Dear sammy,
Your position always harks back to the same point: "I ignore any effects of gravity in the movement part of the putting stroke. The only significant effect of gravity is on the torque created on the hands by the lie angle of the putter in the sagittal plane, which is rather constant during the stroke because it's the cantilevered weight of the putter subject to gravity. ... I advocate a torqued putting stroke, and ignoring any slight pendular gravitational effect, other than the dead weight of the putter. In this way you have constant control of the putting stroke, and you need not bother with switching between muscular force and gravity in the stroking phase of the putt.... Forget the influence of gravity in making a pendular-like putting stroke, and you need not maintain a 'dead hands' grip on the putter ... you are in full control, provided you know how to maintain control. Gravity is not a major force to move the putter in the forward stroke and should be ignored for simplicity."
Apparently, you have some notion about the magnitude of "gravitational effects" in a pendular stroke, that these effects are very minor for typical sizes of putting strokes, and that therefore these effects are insufficient to power the stroke. Let me take a stab at clearing up the confusion between us on this underlying basis for your position.
The "force" that gravity exerts upon an object is called its "weight". The formula from Newton is F = ma, of Force equals Mass time Acceleration, and "a" in the case of gravity is called "g", 9.8 m/sec^2. When the mass is kg, g is m/sec^2, then F is in units of Newtons. But when the mass is in the English system it is "slugs", g is in feet/sec^2, and weight or force is in "pounds". The slug is an English unit of mass. It is a mass that accelerates by 1 ft/s2 when a force of one pound-force (lbf) is exerted on it. Therefore a slug has a mass of about 32.17405 pound-mass or 14.5939 kg. A putter head "weighing" 375 grams has a mass of 375 grams, which is equivalent to 13.2 ounces or 0.83 pounds.
So, as you say, the "force" that gravity exerts by virtue of the putter head is slight or minor, but your point is not relevant to the timing. The timing of the swinging velocity of the putter head at impact is what determines the "force" or "momentum" of impact between putter head and ball, not the force of gravity acting on the putter head.
A ball bearing with a mass of 1 gram swings at the end of a 35-inch string exactly the same (same acceleration rate, same velocity, same total time from top to bottom or from top to top) as does a 1 kg bowling ball, wind resistance effects aside. The masses cancel out in the motion of freely falling bodies in gravity. This is basic Gallilean physics from the Leaning Tower of Pisa in the 16th century. The basic reason is because gravity's force (of attraction between two masses, e.g., one being the "big ball" earth) is less for a light mass and greater for a heavy mass, so that the force of gravity in the free fall of all objects of all different masses on earth is always scaled to result in just the same falling motion (i.e., all masses accelerate smoothly at the exact same rate). Here's a
good, simple discussion of the physics. Mass and weight of objects and the "force of gravity" on an object are irrelevant to the timing pattern of free-fall in gravity.
The above is true with respect to a "simple" or "ideal" pendulum -- masses are irrelevant, and only length matters to the timing and motion pattern. Gravity's "force" is from Newton's Law: Force (F) = Mass (m) times Acceleration [of Gravity] (g), generally F=ma and gravitationally F=mg. Since masses are irrelevant, that means that the "force of gravity" is also irrelevant when comparing small swings and big swings, small masses and large masses. Your statement that the force of gravity is so small etc. is not relevant to an understanding of what happens in a simple pendulum.
Granted, a "simple" or "ideal" pendulum is not the same as a "real" or "physical" pendulum such as the arms, hands, and putter swinging beneath a pivot in the neck. In that case, the masses matter because the system of the "triangle" has a center of mass and a moment of inertia a certain distance out along the length, and this matters in the equations. But looking at exactly how it matters for a typical adult male human body shows again that your position is incorrect.
Whereas for a simple pendulum, the timing of the swing is from the formula:
T = 2*pi*SQR[L/g], with T being the time from top to top and back again to starting top and L is the total length from pivot to bob,
in the case of a physical or real pendulum, the formula includes a "mass" term:
T = 2*pi*SQR[I/MgD], where I is the moment of inertia of the system, M is the total mass of the system, g is the acceleration of gravity, and D is the distance out from the pivot to the center of mass of the system. I = T^2MgD/4*pi^2.
What does this mean for a 35-inch putter swinging from the pivot at the top of its handle?
The simple T would be:
T = 2*pi*SQR[L/g]
T = 2*pi*SQR[(35 in * 2.54cm/in / 100cm/m) / 9.8 m/sec^2]
T = 2*3.14* SQR[0.889m / 9.8 m/sec^2]
T = 6.28 * SQR{0.0907 sec^2}
T = 6.28 * 0.301 sec
T = 1.89 SEC Top to Top to Top
T = 0.946 sec Top to Top (946 ms)
T = 0.473 sec Top to Impact (473 ms)
How fast is the putter head moving at its peak speed at the bottom of the arc? Assuming the angle of the backstroke was 20 degrees, this is 20/360 = 1/18th of the circumference of the circle with radius of 35 inches. Circumference = 2*pi*R, so C = 6.28 * 35 in = 220 inches. 1/18th of this distance is 12.2 inches, so the putter head in this stroke covers 12.2 inches from top to impact in 0.473 sec, so that is an "average velocity" of 25.8 in/sec (12.2 / 0.473). Since the starting velocity at the Top is zero and the rate of acceleration is steady and smooth, the peak velocity that corresponds to this average velocity must be twice as fast as the average, so the peak velocity at impact = 51.6 in/sec.
What does that velocity do to a ball with that putter head? The "momentum" is mass times velocity, so here: 0.375 kg times 51.6 inches/sec. (converted to m/sec = 1.31 m/s), or 0.375 kg * 1.31 m/s = 0.49125 kg*m/s. The Law of the Conservation of Linear Momentum means simply that no energy gets lost and all is evenly accounted for between Time 1 and Time 2. Mathematically, this means that the total momentum of putter head and ball before collision is the same as the total afterwards: mv1(putter) + mv1(ball) = mv2(putter) + mv2(ball). The momentum of the ball before collision is zero, so mv1(putter) [before collision] = mv2(putter) + mv2(ball) [after collision].
To determine the velocity after collision of the putter head and ball, the putter head velocity after collision is the velocity before collision multiplied by the multiplier ratio: (putter head mass - ball mass) / (putter head mass + ball mass). The mass of a golf ball is 45 grams, so the multiplier is (375 g - 45 g) / (375 g + 45 g) = 330/420 = 78%. So the putter head velocity after collision is 1.029 m/s. The velocity of the ball off the putter face is then the velocity of the putter head before collision time the multiplier ratio: 2 * putter head mass / (putter head mass + ball mass), 2 * 375 g / (375 g + 45 g), or 750 g / 420 g, so the multiplier is 1.785 (178.5%). The ball velocity off the face is then 1.31 m/s * 1.785 = 2.34 m/s (equivalent to 92.1 inches/sec or 7.675 ft/s). [As a check, from the law of conservation of linear momentum: mv1 = mv1' + mv2', 0.375 kg * 1.31 m/s = 0.375 kg * 1.029 m/s + 0.45 g * 2.34 m/s, 0.4913 = 0.386 + 0.1053, 0.4913 = 0.4913.] There are 5.28 inches per one roll of the ball, so this is 17.4 revolutions/sec. That's about a 10-foot putt on a fast green.
What does this mean for a real pendulum consisting of a normal adult male and a 35 inch putter swung with a shoulder stroke?
The Moment of Inertia I = T^2MgD/4*pi^2, with M being total mass of two arms and hands and the putter and D being the distance of the effective center of mass of all segments out from the pivot. The typical human adult male arm from shoulder to finger tips has a mass of about 5.7% of body mass in each arm+hand (e.g., 140 pound male has 7.98 pounds in each arm+hand or a total of 15.96 pounds for both arms+hands, 7.3 kg), and a conventional putter has a mass of about 0.490 kg, so the total mass of the system for this male is 7.79 kg. From
anthropometric studies done by NASA (Clauser 1969), the typical D distance of the center of mass of the human arm from shoulder to finger tips along the length from shoulder out to end of hand is about 42% of the entire length. If the normal adult male shirt length is 33 inches (shoulder to wrist) and the hand is another 7 inches long, the total from shoulder to finger tip is 40 inches, and D is 42% of this, or 16.8 inches from shoulder to center of mass (roughly 0.425 meter).
upper arm 3.1% mass 17.9% length
forearm 1.8% mass 16.3% length
hand 0.9% mass 13.0% length
arm+hand 5.7% mass 54.2% length
Dempster 1955:
Dempster, W. T. Space requirements of the seated operator. WADC Technical
Report (TR-55-159), Wright-Patterson Air Force Base, OH, 1955.
upper arm MOI 0.29 Nm^2
forearm 0.037-0.082 Nm^2
International Encyclopedia of Ergonomics and Human Factors By Waldemar Karwowski, p 317.
1 Newton = 0.101 971 621 kilogram-force
1 kilogram-force = 9.806 650 028 6 Newton
The usual geometric model for human arms is the form of a cylinder with uniform density and a given radius and length. From this model, with a COM location, the Moment of Inertia can be calculated, and from that, the timing of the swing can be calculated. Using anthropometric data about human segment masses and sizes and proportionality, a reasonable model that approximates the real human can be defined for calculations.
Cylinder Moment of Inertia Calculations.
Physical Pendulum Calculator per Parallel Axis Theorem.
FIRST MODEL OF HUMAN MAKING PUTTING STROKE, AFTER CLAUSER 1969:
MODEL OF ARMS+HANDS+PUTTER USING 6.7 kg FOR ARMS+HANDS + 0.49 kg FOR PUTTER
Cylinder I = 1/4 MR^2 + 1/3 ML^2.
[N.B.: In the above equation for the Moment of Inertia of a solid cylinder, L is the total length of the "object" or "body segments" or body segments plus object, as the case may be, rather than the D distance from pivot to COM. When using a calculation involving D, the figure 0.7 meter has been guesstimated as the approximate location of the center of mass (COM) of the arms+hands+putter of 1.35 meters in total length, which is roughly half way. For only the arms, the COM is about 42% of the way out from the pivot, so incorporating the thin, relatively light mass of the putter into the structure of the arms and hands with the big blob of the putter head out at the end of the stick probably moves the COG out to about 50% of the total length. The FIRST MODEL uses L and D, and the SECOND MODEL uses D.]
Arms+Hands:
1/4 * 6.7 * .04^2 + 1/3 * 6.7 * 1.01^2 = 2.2809 kg*m^2
Putter:
1/4 * 0.490 * 0.005^2 + 1/3 * 0.490 * 0.635^2 = 0.065863 kg*m^2
Arms+Hands+Putter:
(Model= cylinder radius 1 cm 1.35 m length 7.19 kg mass.)
1/4 * 7.19 * .01^2 + 1/3 * 7.19 * 1.35^2 = 4.3681 kg*m^2
Conversions: above Model
inertia moment or mass center (I) = 4.3681 kilogram-meter^2
mass (M) = 7.19 kilogram
acceleration of gravity (g) = 9.8 meter/second^2
distance from moment of inertia to pivot (D) = .7 meter
Solution:
period (T) = 1.86981935697 second
Backstroke T = 0.935 second
SECOND MODEL OF HUMAN MAKING PUTTING STROKE, AFTER DEMPSTER 1955:
MODEL BASED UPON 5.7% BODY MASS FOR ARM+HAND * 2 + 0.49 kg FOR PUTTER
Various adult male body masses all with the same stature and arms+hands length:
140 lbs 5.7% arm+hand mass * 2 + putter
inertia moment or mass center (I) = 5.61918 kilogram-meter^2
mass (M) = 7.49 kilogram
acceleration of gravity (g) = 9.8 meter/second^2
distance from moment of inertia to pivot (D) = .7
Solution:
period (T) = 2.07784608199 second
Backstroke T = 1.038 second
180 lbs 5.7% arm+hand mass * 2 + putter
inertia moment or mass center (I) = 7.3447027 kilogram-meter^2
mass (M) = 9.79 kilogram
acceleration of gravity (g) = 9.8 meter/second^2
distance from moment of inertia to pivot (D) = .7 meter
Solution:
period (T) = 2.07784704558 second
Backstroke T = 1.038 second
200 lbs 5.7% arm+hand mass * 2 + putter
inertia moment or mass center (I) = 8.16995025 kilogram-meter^2
mass (M) = 10.89 kilogram
acceleration of gravity (g) = 9.8 meter/second^2
distance from moment of inertia to pivot (D) = .7 meter
Solution:
period (T) = 2.07784705266 second
Backstroke T = 1.038 second
250 lbs 5.7% arm+hand mass * 2 + putter
inertia moment or mass center (I) = 10.12053525 kilogram-meter^2
mass (M) = 13.49 kilogram
acceleration of gravity (g) = 9.8 meter/second^2
distance from moment of inertia to pivot (D) = .7 meter
Solution:
period (T) = 2.07784705266 second
Backstroke T = 1.038 second
The above two models of the human arms and hands holding a putter in a fixed triangle shape during a shoulder stroke with a pivot-to-sole length of about 1.35 meters (54 inches from top of sternum at address posture to bottom of putter) and a center of mass about 0.7 meters from the pivot agree that the pendular timing of the backstroke is about 1 second and the time from top of backstroke to impact is half that.
How fast is the putter head moving at its peak speed at the bottom of the arc? Assuming the angle of the backstroke was 20 degrees with the pivot 54 inches from the putter head's sole, this is an arc 18.8 inches long, so the putter head in this stroke covers 18.8 inches from top to impact in 0.5 sec, so that is an "average velocity" of 37.6 in/sec (18.8 / 0.5). Since the starting velocity at the Top is zero, the peak velocity that corresponds to this average velocity must be twice as fast as the average, so peak velocity = 75.2 in/sec.
What does that velocity do to a ball with that putter head? The "momentum" is mass times velocity, so here: putter head mass 0.375 kg times 75.2 inches/sec. (converted to m/sec = 1.91 m/s), or 0.375 kg * 1.91 m/s = 0.71628 kg*m/s. The Law of the Conservation of Linear Momentum essentially says "nothing funny" happens to the total energy of the event between Time1 and Time2 in a collision between two objects. Mathematically, mv1(putter head) + mv1(ball) [before collision] = mv2(putter head) + mv2(ball) [after collision]. Since the ball before collision is not moving, it has zero momentum, and the equation is then mv1(putter head) [before collision] = mv2(putter head) + mv2(ball) [after collision].
To determine the velocity after collision of the putter head and ball, the putter head velocity after collision is the velocity before collision times the multiplier: (putter head mass - ball mass) / (putter head mass + ball mass). The mass of a golf ball is 45 grams, so the multiplier is (375 g - 45 g) / (375 g + 45 g) = 330/420 = 78%. So the putter head velocity after collision is 1.49 m/s (78% of 1.91 m/s).
The velocity of the ball off the putter face after collision is the velocity of the putter head before collision time the multiplier: 2 * putter head mass / (putter head mass + ball mass), 2 * 375 g / (375 g + 45 g), or 750 g / 420 g. So the multiplier is 1.785 or 178.5%. The ball velocity off the face is then 1.91 m/s * 1.785 = 3.41 m/s (equivalent to 134.2 inches/sec or 11.2 ft/s). There are 5.28 inches per one roll of the ball, so this is 25.4 revolutions/sec. That's about an 11-foot putt on a medium speed green.
The velocity of a rolling ball off the bottom of the Stimpmeter is about 76 inches per second or 14.3 revolutions per second or 6.3 feet per second. [N.B.:
Weber's calculations for the USGA seem to be mistaken about the height of the release notch, treating it at 36 inches up the ramp and 12.6 inches high when in fact it is 30 inches up the ramp and 10.8 inches high. (See
USGA Stimpmeter Instruction Booklet, notch at 30 inches up ramp.) Weber's 95.5 inches per second ball speed off the ramp is too high -- see my
Flatstick Forum discussion of this issue.] Based on my putter head mass and stroke, a backstroke that generates a ball speed of 76 inches per second has a putter head speed of 76 / 1.785 = 42.6 inches per second (for a putter head mass of 375 grams). For this to be the peak speed at impact, the downstroke "average" speed of the putter head will be half that or 21.3 inches per second. Since the downstroke is 0.5 seconds or thereabouts, the distance the putter head travels is 21.3 inches/sec * 0.5 sec. = 10.65 inches of stroke arc (see
the Circle Calculator for arc). This backstroke has an angle of 11.3 degrees. The backstroke measured linearly along the ground is 10.8 inches.
CONCLUSION
In summary, a gravity-sponsored stroke with no torquing of the putter into impact but allowing the putter and arms and hands to free-fall in a pendular manner in gravity results in a stroke with a backstroke timing of about 1 second, a downstroke-to-impact timing of about 1/2 a second, and a stroke timing from top to top of about 1 second. Modelling the human "triangle" of a normal adult male with a conventional putter using the physics of a real, physical pendulum predicts the same timing. the swinging peak speed of the putter at the bottom of the arc is sufficient with an 11.3 degree backstroke to mimic the "force" of the Stimpmeter, and therefore roll a ball to the same distance that the Stimpmeter measures on that green. Hence, my backstroke of 10.8 inches should match the distance given in terms of the green speed, rolling the ball 10 feet on a Stimp 10 green and 11 feet on a Stimp 11 green.
In trying to understand why you don't seem to "get" the physics here (or agree that this IS the relevant physics), I note your frequent references to the "force of gravity" in the free-body diagram sense of the top of the backstroke. This is irrelevant to the force of the blow of the putter head as it swings freely thru impact. I also suspect you may be using the static torque of the putter and arms at the top of the backstroke as an indicator that this same dynamic "torquing" action is required to generate the stroke force necessary to roll the ball across the green, when in fact these two senses of "torque" are completely different. The putter and arms and hands will swing down and generate plenty of momentum and force to roll the ball across the green, and it's not a matter of the "force" of gravity being small or of the apparent "need" to "torque" the putter down and thru impact in order to generate force: it's a matter of the peak speed of a pendular swinging as determined by the angle or size of the backstroke.
Another possible explanation for why you don't seem to agree with the physics here is that a gravity-sponsored backstroke typically has backstroke sizes that are larger than your torquing stroke sizes. Perhaps if you really tried the gravity-sponsored stroke, you would not think the forces are so minor. If you roll a ball 20 feet with your stroke, I roll a ball also twenty feet with my larger, slower stroke, so obviously there is no difference in the energy generated.
Finally, I don't see the detailed explanations that should accompany your use of the terms "torque" and "gravity" force. perhaps you are a bit vague about all this, and if you dug deeper into the precise meaning of these terms, you would not use them in senses that strike me as non-scientific because off the mark in the relevant physics.
If you disagree with the physics outlined in this post, that would be interesting. But if you don't disagree with the physics, then I fail to see the logic of your constantly repeated position.
Cheers!
Geoff Mangum
Putting Coach and Theorist
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