jbeam435 --stinger

[supo]

i just get my builder d to do mine, best on the planet this fella., im utterly amazed how i give him a driver one day ,say,hmm think i like the head but it goes high right too mch. or its too light then i get it back , next week, im like ok what on earth did u do to this???

inject it with bulgarian horse growth hormones or something ?

just un real.

bit of wizardry here and theres. and western specs on JDM pieces materialise.

i have beenaskign him to go heaver iSW thos form D2 normamly im flooring things at d3 now.

although................ the steel egg7 driver is only c9 and its 350+ grams , i cant feel any differnce. it must be the shape of that head..

that gets used tomorrow at longy!

[chiromikey]

I got tired if the look on my builder's face when I tell him what I want to experiment with...

[sbeaz]

I'm talking...HEAVY! My current 435 sw isn't in the D's...or even the E's...it's F3! :)

Just curious chiro...If you play your driver @ F3, what are your wedges?

[chiromikey]

Depends on the wedges but typically D6-D8.

[kaaayelll]

No place to stuff cotton, and wrapping them in it tends to kill spin. ?

[sbeaz]

Depends on the wedges but typically D6-D8.

Cool. I thought you were going to say Q8 or something. Anybody that's ever held Hogan's clubs say - to a person - they were the heaviest clubs they'd ever picked up. Not sure what his SW was.

[chiromikey]

Cool. I thought you were going to say Q8 or something. Anybody that's ever held Hogan's clubs say - to a person - they were the heaviest clubs they'd ever picked up. Not sure what his SW was.

I never knew that. I do think he was the master of MB iron design. There's not a blade iron today that isn't influenced by his designs.

[chuck4golf]

I think one of the things I like best about the Epon 503's I bought from someone is he had the swingweight at about D5. I really like this.

[+TourSpecGolfer]

Simple F=MA dictates that the heavier head delivers more force on the ball and your formula agrees. Knowing that the ball weight and COR are constant, if one maintains head speed, the heavier head equals higher initial ball velocity.

The ONLY way a lighter head can yield higher initial ball velocity is if it allows you to gain head speed (but I addressed that in my initial post).

This equation is from sevendreamers

I think the bottom line with the equation is that a lighter club allows you to swing it faster which means higher ball speeds.

I don't know F=MA, how does a heavier head/balance increase ball speed if the opposite is true?

[RIduffer]

Hogan used butt weights in the 40+ gram range, from what I have been told. If he was trying to get to a particular SW then he would need to add tungsten down the shaft or lead tape on the head to get into the low mid D range

[chiromikey]

This equation is from sevendreamers

I think the bottom line with the equation is that a lighter club allows you to swing it faster which means higher ball speeds.

I don't know F=MA, how does a heavier head/balance increase ball speed if the opposite is true?

F=MA is simply Force = Mass x Acceleration. Acceleration is actually more accurate than velocity to describe head speed and ball speed but that's getting nitpicky.

Yes, theoretically one should be able to swing a lighter club faster but that's not always the case and it's something I ruled out by saying to swing the heaviest head weight you can that doesn't cause more than a mph loss of ss.

But regarding the head weight vs ball speed equation, the opposite ISN'T true...no matter which equation is used. Plug in numbers to the equation from Sevendreamers and you'll find that their equation also concludes the same thing...as long as head speed doesn't change, a heavier head weigh yields a higher initial ball speed. So whoever told you that a lighter club equals higher ball speed was either incorrect or failed to mention they manipulated head speed in their calculations.

:)

Edited August 7, 2014 by chiromikey

[chuck4golf]

There are a few variables at play, seems to me. Speed and mass are the big ones - or acceleration (speed squared if I recall my physics). I know people say there is an optimal weight to maximize speed, which I assume to be true. I am thinking that at the same overall weight, shaft weight, head weight, swing weight, and swing speed at impact, that a ball hit on the sweet spot would have more initial speed if the head were more compact because there is more mass being delivered into the impact (decreasing its deflection), but perform worse on off-center hits because the MOI is lessened and more deflection of the clubhead occurs. I would think smaller heads for good ball strikers makes some sense, but I am just guessing.

The other thing I wonder is, lets say a head and shaft is 300 grams overall. It could be shaft 80 and head 220 grams or shaft 40 and head 260 grams (theoretically, just for illustration points). Will these two be swung at the same speed? I don't know. But if so, I would assume better performance from the 40/260 because greater mass is located at the point of impact.

This is really interesting to me.... as I contemplate a new driver, waiting in an airport....

[kaaayelll]

There should an optimal point that opens the possibility that a heavier head (mass) AND a slower swing speed could produce higher ball speeds.

Then factor center of gravity, shaft characteristics, and other factors impacting launch angle, spin, stability, etc and we're starting to have fun. ;)

Thanks for bringing up an easily overlooked point, Mikey.

[chiromikey]

None of us actually swing at our max and I've found that I have what I call my "comfort speed". It's the ss I maintain comfortably (and it fluctuates day to day). However, that speed tends to stay the same for me if it's a light head or super heavy one...my body just tends to find it's comfort zone. Because of that I can swing an overly heavy driver, not lose any head speed, and gain distance. Obviously if I was swinging at 100% effort all the time, then yes, I'd swing a lighter driver faster than a heavy one. But that's not the case for me, nor for most people I'd be willing to bet.

I believe you're right about the concentration of mass at impact. The more mass directly behind the impact point, whether it's a compact head or a heavy head/light shaft combo, will increase force, therefor distance (all other variables remaining constant and assuming you always hit the sweet spot).

[chuck4golf]

That assumption that 'you always hit the sweetspot' is a mouthful.

That's an interesting point about your ss remaining constant because you're not at full tilt... I know that most of my swings are not. at full tilt - they kind of mesh with my psychology of the day, I suppose. It takes are real fearlessness to let it rip, all the way. So does that mean I will swing a heavier club at the same speed? I do know that I play the swing-weight heavy Epons very well with all the distance I expect from the loft of each club. And with great accuracy, best ever for irons.

Kaaayelll, I think physics with F=MV**2 says losing speed over an increase of mass is a losing proposition. I just don't see how losing head speed, all other things like how centered the strike is being constant, could be compensated for by greater mass. But maybe my limited physics is just enough to make me really stupid about all of this.

[kaaayelll]

F = MA is a simple multiplication formula, so it stands that a decrease in Acceleration can be offset by an increase in Mass.

Whether or not Force increase or decrease with changes in M and A has to do with the magnitude of the changes in M and A.

Simple example:

3 x 2 = 6

3.25 x 1.9 = 6.175

3.1 x 1.9 = 5.89

[chuck4golf]

I understand that logic. The thing that stands out in my thinking is that acceleration = velocity squared. So it's not the same thing. To use your example, if assume your speed variable is speed and not acceleration, for illustration purposes:

3 x 2**2 (squared) = 12

3.25 x 1.9**2 = 11.73

So the decrease of speed has the impact of that loss squared on the force imparted. Thus, it's not a linear thing. At least that's how I understand it.

So, I think what this means is, let's go back to the example of 3 x 2**2 = 12. Let's increase the mass by 5%, so now it's 3.15.

To break even on the force, we'd have to lose less than 2.5% of the speed. (3.15 x 1.95**2 = 12)

At least, that's how I understand it. If the increase in mass causes less speed loss than this, then it's a helpful thing. But it has to have a very limited impact on the speed, because the force is the result of the speed squared.

[chiromikey]

Acceleration doesn't equal velocity squared. It's the change in velocity over time.

Or, if we're talking about head spead velocity squared divided by the radius of our swing arc.

Edited August 7, 2014 by chiromikey

[chiromikey]

Note to self...write apology letter to physics professor and admit I did one day finally use physics outside of school. :)

[RIduffer]

Note to self...write apology letter to physics professor and admit I did one day finally use physics outside of school. :)

Scary what stupid tidbits come back to haunt us...

[chuck4golf]

Well, that will teach me not to pretend I remember physics! But forever I have had the notion that F=Mv**2.

I know what the problem is. I just remembered.

I made a C in physics. Guess that was about the right grade.....

Edit # 2: But wait... something doesn't make sense. If a tractor-trailer hits you going a constant 60 mph, it sure as hell is going to deliver more than 0 force. But F=MA, where A is the rate of change of speed over time, in this example is 0 and thus, in this equation, F would equal 0. Except that you'd be totally flattened by the truck hitting you.

From some site (and if it's on the internet, especially in pdf format, well, we know that it is true) - except I can't copy and paste into this space for reasons that escape me, so I will paraphrase - that Newton's 2nd Law equates to F = MV**2.

Anyway, I don't really know. I am just killing time in a layover in Vienna airport where they have GREAT free internet. But somehow, I just don't think acceleration is the key thing. I actually would like to be wrong, because I'd definitely swing a heavier head if I believed the relationship in the tradeoff was close to linear rather than exponential.

Edited August 8, 2014 by chuck4golf

[Mob]

I have narrowed my driver selection down to essentially 3 that I will play.

Crazy 435 with CB80 LS Noir shaft

Ryoma Maxima D with Quadra Fire Express Max WBQ shaft

SLDR 430 with Veylix Alpina 673 shaft

Just started using the Ryoma for the first time this past week and really like it. The feel is great and it is plenty long for me. Had to buy the weight kit to get the weight up as I cut the shaft down to 44 1/4 inches. I can't comment on consistency yet, as I haven't hit it enough. The SLDR is getting benched temporarily for a lack of cooperation in my club championship. I have a feeling that I will be coming back to the 435 within the next month. It always seems to call to me when I am grabbing my clubs to go play. It is like the Sirens in Greek mythology.

[chiromikey]

Well, that will teach me not to pretend I remember physics! But forever I have had the notion that F=Mv**2.

I know what the problem is. I just remembered.

I made a C in physics. Guess that was about the right grade.....

Edit # 2: But wait... something doesn't make sense. If a tractor-trailer hits you going a constant 60 mph, it sure as hell is going to deliver more than 0 force. But F=MA, where A is the rate of change of speed over time, in this example is 0 and thus, in this equation, F would equal 0. Except that you'd be totally flattened by the truck hitting you.

From some site (and if it's on the internet, especially in pdf format, well, we know that it is true) - except I can't copy and paste into this space for reasons that escape me, so I will paraphrase - that Newton's 2nd Law equates to F = MV**2.

Anyway, I don't really know. I am just killing time in a layover in Vienna airport where they have GREAT free internet. But somehow, I just don't think acceleration is the key thing. I actually would like to be wrong, because I'd definitely swing a heavier head if I believed the relationship in the tradeoff was close to linear rather than exponential.

I just saw your edit...I'm not at an airport but I'm bored and can't sleep.

In your example, you're correct in that there is no acceleration...BEFORE contact. Knowing Newton's 3rd law we know there's an equal but opposite force applied to the truck and your body when they do contact. We can either multiply the mass of the truck by it's minimal negative acceleration as it smashes you into it's radiator or we can multiply your mass by your nearly instant acceleration from 0-60mph. Either way, it's going to be a lot of force due to significant mass or significant acceleration, respectively.

Newton's 2nd law is definitely F=ma, not sure why the site you found shows something different.

[+TourSpecGolfer]

This is getting complicated!

To simplify this, what exactly are we trying to solve here, shaft specs for a player or a basic philosophy of club design?

[nobmontana]

Former Physics major here from a not so good school :) ... but I think there are far more human produced variables in a golf swing

that impacts how far or straight a ball goes.

A note on the stability of steel shafted drivers. I have had these combos in the past and they definitely do go straighter.

I always thought this was because of lower torque of the shaft relative to graphite. In addition, having shorter club length for the steel shafted drivers probably allows hitting the sweetspot more often and possibly providing more confidence in swiniging closer to an all out full swing?