design a golf ball based on icosahedrons
IF it has 300 dimples.
Not in the other cases I suspect though you might try using both
faces & vertexes .... 32 = 2^5 .... Nope.

Or am I wrong <G>?

 

Yes you are, indeed. Did you try to model it? I modeled the left golf ball
depicted in
http://content.answers.com/main/content/wp/en/f/f0/Two_similar_icosahedron_golf_ball_designs.jpg

It has one dimple per each icosahedron vertex, four per edge and six dimples
per face, eg. 12*1 + 20*6 + 30*4 = 252 dimples.

I don't know how to arrange the dimples to get exactly 300 dimples.
-h-

 

To begin with, 252 was not on the list <G>.

252=(2^2)*63 .....

To recap:
12=2*3*2^2
20=5*2^2

But you also used 6 per face and 4 per edge .. does that include
the vertex ones?
Probably not very uniform in spacing either ....

Someone might try lesser Platonic solids too ...