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You may recall hearing Dave Pelz tell you that the optimal distance a putt has to roll past the hole is 17 inches. How big might you guess the hole is at 17 inches? 4 inches wide? 3 ½? What if I told you that the hole, at 17 inches, was only about 2¼ inches wide? What? Am I crazy? No - it's just math. Consider a well cut hole, and what's required for the putt to go in. For a putt to be holed, it has to have enough time for the ball - 1.68 inches in diameter - to fall half that distance, or 0.84 inches, or more. Gravity is a constant force (9.8 m/s 2 ) and 0.84 inches is about 0.021336 meters. Given that distance (d) = 0.5at 2 we can solve for t time: 0.021336m = 0.5 * 9.8 m/s 2 * t 2 . 0.021336m = 4.9 m/s 2 * t 2 . 0.021336m / 4.9m/s 2 = t 2 . 0.004354s 2 = t 2 . t = 0.065987 seconds. In other words, a golf ball needs approximately 0.066 seconds to fall 0.84 inches, striking some part of the back of the cup at the equator for the ball to be holed. As you all know, d=rt. Distance equals rate times time. For example, in two hours going 60 MPH you'll travel 120 miles. 120 miles = 2 hours * 60 miles/hour. Simple stuff. I don't know what the "rate" of a ball rolling 17" past the hole is, but smart people do and they've plugged that in to the equation. t is always going to be 0.066 seconds, and thus they can solve for d. That "d" is effectively the amount of air a ball needs to have under it's path in order to fall into the cup by dropping 0.84 inches (or more). A putt that rolls over the very very outside edge of the hole might only have a quarter of an inch of "air during which it can fall. If the putt is barely rolling at any speed, that'll take 0.66 seconds or longer and the putt will drop. If the ball is rolling really fast, it'll cross that 1/4 inch in no time at all, and not fall in. Here are the numbers, in graphical form: In other words, a putt that rolls six inches past the hole (0.5 feet, or 0.5') is effectively 3.8 inches wide. We lose 0.45 inches off the sides of the cup (0.225 off each side, the left and the right) and are effectively putting to a hole that's 3.8 inches wide. If this graphic shows you anything, it's that the cup gets really small, really quickly. The hole is less than an inch wide for a putt that rolls four feet past the hole. Heck, even a putt that rolls only three feet beyond the hole - leaving what would amount to a tap-in for most people - is only 1.4 inches wide. Here's that same graphic in a form that makes the distance a bit more obvious. Notice how at "A" the distance the ball needs to travel is much shorter than the distance needed in "B". Putt a ball about eight feet past the hole and there's almost no chance of it going in the hole - the target is virtually 0 inches wide. These are the putts that hit the back of the hole, fall 0.7 inches instead of 0.84, and pop up and sit behind the hole. (These numbers are for a flat green at about 8 on the stimp. On faster greens, these holes are a little larger because a ball can be rolling more slowly and still roll six inches past the hole, or two feet past the hole. On downhill putts, the hole can be a little larger too for the same reason - the ball is rolling more slowly - but this is often offset by the fact that the far edge of the cup is a little lower, so the ball has to fall more than 0.84 inches. On an uphill putt, a similar thing occurs - the ball will be traveling faster to roll out three feet past the hole (or whatever distance), but the back of the cup is a little higher so the ball only has to fall perhaps 0.8 inches.) So, what's the point of all of this? If you like putting to large targets, strive to hit your ball about six to twelve inches past the cup. 17 inches, two feet... three feet... they're all too firm. The ball will be rolling too fast at the hole and it will be rolling too fast to have a very large target - 2.25 inches, 1.9 inches, and 1.4 inches respectively. Okay, so why not strive for "dead at the hole" capture speed? After all, the hole becomes 4.25 inches wide then, right? Well, three reasons. If you're off by an inch in terms of your speed, or two inches, you're virtually 100% certain not to make the putt. Never up, never in. The lumpy donut. It's not a huge effect, but it can play a small role when the ball is moving that slowly. Wobble. As a ball comes to a stop, the last six inches of its roll are really, really affected by small imperfections in the green. The ball will sharply break left and right on these small things (spike marks, a tuft of grass, a heel print) and nobody wants to see their putts swerve half an inch right because of a tiny little bump and miss the hole. 6-12 inches gives you a pretty darn big hole to putt into and it avoids the "wobble" problem. You've probably heard about golfers hitting the ball "firmly" and "taking out the break." Well, guess what? You know all those putts that lip out? Capture speed problems. The hole was too small, the ball didn't have enough time to drop, and it hit the far side of the hole with most of the ball still above the edge of the hole, veered hard sideways off the back of the cup, and climbed out. Give yourselves the biggest hole to putt to: strive to roll the ball six to twelve inches past the hole. You might leave one short now and then, but I guarantee you'll make more than enough additional putts to offset the occasional ones left short. Update 2017-11-09: Here are the numbers for other stimp speeds: Effective Capture Width at Various Roll Speeds and Stimps Speed Stimp 8 Stimp 10 Stimp 12 Die 4.2" 4.2" 4.2" 6" 3.8" 3.9" 3.9" 1' 2.6" 2.7" 2.8" 2' 1.9" 2.1" 2.3" 3' 1.4" 1.7" 1.9" 4' 0.9" 1.3" 1.5" 5' 0.5" 0.9" 1.2"
I found two amazing PDFs that I would like to share with you. https://arxiv.org/pdf/1502.03501.pdf In this paper, every number from 1 to 1000 is written out using each of the nine digits available and only the five basic functions (+, -, x, /, and powers), plus parentheses for grouping. For example: To get the number 422 using only the number 7: 422 = 7 × 7 × 7 + 77 + (7 + 7)/7. To get the number 814 using only the number 2: 814 = 22 × (2 + 2 + 2)2 +22. https://arxiv.org/pdf/1302.1479.pdf In this paper, every number from 1 to 11,111 (except 10,958 in ascending order) is created using each digit from 1-9 in both ascending and descending order. For example: 2785 in ascending order = 1 + 23 × 4 × 5 × 6 + 7 + 8 + 9. 5296 in descending order = (98 × 7 + 654 × 3) × 2 × 1. Pretty nifty, I thought.