Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Golfer Needs Competing Laws of Physics Explaining Please

  1. Jan 24, 2009 #1
    I do not even have a schoolboy knowledge of physics and so I need help - preferably in simple terms please:

    I compete internationally in an extreme sport known as 'Long Drive Golf.' Essentially one succeeds by hitting a golf ball the longest distance.

    For many decades the manufacture of drivers (the golf clubs which are the longest / most powerful) has been solely based on the premise that if one reduces the head weight then one can swing the club faster and by swinging the club faster then one can hit the golf ball further. (Let us for the moment ignore other factors such as moment of percussion, coefficient of restitution etc.)

    As a rough and ready guide:

    If one has a swing speed of 130 mph, then using a standard driver (whose head normally weighs 200g) one would expect to hit the ball to a landing point 325 yards away.

    For each 1mph increase in swing speed one might reasonably expect to achieve a further 2-3yards in distance. There are adverse effects on head strength etc. if head weight is reduced much below 200g.

    Let us imagine that by increasing head weight swing speed was adversely affected. The question nevertheless arises is:

    'Would the benefit gained in distance under the formula of FORCE = MASS x ACCELERATION outweigh the distance lost through making swing speed fall by increasing head weight?'

    I can measure on my radar meter any drop in swing speed due to increased head weight.

    What I cannot assess is the ratio of distance improvement from increasing the headweight based on swing speed remaining constant.

    Is it possible to calculate such please? What increased distance could one expect to achieve by increasing head weight to 205g / 210g / 215g?

    Ivan Sanders
  2. jcsd
  3. Jan 25, 2009 #2


    User Avatar
    Science Advisor

    will give you ball velocity for different head mases, head velocities. This of course assumes a short contact collision, and not pushing the ball along with the swing.

    The range is more complicated and depends on drag and take-off angle
    But you seem to have some empirical estimates for this already.
  4. Jan 25, 2009 #3
    Thanks A.T. After reading the above two links a line from a film immediately flashed into my fried brain:

    "What we have here is a failure to communicate." (Cool Hand Luke).

    The articles were written for scientists and, for a layman, as communicative as me speaking in Russian to a British audience.

    I do not understand how I could 'push' the ball rather than hit it when swinging a club at 130mph?

    Also, I do not understand the reference to 'drag.' Is such wind resistance as the club head moves through the air? If so then surely such was only one of over a hundred factors which resulted in club head speed / acceleration and therefore irrelevant for the purpose of the formula which I seek? Newton was surely only concerned with the rate of acceleration, not how it was achieved? Forgive me if I am 'speaking through my hat.' My intellect is based on elementary logic, not science.

    For the purposes of my question one should simply assume an optimum take-off angle. This is normally around 14 degrees.
  5. Jan 25, 2009 #4
    Hello Ivan,

    I presume that 'follow through' is carried out, by which I mean that I suspect that you are applying a torque to the golf club during the duration that the head is in contact with the ball. In other words, because of the human applying a force the speed of the club is not greatly reduced after hitting the ball, this is what I mean by 'follow through'.

    As long as follow through is achieved, then the faster the head is moving the further the drive will be. The mass of the head is irrelevant, as long as you can get it up to speed before it makes contact with the ball and keep it at speed using follow through.

    I suspect that the optimal head weight depends on biophysics i.e. at some point it becomes easier to swing a heavy head than a light one, because the light one has less tactile feedback.

    This is only meant to be a preliminary analysis (considering only the main effects without including smaller corrections that could still have an impact), if you share more relevant information then I will take that into consideration.
  6. Jan 25, 2009 #5


    User Avatar
    Science Advisor
    Homework Helper

    The mass of the head should be the same as the mass of the ball for most efficent energy transfer - but a club this light would be difficult to swing.
    Last edited: Jan 25, 2009
  7. Jan 25, 2009 #6
    Hello Confinement,

    Thanks for your input.

    Yes indeed, it is absolutely essential that golfers have a full follow through. But, I could not technically say why.

    The club head is reputed to slow down by approximately one third due to the collision.

    Research on work and power analysis of the golf swing is incomplete but the best article I have found 'Work and Power Analysis of the Golf Swing' by Steven M. Nesbit and Monika Serrano appeared in the Journal of Sports Science and Medicine (2005) 4, 520-533www.www.jssm.org

    The sport of golf is riddled with ancient myths, and often these have absolutely no scientific basis and my own empirical research, using numerous golfers, has proved many to be ill-founded and patently wrong.

    My fringe sport of Long Drive is only 30-40 years old, and for the first time in the history of the sport of golf competitors have had to find ways in order make / adapt equipment to produce maximum possible distances subject to construction limit rules eg. C.O.R. limit of 0.830 www.longdrivers-eu.com[/url] [url]www.longdrivers.com[/URL]

    (Apologies for digressing.)

    The issues concerning control are complex, and I have not seen any research data showing at what point the weight of a club slows the swing down. It is obviously not a straight line progession as one can swing six feathers as quickly as one feather.

    I have to confess to be stunned by your advice that head weight is irrelevant within the confines of pure physics, i.e. ignoring the physical abilities and attributes of each player. Such shows that I have totally misunderstood the meaning and application of Newton's second law.

    I hit the ball further with a 211g driver head than I do with a 203g driver head but clearly not due to the F = M x A formula!

    The question I raised is a relative simple one in golfing terms. There are many more complex questions yet to be answered once the basic questions have been fully dealt with.

    One such advanced question is why a competitor's swing speed usually increases by between 3 and 5 mph when a golf club is 'counter-balanced.'

    The process of counterbalancing involves inserting weight into the handle (grip) of the club. The overall weight of the club increases but it feels lighter. The results are consistent irregardless of whether or not a subject knows that his club has been counter-balanced. Strange!
    Last edited by a moderator: Apr 24, 2017
  8. Jan 25, 2009 #7


    User Avatar
    Science Advisor
    Homework Helper

    Is the follow through anything to do with the ball being in contact with the club or is it just to give you the proper form in the swing?
    I seem to remember an experiment that showed that if a baseball bat was dropped at the moment it hit - it made no difference to the ball.
  9. Jan 25, 2009 #8
    mgb phys,


    Thanks to you too for your input.

    The weight of a golf ball is 45.50g. I have provided the range of driver head weights.

    Drivers of course vary in weight. One of my typical driver's weight in total is approximately 370g. Such of course includes the weight of the driver head.


    I honestly do not know. What I can say with certainty is that if one has an abbreviated follow through the ball fails to elevate properly and may have other flight faults. This is true for every golfer.

    This is strange is it not because the ball has left the face of the club by that time and one would imagine that the follow through would be a superfluous operation!

    I really cannot get my head round why a good follow through is absolutely vital.

    P.S. Sorry. But I must go to bed now. We athletes must get our sleep. (It's now 1.03 am in the UK)
    Last edited: Jan 25, 2009
  10. Jan 25, 2009 #9
    After reading the jssm article, I programmed a simulation based on the measured forces exerted by the golfer eg figure 11. The input into the model was the following:

    a. The Force vs Time measurments, which is a profile that is unique for each golfer.
    b. The mass of a golf ball, 0.046 kilograms.
    c. The drag coefficient of air resistance acting on the club head.
    d. The coefficient of restitution i.e. a parameter that summarizes the rigidity of the collision between the club and ball.

    For (a) I used a qualitatively similar estimate, but in a more detailed simulation the measured data could be used instead. Obviously (b) is exact, and I estimate (c) to be about 0.02. For (d) I have assumed for now a perfect collision, since I don't know what value might be realistic.

    The simulation was a success in two ways:

    1) Without the consideration of air resistance acting on the club during the swing (set (c) to zero), I found that the speed of the ball after it had been struck was independent of the mass of the club.

    2) With the effect of air resistance on the club taken into account, the simulation indicates a well-defined optimal mass value in a reasonable range.

    I have found that the optimal club mass depends strongly on the air resistance, and to a lesser extent on the player's power profile. Specifically, the simulation indicates that a lighter club should be used when the air has a lower density, relative to the optimal club mass in normal humidity.

    Attached Files:

    • golf.bmp
      File size:
      89.4 KB
  11. Jan 25, 2009 #10
    A good follow-through is "necessary" because it encourages a smooth, linear, strike. Also, I suspect that the push of the club on the ball also lasts longer than appropriate for assuming a single "instantaneous" collision or energy transfer.


    First note that, the maximum speed of the club isn't only reliant on the "strength" of your muscles, but also the "speed" of their contraction. This is important for the reason why your able to hit further with a heavier club despite the fact that energy transfer is most efficient at equal weights.

    So you have the initial collision which imparts a velocity on the ball, then you have a momentary (but non-negligible) push on the ball from the club. I believe the dynamics of this push will rely on the trajectory of the head and the golfer's muscular characteristics.

    The swing should be elliptical with the ball strike occurring at the outer most point on the ellipse so that conservation of angular momentum will accelerate the head of the club.

    In addition to that, the golfer's muscles should also be able to bring the club back up to maximum contraction speed.

    The sustained push will be stronger if the mass of the head is as large as possible without greatly affecting your maximum swing speed.

    Main Points

    -Hit the ball when the head of the club is as far from the "center" of rotation as possible.

    -Your "ideal" club should have the largest mass that allows you to achieve your maximum swing speed.

    In your case Ivan, you probably reach the same maximum speed with both clubs, but the larger lets you bang out a few extra yards because F=MA makes that sustained push a little stronger. (Also, you may consider attaching a 40.5 gm weight to your club when you're checking your swing speeds to account for the ball)

    -If your swing speed IS limited by strength rather than muscle contraction, then use a longer club to increase the speed of the head. (just an added thought)

    -Try to hit to the ball so it takes off at a 35 degree angle relative to the ground (This is the angle that has been experimentally shown to maximize projectile range in earth's atmosphere)

    Fairly obvious observations (to be redundant). The numbers would be tedious to crunch.
  12. Jan 26, 2009 #11
    I am amazed at the time, trouble and consideration invested in a stranger (or for science, or both) for which I am greatly appreciative. Unfortunately I could not open Confinement's attachment. I would very much like to see this and wonder if such could possibly be e-mailed to me please at sanderslongdrive@yahoo.com ?

    Please excuse my ignorance, but references to 'mass' presumably means the SIZE of the head rather than its WEIGHT? In other words perhaps choose the maximum allowed head size of 460cc, rather than the 360cc heads which I presently use because of their aerodynamic qualities? Even though the COR ratings are similar? (I think that 'the penny has finally tumbled,' please see my final paragraph).

    COR is a fascinating subject, which I barely understand. Although I do manage to secure the gist of some science articles. MackBlanch advises hitting the ball as far away from the centre of rotation as possible. I think that I understand this: In other words do not bring the head in square to the ball. This point was touched upon at www.en.wikipedia.org/wiki/Coefficient_of_restitution[/URL] where it was stated that COR can take a value greater than one in a special case of oblique collisions. Incredible!

    Mack appears to make a different point though, namely that by hitting towards the side of the golf ball the club head speed is not slowed down so much and reducing this factor is key to flight distance. I hope that my interpretation is correct?

    Next, the words 'sustained push' appear to be key. However, I am unsure as to what this means. As a layman I either hit the ball, or in putting terms cheat by pushing the ball into the hole with the putter club head rather than cleanly striking it.

    For a scientist though there appears to be a more technical meaning. Apparently all balls are pushed when struck and one has to push as well as possible. I do not think that MOI is being referred to, but rather one simply has to keep the ball and club head in contact with each other as long as possible? No, that can't be correct either. Please explain.

    Hitting a ball long involves numerous factors, but yes speed and strength are key, as is technique. Muscles have to remain loose for speed and efficiency. Muscles are not contracted in the same way as in say power lifting. There is a raging debate between competitors as to the importance, or otherwise, of strength and all that can be agreed is that flexibility is paramount as power is loaded into the lower back / core. Most regular golfers simply cannot stretch their bodies into a vital body power position. Most golfers essentially play golf by swishing their arms to and fro which, by itself, could never secure Long Drive competition distances.

    What is also key is the width of the swing arc which can be achieved in three principal ways but being six feet eight inches tall (as was last year's World Long Drive Champion) certainly helps.

    Speed is limited by everything, but likewise increased by everything. It would be a fascinating exercise to take a competitor and try to assess which his key limiting factor to distance was as muscle speed surely relies on strength and so there are no independent variables.

    So far, to summarise, I believe that I am being advised to use large heads (although I am unsure why - but please see below) and that head weight is almost immaterial: Use as light in weight head as possible, but in certain atmospheric conditions a slightly increased head weight could be beneficial.

    In conclusion, I think that the essence of the advice which I have been kindly given is that in the equasion F = M x A mass means SIZE not WEIGHT. There appears to have been an elemental fundamental misunderstanding on my part.

    Have I summed up correctly?

    You chaps are BRILLIANT! Thanks again.
    Last edited by a moderator: Apr 24, 2017
  13. Jan 26, 2009 #12
    Hmm, I'm sorry for not being articulate. One of my biggest problems is explaining my ideas clearly.

    You may think of mass as weight since you use the metric system. However, that is technically wrong since 'weight' refers to the force of gravity on an object (1kg of iron weighs less on the moon, but there is still 1kg of iron). Your 211 gm club has more mass than your 203 gm club.

    I did not account for COR at any point in my thought process. It seems to me that when dealing with a sphere, you will want to strike it so that the motion of the club is directed through the center of the ball. That said, I know it is good practice to put top-spin* on the ball to generate lift.

    Now, thinking about it some more, I found a slow-motion video of Tiger Woods driving a ball and I realize my previous analysis is flawed.

    There is, in Tiger's swing, a single instant of contact for a transfer of momentum. After the contact, you can see the shaft of the club bend back so that the head is no longer in touching with the ball. You can see, however, that the club's head does travel in an ellipse and if we assume the point between his shoulders is the center of rotation, then the head of the club is closest to that point when it strikes the ball.

    Since it -is- only an instant of contact, there is no sustained push in addition to the transfer of momentum. I can only assume then, that you would want your club to have a mass close to that of the ball's in order to maximize the amount of momentum transfered. Your next task would be to maximize the speed of the head at the moment of contact.

    Ivan, have you measured your swing speed at the point of contact with both clubs? I suspect that if you really are hitting it further with the 211gm club then you must be swinging that club notably faster.


    1. A lighter club will have transfer energy to the ball more efficiently. However, you need to still be able to maintain control of the club while you swing it. If you can't control your swing well enough to strike the ball properly you should use a heavier club.

    2. Swing your club in an ellipse with the the point of contact being the moment at which the head of the club is closest to your chest. This will help you hit the ball when the head is moving at its fastest.

    3. Still try to drive the ball off at a 35 degree angle from the ground. (Maybe less if you can consistently add top-spin to the ball as the generated lift will carry the ball further)

    - Tiger's Swing

    A little bit about the "power" of muscles. Power is defined as force*speed. Think of two people lifting weights. A boxer can bench 100kg and do it very quickly, say in 2 seconds. A body builder can lift 200kg, but he has to inch it up slowly and it takes him 7 seconds. the boxer's power is 100kg *10 * (1meter /2seconds) = 1000. The body builder's power is 200kg * 10 * (1 meter/ 7seconds) = 285.7 . So despite being able to lift more and, in the traditional sense, being stronger, the Body Builder is less powerful. The boxer would probably be able to drive a golf ball further using the same club. However, the body builder could mitigate the detriment of his lack of muscle speed by using a longer club.

    *you would want the ball spinning they same direction a ball rolling towards you would spin. It increases the speed of the air passing over the top of the ball while decreasing the speed of the air passing under the ball making it subject to Bernoulli's principle.
    Last edited by a moderator: Sep 25, 2014
  14. Jan 26, 2009 #13

    You have put a great deal of though, effort, and explanation into this for which I am greatly appreciative.

    I believe that we have both digressed somewhat from the question in hand namely:

    "Will adding weight to a 200g golf head increase distance assuming that the swing speed and everything else remains constant?"

    I believe that the consensus of scientific expert opinion here is "No it will not, but a slight increase in head weight may be beneficial in heavy atmospheric conditions."

    It would not be realistic to reduce head weight much below 200g as such would weaken the structural integrity of the head.

    The physics are human bodily movement necessary to achieve good results are immensely complex and I prefer not to discuss them here. They are set out in the scientific research article which I mentioned previously. Indeed the flexing and contra flexing of the shaft in the downswing and follow through is in itself complex and occurs several times. Such would possibly be a good topic for a different debate.

    Whilst it is unnecessary for me to understand most of you, my friends', reasoning I feel impelled to solve the mystery (for me) of exactly what 'mass' amounts to. I initially thought 'weight' then felt that I had misunderstood and the real meaning was 'volume.'

    You inform me that a 203g head has a lesser mass than a 211g head, even though they can both be of the same size.

    Let me explain how my uneducated mind works:

    If F = M x A

    I thought that the greater the force the further I would hit a golf ball.

    If I increase M (whatever such is) then I thereby increase F, and thereby increase distance.

    Yet, I am informed that increasing M (by increasing weight) will have virtually no effect except in heavy atmospheric conditions.

    Whilst I trust you scientists implicitly you will hopefully nevertheless see why I am confused.
  15. Jan 26, 2009 #14

    "If I increase M (whatever such is) then I thereby increase F, and thereby increase distance."

    There are actually three possible results of increasing M.

    1. You increase M, you maintain the acceleration A, so the Force F will increase.

    2. You Increase M, but you are unable to maintain acceleration A, so A decreases, and F stays the same.

    3. You increase M. F and A will alter in some dynamic combination as a result that matches the formula: M = F/A.

    So, long story short, increasing M does not necessarily increase F. you have to worry about your ability to accelerate the club as well.

    Further confounding your understanding is your misconception that a greater force will result in a greater distance.

    An object, A, can only exert a force on another object, B, that is equal or less than the capacity of object B to exert a force on object A. That is, Every action has an **equal** and opposite reaction.

    So, while you can increase the capacity of the club to exert force, the actual force affecting the golf ball is limited by the golf ball's inertia (how resistant it is to having its momentum changed).

    I will try to think of a good way to explain this more clearly, right now I have to go to the gym. Someone else, feel free to pick up where I left off.
  16. Jan 26, 2009 #15

    OK. In the absence of a definition of 'mass' everything else is non-understandable. What is clear though is that the F = M x A equasion does not operate like simple algebra: If 8 = 4 x 2 then if one increases 4 to 6 then 12 is not the answer. Least not for Mr. Newton.

    I think that I prefer to abandon my attempt at learning science as unpredictable dynamic combinations are of no use to a simple man who just wants to know if adding a bit of weight to his club makes it more powerful.

    I think I'll walk now knowing that adding weight is basically no help unless perverse golfer attributes kick in, or the weather is inclement. I'll wet my finger and stick it in the wind and swing true.

    Please excuse my 'humour.' I tend to laugh at many situations, but never at people.

    No scientist has herein challenged the conclusions of my new friends (above), namely that adding weight does not necessarily add distance when acceleration is constant. Apparently, it usually will not.

    Thank you all so much. I know that I have been a pain, but such was not my intention.

    Goodbye and good luck to my friends!
    Last edited: Jan 26, 2009
  17. Jan 26, 2009 #16
    hmmm it DOES operate the same way. You're thinking kind of like a calculator. Where you can only change one side of the equation. Try this:

    8 = 4 * 2 Now, if you increase 4 to 6 you can either change 8 to 12 to maintain the integrity of the equation OR you can change the 2 to 8/6ths.

    that is:

    12 = 6 * 2


    8 = 6 * 8/6 = 6*1.33333...

    Both are true statements. It's just a matter of which number you want to change in order to maintain the truth of the equation.


    F = M * A

    If you add to M, you can either lower A and F will stay the same, or you keep A the same and F will increase.

    The question is, "Can you swing a heavier club at the same speeds you swing a lighter club?" or, "Is your acceleration the same despite the fact that you have changed the mass?" If you can, then your swing will have more force. However, you may not be able to, so the force will likely be the same and your club will be slower.

    See how fast you can pick up a rock. Then see how fast you can pick up a much bigger rock using the same amount of force. If you can lift both rocks, then you will lift the heavier rock a lot slower than the lighter rock. That reduced speed for the more massive rock reflects that the rock is accelerating slower.

    Mass is hard to define. An operational definition that should suit any purpose you may have is that it is the amount of material (Not the Volume which is how much SPACE that material takes up). A large rock has more MASS than a Balloon of the same size. There is more stuff in the rock than in the balloon.
  18. Mar 17, 2009 #17

    Perhaps it might be easiest to simplify and eliminate the effects that can't be controlled, such as wind speed and air density. Assuming that all unaccounted for effects are equal, when you increase the mass, F = MA guarantees an increase in the amount of force. However, a golf-ball collision with a golf club head is NOT an inelastic collision, as might be concluded from a youtube video. The only way to see the elasticity of such a collision would be with a VERY HIGH SPEED camera, probably requiring at least a thousand frames per second, or possibly more, so as to allow you to see it actually happen. As you're experience and judgement tell you, roughly 2/3 of the club head's force is transferred to the golf ball, and the COR is a measurement of the efficiency of such impacts. For simplicity, I would choose to multiply the COR by MA, and then compare those results based on the change in club-head weight as well as COR. You may find that the overall differences might be too small to not be overshadowed by the unaccounted for effects, such as a poor angle of impact, or other club-swing faults, or by atmospheric effects such as wind or air density (aka humidity). My gut says anything less than 5 yards difference is probably statistically insignificant due to the potentially large differences in two consecutive human swings of a golf-club. However, if you could get one of those golf-club swing testers... you might then be able to at least remove the human element. Of course, since the entire objective is to improve the human performance, you have to take the human element into account. Thus, my conclusion is that a previous posters comment about maximizing club-head speed within the parameters of an accurately swingable club and the maximum allowable weight at which that can still occur, is probably the best advice.

    (aka sgmunson)
  19. Mar 17, 2009 #18


    User Avatar
    Gold Member

    The OP, have you considered physiology to improve your distances? I gained impressive yardage switching to left-handed clubs (at the cost of short-game accuracy). Lots of problems that plague right-handed golfers stem from the dominance of the right hand in gripping/controlling the club when the right hand needs to be "along for the ride". Turn this around and try driving left-handed. You might be surprised. I was. I can easily get 200 yards out of a 5-iron when the ball is back in my stance (effectively de-lofting the club), but sometimes had spotty results driving for distance. Switching to lefty improved my long game, but I had trouble with the short game, which is where the points are.
    Last edited: Mar 17, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook