| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
12 Jan 2005 08:16:40 AM |
| Object: |
Coefficient of Restitution (COR) for Driver |
Looking for simple experiment for measuring the COR of a golf club
surface. Any ideas?
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| User: "PD" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 09:27:13 AM |
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wrote:
Looking for simple experiment for measuring the COR of a golf club
surface. Any ideas?
I can measure a *relative* COR, compared to a heavy concrete surface.
1. Drop a golf ball on a thick concrete slab and measure the ratio R1
of the bounce height to the drop height.
2. Clamp a golf club to the slab. If you are interested only in the COR
of the club surface, then clamp the head tightly to the slab. If you
are interested in the COR of the club as a whole, including the shaft,
then it is only necessary to clamp it in a vise where the hands would
grip.
3. Drop a golf ball on the golf club surface and measure the ratio R2
of the bounce height to the drop height.
4. The COR of the golf club surface will be R2/R1.
There is a caveat and it is a warning about the care with which you
clamp the club to the floor. Exercise: Hold a basketball and a golf
ball above the ground, with the golf ball resting directly on top of
the basketball. Release both at the same time, letting them fall to the
floor together. What is the observed COR of the floor/basketball
combination in this case? (Safety tip: wear goggles, a mouthguard, and
a groin cup.)
PD
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| User: "Franz Heymann" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 01:30:44 PM |
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"PD" <pdraper@yahoo.com> wrote in message
news:1105543633.050066.128310@c13g2000cwb.googlegroups.com...
cnctutwiler@wmconnect.com wrote:
Looking for simple experiment for measuring the COR of a golf club
surface. Any ideas?
I can measure a *relative* COR, compared to a heavy concrete
surface.
1. Drop a golf ball on a thick concrete slab and measure the ratio
R1
of the bounce height to the drop height.
2. Clamp a golf club to the slab. If you are interested only in the
COR
of the club surface, then clamp the head tightly to the slab. If you
are interested in the COR of the club as a whole, including the
shaft,
then it is only necessary to clamp it in a vise where the hands
would
grip.
3. Drop a golf ball on the golf club surface and measure the ratio
R2
of the bounce height to the drop height.
4. The COR of the golf club surface will be R2/R1.
There is a caveat and it is a warning about the care with which you
clamp the club to the floor. Exercise: Hold a basketball and a golf
ball above the ground, with the golf ball resting directly on top of
the basketball. Release both at the same time, letting them fall to
the
floor together. What is the observed COR of the floor/basketball
combination in this case? (Safety tip: wear goggles, a mouthguard,
and
a groin cup.)
Your measurement has little to do wiith reality. In practice, the
club is held at the far end of the handle. In that case, the
behaviour of the whole handle is part of what determines the behaviour
of the club..
Franz
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| User: "PD" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 03:41:49 PM |
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Franz Heymann wrote:
"PD" <pdraper@yahoo.com> wrote in message
news:1105543633.050066.128310@c13g2000cwb.googlegroups.com...
cnctutwiler@wmconnect.com wrote:
Looking for simple experiment for measuring the COR of a golf
club
surface. Any ideas?
I can measure a *relative* COR, compared to a heavy concrete
surface.
1. Drop a golf ball on a thick concrete slab and measure the ratio
R1
of the bounce height to the drop height.
2. Clamp a golf club to the slab. If you are interested only in the
COR
of the club surface, then clamp the head tightly to the slab. If
you
are interested in the COR of the club as a whole, including the
shaft,
then it is only necessary to clamp it in a vise where the hands
would
grip.
3. Drop a golf ball on the golf club surface and measure the ratio
R2
of the bounce height to the drop height.
4. The COR of the golf club surface will be R2/R1.
There is a caveat and it is a warning about the care with which you
clamp the club to the floor. Exercise: Hold a basketball and a golf
ball above the ground, with the golf ball resting directly on top
of
the basketball. Release both at the same time, letting them fall to
the
floor together. What is the observed COR of the floor/basketball
combination in this case? (Safety tip: wear goggles, a mouthguard,
and
a groin cup.)
Your measurement has little to do wiith reality. In practice, the
club is held at the far end of the handle. In that case, the
behaviour of the whole handle is part of what determines the
behaviour
of the club..
I allowed for that in Step 2.
I'm actually not sure how much part that plays, and how much of a
contribution to the COR (or rather, to the deviation of the COR from 1)
it makes.
I would think that a good designer would choose a shaft material and
shape that would flex elastically but return virtually all of its
stored energy to the ball before the ball left the club face. I'm sure
the designer knows how much the shaft contributors to the COR.
Incidentally, keep in mind that where the hand grips the shaft is
neither a free or rigid joint...
But what do I know? I'd just hit the golf ball with a baseball bat and
be done with it. :>)
PD
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| User: "Kenneth Doyle" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 08:05:57 PM |
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"PD" <pdraper@yahoo.com> wrote in
news:1105566109.148236.119230@c13g2000cwb.googlegroups.com:
I would think that a good designer would choose a shaft material and
shape that would flex elastically but return virtually all of its
stored energy to the ball before the ball left the club face.
But that's mostly a function of how the golfer swings the club. A smoothly
accelerated swing won't allow the handle to straighten up until well after
the ball has left the club-face. Think of a fishing rod. It stays bent
all the way throught the casting action and only straightens up after the
rod has stopped accelerating.
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 08:16:45 PM |
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Good point--how far can one cast if the lure (head) is to heavy for a
flimsy pole--or if the lure is to light for a stiffer pole. You are
correct when viewing the entire system (man, shaft, head,ball)--there
is a balancing act--my question though is trying to focus on the
interaction between the head and ball but not at the more simplistic
level where the balls mod of e is 1. What are your thoughts?
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| User: "Kenneth Doyle" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 09:06:40 PM |
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wrote in
news:1105582605.571042.188160@f14g2000cwb.googlegroups.com:
Good point--how far can one cast if the lure (head) is to heavy for a
flimsy pole--or if the lure is to light for a stiffer pole. You are
correct when viewing the entire system (man, shaft, head,ball)--there
is a balancing act--my question though is trying to focus on the
interaction between the head and ball but not at the more simplistic
level where the balls mod of e is 1. What are your thoughts?
I saw a hint of an experiment involving a pendulum and a ball-bearing which
seemed worthy of some study.
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 11:30:19 PM |
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I read somewhere that a club manufacturer said--"the difference between
the clubs a pro golfer uses and those that you buy in the store are
about the same as the difference between a NASCAR engine and the same
engine that has been hopped up and setting in the car in your
driveway...." Just trying to get to the point where I can say I've at
least got "the hopped up engine setting in the drive"--Thanks
again--your comments have moved me up the ladder a couple of more rungs.
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| User: "Franz Heymann" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
13 Jan 2005 01:18:50 AM |
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"PD" <pdraper@yahoo.com> wrote in message
news:1105566109.148236.119230@c13g2000cwb.googlegroups.com...
Franz Heymann wrote:
"PD" <pdraper@yahoo.com> wrote in message
news:1105543633.050066.128310@c13g2000cwb.googlegroups.com...
cnctutwiler@wmconnect.com wrote:
Looking for simple experiment for measuring the COR of a golf
club
surface. Any ideas?
I can measure a *relative* COR, compared to a heavy concrete
surface.
1. Drop a golf ball on a thick concrete slab and measure the
ratio
R1
of the bounce height to the drop height.
2. Clamp a golf club to the slab. If you are interested only in
the
COR
of the club surface, then clamp the head tightly to the slab. If
you
are interested in the COR of the club as a whole, including the
shaft,
then it is only necessary to clamp it in a vise where the hands
would
grip.
3. Drop a golf ball on the golf club surface and measure the
ratio
R2
of the bounce height to the drop height.
4. The COR of the golf club surface will be R2/R1.
There is a caveat and it is a warning about the care with which
you
clamp the club to the floor. Exercise: Hold a basketball and a
golf
ball above the ground, with the golf ball resting directly on
top
of
the basketball. Release both at the same time, letting them fall
to
the
floor together. What is the observed COR of the floor/basketball
combination in this case? (Safety tip: wear goggles, a
mouthguard,
and
a groin cup.)
Your measurement has little to do wiith reality. In practice, the
club is held at the far end of the handle. In that case, the
behaviour of the whole handle is part of what determines the
behaviour
of the club..
I allowed for that in Step 2.
My apologies. I read too fast.
Franz
I'm actually not sure how much part that plays, and how much of a
contribution to the COR (or rather, to the deviation of the COR from
1)
it makes.
I would think that a good designer would choose a shaft material and
shape that would flex elastically but return virtually all of its
stored energy to the ball before the ball left the club face. I'm
sure
the designer knows how much the shaft contributors to the COR.
Incidentally, keep in mind that where the hand grips the shaft is
neither a free or rigid joint...
But what do I know? I'd just hit the golf ball with a baseball bat
and
be done with it. :>)
Me, I've never hit, thrown or kicked a ball except in anger.
{:-))
Franz
Franz
PD
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
13 Jan 2005 08:46:22 AM |
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Kenneth Doyle: Ref your smooth swing comment. I have a radar detector
here at home and I use it experimentally to check for speed variances
after equipment mods. What radar shows for me is that a smooth swing
increases clubhead speed by about 8-10% over going for the grandstand
at Wriggly Field. Watching my body movement in a mirror has lead me to
believe the extra power comes more from a pause at the top (which one
does if swinging smoothly)--this allows the lower body to begin turning
ahead of the upper section--providing more of a whipping effect. The
longer the pause--the higher the swing speed according to my tests--and
"Search for the Perfect Swing" 1968 (written by group of UK engineers)
suggests a 1 mph increase in clubhead speed results in aprox 2yds extra
distance.
So in sum--that's why I'm looking at the club face interation with the
ball--a trampoline works pretty well.
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
13 Jan 2005 05:27:12 PM |
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PD--tried your Bball experiment as you requested. Although the bball
was half deflated (low CRO), when dropped from about 3 feet the golf
ball came flying off the bball and rose to 6 or 7 feet. Here's my
analysis--the bball and gball hit the ground with Mgh1 energy (m of
bball + m bball). At v=o for the system, stored energy of the bball is
released propelling the gball to a height equililent the same height it
would have gone if dropped to the surface from mgh2 where h2=Mh1/m.
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| User: "PD" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
14 Jan 2005 09:58:27 AM |
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wrote:
PD--tried your Bball experiment as you requested. Although the bball
was half deflated (low CRO), when dropped from about 3 feet the golf
ball came flying off the bball and rose to 6 or 7 feet. Here's my
analysis--the bball and gball hit the ground with Mgh1 energy (m of
bball + m bball). At v=o for the system, stored energy of the bball
is
released propelling the gball to a height equililent the same height
it
would have gone if dropped to the surface from mgh2 where h2=Mh1/m.
Again, you're letting the algebra trip up your pant cuffs. You're
trying to apply one (energy conservation) or the other (momentum
conservation) to explain it, when BOTH are required in a collision,
even in one where COR < 1 (provided it's accounted for).
Here's another, simpler way to look at it, using the relative velocity
conservation argument appropriate to elastic collisions. (This is what
you would have seen if the basketball had been inflated.) Since
momentum and energy conservation seem to trip you up, I suggest you
stick with this method of analysis for now.
Dropping from a height of about 1 m, both balls would have a speed of
about 10 m/s by the time they hit the ground. Basically what happens is
that, because the basketball compresses on impact, the golf ball
doesn't feel the impact right away. The basketball rebounds from the
floor before the golf ball starts to really impact with the basketball,
and then the golf ball is bouncing off an upward-moving basketball. So
we'll take it a step at a time.
Basketball-floor impact. Just before impact, the relative speed of
floor and ball is 10 m/s. Since this remains constant in an elastic
collision, the relative speed of floor and ball is still 10 m/s after
the collision. Assuming the mass of the floor is big enough that its
speed is zero after the collision, the basketball bounces upward with
speed 10 m/s. (That seems intuitive.)
Golfball-basketball impact. Just before impact, the golfball is headed
downward at 10 m/s and the basketball upward by 10 m/s, meaning a
relative speed of 20 m/s. The relative speed will still be 20 m/s after
the collision. Assuming the basketball is massive enough that it
doesn't slow down appreciably after impact with the golfball, it is
still going up at 10 m/s after the impact. This means the golfball must
bounce upward at 30 m/s after impact.
After the second impact, the height the golfball goes is strictly a
matter of energy conservation. Since the speed on the rebound is three
times higher than the speed it got going down, and since the kinetic
energy is proportional to v^2, its kinetic energy after impact goes up
by a factor of *nine*. Since the energy is proportional to the height
it will go (mgh), the golfball ideally would rebound to a height of
nine meters, or 28 ft. The COR of the basketball/floor combination is
*nine* in this analysis. This is why I warned you to wear a cup. I'm
glad your basketball was not elastic, but I hope that this inspires you
to inflate the basketball and try it again. With a cup.
Why is this important? If you only half-heartedly clamp the golf club
to the floor in the test I proposed to you, it will rebound off the
floor when the golfball hits it and then hit the golfball with an
upward speed. The golf club head will play the role of the basketball
in the experiment I had you perform. Obviously, you can see how that
would affect your measurements of the COR. You must *seriously* clamp
the golf club head to the floor -- make it one with the floor, so to
speak.
Isn't physics fun?
PD
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
14 Jan 2005 11:29:12 AM |
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I thought I had the energy conservation covered by Bball rebound v=0
and gball energy equal to the total Mgh factor of the bball and gball.
Since gball has all the energy now it rebounds to a height equal to
energy of Mgh at v=0 prior to rebound. Its velocity would be determined
by solving for v where Mgh=1/2mv2. Right? The height would be equal to
h2=1/2gt2 and t would be t=v/g.
Appreciate your time--can't thank you enough for your simple solution
for solving for COR.
Thanks--and physics is fun but so are chem and EE.
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| User: "PD" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
14 Jan 2005 11:49:26 AM |
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wrote:
I thought I had the energy conservation covered by Bball rebound v=0
and gball energy equal to the total Mgh factor of the bball and
gball.
Since gball has all the energy now it rebounds to a height equal to
energy of Mgh at v=0 prior to rebound. Its velocity would be
determined
by solving for v where Mgh=1/2mv2. Right? The height would be equal
to
h2=1/2gt2 and t would be t=v/g.
Not quite. Note your analysis requires knowing M and m to calculate the
height. That is, solving for v will involve a ratio of M/m. You should
observe that my answer doesn't involve any ratio of M/m. The golf-ball
rebounds to nine times its original height regardless of M/m, as long
as M/m is "big". Notice that I'm talking about qualitative features of
the solution, not quantitative -- enough to tell you that something
can't be right.
So where did you go wrong?
Well, you assume the basketball stops dead on the ground, so that all
of its kinetic energy goes into the golf-ball. That's not what happens.
Most of basketball's KE stays in the basketball. Just enough goes into
the golf ball to get it to go to nine times its original height.
Of course, the basketball is not infinitely heavy, and it doesn't
bounce off the floor with exactly the same speed with which it hit the
floor. If it did, then the final energy would be more than the initial
energy -- and that can't be right. But that approximation is darn close
to being right.
PD
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
14 Jan 2005 11:55:22 AM |
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One final thought--seems to me that energy is always conserved--so is
momentum when COR=1. When COR is not 1, we have to trump up the
equations to make theory match reality. Seems easier to me to work the
energy side and let the cards fall where they may. In other
works--integrate F=ma and move towards the energy side (that way one
keeps all the unknown constants in the equation) rather than using
differentiation to move the other direction.
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| User: "PD" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
14 Jan 2005 12:15:55 PM |
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wrote:
One final thought--seems to me that energy is always conserved--so is
momentum when COR=1. When COR is not 1, we have to trump up the
equations to make theory match reality. Seems easier to me to work
the
energy side and let the cards fall where they may. In other
works--integrate F=ma and move towards the energy side (that way one
keeps all the unknown constants in the equation) rather than using
differentiation to move the other direction.
You have it backwards. Momentum is always conserved, and independently
for translational and angular momentum.
Total energy is always conserved as well, but mechanical energy is not.
What makes energy different is the transformation of one kind of energy
into another -- kinetic into rotational, thermal into kinetic,
potential into thermal, etc -- that is not the case for momentum.
What makes a collision inelastic is the conversion of some mechanical
energy into forms of energy that are not completely recoverable, like
deformation energy or thermal energy. In a completely elastic
collision, that doesn't happen and mechanical energy is conserved.
For *completely* inelastic collisions (COR = 0), we can't use
mechanical energy as an extra "handle" to solve the problem, but we're
given another gift instead. This is the case where the colliding
objects are stuck together in the final state (or the initial state --
an exploding firecracker is an example of a completely inelastic
collision), and that extra constraint provides the handle needed to
solve. (Note: this does NOT mean that ALL the mechanical energy is
"used up" in a completely inelastic collision.)
Unfortunately, completely inelastic or completely elastic collisions
aren't as common as partially inelastic collisions, where momentum is
conserved but mechanical energy is not, but you also don't have the
extra constraint of sticking together. This leaves the problem
insoluble -- unless you happen to know experimentally what the COR is,
which tells you *exactly* how much mechanical energy is lost in the
collision and gives you back the energy equation as a handle to work
with.
In other words, the theory is never trumped up. The issue is whether
you have enough information to solve for what you want to find.
This was fun. Thanks a lot.
PD
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
14 Jan 2005 12:23:17 PM |
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Your are correct--please forgive my monentary brain short. Thanks again
for all your help--and it was fun.
VR
C Tutwiler
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
14 Jan 2005 01:15:52 PM |
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Did have a question about compressible fluids (compressed air) and it's
interaction with a pellet guns projectile velocity. When I have time
to get smarter on the subject perhaps we could discuss.
Thanks again.
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
13 Jan 2005 05:51:11 PM |
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Sorry--Mgh1---> M=mass of bball +gball ; g=accel of gravity; h1=3
feet
mgh2---> m=mass of gball, h2=Mh1/m
h2 is good number for COR 1 but could be calculated for CORs less than 1
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 07:57:35 PM |
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PD don't give up on me so easily--I love your idea on measuring COR.
What about my second question: Is it an MV relationship or a 1/2MV2
one--or are they the same--we do just do an intergration of MV to get
1/2MV2. One is conservation of momentum the other conservation of
energy. What do you think? So do I attempt to lighten the club in
areas where swing speed increases to a max i.e.where I can't swing any
faster--then add weight to it until swing velocity begins to decrease
slightly--velocity max and mass max--Would that give max ball velocity
off the club face for the power I'm able to direct to the shaft/club
head? I could give Calloway golf 5 grand to computer model the whole
thing--but I'm not that excited about golf.
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| User: "PD" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 08:26:08 PM |
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wrote:
PD don't give up on me so easily--I love your idea on measuring COR.
What about my second question: Is it an MV relationship or a 1/2MV2
one--or are they the same--we do just do an intergration of MV to get
1/2MV2. One is conservation of momentum the other conservation of
energy. What do you think?
In an elastic collision, both conservation laws are in effect.
So do I attempt to lighten the club in
areas where swing speed increases to a max i.e.where I can't swing
any
faster--then add weight to it until swing velocity begins to decrease
slightly--velocity max and mass max--Would that give max ball
velocity
off the club face for the power I'm able to direct to the shaft/club
head? I could give Calloway golf 5 grand to computer model the whole
thing--but I'm not that excited about golf.
It is not a simple optimization between momentum and energy. Since the
swing is a rotation, you have to consider center of mass, moment of
inertia. Since the collision should be elastic you have to consider the
natural frequency of both the shaft and the head surface, which depend
on material, structure, and mass-distribution. You also *do* want to
impart spin, and the transfer of energy to rotation of the ball also
needs to be considered. The optimization is difficult and complicated.
I don't have a simple answer. Heck, I don't have an answer at all. I'm
not a golf-club engineer.
PD
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| User: "Gregory L. Hansen" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 01:47:14 PM |
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In article <cs3tt3$qd8$1@sparta.btinternet.com>,
Franz Heymann <franz.heymann@btopenworld.com> wrote:
"PD" <pdraper@yahoo.com> wrote in message
news:1105543633.050066.128310@c13g2000cwb.googlegroups.com...
cnctutwiler@wmconnect.com wrote:
Looking for simple experiment for measuring the COR of a golf club
surface. Any ideas?
I can measure a *relative* COR, compared to a heavy concrete
surface.
1. Drop a golf ball on a thick concrete slab and measure the ratio
R1
of the bounce height to the drop height.
2. Clamp a golf club to the slab. If you are interested only in the
COR
of the club surface, then clamp the head tightly to the slab. If you
are interested in the COR of the club as a whole, including the
shaft,
then it is only necessary to clamp it in a vise where the hands
would
grip.
3. Drop a golf ball on the golf club surface and measure the ratio
R2
of the bounce height to the drop height.
4. The COR of the golf club surface will be R2/R1.
There is a caveat and it is a warning about the care with which you
clamp the club to the floor. Exercise: Hold a basketball and a golf
ball above the ground, with the golf ball resting directly on top of
the basketball. Release both at the same time, letting them fall to
the
floor together. What is the observed COR of the floor/basketball
combination in this case? (Safety tip: wear goggles, a mouthguard,
and
a groin cup.)
Your measurement has little to do wiith reality. In practice, the
club is held at the far end of the handle. In that case, the
behaviour of the whole handle is part of what determines the behaviour
of the club..
Ooh, wait, we've done this before. A pendulum swings down and hits a
ball bearing, which flies through the air and hits a peice of carbon paper
atop white paper, making a mark. The student verifies that energy is
conserved and leaves with no idea that such a thing as a coefficient of
restitution exists.
Let pendulum -> golf club and ball bearing -> golf ball.
--
"Beer is living proof that God loves us and wants us to be happy."
-- Benjamin Franklin
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 09:56:32 AM |
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Great idea. I've been trying to to quantify for sometime the relative
Mod of Elas for the golf ball and happened to stumble on to sci.physics
sight.
Many thanks!
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 11:38:33 AM |
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I've been thinking about your answer and another question has come up.
In your basketball example--are you saying there is a optimal period of
time that a golf ball should sit on the face of the golf club for max
distance? Also, is the reaction a dP/dT thing or more a 1/2mv2 thing?
Would a ball hit by a train (high mass) at 100mph go further than hit
by me swinging a 100 mph--I suspect it would by alittle but I wouldn't
expect it to go into orbit. What do you think? So in tuning my club for
max distance--do I build for max velocity (lighter swings faster up to
a point)--then after reaching max v weight the head a much a possible
until max v begins to decrease? I know launch angle and spin rate
impact--but just in general terms what do you think?
Thanks again
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| User: "PD" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 12:08:28 PM |
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wrote:
I've been thinking about your answer and another question has come
up.
In your basketball example--are you saying there is a optimal period
of
time that a golf ball should sit on the face of the golf club for max
distance? Also, is the reaction a dP/dT thing or more a 1/2mv2
thing?
Would a ball hit by a train (high mass) at 100mph go further than hit
by me swinging a 100 mph--I suspect it would by alittle but I
wouldn't
expect it to go into orbit. What do you think? So in tuning my club
for
max distance--do I build for max velocity (lighter swings faster up
to
a point)--then after reaching max v weight the head a much a possible
until max v begins to decrease? I know launch angle and spin rate
impact--but just in general terms what do you think?
As to the basketball example, I invite you to actually do the
experiment in your garage and then think about what you see. (I.e. how
on earth can THAT happen?) If you can tell me what you observe, then we
can talk about what's going on and why that has any bearing on the
golf-club measurement.
As for your general question, for a perfectly elastic collision, the
relative velocity of the golf ball and the club will be the same before
and after the collision. This means that, assuming the club doesn't
slow down after impact, the ball will take off with twice the swing
speed of the club. Of course, the approximation that club doesn't slow
down only works when the club is much heavier than the ball. Of course,
the heavier the club, the harder it is for the golfer to accelerate it,
especially if the mass is all concentrated at large radius from the
hands. And the assumption of perfectly elastic collision means a COR of
1. This doesn't require that the club be non-deformable, but it does
mean that the natural frequency of oscillation of the deformation of
the club face has to be matched to the natural frequency of oscillation
of the ball's deformation, or the ball will leave the club before the
club face rebounds completely. All of this should perhaps explain why
club manufacturers spend months computer-modeling new designs and
months testing and tweaking designs.
PD
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| User: "" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 01:24:54 PM |
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PD--thanks for your input--I guess the golf ball shoots up off the b
ball because MVtotal(B ball and golf ball and earth)=MV (B ball) + MV
(golf ball):(earth V=0)and M for the golf ball is very small, meaning V
must be very large to keep the equation balanced.
Are you saying the collision then is strickly MV type with no
consideration for V2 energy term?
Note: 71 EE grad who retired from flying jets in 97--not an ed student,
contractor, and etc. with nothing personal at stake here other than
attempting to get another 5 or 6 yards out a golf shot.
Thanks for you help
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| User: "PD" |
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| Title: Re: Coefficient of Restitution (COR) for Driver |
12 Jan 2005 03:34:41 PM |
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wrote:
PD--thanks for your input--I guess the golf ball shoots up off the b
ball because MVtotal(B ball and golf ball and earth)=MV (B ball) + MV
(golf ball):(earth V=0)and M for the golf ball is very small, meaning
V
must be very large to keep the equation balanced.
No, and you've let algebra tangle your pant cuffs. Recall what I said
earlier: In an elastic collision, the relative velocity is the same
before and after the collision. Consider a baseball thrown at 90 mph at
a batter who swings a bat a 60 mph, so that the relative velocity is
150 mph. After impact, suppose the bat continues on in the
follow-through at 60 mph forward. This requires that the ball will
leave the bat at 210 mph forward. This is why a batter can hit a
pitched ball out of the park even if he can't hit a ball on a T-stand
into the bleachers.
You haven't told me what you *saw* with the basketball-golfball
experiment. I presume you can get your hands on both -- certainly the
latter. There is no replacement for experimental tinkering for complete
understanding. When you do the experiment and you tell me what you see,
I'd be happy to talk with you about why that's what you see.
Are you saying the collision then is strickly MV type with no
consideration for V2 energy term?
Not at all. In a completely elastic collision, both momentum ("mv") and
kinetic energy ("(1/2)mv^2") are conserved. However, that's the
limiting case when COR = 1. Most real collisions have 0 < COR < 1,
which means that kinetic energy is only partially conserved -- the
fraction that is conserved is precisely the COR.
Note: 71 EE grad who retired from flying jets in 97--not an ed
student,
contractor, and etc. with nothing personal at stake here other than
attempting to get another 5 or 6 yards out a golf shot.
Thanks for you help
Yeah, I just thought I'd remind you there are engineers who earn a
living designing expensive products to get another 5 or 6 yards out of
a golf shot. If it were easy, there'd be a better club for half the
price at Costco.
PD
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