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Sports » rec.sport.golf » Smash factor
| Smash factor [message #1033495] |
Fr, 19 Mai 2006 23:39 |
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Can someone explain the physics of the so_called "smash factor". The
'formula' states that it is ball speed over club head speed (and result in
indices in the 1.2 ish to 1.4 ish range). That is all well and good ......
but how is that ball speed exceeds club head speed ???? Is that possible
???? I assume that compression must be adding velocity ...... but how is
that possible (does that not defy the "to every action there is an opposite
an equal reaction" ?? Or is this (hitting a golf ball) some weird / complex
transfer of energy concept ???? It takes energy to accelerate the golf ball
..... so this must have something to do with the masses involved. i.e. large
club head .... some of that energy (of swing) at impact is used to compress,
.... and the compression release "ADDS" velocity to the ball as it leaves the
driver (club).
The smash factor in itself is all well and good and yet another set of data
for us to get all hung up on ........ but can ANYONE explain the physics of
the ball speed being greater than the club speed ???
tia
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| Re: Smash factor [message #1033518 ] |
Sa, 20 Mai 2006 01:13 |
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You've got a very massive object colliding at high
speed with a light object. The massive object,
the golfer plus club spinning with all his might,
can impart a fraction of his total energy
and momentum to the golf ball, and the golf ball
because it so light has to take off at a higher
speed than the incoming clubhead. It's all
according to the laws of physics, and I don't
think you have to worry too much about C.O.R.
and the ball technology. High C.O.R. clubs and hot
balls just help the experiment approach Newton's
ideal.
With perfect clubs and balls,
v (golf ball) = the sum of the incoming and outgoing
clubhead speed.
So if the clubhead has any speed
at all after impact, the golf ball's exit speed
will exceed the clubhead's incoming speed.
I think that means the max. smash factor with perfect
everything would be 2.0.
If you just fired a 1.6 oz clubhead (with no golfer
attached) at the ball, then you would be right, the
outgoing velocity of the golfball could be no higher
than the incoming velocity of the clubhead.
Good compression technology in
the golf ball doesn't really add any
energy. It preserves it. You know that if you
drop a golf ball on a cement floor, it will bounce
back up lower than the height you dropped it from.
So there is lost energy even with a hot ball.
When you assume that the golf ball can't fly out
faster than the incoming clubhead, you are really
making a very natural assumption, "Speed is
conserved.", ie the sums of all the speeds before
and after impact must be the same. This is close,
but the real thing that is conserved (mostly) is momentum,
the product of mass and speed. So with a big
difference in masses between the clubhead/golfer and
the ball, the incoming and outgoing speeds of the club
and ball may not follow your intuition.
Rod's news wrote:
> Can someone explain the physics of the so_called "smash factor". The
> 'formula' states that it is ball speed over club head speed (and result in
> indices in the 1.2 ish to 1.4 ish range). That is all well and good ......
> but how is that ball speed exceeds club head speed ???? Is that possible
> ???? I assume that compression must be adding velocity ...... but how is
> that possible (does that not defy the "to every action there is an opposite
> an equal reaction" ?? Or is this (hitting a golf ball) some weird / complex
> transfer of energy concept ???? It takes energy to accelerate the golf ball
> .... so this must have something to do with the masses involved. i.e. large
> club head .... some of that energy (of swing) at impact is used to compress,
> ... and the compression release "ADDS" velocity to the ball as it leaves the
> driver (club).
>
> The smash factor in itself is all well and good and yet another set of data
> for us to get all hung up on ........ but can ANYONE explain the physics of
> the ball speed being greater than the club speed ???
>
> tia
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| Re: Smash factor [message #1033523 ] |
Sa, 20 Mai 2006 02:17 |
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>I assume that compression must be >adding velocity .....
=3D=3D=3D=3D=3D
my assumption would be that the ball's spinning adds velocity..
>mho
>v=83e
>drive 10% less,
>don't discover safety by accident
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| Re: Smash factor [message #1033566 ] |
Sa, 20 Mai 2006 16:37 |
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<fiveiron [at] webtv.net> wrote in message
news:4225-446E6011-950 [at] storefull-3317.bay.webtv.net...
>I assume that compression must be >adding velocity .....
=====
my assumption would be that the ball's spinning adds velocity..
____________________________________________________________
I don'tthink so. I may be inclined to buy into 'blueposts' explanation about
preservation of momentum ...... but I can't believe that 'spin' adds
velocity (spin can add lift ..... or enhance that cursed slice :(
And I'm not completely sold on blueposts momentum explanation. Surely there
must be some physics experts in here :) The explanation must lie in the
momentum transfer (club to ball). i.e. not ALL of the club and golfers
momentum are transferred to the ball as velocity and mass of the ball
(obviously that is the case because the club head continues to 'swing' AFTER
impact). It just seems to me that some of that club head energy MUST be
taken up in compressing the ball ..... and as the ball 'pops' / returns to
round - if you like .... it adds acceleration forces .. and thereby
explaining a ball speed greater than that of the club head.
>mho
>vfe
>drive 10% less,
>don't discover safety by accident
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| Re: Smash factor [message #1033570 ] |
Sa, 20 Mai 2006 17:24 |
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Ok, so my physics degree is almost 30
years old. I'm not an expert in mechanics,
but who is anymore?
For the case where momentum and energy are
conserved, that is perfectly elastic balls
and clubfaces, you write down two sets
of equations, divide, and get for the
velocities just before and after impact:
v(ci) + v(cf) = v(bf) + v(bi),
Where c stands for clubhead, b = ball,
i= initial, and f = final
This follows "Physics" by Halliday
and Resnick, 1967, pg 216.
Since the ball is stationary before
impact, v(bi) = 0, so that's how you get
the result that v(bf) = v(ci) + v(cf),
i.e. the ball can fly away with greater
than the incoming clubhead velocity.
I just dropped a hard golf ball from
60" onto cement. It rebounded back up
to 45". That 15" is "lost" energy that
went into compressing the ball. The
standard explanation is that this type
of "lost" energy ends up in the ball as
heat which then gets radiated out into
the air (if the ball is warmer than the
air).
So the golf ball isn't a perfect receiver
of the clubhead's energy -- that means
you'll never get a smash factor of 2.0,
the theoretical maximum.
Regarding spin, as I understand it
the right amount of spin
and dimples help to *preserve* the velocity
of the golf ball. Presumably the golf
ball velocity is measured just after
impact, so spin wouldn't have had time
to do much anyway.
I did another post where I tried to estimate
the relative energy of the golfer/club and
the ball.
http://groups.google.com/group/rec.sport.golf/browse_frm/thr ead/b2220fcf237f8b79/4c2f452328ae67bb?lnk=st&q=ft-lbs+gr oup%3A*golf*&rnum=1&hl=en#4c2f452328ae67bb
The title of the thread was "overcoming inertia".
The answer I got was that the golfer/club has
about 10x the kinetic energy of the exiting
golf ball. (675 ft-lbs vs 75 ft-lbs). So there
is plenty of excess energy to power the ball
and still keep the clubhead swinging on
to follow-through. Only a fraction of the
golfer's incoming energy is transferred to
the ball.
In our everyday lives we don't see
collisions that are elastic enough to demonstrate
the fact that the less massive object can
exit with higher velocity than the incoming
more massive object. In most such collisions
we see, something crumples, ruining the
elasticity.
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| Re: Smash factor [message #1033590 ] |
Sa, 20 Mai 2006 22:28 |
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Don't hold your breath waiting for a response from tire iron.
"Rod's news" <rod.gram [at] shaw.ca> wrote in message
news:tQFbg.176699$P01.67983 [at] pd7tw3no...
>
> <fiveiron [at] webtv.net> wrote in message
> news:4225-446E6011-950 [at] storefull-3317.bay.webtv.net...
>>I assume that compression must be >adding velocity .....
> =====
> my assumption would be that the ball's spinning adds velocity..
>
> ____________________________________________________________
>
> I don'tthink so. I may be inclined to buy into 'blueposts' explanation
> about preservation of momentum ...... but I can't believe that 'spin' adds
> velocity (spin can add lift ..... or enhance that cursed slice :(
>snip
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| Re: Smash factor [message #1033599 ] |
So, 21 Mai 2006 00:07 |
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"Rod's news" <rod.gram [at] shaw.ca> wrote in message
news:2Wqbg.175356$7a.44415 [at] pd7tw1no...
> Can someone explain the physics of the so_called "smash factor". The
> 'formula' states that it is ball speed over club head speed (and result in
> indices in the 1.2 ish to 1.4 ish range). That is all well and good ......
> but how is that ball speed exceeds club head speed ???? Is that possible
> ???? I assume that compression must be adding velocity ...... but how is
> that possible (does that not defy the "to every action there is an
opposite
> an equal reaction" ?? Or is this (hitting a golf ball) some weird /
complex
> transfer of energy concept ???? It takes energy to accelerate the golf
ball
> .... so this must have something to do with the masses involved. i.e.
large
> club head .... some of that energy (of swing) at impact is used to
compress,
> ... and the compression release "ADDS" velocity to the ball as it leaves
the
> driver (club).
>
> The smash factor in itself is all well and good and yet another set of
data
> for us to get all hung up on ........ but can ANYONE explain the physics
of
> the ball speed being greater than the club speed ???
>
> tia
>
>
If you model the impact of a golf club on a golf ball as a free body
collision between the clubhead and the ball (a reasonable assumption
according to Jorgensen in "The Physics of Golf") and make it a 'no
loft/square' impact (for simplicity), you'll find the following.
1) The quantity that is conserved is momentum (mass*velocity) before/after
impact
2) The velocity of separation between the club and the ball (speed of the
club minus the speed of the ball) after impact divided by the velocity of
separation before impact is the COR (coefficient of restitution). Assuming
that the masses, the velocity of the clubhead before impact, and the COR are
known, these two equations are all that is required to solve for the
velocity of the ball and clubhead after impact. Note that if the COR =1 then
energy will be conserved, FWIW. The ROG COR limit of 0.83 is for a clubhead
and some kind of a steel ball. I have no idea what the COR between a
clubhead and a golf ball is. It is my understanding that the collision of
two perfectly round diamonds is very close to a COR of 1. I'd be happy to
run the experiment if someone will supply me with the required materials :-)
3) For the case of the COR being relatively high and the mass of the
clubhead being a good deal larger than the mass of a golf ball, the clubhead
and golf ball both go in a forward direction after impact. Assuming that the
clubhead wasn't broken by the ball, then the ball must end up going faster
than the clubhead is going after impact. The math says that for most cases
the ball will end up going faster than the clubhead was going before impact
as well.
4) For the interesting case of COR=1 and the mass of the ball = mass of the
clubhead, the clubhead will come to a complete stop after impact with all
the energy being transferred to the ball.
5) For high COR's and the ball having a good deal more mass than the
clubhead, the clubhead will actually bounce back after impact (note the
beginning assumption of free body collision).
Why does it work this way? That is what the mathematics says that it does
and it is all part of what Eugene Wigner (early 20th century physicist)
called the "incomprehensible success of mathematics in explaining physics".
dave
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| Re: Smash factor [message #1033622 ] |
So, 21 Mai 2006 04:27 |
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ok ....... I'm starting to believe :) (regarding the preservation of
momentum ..... otherwise shoulders and shotguns couldn't mix :)
But there remains the unknown of "how much of the 'clubs initial' (i.e.
prior to collision) momentum remains in the club ???
Is the 'sweet spot' technology all about maximizing that transfer of
momentum ??????
...... and finally (I promise :) .... can compression of the ball (well,
actually the re_forming of the ball to spherical AFTER contact) "add"
velocity (to the ball's vector) ...... or does this entire discussion boil
down to two bodies, a collision, .... and preservation of momentum ???
<bluepost22 [at] yahoo.com> wrote in message
news:1148138688.293533.63880 [at] u72g2000cwu.googlegroups.com...
> Ok, so my physics degree is almost 30
> years old. I'm not an expert in mechanics,
> but who is anymore?
>
> For the case where momentum and energy are
> conserved, that is perfectly elastic balls
> and clubfaces, you write down two sets
> of equations, divide, and get for the
> velocities just before and after impact:
>
> v(ci) + v(cf) = v(bf) + v(bi),
>
> Where c stands for clubhead, b = ball,
> i= initial, and f = final
>
> This follows "Physics" by Halliday
> and Resnick, 1967, pg 216.
>
> Since the ball is stationary before
> impact, v(bi) = 0, so that's how you get
> the result that v(bf) = v(ci) + v(cf),
> i.e. the ball can fly away with greater
> than the incoming clubhead velocity.
>
> I just dropped a hard golf ball from
> 60" onto cement. It rebounded back up
> to 45". That 15" is "lost" energy that
> went into compressing the ball. The
> standard explanation is that this type
> of "lost" energy ends up in the ball as
> heat which then gets radiated out into
> the air (if the ball is warmer than the
> air).
>
> So the golf ball isn't a perfect receiver
> of the clubhead's energy -- that means
> you'll never get a smash factor of 2.0,
> the theoretical maximum.
>
> Regarding spin, as I understand it
> the right amount of spin
> and dimples help to *preserve* the velocity
> of the golf ball. Presumably the golf
> ball velocity is measured just after
> impact, so spin wouldn't have had time
> to do much anyway.
>
> I did another post where I tried to estimate
> the relative energy of the golfer/club and
> the ball.
>
> http://groups.google.com/group/rec.sport.golf/browse_frm/thr ead/b2220fcf237f8b79/4c2f452328ae67bb?lnk=st&q=ft-lbs+gr oup%3A*golf*&rnum=1&hl=en#4c2f452328ae67bb
>
> The title of the thread was "overcoming inertia".
>
> The answer I got was that the golfer/club has
> about 10x the kinetic energy of the exiting
> golf ball. (675 ft-lbs vs 75 ft-lbs). So there
> is plenty of excess energy to power the ball
> and still keep the clubhead swinging on
> to follow-through. Only a fraction of the
> golfer's incoming energy is transferred to
> the ball.
>
> In our everyday lives we don't see
> collisions that are elastic enough to demonstrate
> the fact that the less massive object can
> exit with higher velocity than the incoming
> more massive object. In most such collisions
> we see, something crumples, ruining the
> elasticity.
>
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| Re: Smash factor [message #1033623 ] |
So, 21 Mai 2006 04:44 |
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See if you can scare up a copy of "Search for the Perfect Swing" by
Chrochron et al. The authors were Brit physicists and wrote clearly
about the physics of swing and impact.
About balls moving faster than what hits it: think of ping pong balls,
or tennis balls in play. Better than that, think about any kind of ball
bouncing. The 'paddle' which is the floor, isn't moving very fast!
The thing that's conserved during impact is of course momentum, not
kinetic energy. The energy stored in the clubhead by its speed before
impact goes into the velocity of the ball and heat.
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| Re: Smash factor [message #1033630 ] |
So, 21 Mai 2006 05:59 |
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one guy sez, this is really a measure of your elasticity at impact. it's
a measure of
how much energy you transfer from the club head to the ball when you
swing.
>mho
>v=83e
>drive 10% less, help create a gas glut
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| Re: Smash factor [message #1033631 ] |
So, 21 Mai 2006 06:28 |
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"smash factor" - who come up with that term? just something else to
"kick around",
meaning nothing unless the contributing factors change,
and then you'll only see it in a more or less distance situation. there
are so many
variables in the hardware, and swings that it would be hard to get a
standard.
the smash factor read could be determined, but varying implementations
could and would give different results.
"a" ball dropped from a 5' height bounces back to within 15" of it's
starting point is not
and experiment for ascertaining facts, some balls when dropped will
return to their starting point.
and of course some balls never return to their starting point.:--)
>mho
>v=83e
>drive 10% less, help create a gas glut
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| Re: Smash factor [message #1033634 ] |
So, 21 Mai 2006 13:31 |
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"Tony" <ajw27703 [at] yahoo.com> wrote in message
news:1148179440.234948.241330 [at] i39g2000cwa.googlegroups.com...
> See if you can scare up a copy of "Search for the Perfect Swing" by
> Chrochron et al. The authors were Brit physicists and wrote clearly
> about the physics of swing and impact.
>
> About balls moving faster than what hits it: think of ping pong balls,
> or tennis balls in play. Better than that, think about any kind of ball
> bouncing. The 'paddle' which is the floor, isn't moving very fast!
>
> The thing that's conserved during impact is of course momentum, not
> kinetic energy. The energy stored in the clubhead by its speed before
> impact goes into the velocity of the ball and heat.
>
I found a copy of "Search for the Perfect Swing" at a used book store for
$3. A great book.
dave
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| Re: Smash factor [message #1033688 ] |
So, 21 Mai 2006 22:23 |
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On Fri, 19 May 2006 21:39:10 GMT, "Rod's news" <rod.gram [at] shaw.ca>
wrote:
>
>The smash factor in itself is all well and good and yet another set of data
>for us to get all hung up on ........ but can ANYONE explain the physics of
>the ball speed being greater than the club speed ???
Inertia has to be conserved. Inertia is mass times velocity. The
effective mass of the club head times its velocity + the mass of the
ball times its velocity has to be the same before and after the ball
strike.
If the club head had the exact same mass as the ball, then it is
possible for it to hit the ball, stop short, and the ball continues at
the speed the club was moving. This is "effective" mass, as the
club head is attached to some other masses that contribute a bit.
Hitting the ball doesn't have to stop the club head, slowing it down
still passes momentum to the ball. Since the ball is so light, we
see that momentum with a high velocity component.
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| Re: Smash factor [message #1033689 ] |
So, 21 Mai 2006 22:24 |
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On 19 May 2006 16:13:12 -0700, bluepost22 [at] yahoo.com wrote:
>With perfect clubs and balls,
>
>v (golf ball) = the sum of the incoming and outgoing
> clubhead speed.
Velocity isn't conserved, momentum is.
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| Re: Smash factor [message #1033702 ] |
So, 21 Mai 2006 23:59 |
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On Sun, 21 May 2006 20:23:36 GMT, Howard Brazee <howard [at] brazee.net>
wrote:
>
>Inertia has to be conserved. Inertia is mass times velocity. The
>effective mass of the club head times its velocity + the mass of the
>ball times its velocity has to be the same before and after the ball
>strike.
My bad. Replace "Inertia" with "Momentum". I had a brain fart
while proof-reading.
Sorry.
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