1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Golf ball rolling coefficient

  1. Dec 20, 2007 #1
    Hello to all.

    I am a golf course designer and am researching information on the slope and the roll of a golf ball on different greens speed.

    I am attempting to determine a golf ball rolling on "X" slope will not stop on "Y" surfaces(friction coefficient)

    My interest is in a golf ball rolling on different speed greens (surfaces with different rolling friction coefficient) and measuring that info and then determining the slope that a ball will not stop rolling.

    We use a stimpmeter now but it is primitive at best.

    I intend to use this for determining the maximum slope for putting greens based on speed of greens.

    A few tidbit of info:

    The weight of a ball is 1.62 ounces
    The stimpmeter we us to determine friction coeffient is a 3' bar raisedto a 20.5 degree slope

    The distance we call "green speed" is the distance the ball travels after leaving the 3' bar on a level surface.

    This Green speed number is a industry standard.

    Initial work I have done in the field shows a ball will not stop rolling at differning speeds and slope and can be plotted.

    I wish calculate the the graph using physics and have no way to deterime rolling friction coefficients with a golf ball. (which mustallso have a friction coefficient?)

    Any one willing to advise me (remember I am a layman not educated in physics enough to know much more than what goes up must come down :)

  2. jcsd
  3. Dec 20, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You might find this article of interest: http://turf.lib.msu.edu/1990s/1997/970312.pdf" [Broken]
    Last edited by a moderator: May 3, 2017
  4. Dec 20, 2007 #3
    I have it and it is not accurate and cannot be duplicated in field. I have spoke to Mr Weber and he is reviewing it. He showes a ball will not stop rolling at a higher angle than can be replicated in the field.

    This is close but misses. Did you look at the article Doc? Did he not assume the Coefficient of the ball also ?
    Last edited by a moderator: May 3, 2017
  5. Dec 20, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I have to admit that I didn't read it closely before posting the link. (D'oh! :redface:) But what do you mean by "assume the coefficient of the ball"? Looks to me like he's derived the relationship between stopping distance and coefficient of friction, then he uses his measured distance values to calculate the coefficient of friction. (I'm not sure I buy his derivation, for one thing.)
  6. Dec 20, 2007 #5
    What I am asking is each the green(surface) and ball have a friction coefficient don't they? What I have read is rolling coefiient is based on a steel ball (smooth) on a surface.

    What do you mean about his devrivation? I can email you my field data in an xls if you liek to see it.
  7. Dec 20, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    That's part of the problem. I would say that what he calls "rolling coefficient" would strongly depend on the particular green surface and perhaps speed as well. Seems like a bit of a bogus model to me.

    Another thing that bugs me is his equation 3; that's not obvious to me. He uses the same logic for equation 8. Not sure I buy it.

    I take it that you are measuring the stopping distance, then are trying to use this guy's equations to determine the slope at which a ball would keep rolling (when it leaves the stimpmeter)?
  8. Dec 20, 2007 #7
    The green suraces will have differing rolling coef. or green speeds as we call them

    Last edited by a moderator: May 3, 2017
  9. Jan 19, 2009 #8
    Dear jlemon and Doc,

    About five years ago I corrected Weber's Stimpmeter calculations and posted the new calculations on my http://www.network54.com/Forum/52812/thread/1088599618/last-1110805824/Stimpmeter+Physics". His basic mistake is in using the top of the ramp (12.6" high at top of 36" ramp) as the release height for a release angle of 20.5 degrees, when in fact the release notch is 30" up the ramp at a height (from basic trignometry) of 10.5" (sin[20.5] * 30").

    This is what I then posted:

    Arthur Weber has written for the USGA Green Section an article entitled Green Speed Physics (pdf document) that calculates a speed of the golf ball off the bottom of the ramp of 95.5 inches per second. His otherwise very nice paper, however, has a mistake embedded in the calculations, in that he uses 12.6 inches for the beginning height of the ball, and the actual height is only 10.5 inches. Let me re-do his calculations.

    First, what exactly is a Stimpmeter?

    A Stimpmeter is an inclined plane or ramp that is 36 inches long, with a notch for the golf ball 30 inches from the bottom end, such that raising the back end of the ramp to a height of 12.6 inches and an angle of 20.5 degrees causes the ball to tip out of the notch and start rolling down the ramp. See USGA Stimpmeter Instruction Book. The golf ball notch is 10.5 inches above the ground at release, from Pythagorean geometry, 30 inches * sine(20.5 degrees) = 10.5 inches.

    An inclined plane is simply a mechanism for spreading out the WORK of raising a WEIGHT to a HEIGHT. The milder the angle of the ramp, the more spread out the WORK and the less FORCE required to be applied over time to move the WEIGHT up to the increased HEIGHT. If you want to raise a bowling ball five feet off the ground, its a lot easier to roll it up a gently sloped ramp than it is to hoist it straight up, and carrying a suitcase up a flight of stairs is harder on steep stairs than on a long, more gently sloped flight of stairs. The point is that the ENERGY to raise the WEIGHT a given HEIGHT does not at all depend upon the angle of length of the ramp, but only on the HEIGHT. Galileo sorted all this out about 400 years ago. Students at Rice University in Texas have reproduced Galileo Inclined Plane Experiment.

    Since from basic Pythagorean geometry we know that any one angle and any one length of a hypotenuse of a right triangle corresponds to one and only one triangle with adjacent and opposite sides (and the opposite side is the same as the back-end HEIGHT of a Stimpmeter), then it is obvious that a Stimpmeter is just a device for always giving a golf ball the same ENERGY on every roll, since that LENGTH and ANGLE always corresponds with the same HEIGHT.

    The relevant physics formulas are the following:

    1. PE = W*H

    The Potential Energy (PE) is the ENERGY stored in the golf ball's WEIGHT (W) by raising it higher than it is on the ground a certain HEIGHT (H).

    For the Stimpmeter and a golf ball, the numbers are PE = 1.62 ounces (oz) * 10.5 inches. Thus PE = 1.62 * 10.5 = 17.01 in-oz of ENERGY with the Stimpmeter raised.

    2. KE=PE*cosineANGLE, or KE=W*H*cosineANGLE

    The Kinetic Energy of the ball off the bottom of the ramp is the PE reduced by part of the ANGLE of the ramp's tilt.

    For a Stimpmeter and ball, the numbers are KE = 17.01 * cosine(20.5 degrees). This KE = 17.01 in-oz * 0.9367 = 15.93 in-oz. The difference between PE and KE (17.01 - 15.93) of 1.08 in-oz is just the bounce at the bottom that dissipates or wastes some of the PE.

    3. KE = 0.5 W V^2 / g and V = SQR[2*KE*g/W]

    The VEOLICTY (V) of the ball at the bottom of the ramp is determined by the KE, which in turn depends on the WEIGHT, HEIGHT and ANGLE of the ramp, as seen in formulas 1 and 2 above. The variable "g" is earth's constant of acceleration form gravity's force, and is usually taken to be 32.2 feet per second per second (velocity increases each second by this amount of added velocity). In units of "inches per second per second," this is 32.2 * 12, or 384 inches per second per second (in/s/s).

    For the Stimpmeter and golf ball, the numbers are V=SQR[2*15.93*384/1.62], or SQR[7,552] = 86.90 inches / second velocity at the bottom of the ramp. [Weber gets 95+ inches per second, too high, since he erroneously uses 12.6 as the release HEIGHT.]

    To figure how long in TIME it takes to roll down the ramp, which is always the same, the formula is simply DISTANCE = SPEED * TIME. Here, SPEED is the average VELOCITY while rolling down the ramp. Since the ball starts at 0 inches per second velocity in the notch, and smoothly accelerates to a peak VELOCITY at the bottom or end of the roll off the ramp of 86.9 inches per second, the average velocity or SPEED is one half 86.9, or 43.45 inches per second. The DISTANCE is 30 inches. So the TIME = DISTANCE / SPEED, or T = 30 in / 43.45 in/s, for a total roll time of 0.69 seconds. If you silently mouthed "one mississippi" after the ball starts rolling, then ball would exit the bottom of the ramp after about "one missip..."

    So the ball rolls for 0.69 seconds and leaves the ramp traveling at 86.9 inches per second (which is about 16.5 revolutions per second).

    To convert the Stimpmeter reading into the green's "coefficient of friction," where all Stimp 9 greens have the identical "coefficient of friction," the formula is:

    5. KE = W*S*f

    KE is Kinetic Energy of the ball off the Stimpmeter, W is WEIGHT of the ball, S is Stimpmeter of the green (e.g., 9, 10.5, etc.), and f is the "coefficient of friction" of the green.

    The numbers are 15.93 in-oz = 1.62 oz. S f. Thus, rearranging to see the relationship between Stimp and f,

    f = 15.93 in-oz / 1.62 oz * S (*12 in/ft to convert Stimp in feet to Stimp in inches)

    f = .82 ft / S (back to S in feet), or f = .82 ft / S ft

    So, for a Stimp of 10, the green has a "coefficient of friction" of 0.082. For a Stimp 5, f = 0.165 or twice as high.

    The above calculations corrected for the effects of rolling friction down the ramp predict an off-ramp velocity a little lower: 76.9 in/s.

    The rolling of a ball down the Stimpmeter is slower than the frictionless sliding of a block down the same ramp due to some energy going into making the ball roll. The TRANSLATIONAL velocity of the ball at the bottom of the ramp is not as fast as the above calculations suggest in ITEM 5. The actual formula that accounts for this difference generates a predicted speed at the bottom of the ramp of 76.9 inches / second or 6.35 feet / second, so the AVERAGE velocity down the ramp is 1/2 this, or 38.45 inches / second or 3.175 feet / second over 30 inches. Thus the time down the ramp is 0.78 seconds. It takes the ball about 8/10th of a second to roll from the top of the ramp to the bottom.

    Here is the correction in another http://www.network54.com/Forum/5281...805003/Ball+Velocity+at+Bottom+of+Stimpmeter":

    You can't be too careful with getting physics right! The above calculations really assume the ball just drops in free fall, but in actuality it rolls. When the rolling (versus dropping) is observed, the distinction between rotational speed and translational speed of the ball's center is critical. At the bottom of the ramp, the translational speed is what you are asking about.

    The rolling of a ball down an inclined plane does not accelerate the translational speed the same way that a block (for example) would slide without friction down a ramp. The rolling reduces the friction between the ball and the ramp so that friction does not play much of a role in affecting the acceleration (and this is why Gallileo used a ball and ramp to isolate the effect of gravity from that of friction). But if you compare a frictionless block sliding down a ramp and a ball rolling down the same ramp, the translational acceleration of the block is faster than that of the ball, and the resulting velocity at the bottom of the ramp is accordingly faster for the block than the ball.


    This SparkNotes physics website explains the distinctions and gives the correct formula. The upshot is that the velocity at the bottom of the ramp of a rolling DISK is given by:

    v = SQR(g h 4/3)

    where g = acceleration force of gravity, 32.2 feet per second per second, or 384 inches per second per second;

    where h = height of ball at release (10.5 inches)

    The explanation is given:

    "At the top of the incline, the disk has no kinetic energy, and a gravitational potential energy of mgh. At the bottom of the incline, all this gravitational potential energy has been converted into kinetic energy. However, in rolling down the hill, only some of this potential energy becomes translational kinetic energy, and the rest becomes rotational kinetic energy. Translational kinetic energy is given by 1 /2 mv2 and rotational kinetic energy is given by 1 /2 I2. We can express in terms of v and R with the equation = v/R, and in the question we were told that I = 1/2 mR2. We now have all the information we need to solve for v:


    My calculator gives the answer as 76.6 inches per second at the bottom of the ramp, and in feet per second that is 6.38 ft/s, for a rolling DISK.

    The MOMENT OF INERTIA (MOI) of a disk or rolling cylinder or wheel is I = 1/2*M*R^2

    where R = radius.

    But in the case of a BALL or SOLID SPHERE, the MOI is I = 2/5*M*R^2. See Eric Weinstein's Wikpedia for the MOI of a Solid Sphere versus the MOI for variously shaped objects.

    Plugging in the sphere MOI in the basic equation,

    1/2 * Mv^2 + 1/2 * Iw^2 = Mgh

    where w [omega] = v/R; where I = 2/5 M R^2; were R golf ball = 0.84 inches (but which can be ignored since the term cancels out); where g = acceleration of gravity (384 in/sec^2); h = height of release (10.6 in) and M = mass of ball (which can be ignored, since the term cancels out), the formula reduces to:

    1/2 * Mv^2 + 1/2 * [2/5 M R^2] * [v/R]^2 = Mgh

    [1/2 * Mv^2] + [2/10 M R^2 *v^2 / R^2] = Mgh

    1/2 * Mv^2 + 2/10 * Mv^2 = Mgh

    5/10 * Mv^2 + 2/10 * Mv^2 = Mgh

    7/10 * Mv^2 = Mgh

    v^2 = Mgh / M * 10/7

    v^2 = gh * 10/7

    and finally:

    v = SQR(gh * 10/7)

    Plugging in the numbers:

    v = SQR(384 in/sec^2 * 10.5 in * 10/7)

    Thus the velocity of the ball at the bottom f the ramp (v) released from a height of 10.5 inches above the ground is 75.9 inches / second. In other unit terms, this is 6.32 ft/sec. or 1.93 m/s.

    Any sliding or rolling friction will reduce this slightly, as will twisting the ramp out of vertical or the ball's bouncing at the bottom of the ramp. This also assumes an uneventful release that does not add or subtract force in the initial start of the ball's motion down the ramp.

    The coefficients of friction for various (level) green speeds are calculated as follows:

    PE = WH
    PE = (1.62 oz.)(10.5") = 17.01 in-oz
    KE = PE cosΘ
    KE = 17.01 in-oz cos(20.5)
    KE = 17.01 in-oz 0.937
    KE = 15.94 in-oz
    DissipatedE = PE - KE
    DissipatedE = 17.01 in-oz - 15.94 in-oz
    DissipatedE = 1.06 in-oz
    KE = 1/2 * W/g * V(f)^2
    KE = 15.94 in-oz = 1.2 * 1.62oz / (32.2ft/s^2*12"/ft) * V(f)^2
    V(i) = SQR [15.94 * 2 * 32.2 * 12 / 1.62]
    V(i) = 87.2 in/s velocity of ball off bottom of Stimpmeter
    KE = W*St*CF
    CF = KE / St * W
    CF = [15.94 in-oz / 1.62 oz] / St ft * 12 in/ft
    CF = 0.8200 / St

    For level Stimpmeter measurements between 6 ft and 13 ft, the coefficient of friction is:

    6.0 ft, CF = 0.8200 / 6.0 = 0.137 (Weber 6.0 ft: 0.164)
    7.0 ft, CF = 0.8200 / 7.0 = 0.117
    8.0 ft, CF = 0.8200 / 8.0 = 0.102
    9.0 ft, CF = 0.8200 / 9.0 = 0.091 (Weber 8.5 ft: 0.116)
    10.0 ft, CF = 0.8200 / 10.0 = 0.082
    11.0 ft, CF = 0.8200 / 11.0 = 0.075 (Weber 11.0 ft: 0.089)
    12.0 ft, CF = 0.8200 / 12.0 = 0.068
    13.0 ft, CF = 0.8200 / 13.0 = 0.063 (Weber 12.5 ft: 0.079)

    M. Hubbard and L. Alaways, Mechanical interaction of the golf ball with putting greens, in M. Farrally and A. Cochran, eds., Science and Golf III: Proceedings of the World Scientific Congress of Golf (Champaign IL: Human Kinetics, 1999), pp. 428-439, at p. 438, Table 54.2, using 1.83 m/s [72 in/s] as the exit velocity of the ball off the Stimpmeter, calculate the following lower coefficients of friction indicating substantially faster green surfaces with less friction than does Weber:

    4.5 ft, CF = 0.125
    5.5 ft, CF = 0.102
    6.5 ft, CF = 0.0862
    7.5 ft, CF = 0.0745
    8.5 ft, CF = 0.0659
    9.5 ft, CF = 0.0589
    10.5 ft, CF = 0.0533

    A. Penner, The physics of putting, Can. J. Phys. 80: 1-14 (2002), at 2-3, derives the green friction coefficient from the inherent deceleration of a rolling ball's inertia, assuming immediate rolling upon impact without any skidding, using data from the distance a ball rolls on a "slow" green (4 ft, 1.22 m) versus a "fast" green (12 ft, 3.66 m) according to the formula:

    α = - 5/7 * ρ * g

    where ρ = vertical distance from contact point on ball to ball center (how far above or below the ball's equator the contact is made, vertically) / radius of the ball. Penner equates ρ with the coefficient of friction (CF), and calculates CF for a Stimp 4 ft green as 0.196, for a Stimp 8 ft green as 0.131, and for a Stimp 12 ft green as 0.065, using the initial velocity of the ball of the end of the Stimpmeter as 1.83 m/s (72 in/s, 13.6 rps). These CF figures are in general accord with those of Weber.

    The off-ramp velocity used in the above is 1.83 m/s (72 in/s) as calculated by B. Holmes, Dialogue on the stimpmeter, Phys. Teacher 24(7), 401 (1986). My calculation is a little faster (75.9 in/s or 1.93 m/s), so I would predict coefficients of friction slightly larger than the Weber-Penner figures above and slightly steeper limiting slopes as well. (The greater CF slows my faster ball down to the stopping distance of the Holmes 1.83 m/s slower ball.)

    Maximum slopes above which the ball will not stop given the Stimpmeter coefficient of friction are, according to Weber, from the formulae:

    St(dn) = (10.5 in * 0.937) / [CF - sin Φ cos Φ]
    St(dn) = 9.84 in / [{CF - sin Φ cos Φ} * 12 in/ft]
    St(dn) = 0.82 ft / [CF - sin Φ cos Φ]

    A ball will not come to rest when the slope and Stimpmeter combination is such that the denominator in the equation

    St(dn) = 0.82 ft / [CF - sin Φ cos Φ]

    goes to 0, that is, when sin Φ cos Φ > or = CF. The ballpark formula is Maximum slope (Limit) for St(lv): Limit = arcsin(CF). Each 1.74555 % of slope is 1 degree slope; each 0.5729 degree is 1 % of slope. These "limit" combinations are:

    St(lv) 6.0 ft, CF 0.137, Limit 7.87 degrees 13.73% slope
    St(lv) 7.0 ft, CF 0.117, Limit 6.72 degrees 11.73% slope
    St(lv) 8.0 ft, CF 0.102, Limit 5.85 degrees 10.21% slope
    St(lv) 9.0 ft, CF 0.091, Limit 5.22 degrees 9.11% slope
    St(lv) 10.0 ft, CF 0.082, Limit 4.70 degrees 8.20% slope
    St(lv) 11.0 ft, CF 0.075, Limit 4.30 degrees 7.51% slope
    St(lv) 12.0 ft, CF 0.068, Limit 3.90 degrees 6.81% slope
    St(lv) 13.0 ft, CF 0.063, Limit 3.61 degrees 6.30% slope

    Jerry Lemon's calculations of limiting slope plot data points (read from his chart, J. Lemons, Putting green speeds, slopes, and "non-conforming" hole locations, USGA Green Section Record (Jul-Aug 2008), 21-25, at p 22, Fig. 1):

    Stimp. Limit deg. Limit %

    7.0 ft: 4.7 deg., 8.20%
    8.0 ft: 4.0 deg., 6.98%
    9.0 ft: 3.6 deg., 6.28%
    10.0 ft: 3.3 deg., 5.76%
    11.0 ft: 3.0 deg., 5.24%
    12.0 ft: 2.8 deg., 4.89%
    13.0 ft: 2.55 deg., 4.45%
    14.0 ft: 2.4 deg., 4.19%

    These slopes represent much milder slope when the limit is reached than in Weber. The USGA notes: "Based on current information, any sliope 3% (1.7 degrees) or greater on a 10 Stimpmeter reading is too steep for hole use." Turf twisters, USGA Green Section Record (Mar-Apr 2005).

    M Sweeney and M Carroll, The fine line of fair competition: How to avoid hole location disasters, www.aimpointgolf.com, write:

    "A ball on a perfectly flat putting surface will not roll on its own, however, as the slope or grade of that surface is increased, there will come a critical point at which the frictional force is overcome and the ball will begin rolling. That critical point is dependant on the green speed, or stimpmeter reading. For typical tournament green speeds of 12, the grade at which the ball will roll off it position, or accelerate while rolling over, is 5.9%."

    Another indication of how increasing slope alters distance control is the comparison of a one-foot putt's force expressed on different slopes and different green speeds. This data is from Sweeney and Carroll, loc cit., p. 2:

    3% slope
    10: 1.1
    11: 1.3
    12: 1.5
    13: 18
    14: 2.2

    4% slope
    10: 1.3
    11: 1.6
    12: 2.1
    13: 2.8
    14: 3.8

    5% slope
    Stimp: Roll (ft)
    10: 1.7
    11: 2.4
    12: 3.9
    13: 7.9
    14: ball does not stop


    Geoff Mangum
    Putting Coach and Theorist
    http://puttingzone.com" [Broken]
    Golf's most advanced and comprehensive putting instruction -- you're either in the PuttingZone, or not.
    Over 2.6 million visits and growing strong.

    Last edited by a moderator: May 3, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook