The only thing constant about golf is its inconstancy.
—Jack Nicklaus
Many golfers, including Sam Snead and Lee Trevino, considered Moe Norman to be the best striker of the ball in golf history.1 Tiger Woods has said that only two players have ever truly “owned their swing”—Moe Norman and Ben Hogan.2 Yet, many people have never heard of Norman, whose story is fascinating. Raised in Ontario in the 1930s, he latched onto golf as an escape from social and family difficulties. He taught himself a unique, highly repeatable swing that utilized heavy clubs, a very wide stance, a grip anchored in the palm of his right hand, and a single plane for backswing and downswing. It is closely related to the Natural Golf method.3
In spite of amassing staggering statistics—like 33 course records, 17 holes-in-one, and three rounds of 59—Moe Norman never made it on the PGA Tour. He had a number of, to put it nicely, eccentricities. Opponents carefully reaching out to mark their balls on the greens might find Moe’s putt scurrying underneath their arms. Moe liked to play fast. A playing partner bemoaning his drive of a brand new ball into a creek was startled to find Moe knee-deep in the water trying to retrieve the ball. Moe had finely tuned ball-hawking skills from his days as a caddie, appreciated the value of a new ball, and was a very loyal friend; however, he dressed poorly and was ill-at-ease socially. While some reporters loved having a character on tour, others thought he was not enough of a gentleman. He was made to feel unwelcome on the Tour and returned to Canada.4
If we say that Moe Norman was the best ball striker, what exactly do we mean? To a mathematician, the word “best” is not meaningful without a precise definition. Are there objective criteria by which we can say that Moe was better than Ben Hogan* or Jack Nicklaus? While distance and height, and the ability to hit controlled draws and fades, are part of ball striking, the primary criterion for ball striking is consistency.
Don Wade tells the story of Moe Norman arriving at the site of a tournament and finding the course blanketed in fog. Instead of feeling his way to the driving range, Moe grabbed a handful of balls and drove them down the first fairway. The caddie who went to retrieve the balls was astonished to find two balls touching each other and all six balls within a 10-foot circle.5
This story gives us a place to start. Imagine two golfers hitting six balls each at the driving range. To keep things simple, let’s assume that all of the balls end up in a line, as illustrated in figure 2.1. Obviously, golfer A is more consistent than golfer B, since B’s shots are more spread out, but how much more consistent is golfer A? A basic measure used by statisticians to quantify spread is standard deviation.
The formula for the standard deviation σ of a data set is often written in the form
which is a little like legal fine print: you know that it’s good, but it doesn’t seem to be in a recognizable language. Taken in the right
Figure 2.1 Measuring consistency: (a) golfer A, more consistent; (b) golfer B, less consistent
order, the standard deviation calculation is quite simple. First, find the mean (average) of the data; this is denoted 
. Golfer A has distances 143, 144, 145, 146, 148, and 150, and the mean is
Golfer B has distances 141, 142, 144, 147, 149, and 153, and the mean is
The next step is to subtract the mean (146) from each data point.
A: 143, 144, 145, 146, 148, 150 → −3, −2, −1, 0, 2, 4
B: 141, 142, 144, 147, 149, 153 → −5, −4, −2, 1, 3, 7
Square these numbers and add them up.
A: 9 + 4 + 1 + 0 + 4 + 16 = 34
B: 25 + 16 + 4 + 1 + 9 + 49 = 104
Divide by n − 1, which is one less than the number of data points (in this case, divide by 6 − 1 = 5), and then take the square root.
The standard deviations are approximately 2.6 for golfer A and 4.6 for golfer B. This gives us a measure of how much more consistent golfer A is than golfer B, as the spread in golfer B’s shots is nearly twice the spread in golfer A’s shots.
Notice that, for each golfer, four of the six data points are within one standard deviation of the mean. That is, for golfer A there are four balls in the distance range 146 ± 2.6, between 146−2.6=143.4 and 146 + 2.6 = 148.6. For golfer B, four of the six distances are in the range 146 ± 4.6. This illustrates a guideline that can be made precise: for samples from a normal (bell curve) distribution, about 68% (roughly two-thirds) of the data should be within one standard deviation of the mean; about 95% of the data should be within two standard deviations of the mean; and 99.7% should be within three standard deviations of the mean.
Standard deviation is a quantity that will resurface in this book several times. In golf and numerous other situations, it is a simple and informative measure of the amount of variability in a set of data.
In figure 2.1, we ignored the left/right movement of the ball. We will now factor in this second dimension of motion. The question is what the result would be if a golfer dropped several balls and hit them all with the same club at the same target. If the golfer is Moe Norman, you might imagine a neat pyramid of balls stacked on top of the target. By contrast, a large park would be required to hold my shots. What would the scatter of balls look like?
All mathematical investigations begin with assumptions. In what follows, I assume that every swing is the same except for random variations in swing speed, swing angle, and face angle. These variations will be assumed to be independent—that is, the size of the variation in swing speed has no relationship to the variation in swing angle or face angle. They are not really independent—a particular twitch in the downswing could simultaneously pull the swing plane to the left and open up the club face—but the calculations are easier with this assumption. The remaining question is how much variation to allow. The results of 5° misalignments are shown in figures 1.7 and 1.9 of the previous chapter. Is 5° too large or too small?
Researchers Werner and Greig estimate a standard deviation in swing angle for scratch golfers of 2.314°.6 Thus, an error as large as 5° is greater than a two-standard-deviation error and would occur less often than 5% of the time. For non–scratch golfers, the standard deviation increases by 0.0673 times the handicap. For example, a 20-handicapper would have a standard deviation estimated at 2.314 + 20 * 0.0673 = 3.66°. For this golfer, a 5° error is between one and two standard deviations and would be likely to occur every few holes.7
The standard deviation in clubface angle for scratch golfers is estimated at 1.42°, with an adjustment of 0.0412 times handicap for other golfers. For example, a 10-handicapper would have a standard deviation of approximately 1.42 + 0.412 = 1.832° for clubface angle. The standard deviation in swing speed is estimated as 2.984% plus 0.0868 times handicap.
The computer experiment simulates shots by a scratch golfer. The starting point is a basic swing with clubhead speed equal to 161 ft/s (about 110 mph, the Tour average) and both angles set to 0. The landing point is found and then an estimate is made of the stopping point after roll.8 Two more speeds are generated: 161 + 9.608 ft/s and 161 − 9.608 ft/s. The change of 9.608 corresponds to two standard deviations, where one standard deviation is 2.984% of 161, or 161 * .02984 = 4.804 ft/s. Similarly, swing angles of ±4.628° and face angles of ±2.84° are generated. There are 27 combinations of one of the three swing speeds, one of the swing angles, and one of the face angles. Stopping points for each of the 27 sets of initial conditions are estimated, and they are shown in figure 2.2.
The dots in figure 2.2 represent the computed stopping points. A solid polygon has been drawn around the points to help visualize the region where all good shots would be. In this case, a “good” shot means that each of the three variables is within two standard deviations of its base value. Assuming that the variables are independent, about 86% of the shots would be inside of this polygon. (This is obtained by using the two-standard-deviation rule on each of the three variables: 0.953 ≈ 0.86.) The shaded region represents a 60-yard-wide fairway and is included for reference.
Figure 2.2 Stopping points for shots from 27 swings of different speed, face angle, and swing plane
Loosely speaking, the points in figure 2.2 form three arcs corresponding to the different swing speeds. A 20-handicap golfer would show a similar pattern but with more scatter and, therefore, fewer balls in the fairway.
Figures 2.3a and 2.3b show the corresponding scatters with the basic shot being a fade9 and a draw, respectively. There are fewer balls in the fairway in these cases compared to the “straight” drives in figure 2.2. This result depends strongly on the orientation of the fairway. A straight fairway calls for a straight drive.
Suppose that the fairway curves to the right or left. You can fit more of the dots in figure 2.3b (from a draw) into a fairway that curves from right to left than a fairway that curves left to right. This illustrates the standard wisdom of shaping the shot to the shape of the fairway. However, the difference in the number of balls in the fairway is not large. For hitting fairways, the most important factor by far is the size of the spread, which is determined by the consistency of the ball striker.
Figure 2.3 Stopping points for 27 shots with (a) a basic fade swing and (b) a basic draw swing
The shot pattern for irons will have the same general shape as that seen in figures 2.2 and 2.3. Instead of staying within a strip of fairway, of course, the goal is to be on the green. Clearly, the shape of the green will have a large effect on how many balls stay on the green.
We next examine what happens once you’re on the green and ready to putt.
When golfers think about putts, there are usually two properties that come to mind: speed and line.10 That is, we have control over how hard we hit a putt and in which direction we hit it. If we start the ball on its way with the correct initial speed and direction, the putt will go in ... unless the golfing gods are angry with us (the topic of the next chapter). Mathematically, we describe putts in terms of polar coordinates.
The mathematical position of an object in two dimensions is usually described in relation to a fixed location called the origin. In rectangular (two-dimensional) coordinates, we determine the object’s horizontal distance x and vertical distance y from the origin and call the location (x, y). In polar coordinates, the object is defined by its distance r from the origin and the angle θ from the positive x-axis, as shown in figure 2.4. Mathematically, the x and y version of a point is so familiar that polar coordinates can seem odd. For putting, though, polar coordinates are natural, since they identify distance and direction.
For this reason, Werner and Greig use polar coordinates to describe some important results from their research.11 To gather data
on putting accuracy, they recruited numerous golfers of varying handicaps. Each golfer was given a small target (not a hole) at which to aim, on a level green from varying distances. The actual stopping points of the ball were recorded for a large number of putts. Averaging over numerous golfers, Werner and Greig report that the standard deviation for distance is
σr = (0.0536 + 0.0017 * HCP) * D,
where D is the target distance and HCP is the player’s handicap.
To put some numbers on this, start with a 15-foot putt by a 10-handicapper. Then D = 15 feet, HCP = 10, and you can compute σr = (0.0536 + .017) * 15 ft = 1.059 ft. This says that about two-thirds of the putts will roll a distance in the interval 15 ± 1.059 feet and, therefore, end up in the range from 14 feet to 16 feet. About 95% of the putts will be within two standard deviations of the mean, roughly between 13 feet and 17 feet. Remember that this calculation applies to flat putts that the golfer is trying to hit exactly 15 feet. The final 5% of putts are either more than 2 feet short of or more than 2 feet beyond the target. This is part of what separates a 10-handicapper from a scratch golfer.
The standard deviation for angles is estimated as
σθ = (1.718 + 0.0039 * HCP) deg.
Applying this formula to a 10-handicapper gives σθ = 1.757°. About one-third of the putts will have an angle that is at least 1.757° off. Over 15 feet, this translates to at least 5.5 inches off-line, more than enough to miss the putt.
Combining this information gives a picture of where the putts finish. Figure 2.5 shows the stopping region for a putt of 15 feet by a 10-handicapper. A circle is drawn at the 15-foot mark to add perspective on the size of the hole. The shaded area covers ±1 standard deviation for distance and ±1 standard deviation for direction. Assuming that the errors are independent (unfortunately they are not), that means only about 46% (0.4624 = 0.682) of the putts will be inside the shaded area, and over half of the putts will be outside of the shaded region.
Figure 2.5 Stopping points for 15-foot putts by a 10-handicapper, assuming that distance and angle are within 1 standard deviation of the target. The ball moves left to right, and the hole is shown as a circle.
This raises an interesting question: which of the stopping points in figure 2.5 corresponds to putts that would go in? Certainly every stopping point inside the hole would count. However, what about a point just to the right of the circle? The figure corresponds to a putt that travels from left to right. A putt that is at the center of the hole but hit an inch too far will go in. So, many of the points that are outside of the circle represent putts that would go in. The “make zone” examined next provides more details.
The most important question about a putt is whether the ball goes in the hole or not. As all golfers know, having the path of the putt intersect a portion of the hole does not guarantee that the ball goes in. If the ball has too much speed, it will roll right over the hole, or “lip out.” On the other hand, the putt does not have to be perfect. Some putts pound the back of the hole and drop in, and there is the “all-around good” putt that does a lap or two around the hole before dropping. The make zone is a way of illustrating which putts go in and which do not.
One make zone is based on where the putt would stop if there was no hole. Werner and Greig have developed formulas for the make zone for different speed greens, based on extensive data collection.12 Hoadley found a similar shape.13 In both cases, the technique is to roll balls on flat putts from a fixed distance with a variety of initial speeds and lines. For the putts that go in, the same speeds and lines are repeated on a flat putt with no hole present, and the stopping point is determined. The make zone is an idealized region showing where made putts would have stopped if the hole had not been in the way.
Figure 2.6 shows the make zone for a flat green with the putt moving from left to right and the hole centered at the crossing of the axes. If you hit a putt at the center of the hole, it will drop even if it has enough speed to roll 52 inches (a little over 4 feet) past the hole. However, a putt that is just inside the edge could have only enough speed to roll 10 inches or fewer past the hole.
The portion of the make zone to the left of the vertical axis in figure 2.6 is a semicircle, representing the front half of the hole. Its radius is larger than the radius of the hole. To see why, you need to know that in this analysis the “location” of the ball is actually the location of the center of the ball. So, what should happen if the ball stops just outside the edge of the hole? The question refers to the center of the ball being just off the edge, which means that 45% or so of the ball is hanging over the hole. I would certainly want that putt to drop. And, in fact, it usually does. Werner and Greig estimate the radius of the make zone semicircle to be 2.125 + 1.944/S inches, where S is the Stimp number14 of the green in feet. The actual radius of the hole is 2.125 inches, so the expression 1.944/S gives the leeway for a ball dropping in. For a fast green with Stimp 10, you get only an extra 0.2 inches. For a slower, hairier green of Stimp 5, the leeway is close to 0.4 inches, which is about half the radius of the ball. For a putt dying at the hole, this means that slower greens are more forgiving.
Figure 2.6 Make zone for a flat putt rolling from left to right
A different way of constructing a make zone is to show the combinations of initial speed and direction that produce a made putt—that is, the variables that the golfer directly controls. However, this version of the make zone is highly dependent on the length of the putt, whereas figure 2.6 is valid for any length of putt. The make zone in figure 2.7 is for a flat 15-foot putt.
The shaded region in figure 2.7 indicates the values of the initial speed (in ft/s) and initial angle (where 0 is directly at the hole) which will produce a made 15-footer, based on the make zone in figure 2.6. Notice that there is less than a 1° margin of error on either side of center. Reading off of the vertical axis, you can see that even for a putt hit in the center of the hole, the margin of error for speed is less than 1.5 ft/s. It takes a precise stroke to make a 15-footer!
Figure 2.7 Combinations of speed and initial direction to make a flat 15-foot putt
Figure 2.8 Combinations of speed and initial direction to make a downhill, right-to-left, 15-foot putt
The symmetry in figure 2.7 depends on the putt having no break. Figure 2.8 shows the analogous make zone for a putt with break. The ball is still 15 feet from the hole, and the speed of the green has not been modified. The green is assumed to be a flat plane tilted at an angle of 2°, such that the putt is downhill and breaks right-to-left.
To read this graph, you can start with an angle. For example, 3.5° means that the putt is aimed 3.5° to the right. This converts to allowing for 11 inches of break.15 Finding 3.5 on the horizontal axis, you see that the shaded region extends from just below 9.6 (the actual value is 9.57) to about 10.2 (more precisely, 10.19). The putt will be made if the initial speed is between 9.57 ft/s and 10.19 ft/s. An initial speed of 9.57 ft/s barely gets the ball to the hole, whereas an initial speed of 10.19 sends the ball to the back of the hole, almost producing a lip-out. The graph can give an answer to the question of what the best line is. There are numerous ways to answer this question, depending on how you define “best.” One goal might be to have the maximum margin of error in speed. At angle 3.2°, speeds between 9.67 ft/s and 10.35 ft/s are all good. The difference of 0.68 ft/s is the margin of error, which is the largest such margin of error possible. By contrast, notice that the angle of 3.5° only produces a margin of error of 10.19 − 9.57 = 0.62 ft/s.
Unfortunately, it is not helpful to tell a golfer to use an angle of 3.2° and any initial speed between 9.67 ft/s and 10.35 ft/s. Protractors are not standard golf equipment, even in the well-accessorized golf bag. However, this result can be translated into golfing terms. At the angle of 3.2°, all made putts enter the hole with a fair amount of speed. The extreme speeds of 9.67 ft/s and 10.35 ft/s are right at the values for lip-outs on the low and high sides of the hole, respectively. The conclusion is this: if you want the largest range of initial speeds possible, use a relatively small angle and hit the putt firmly. The downside is that the high speed will take the putt well past the hole if you miss. If you do not want the agony of a long second putt, then you should choose a larger angle and slower speed. Figure 2.8 shows that this safe play gives a slightly smaller margin of error. We will revisit this trade-off between aggression and safety in the next chapter.
Should you aim higher or lower on a breaking putt? I have never felt terribly comforted by the comment, “At least you missed on the high side. That’s where pros miss.” However, there are a couple of factors that point to “aim higher” being good advice. First, most golfers underestimate the break.16 (This is discussed further below.) Plus, if you lag the putt to finish near the hole, you will need to play much more break. In figure 2.8, the angles range from 2.5° to 4.9°. These angles correspond to breaks of about 8 inches and over 15 inches, respectively. The amount of break you play can change drastically with the firmness of the putt.
The ability to read the amount of break on a putt involves an interesting mental illusion. Golfers talk about having a “feel” for the green, and this vague term is appropriate. It turns out to be difficult to quantify the ability to read greens. For example, if you say that a putt will break 5 inches, what exactly does that mean?
Suppose that the bottom of the box in figure 2.9 represents a distance of 5 inches below the center of the hole. Eyeballing this graph, would you say that this putt breaks more or less than 5 inches? What most golfers mean when they say a putt breaks 5 inches is that the ball should initially move directly at a point even with the hole and 5 inches to the side. By this measure, the putt in figure 2.9 breaks nearly 8 inches. Surprised? The graph and calculation in the notes may help.17
Short-game expert Dave Pelz has found that most people severely underestimate the break of a putt. However, this may be partially due to the optical illusion demonstrated in figure 2.9. Translating Pelz’s data to a putt that breaks 20 inches, on the average golfers report that they think the putt will break a mere 5 inches but line their putter head up for a 15-inch break and then “mishit” the putt on the line for an 18-inch break.18 Some of these putts go in, so in this sense the read is not far off. Much of the error is in the mental measurement of where the ball is actually being aimed. The unusual, and apparently intentional, mishit of the putt may be part of why sharp-breaking putts are so difficult for amateurs to make.
For the average golfer, then, reading a green is more of an intuitive visualization than a precise calculation. This is why getting the feel of a green and visualizing the path of a putt going in are so important.
Figure 2.9 Overhead view of a sidehill putt: how much did it break?
The graphs in this chapter have shown some of the effects of variation in golf. But where does the spread come from, and how can you control it?
For drives and iron shots, most of the variation comes from our swings. We have seen how misaligned swing planes and face angles produce pulls, hooks, pushes, and slices. Also, variations in swing speed affect ball speed and spin. Similarly, there are three main sources of error in a putting stroke. The swing plane can be misaligned, the face angle can be misaligned, and the ball can be hit off-center. Dave Pelz has carefully researched the effects of these errors.19
Only about 20% of the swing plane error is transmitted to the ball. That is, if the swing plane aligns 10 inches to the left of target, the ball will go 2 inches left of the target. This is good news for those of us with outside-in putting strokes and explains why Billy Mayfair could win tournaments with a stroke that followed no known path.
On the other hand, a full 90% of face alignment errors are transmitted to the ball. If the putter face points 10 inches left of the target, the ball will go 9 inches left of target. If you have ever helped someone line up a putt, you probably focused on face alignment, which is the single most important factor in lining up correctly.
If you think that hitting putts in the center of the club should not be difficult, you are not alone. However, Pelz finds that this is a common and devastating error to make. When a putt is hit off center, the putter head twists. The twisting affects the initial direction and speed of the putt so that the actual putt bears little resemblance to the intended putt.
The primary sources of our golfer’s spread are our own swing errors. Our swings can be improved with practice and good instruction. Along with practicing more, we can reduce much of the spread with better clubs. The principal ingredient in this technological diet pill is MOI, or moment of inertia. Without getting into technicalities, the moment of inertia for a particular object and a particular rotation axis is a measure of the object’s resistance to rotation about that axis. The larger the MOI, the less likely the object is to rotate.
Golf club manufacturers can now make clubs that, while light, have large MOI’s for common rotations. The gigantic 460-cc driver clubheads, perimeter weighting on irons, and branding-iron shapes of putters are all intended to increase MOI. With less twisting of the club on off-center hits, we get more consistency on our shots.
The graphs in this chapter illustrate the importance of consistency for all golfers. The more you can reduce the spread of your shots, the better you will score. How important is consistency at the top level of the game? The following study, conducted by Richard Goeres,20 examines one aspect of consistency in the professional game. In this discussion, we will be looking at consistency in scoring for a round, as opposed to consistency on a particular shot.
From 1998 to 2005, Tiger Woods made the cut in every tournament he played. His streak of 142 consecutive cuts made is the official PGA record.21 Goeres wanted to get a handle on how impressive this record is. In the study, a field of 110 golfers was generated randomly by a computer. Each golfer was assigned a “talent level” chosen from a normal distribution with mean 70.8 (the Tour average for the 1998 through 2005 seasons) and standard deviation 0.7. Scores for two rounds for each player were then simulated, from a normal distribution with mean equal to the player’s talent level and standard deviation of 2.5 (the standard deviation in scores for a typical round in a PGA tournament). The cut was set at the top 70 plus ties.
Tiger’s mean was first set to 68.5 (his scoring average from 1999 through 2003), meaning that he would be more than 2 strokes per round better than average and a full stroke better than 97% of the golfers he might face. With the same standard deviation of 2.5 strokes for a round as the rest of the field, the simulated Tiger missed the cut 3.84% of the time.
With this percentage, the likelihood of making 142 straight cuts computes to about 1 in 260. To be precise, the probability of Tiger making a cut is assumed to be 0.9616. Assuming that each tournament is an independent event, the probability of making 142 out of 142 cuts is 0.9616142, or approximately 0.004. The fraction evaluates to approximately 0.004. The fraction 
evaluates to approximately 0.004.
To get a handle on this, some graphs may help. In figure 2.10a, the curve illustrates the distribution of talent levels for the simulation.22 Tiger’s talent level of 68.5 is marked for reference. Figure 2.10b shows the distribution of Tiger’s simulated scores, along with the distribution of simulated scores for an average golfer in the field. Since making the cut in this study only required being in the top 70 of 110 (many fields are larger), the advantage given to the simulated Tiger would seem to guarantee a made cut. However, it turns out that there is enough overlap in the distributions to produce missed cuts nearly 4% of the time.
In a different simulation, Tiger’s mean for the week was set equal to the best of the randomly generated means for the field of 110. His standard deviation was lowered by a full stroke to 1.5. Examining figure 2.10a, we can see that, if one of Tiger’s competitors drew a talent level in the left tail of the distribution (in particular, more than 3 standard deviations to the left), Tiger’s talent level could be the same 68.5 as in the first simulation. In many cases, Tiger’s talent level was higher in these tournaments than in the first simulation. Being rated as the best player in the field and being far more consistent than anybody in the field, Tiger would see his likelihood of making 142 straight cuts increased to about 1 in 7—still not very likely, but not all that shocking.
Figure 2.10 Consistency in scoring for a sound: (a) talent levels (average scores) of simulated field; (b) score distributions for simulation
Notice that when Tiger’s mean was raised (slightly, on the average) and his standard deviation was lowered by a stroke per round, his odds of making 142 straight cuts went up dramatically. The consecutive cut streak is not likely if Tiger is just better than everybody else. Along with being the best player in the field every week, he must play closer to his average than everybody else to keep making cuts. The graphs illustrate this result. Figure 2.11 shows the distribution of Tiger’s simulated scores with an average of 68.5 and a lower standard deviation, along with the distribution of simulated scores for an average golfer in the field. Comparing figures 2.11 and 2.10b, you may not immediately conclude that the streak is more likely in the situation of figure 2.11. The key is to realize that only a bad round by Tiger would put the streak in jeopardy. In figure 2.10b, Tiger’s bad rounds extend out to 75 and beyond. In figure 2.11, a score higher than 72 is very unlikely. Therefore, with the lower standard deviation there was little chance that Tiger would have two rounds bad enough to put him in danger of missing the cut.23
Figure 2.11 Score distributions for second simulation
The simulations point to Tiger’s consecutive cut streak being very special. There are other ways of reaching the same conclusion. As of October 2010, the longest active streak of made cuts was Steve Stricker’s 23 straight cuts. He was the only player who had made cuts in more than 20 consecutive tournaments and one of only three who had made 15 consecutive cuts. The record of 142 looks very, very good.
