Hi Geoff,

Have you determined what the effective hole diameter is for a putt that would
roll past the hole with correct speed (~12 inches past)?

Thanks
David

Dear David,

Although Dave Pelz made widespread the notion of judging speed of putts by how far past the hole a miss rolls, this is in fact a wrongheaded, incorrect, harmful and confusing approach that the golf world would be a lot better off without. The physics of the ball-hole interaction is separate and apart from the ball-green interaction. The effective width of the hole basically has next to nothing at all to do with go-by distance. The distance a ball rolls past the hole in case of a miss depends upon the speed of the ball at the front lip of the cup, the green's speed, and uphill-downhill effects. The capture of the ball in the ball-hole interaction depends upon the length of the path across the hole and the ball's lateral speed over the hole as the bottom of the ball crosses over the lip, and sometimes whether the back rim is higher or lower than the front rim.

The "best" ball speed at the front lip (in light of usual green conditions today) is about two revolutions per second (about 10.5 inches per second, just less than 1 foot per second). This speed makes just under 87% of the hole available for capture.

From Ptolemy's Table of Chords in the Almagest (Syntaxis Mathematica, 2nd century AD), the case where the "chord" across the hole is one half the full diameter is known from the expression sin (theta/2) = 0.5 diameter / diameter, with theta being the angle at the center of the cirle with radii out to the ends of the half-diameter chord. When the diameter of a golf hole is 4.25", sin (theta/2) = 2.125/4.25, of 1/2. So theta/2 = arcsin(0.5), and theta is therefore 60 degrees, and theta/2 is 30 degrees.

This diagram illustrates the general idea:



From the trigonometry of the smaller triangle OAM, with angle OAM being 30 degrees, the length O-M is 1/2 of the radius OM, and 1/4 the diameter P-P'. The distance A-M is then 1.84" (A-M = O-M / tan(30)). The "capture profile" A-B is twice this or 3.68". This is 86.6% of the full width of the hole.

Now, if a ball rolling 2 revolutions per second at the front lip misses, and the green speed is a "usual" Stimp 10 and the surface level, this ball will roll probably for between 1 and 2 seconds more, with its speed decreasing from 2 rps to 0 rps, so the ball averages 1 rps for 1 or 2 seconds. From the front lip, this is a roll distance between 5.28" and 10.56" from the front lip. Subtracting the width of the hole (4.25") to see "how far PAST the HOLE" the ball would roll, the answer is between 1.03" and 6.28" past the hole. If the green is slower (say, Stimp 9), the ball will roll for about 1 second with an average speed of 1 rps, for 5.28" distance, and merely 1.03" "past the hole". If the green is significantly faster (say Stimp 12), the ball might roll 2-3 seconds past at an average speed of 1 rps or 10.56" to 15.84" inches past the lip and about 6.28" to 11.56" past the hole. If the surface past the hole has much downhill slope, these distances get much larger.

But the "usual" green will see a 2 rps ball go a well less than a foot past the hole, on average.

Cheers!

Geoff Mangum
Putting Coach and Theorist
PuttingZone.com
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Geoff .... I read your treatise on Ball-Hole Capture Physics at http://puttingzone.com/capture.html with great intererst, and I must ask you to review it particularily on the topic of decay patterns and Pelz's 18" past-the-hole standard.

The way I interpret this passage in the study:


The "Decay" phase of putts sets in at about the same point in the slowing speed of the ball, depending on the green speed.

For a given green, with a set green speed, the decay phase ALWAYS sets in at about the same ball speed as it reaches a certain slowness of roll, and the distance from this point to the end of the roll is ALWAYS about the same (ignoring uphill-downhill effects) and ALWAYS takes about the same amount of time.

A 20-foot putt has a decay phase onset at the same ball speed near the end of the putt as does a 5-foot putt, and both putts have a decay phase that lasts about the same length and time after onset.


.. seems to indicate that Pelz may be correct, maybe not in the 18", but that most all putts of equal energy on different stimp greens will decay approximately the same distance, and that putts of different distances on a given green will have the decay phase begin the same distance from where the ball will eventually stop. Does this not offer some credence to Pelz's observations and conclusions?

Could you help clarify this issue, and comment on the reasons why you think the decay phase behaves as it does.

Thanks ....

Dear sammy,

That the decay phase sets in on a given green at a specific point in the slowing speed has been known since at least 1968 in Cochran and Stobbs' Search for the Perfect Swing book. The implication or meaning of this for optimal speed control and ball capture, however, has not previously been elaborated in a thoughtful way. Dave Pelz, in particular, has never mention the "decay phase" in his life, so far as I can determine. So Pelz does not have any "observations and conclusions" on this issue of the relationship of the decay phase onset and optimal speed at the front lip or the so-called go-by speed for maximizing sinks.

The point that Pelz is making is a totally bogus and unscientific point: that the same go-by distance is best for ALL greens in ALL conditions, regardless of grass type or speed. His actual research, published only in 1977, proves exactly the opposite of what he always claims. he always claims (since 1983, and about once every year or two thereafter with clocklike regularity) that he "has proved scientifically that 17 inches past the hole is the optimal go-by distance to maximize sinks for all putt lengths on all grass types regardless of green speed" or words very similar. In fact, Pelz's actual research proved just the opposite, and he said so in print (Larry Dennis, Die your putts at the hole and you're dead, Golf Digest, July 1977, pages 52-55). So Pelz actually KNOWS that his statements since 1983 are not truthful or scientific. In 1977 Pelz said that his research proved that

"there is not any one specific distance past the hole for maximizing sinks," and

"the optimal go-by distance varies from green to green and depends upon the grass type and its condition", and

on bent grass greens in tournament condition "the optimal go-by distance is not a discrete number but a wide range between 5-10 inches," and

on bent grass greens in poor condition "the optimal go-by distance is not a discrete number but a wide range between 10-15 inches," and

on Bermuda grass greens in tournament condition "the optimal go-by distance is not a discrete number but a wide range between 20-30 inches," and

on Bermuda grass greens in poor condition "the optimal go-by distance is not a discrete number but a wide range between 30-40 inches."

or words to this effect (read the article yourself). In light of this, when you ask me "Does this not offer some credence to Pelz's observations and conclusions?", I have to ask whether you are referring to Pelz's numerous bogus statements since 1983 or referring to his actual research report in 1977. I assume, since you ask, that you are taken in by Pelz' false claims since 1983, when he pursued his cock-eyed nonscientific approach to making his statements about optimal go-by speed. His actual science says this approach has no merit to it.

Expert golfers since the 1940s at least had previously known that the go-by speed with optimal touch will vary depending upon grass type and condition (see, for example, Cary Middlecoff's books in the 1950s), so Pelz' actual scientific research merely confirmed this obvious-to-many fact. The research added no lustre to Pelz' personal star. Apparently, he decided sometime around 1983 to bury his research (he has never mentioned the 1977 article in print or in person so far as I can determine) and claim something (false) that he says he "discovered scientifically." Making scientifically false claims does a huge disservice to the golfing community.

Incidentally, you cannot prove a false assertion with sound science. Therefore I can state without any fear of contradicition that NO ONE can possibly prove that 17 inches past the hole is the optimal go-by speed to maximize sinks for all putts on all grasses in all conditions. And no one ever has, and no one can replicate Pelz' experiments from 1976 and prove this assertion either.

Forgetting the confusion caused by Pelz, and returning to your question with the Pelz issue removed, I understand you to ask whether the decay phase onset being the same for a 20-footer and a 5-footer means that the go-by distance will be something like 17 inches past the hole? Not at all. The onset of the decay phase occurs on different green speeds at different points in the slowing speed of the ball, and with different roll-out distances thereafter. The decay phase is really just a noticeable difference in the deceleration causedd by the surface: there is one friction when the ball is rolling "on top of the grass" as they say, and a sharper friction when the ball "settles back down lower into the grass" as they say. The "noticeability" of the difference is not very on fast bent grass greens, and is very noticeable on slow Bermuda greens.

I believe that I am the only person ever to suggest that handling the delivery speed of the ball to the cup so that the ball reaches the cup just before or simultaneously with the onset of the decay phase MIGHT be a beneficial approach. But more precisely, this issue needs more fleshing out with scientific observations of the exact onset speed, the roll-out, and whether it makes any difference if the onset occurs much before the lip is reached.

So, no, Pelz's "observations and conclusions" do not merit any credence at all. His 1977 science added nothing new, and his false claims since 1983 have caused a lot of harm to how people putt.

Cheers!

Geoff Mangum
Putting Coach and Theorist
PuttingZone.com
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Thanks Geoff .... and I would like to take you up on this statement of your's:

" The decay phase is really just a noticeable difference in the deceleration caused by the surface: there is one friction when the ball is rolling "on top of the grass" as they say, and a sharper friction when the ball "settles back down lower into the grass" as they say. The "noticeability" of the difference is not very on fast bent grass greens, and is very noticeable on slow Bermuda greens."

... as it might apply to the practical application of the science to putting. I understand this to mean that as the ball slows down, it's weight has more effect on dynamic friction and the ball tends to settle down ever so slightly into the grass .... thus initiating the decay phase.

If this is correct then the pro golfers in a tournament experience much different putting physics from the beginning of the field to the last golfers of the day. Presumably, the trodden grass around the hole reduces the decay phase, and together with the altered topography around the hole, putting becomes an unequal contest.

The same may apply to recreational golfers playing early or late in the day. Putting is a chaotic situation !!

Dear sammy,

Pros clearly already know that late starting times mean poor greens, so yes "pro golfers in a tournament experience much different putting physics from the beginning of the field to the last golfers of the day. " However, the pros seem to focus most on the poorer conditions that slow the green and make holding the line more problematic, such as spike marks, ball marks, and footprints. The point you make ("Presumably, the trodden grass around the hole reduces the decay phase") is not generally recognized among today's pros. The point has been made in somewhat obscure golf literature of the past, but it is a subtle (and valuable) point not widely appreciated.

Cheers!

Geoff Mangum
Putting Coach and Theorist
PuttingZone.com
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Geoff,
you stated:
"Now, if a ball rolling 2 revolutions per second at the front lip misses, and the green speed is a "usual" Stimp 10 and the surface level, this ball will roll probably for between 1 and 2 seconds more, with its speed decreasing from 2 rps to 0 rps, so the ball averages 1 rps for 1 or 2 seconds. From the front lip, this is a roll distance between 5.28" and 10.56" from the front lip."

This ball will roll "probably"? for between 1 and 2 seconds?
Is this a guess off the top of your head or did you calculate it or read it from a graph?
Roll distance between 5.28" and 10.56" from the front lip?
If the ball has an initial speed of 2 revs per second (10.56"/sec), why would it roll
between 5.28" and 10.56" and for 1 or 2 seconds?
It seems that you are using the average speed of 1 rps and applying it to a "guessed" roll time range of 1 to 2 seconds. You are then projecting that if the speed is 1 rps, it will roll for 1 second and will cover a distance of 1 revolution or 5.28in. For the 2 second roll case, you are saying that if the speed is 1 rps for 2 seconds, it will cover 2 revolutions or 10.56 inches. This logic indicates some confusion. I think you're trying to apply "decay" (from your "scanned" Ball Delivery Speed graph), extrapolated guesses about roll time, and incorrect physics to arrive at these statements. Any graph of speed vs distance for an object under constant uniform deceleration will have a "knee" or decay phase at the end just prior to stop because of the squared relationship of speed to distance. V = sqrt(Vo^2 + 2as). On fast greens, the extra friction of the ball "falling" into the grass blades at its end of roll has little added effect except to make the roll distance and roll time minutely shorter. A ball with an initial speed of 1 rev/sec can't roll for 1sec and cover 1 revolution on a Stimp of 10, nor can the same ball on the same green then roll 2 revolutions for 2 seconds with the same initial speed. It would take a Stimp of 82 for the ball to roll 5.28 inches with a Vo of 5.28 in/sec. The corresponding roll time would be 2 seconds.

If a ball reaches the front lip at 2 rps (.88ft/sec) on a green of Stimp=10, the roll distance would be:
dist = .215ft -> 2.58"
time to roll to a stop = .49sec

dist on green of Stimp=10-> 2.6in, time = .49 sec
dist on green of Stimp=9 -> 2.3in, time = .44 sec
dist on green of Stimp=12 -> 3.1in, time = .59 sec


Dean

Dear Dean,

You make a very interesting point about what actually causes the "knee", but your explanation is not really clear to me. Could you elaborate on that?

My 1-2 seconds is not read from a graph, but is "read from reality" of watching nearly 4 million putts die out. That's a VERY LARGE data sample. It's really closer to just 1 second, but I left it a little open-ended on purpose, so as not to appear overly precise.

The average velocity from an initial velocity at the front lip (2 rps) to a final velocity (0) is simply how the total distance is covered: average velocity times time = distance. I don't at all understand what you are saying is incorrect about this. The only assumption embedded in this approachg is that the deceleration rate is uniform, as it appears to be. For example, to detrmine the distance a ball drops when it starts at zero and at the final velocity is is traveling at 32 feet per second, and the time that elapses is 1 second, the distance = average velocity (32 - 0 / 2) multiplied times the time of travel (1 second). So the distance traveled is 16 feet. And that is exactly what happens in gravity: drop a ball from 16 feet and it takes exactly 1 second to hit the ground, at which time it is traveling at 32 feet per second.

Can you explain a little more clearly what your objection is and what you say is the correct physics?

Cheers!

Geoff Mangum
Putting Coach and Theorist
PuttingZone.com
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Geoff,

You are given the initial speed (2rps or .88ft/sec) and the friction of the green (Stimp=10). From these two values, the distance and the roll time can be calculated.

Instead, you used an "eyeballed" roll time of 1 to 2 seconds (with some sort of innate time measurement ability that you acquired through watching 4 million putts).
You got the roll range of 5.28" to 10.56" by plugging the guestimated time range and average speed into d=rt.

As I listed in the previous post, I got:
dist on green of Stimp=10-> 2.6in, time = .49 sec
dist on green of Stimp=9 -> 2.3in, time = .44 sec
dist on green of Stimp=12 -> 3.1in, time = .59 sec

which varies quite a bit from your distance & time numbers:
dist on green of Stimp=10-> 5.28-10.56", time = 1-2 sec
dist on green of Stimp=9 -> 5.28in, time = 1 sec
dist on green of Stimp=12 -> 10.56" to 15.84", time = 2-3 sec


By the way, 4 million thoughtful putts?

If, for each putt,
1. you take (10 seconds) to choose a target
2. take (10 seconds) for a pre shot routine
3. take the putter back (1 second), strike the ball (1/2 sec) (per the "gravity" stroke)
4. watch the ball roll to a stop (7 seconds)(assuming the 4 million putts average to 10ft on a Stimp of 9)
5. walk to retrieve the ball (10 seconds) and start over to step 1),
the total time per putt would be ~38.5 seconds or .64 minutes

4,000,000 putts * .64 min/putt = 2,560,000 minutes = 42,667 hours
If you putt for 8 hours a day for 5 days a week for 50 weeks a year, that's
21.33 years! (I gave you weekends off and two weeks for Christmas/New Year)

Dean

Dear Dean,

It appears you are using this formula, as quoted form your earlier post: "V = sqrt(Vo^2 + 2as). On fast greens, the extra friction of the ball "falling" into the grass blades at its end of roll has little added effect except to make the roll distance and roll time minutely shorter. A ball with an initial speed of 1 rev/sec can't roll for 1sec and cover 1 revolution on a Stimp of 10, nor can the same ball on the same green then roll 2 revolutions for 2 seconds with the same initial speed. It would take a Stimp of 82 for the ball to roll 5.28 inches with a Vo of 5.28 in/sec. The corresponding roll time would be 2 seconds."

The usual complex of interrelated equations is:

1. x = Vt

2. V = (Vo + V) / 2

3. V = Vo + at

4. x = Vot +_ 1/2 at^2

5. V^2 = Vo^2 = 2ax

(See Nave's HyperPhysics.) Equation 5 combines equations 1 and 2. I assume your "s" is the same as x, a position or distance measure. But what is your "a", the acceleration / deceleration? Equation 2 is the "average velocity" for the case of "constant acceleration / deceleration" as here. So you don't need to know "a" directly -- you just need to know the V at the front lip (Vo) as the other V in parenthesis is really Vfinal (i.e., zero).

The real question is "how long does the ball continue to roll after the onset of the decay phase?" That is a question asking that "t" (time, in seconds) be solved from other, known factors.

Equation 3 allows that, where V (the final velocity after a given time) is zero, Vo is 2 rps, 10.56 inches/sec. or some other units, and "a" is the rate of deceleration from Vo to V.

Admittedly, it is tautological to claim that t is 1 second (or 2 seconds) and then calculate "a", or to claim that "a" is what causes Vo to drop to V (2 rps to 0) in 1 second and hence a = -2 rev./sec.^2.

Although these guys ignore the "decay" phase as separate from the rolling phase, Werner and Grieg gives these formulas based upon empirical examinations of rolls on real greens (as summarized by me in an earlier post):

Even though there is a collection of formulae, a putt can be broken down into phases of 'flight" (Fd) at the very start, "skidding-rolling" (Sd) after that for a short part, and the "pure rolling" (Rd) the rest of the way. The total distance (D) a putt travels is then these three phases added together, D = Fd + Sd + Rd. There are separate formulae for each phase.

The Fd (in feet) = 2* Launch velocity (ft/sec) Squared * cosine of Elevation Angle of launch (in radians, not degrees) * sine of Elevation Angle of launch (in radians) / Accelerating force of gravity (in feet per second per second)

The Sd (in feet) = (Launch velocity * cosine Elevation Angle - [Acceleration of gravity*Coeeficient of friction of sliding over the grass on that green*Time of slide (secs)/2] * Time of slide (seconds)

The Time of sliding or skidding (in seconds) = (Velocity of launch * cosine Elevation Angle of launch - Angular velocity at launch (in radians per second of spin, negative for backspin) * Radius of the ball (i.e., 0.84 inches / 12 inches per foot)) / (Coefficient of sliding friction* (Acceleration of gravity + Ball weight (i.e., 1.62 ounces) + Ball radius Squared / Moment of Inertia of the ball (in slug-feet Sqaured))

The Coefficient of sliding fricition depends upon the Stimpmeter speed of the green. Coefficient = .946 -.021*Stimpmeter reading (in feet). There is no unit for this coefficient.

The Rd (in feet) = .0477*Stimpmeter reading (in feet)*Velocity of ball at point skidding stops and becomes pure rolling, raised to the 1.595th power.

The Velocity when the ball stops skidding (in feet per second) = Velocity at launch *cosine Elevation Angle at launch - Acceleration of gravity Coefficient of sliding friction Time of sliding. In all these matters, there is the "Coefficient of sliding fricition", which Werner and Grieg find by empirical investigation to relate to Stimp reading by the formula: Coefficient = .946 -.021*Stimpmeter reading (in feet). So a Stimp 10 green has a coefficient of "sliding friction" of 0.736.

This is all "in the ballpark", but what is really needed is a sense of the "rolling friction" at the contact between the ball and the surface, especially as the contact area increases in the decay phase. This physics website illustrates the concepts pretty well.

Car tires have a coefficient of rolling friction of 0.02 to 0.06 or thereabouts. Werner and Grieg use the term "sliding friction" (also called "kinetic friction") for the formula to determine when skidding stops and rolling starts. Rubber sliding on asphalt has a coefficient of between 0.5 and 0.8. Here is another table. Rolling friction is much less than sliding friction (see Engineer's Edge).

This experiment by the University of Florida Turfgrass shows how the coefficient is used. Croquet balls and lawns relate to rolling friction as shown in this study by Oxford Croquet. two researchers rolling golf balls across a carpet found the carpet's "deceleration force" to be -0.305 m/s^2. This is 3.1% of gravity (g, 9.8 m/s^2), which corresponds to a "rolling friction" "mu" factor of 0.031. According to Grant Palmer's Physics for Games Developers, the "rolling friction" of greens ranges from about 0.05 to 0.075. If a "fast" green has 0.05, formula

6. Vf = Vi - mu*gt

tells us that if the initial velocity is 2 rps, and the final velocity is 0 rps, and mu is 0.05, and g is 384 inches/sec^2, then time of roll is:

0 = 10.56 in./sec. - 0.05 * 384 in./sec.^2 * t

0 = 10.56 in./sec. - 19.2 in./sec.^2 * t

-10.56 in./sec. = -19.2 in./sec.^2 * t

t = -10.56 in./sec. / -19.2 in.sec.^2

t = 0.55 secs.

The "rollout" distance of roll after 2 rps is reached is from the formula:

x = Vi * t + 1/2 (a * t^2) + 1/2 (Vi^2 / mu * g), where a = deceleration from 2 rps to 0 rps in 0.55 sec., or - 19.2 in./sec.^2 deceleration.

x = 10.56 in./sec. * 0.55 sec. + 1/2 (-19.2 in./sec.^2 * 0.55 sec.^2) + 1/2 (10.56 in./sec.^2 / 0.05 * 384 in./sec.^2)

x = 5.81 in. + 1/2 (-10.56 in.) + 1/2 (5.808 in.)

x = 5.81 in. - 5.28 in. + 2.904 in.

x = 3.4 inches rollout.

For a slower green with mu = 0.075, the time of roll after the ball slows to 2 rps is:

0 = 10.56 in./sec. - 0.075 * 384 in./sec.^2 * t

0 = 10.56 in./sec. - 28.8 in./sec.^2 * t

-10.56 in./sec. = - 28.8 in./sec.^2 * t

t = -10.56 in./sec. / - 28.8 in./sec.^2

t = 0.37 sec.

The "rollout" distance of roll after 2 rps is reached is from the formula:

x = Vi * t + 1/2 (a * t^2) + 1/2 (Vi^2 / mu * g), where a = deceleration from 2 rps to 0 rps in 0.37 sec., or - 28.5 in./sec.^2 deceleration.

x = 10.56 in./sec. * 0.37 sec. + 1/2 (-28.5 in./sec.^2 * 0.37 sec.^2) + 1/2 (10.56 in./sec.^2 / 0.075 * 384 in./sec.^2)

x = 3.91 in. + 1/2 (-3.91 in.) + 1/2 (0.37 in.)

x = 3.91 in. -1.955 in. + 0.185 in.

x = 2.14 inches rollout.

This doesn't seem too out of whack from my experience. (Lord only knows what the real numbers are, though.) I frequently find 40-foot putts that miss stopping within the diameter of the hole front-to-back, so they stop before a 4.25" rollout. Both of the calculations above stop the rollout before passing the rear edge of the hole. I see that a lot -- balls nearly always get as far as the front lip and don't roll past the back lip before stopping.

So, your figures appear more accurate than mine! I'll have to adjust my senses for what is good and bad. I have long known that instinctive is right all the way at least to the front lip and not far past, and now I know that optimal is stopping beside the hole without getting past the back lip. Seems right to me.

As to "nearly" 4 million putts, it is about 650 putts per day for 17 years (6,205 days), an average of about 650 putts per day. I normally putt with two balls, each stroke taking about 10 seconds, and each walking after and resetting for more stroke is 20 seconds tops, so the cycle is 2 putts per 30 seconds, or 1 putt each 15 seconds, and thus 4 putts per minute. So that is an average of 160 minutes a day, 2 hours and 40 minuntes. That's pretty close.

Cheers!

Geoff Mangum
Putting Coach and Theorist
PuttingZone.com
Golf's most advanced and comprehensive putting instruction.

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Geoff,
I used mu=.056 for a Stimp=10 and got .49 sec while you got .55sec using a slightly different mu (.05) so we're very close on time.
The distance can be calculated by conservation of energy or graphically using v vs t
shown below:

Vi
|\
|*\
|**\
|***\ slope = ug = acceleration
|****\
|*****\ area = s = distance
|******\
|*******\
*********V=0
---------->t

rise/run = slope, run = rise/slope, run = t = Vi/ug
area = s = 1/2bh(triangle) = 1/2(Vi)(Vi/ug) = Vi^2/2ug where u = .056 for Stimp=10
Vi = .88 ft/sec
distance = (.88^2)/(2)(.056)(32) = .216ft = 2.6 inches

You got 3.4 inches using x = Vi * t + 1/2 (a * t^2) + 1/2 (Vi^2 / mu * g)
I'm not sure about the extra Vi*t and 1/2(at^2) terms in your equation that lead to a distance of 3.4 inches,
Conservation of energy method doesn't require t to get the distance:
Kinetic energy before = Thermal energy after
(1/2)mv^2 = Fd
F = umg
(1/2)mv^2 = umgd, cancel mass,m
v^2/2 = ugd
v^2/(2ug) = d
d = (.88^2)/(2)(.056)(32) = .216ft = 2.6 inches

Anyway, it looks like we're close.
Dean