Dear Rachel,
Physics applies to golf in a number of ways, naturally. The main way is in how the club and the ball interact in a swing and hit. Then there is the physics of how the ball moves through the air, interacts with the ground (fairway, rough, hazard, or green), and interacts with the hole. You can also think about the physics of the human body in a golf swing.
I. Club-ball interactions
The main book for this is C.B. Daish, The Physics of Ball Games. He also wrote Learn Science through Ball Games (NY: Sterling Press, 1972), also covering the golf swing and flight of the ball through air. Here is some info about
the second book.
The main idea is that the energy (momentum) of the club is transferred to the ball by the clubhead-ball collision. The energy of the clubhead is the mass of the clubhead times the velocity (speed + direction). The heavier the head, the more the energy for a given clubhead velocity. But there is a point where the head gets too heavy to move, and the velocity starts to drop off. Most DRIVER clubheads are near 7 ounces (about 28 grams per ounce, so about 200 grams), and most drivers in total (clubhead, shaft, grip) are around 11-12 ounces. A typical adult male clubhead velocity is about 100 mph at the bottom of the swing. Tiger Woods gets the clubhead up to 125-140 mph. Obviously, the faster the swing at impact, the more energy that transfers. The golfer has to balance clubhead mass against his ability to generate clubhead speed for the best mix for energy.
The typical golf ball itself weighs 1.62 ounces. Its velocity is zero to start with, so its momentum is zero too (mass times velocity).
The physicist looks at the golf swing at two precise moments in time to figure out how the energy/momentum transfers from club to ball -- just before impact, and just after impact.
Just before: The clubhead has all the energy (mass x velocity)
Just after: The clubhead has some and the ball has some.
The BIG IDEA from theory to help figure this out is that the TOTAL energy in the ball and clubhead just before is the SAME as the TOTAL energy in clubhead and ball just after. Nothing funny happens to the energy, and it is all accounted for by physics. This is called the Law of Conservation of Energy, but it just means nothing funny happens to the energy.
To use this idea, set up formula for before and after for the Total Energy, and then put the formulas together, this way:
Just before: Clubhead Energy (mass x velocity) + Ball Energy (mass x velocity, which is zero) = Total Energy
Just after: Clubhead Energy (mass x velocity) + Ball Energy (mass x velocity) = Total Energy
SO THAT:
Clubhead Energy Before + Ball Energy Before (0) = Clubhead Energy After + Ball Energy After
The one you need to know is Ball Energy After. To fugure that out, you have to know the mass of the clubhead and its velocity before impact and after impact. Typically, the mass is 200 grams and the velocity before is 100 mph before and after, the clubhead still has some velocity but a lot less (maybe 60 mph).
UNITS: You have to use the same system of units for physics formulas, and unfortunately the English system is not very convenient (inches, feet, pounds) and the metric System is a lot easier (meters, grams). So you have to convert mph to meters/second. We know that 1 meter has 39.37 inches, and that 1 hour has 60 minutes with 60 seconds each or 60*60 seconds (3,600 seconds in 1 hour). First convert mph to inches per second. 1 mph is 5,280 feet per hour, and in terms of inches is 5,280*12, or 63,360 inches per hour. Covert this to inches per second: 63,360/3,600 = 17.6 inches per second. So, 1 mph is the same as 17.6 inches per second. That means that 100 mph is the same as 17.6*100 or 1760 inches per second. How many meters are covered vy 1760 inches? 1760/39.37 = 44.70 meters per second. So 100 mph is the same as 44.70 meters per second. (A car traveling this fast covers 44.70 meters of roadway every single second. How many feet is that? 1760 inches / 12 inches per foot = 146.7 feet every second!)
How about 60 mph? 17.6 inches/sec/1 mph * 60 mph = 1056 inches / sec. 1056 in/sec divided by 39.37 inches/meter = 26.82 meters per second. (How many feet? 1056/12 = 88 feet every second is covered by a car traveling at 60 mph).
So, back to the formulas:
Before: Total Energy = 200 grams * velocity (use kilograms instead of grams, so use 0.2 kg)... 0.2 kg * 44.7 m/s = 8.94 units of energy (kg-m/s).
After: Ball Energy + Clubhead Energy = 8.94 kg-m/s (by the Conservation Law)
So, Ball Energy = 8.94 - Clubhead Energy After
So, Ball Energy = 8.94 - (0.2 kg * 26.82 m/s)
So, Ball Energy = 8.94 - 5.36
So, Ball Energy = 3.58 kg-m/s
Hoiw FAST is that ball moving?
Ball Energy = mass * velocity
Ball mass = 1.62 ounces. Convert to kilograms: 1 ounce = 28 grams, so 1.62 ounces = 1.62 * 28 = 45.36 grams. I kg = 1000 grams and 1 gram = 0.001 kg, so 45.36 grams */ 1000 g/kg = 0.04536 kg (round to 0.045 kg).
Ball Energy After = 3.58 kg-m/s = mass * velocity = 0.045 kg * Velocity m/s
So, 3.58 kg-m/s = 0.045 kg * Velocity m/s
So, Velocity m/s = 3.58 kg-m/s divided by 0.045 kg
So, Velocity m/s = 79.55 m/s
How fast is that in mph? 79.55 meters * 39.37 inches / meter = 3132 inches; this is 3132/12 feet, or 261 feet per second. There are 3,600 seconds in each hour, so in feet/hour, this is 261*3600 = 939630 feet per hour. There are 5280 feet in each mile, so in mph this is 939630/5280 = 118 mph!
This means that swinging a 200 gram clubhead at 100 mph into a 45 gram golfball on a tee so that after impact the club slows to 60 mph, the ball shoots off into the air at 118 mph! Fore! The ball travels through the air and is slowed by having to knock all the air molecules out of the way, plus the earth's gravity makes it drop downward all the while, so its flight path looks like a rainbow, at the end of which it bounces on the ground a couple of times and rolls to a stop. The AVERAGE velocity during the whole flight is something like halfway between the top speed (118 mph) and the bottom speed (0 mph), so the Average speed would be about 59 mph. Each mph is the same as 1.47 feet per second, so 59*1.47 = 86.7 feet each second of travel (on average). If the ball stays in the air about 6 seconds, it will travel 520.4 feet. That's a drive of 520.4/3 yards, or 175.3 yards. That's probably about right for a lot of people.
At Golf Stores, they have a machine that can measure how fast the clubhead is moving BEFORE impact, but we are really guessing about how fast it is going JUST AFTER impact, so all these numbers require precisie scientific measurement. But you can still gets some numbers on the golf course, and work backwards with the physics to try to figure out other numbers. For example, a man hits a golf ball 200 yards with a 100 mph swing speed and it flew for 6 seconds. How fast was the clubhead moving AFTER impact? 200 yards = 600 feet = 600/6 seconds = 100 feet per second average ball speed, so top speed just after impact of ball is 200 feet per second. 1.47 feet/sec for each mph, so 200/1.47 = 136 mph ball speed just after impact. Total Energy before = Total Energy after, so
Ball Energy after + Club Energy after = Ball Energy before (0) + Club Energy before
136 mph * 0.045 kg + Club speed after * 0.2 kg = 0 + 100 mph * 0.2 kg
6.12 mph-kg + clubhead speed after *.2 kg = 20 mph-kg
Clubhead speed after * 0.2 kg = 20 - 6.12 mph-kg
Clubhead speed after * 0.2 kg = 13.88 mph-kg
Clubhead speed after = 13.88 mph-kg / 0.2 kg
Clubhead speed after = 69.4 mph
This is a little faster than we estimated (60 mph), since the ball went a little further (200 yards vs 173 yards). Nothing funny happens to the energy. The lesson is that if you know the mass of the clubhead and your swing speed, you just count how long the ball is in the air and see how far it goes. These two numbers plus the formulas allow you to figure out the average speed of the ball, to top speed of the ball after impact, and the clubhead speed after impact. Neat!
The same series of calculations have to be done for each club and each swing speed. There are no more than 14 clubs in a set, including the putter, and they all have different masses and swing speeds.
VELOCITY- the concept of velocity is BOTH speed and direction. All along, we have been assuming a STRAIGHT swing with a straight clubface and a resulting straight ball flight. Unfortunately, in golf, swings aren't always straight, or the clubface is a little twisted, or both. Then, all the formulas have to use only the straight part of the hit, for the most part. The straight part of the hit is just that energy that is moving on the straight line of the blow, OR the straight line of the ball flight. They won't be the same in a cockeyed hit with the clubface a little twisted. When the clubface is really square, pointed the same way the swing is moving the clubhead, the ball will go the same direction. But if the face is twisted different from the direction of the clubhead movement, some of the energy of the clubhead is sort of wasted. The ball doesn't go the way the face is pointing or the way the clubhead is moving, but in between the two. To figure out precisely where the ball will go requires separateing the forces into "components" sideways and straight ahead. This makes a triangle (base = left or right, height = straight) of forces, and the one that RESULTS in the ball direction is in between these two directions. The way to see this is to imagine the triangle, and then make a duplicate triangle, flip it, put the two together in a rectangle, and draw a diagonal from corner to corner. The diagonal is the RESULTING direction of the ball flight, adding the two separate "components" back together. If you need to know more about the sidewise and the straight ahead parts, you have to use a little trigonometry (sine, cosine, and tangent) to get the relative LENGHTS of these parts of the forces. The LENGTH is the speed part of velocity (speed plus direction) and the ANGLES tell you direction. Sine, cosine, and tangent just let you start with a known angle and one side to find the other side or the hypoteneus (diagnonal), or start with two sides and find an angle, and so on. That's pretty advanced.
Another assumption has been that the ball and club smack each other and don't change shape from the impact. In fact, they both change some -- the club (metal) practically none, but the ball (plastic and rubber) quite a bit. In slow motion, you can see a club "smushing" a golfball, so that nearly one-third of the ball gets smushed flat by the club. What really happens is that the clubface is angle up a bit (more and more from driver to pitching wedge through the set of clubs) -- straight but up some. When the club smushes the back of the ball flat, it also smushes it a little upward. The ball then rebounds back to shape (from the physics of the ball design) AND slides up the clubface a bit. As it rebounds, it is accepting the energy from the club (and the club is slowing down as it loses or gives up energy to the ball) and the ball "shoots" off the clubface in a straight but upward direction. It also is spinning with backspin, so if you watched from directly opposite the golfer as he swings, the bottom of the ball moves up forward and back over the top towards the golfer.
For all clubs but the putter, the ball flies through the air, bounces, and rolls to a stop. The physics of balls bouncing on the ground take into account how hard the ground is and the angle the ball comes in at and how fast the ball is going when it hits.
II. Ball Flight through the Air
The physics here is all about the ball blasting through all the tiny air molecules. The ball is spinning, and if hit straight, all the spin is backspin in the plane of the flight and none is sidespin. This spin affects how the ball cuts through the air. Also, the dimples on the ball affect the ball-air interaction. Basically, hitting all the air molecules is slowing the ball down (the same way the club hitting the ball slowed the club down), but its pretty gradual because the air molecules are so small (light mass) compared to the ball. The air "flows" around the ball. This means the ball cuts or pushes through the air gas cloud like a bullet fired into water. The air is pushed apart and then behind the ball the general air pressure pushes the air molecules back. There is a bit of a vacuum behind the ball because the ball is dirturbing the generally equalized air pressure as it flies. This vacuum pocket behind the ball is called "drag" because it literally sucks on the ball and slows it down a bit. How do dimples affect this situation? By lessening the drag vacuum so the ball will go farther. How? By chopping up the airflow as it passes around the ball, the airflow at the back of the ball is more mixed up and directed some into the blank area behind the ball. This means SOME of the air pressure is kept close to the ball and there is less of a suck back there.
What does the backspin do? The backspin means the bottom of the ball is hitting the air faster and harder than the top of the ball. In other words, the "pressure" of the ball-air collissions is greater on the bottom and less on the top. This is the same "lift" that an airplane wing creates in cutting through the air, so the ball rises because it spins. The loft of the club starts the ball out at a certain angle upward but spin makes it go even higher upward. If you launch a driver shot at 17 degree (hitting with a 12 degree loft 5 degrees up through the ball), the ball starts out at 17 degreees upward but is also spinning backwards, so it "climbs" an even steeper angle up (maybe 20 degree?).
There is an ideal or optimum combination of launch angle, spin, and length of total flight. It's usually somewhere around 17 degrees. So if you have a 12 degree loft on your driver clubface, hit slightly up through the ball on the tee for maximum distance.
With other clubs, the loft changes. The MOST loft is the wedge. Most wedges have about 56 degrees (while a driver has only 8 to 12 degrees). Combined with backspin, the wedge sends the ball WAY high and not too far out. So for the whole set of clubs, each club is suited to a certain distance shot, as may be called for when playing the course. AND each club has a typical shape of the ball flight -- differently shaped rainbows for different clubs -- and the shapes get shorter and higher going from driver to wedge. Or lower and longer going from wedge to driver.
What does sidespin do? Sidespin comes from hitting the ball with the face twisted and aimed differently than the direction of the clubhead movement. The club smushes a bit of the side of the ball and then when the ball rebaounds, it has a little side spin PLUS a little back spin. These two "components" add up to a RESULTING spin that is tiled out of the vertical plane of the ball flight (just like we did above for the clubface). This spin weakens the ball-air pressure on that side of the ball -- right spin lessens the pressure to the right. So the ball has "lift" off to the side and flies off to the side, depending upon how much sidespin is in the total mix. A slice is a lot of right spin in the mix. A hook is a lot of left spin in the mix. What causes slice spin? Twisting the face to the right in relation to the direction of the clubhead movement. This is said to "open" the clubface. If you swing straight with an open face, you ball spins right and slices. If you swing straight with a "closed" face twisted left, the ball spins left and hooks.
Another assumption we have been making is that the clubface and ball make contact like two ideal points. Two billiard balls can hit head-on or with a glancing blow. We have been assuming a head-on impact. This is all about the center of gravity of the clubhead and ball. The ball is easy - the center of gravity is the center of the sphere. The clubhead is hard. The center of gravity is that one point in the total shape of the clubhead where the different distributions of mass in the weird shape all come together for balance. A round ball with a uniform material throughout has a center of gravity (CG) in the center, but a driver clubhead is more like a bigger ball cut in half and flattened mostly and then stuck on the end of a stick. Depending upon the design of the driver, the CG may be halfway up from bottom to top and halfway back from clubface to back, but not necessarily. In the blow, if the clubhead's center of gravity is moving BELOW the CG of the ball, it will make the club twist upward through impact (the club CG sort of wraps beneath and up around the ball CG). This makes the ball go higher because it alters the loft and launch angle. Golfers who want to hit the ball higher with any club add lead tape to the bottom of the club. To hit the ball lower with that club, do the opposite -- on the back of the club but nearest the top, add some lead tape. The tape moves the CG higher or lower.
The same happens with a clubhead CG off to the side as it moves through the ball. Club CG to the outside of the ball's CG makes the clubface twist around to the inside (left), and this creates hook spin. A club CG inside the ball's CG creates slice spin. This is separate from clubface twist, so you can have a straight swing and a straight clubface but have the clubface CG inside the ball center as it goes through impact and STILL get some slice! Tough game!
Drivers are designed so that the clubface has a "bulge" in front of the club's CG. This way, if you contact the ball with the CG inside, the contact is more glancing than it would be with a flat face, and this reduces the sidespin so the ball doesn't slice so much. It also reduces hook spin the same way.
III. Green-ball interactions.
In putting, the roll of the ball across the green is affected by gravity and the ball hitting the grass blades as it rolls. If the green surface is tilted, then gravity is free to influence the direction of the ball. It is like driving a car straight on a level highway. If the green is tilted, it is as if the steering wheel of the car is given a slight turn away from straight. This constant influence will make the car curve off the road all right, but the exact shape of the curve of the car depends on how fast the car is rolling. The same is true of a ball rolling on a tilted green. Daish covers this pretty well.
The force of the ball is figured just as it is for other clubs, so this tells you that the putterhead speed and mass determines how fast the ball will start out after impact. (And the face angle and direction of the blow matter as well, and so does the CG or sweetspot of the putterhead and ball.) But after the ball starts out, here's what happens:
A "roll" of a ball is not the same as its forward motion. When you first punch a ball with a putter, the ball skids over the grass because it is not rolling yet. The grass friction on the bottom makes the ball start to spin and roll (slows the bottom of the ball down and the top of the ball spins forward) and it takes a little bit before the spinning / rolling matches the forward skidding. When the circumference of the ball rolls along over the grass without skidding, then the grass has added enough spin to the ball that the roll matches the forward motion. When it does, the skidding no longer happens and the ball just rolls. The shaggier the green, the quicker this skid phase is over because the extra grass friction spins the ball up to speed quicker. On really "slick" greens, the skid phase is longer and matters less. Typically, the skid lasts about one seventh of the total putt length. A seven foot putt has a typical skid of one foot, and a 14 foot putt has a skid of about 2 feet.
Also, when the ball is at rest to begin with, it's bottom sits nestled down in the grass probably all the way to the dirt, sort of cradled by the spongy grass blades. When struck with a putter, the ball moves and skids over grass ahead of it and this makes it rise up a bit (the grass blades get shoved down and create a slight incline plane that perpetually lifts the ball up a bit). As this lifting continues and as skidding becomes pure roll, the ball enjoys a state of least interference from the grass and gets a lot of ground covered nicely. Eventually, though, the constant friction of the grass takes its toll and the ball reaches a point where the "inclined plane" effect goes away and the grass is seriously in the way again. This is the "decay" phase of the putt where the ball slows down quicker. For any given green with grass conditions of a certain sort, this phase will set in whenever the ball slows to one specific speed. Then the ball slows quickly to a stop. The interesting point is that on one green the decay phase is always about the same length of roll. A typical length for a typical green is about 1 foot.
So, to putter better with this knowledge, you want to know if you can influence the skid and you want to know what to do about the decay. yes, you can influence the skid by making it more or less, but what difference does it make? Many golfers believe that skidding increases the chance of the ball going off line by hitting something low in the grass. I don't know about that. At any rate, if you want to REDUCE skid you have to either start the ball with some overspin or increase the grass friction to get the spin up to speed. Hitting slightly up on the ball with the putter is thought to add overspin at the start and thus reduce skid. Hitting above the ball's equator is also thought to add overspin. In either event, you are lessening the energy transfer from club to ball unless there is sweespot-to-sweetspot head-on contact between the CG of the putter and the CG of the ball. It's a complicated matter, because less skid makes the ball end up going farther but not having sweetspot-to-sweetspot contact means the ball goes less far. It's a trade-off you have to think about and try out for best effect. Also, hitting up lessens grass friction and so tends to make any skid last a little longer, while hitting down across the top of the ball adds friction. Because of this, it would seem hitting down slightly across the top of the ball would be a little superior to hitting up. On the other hand, perhaps hitting up makes the "inclined plane" effect happen quicker. Also, hitting down makes the ball smush into the spongty turf and then rebound out and it might bounce off line. Hitting up also might "launch" the ball and make it hop along and go offline. Even if bouncing or hopping stays on line, this lessens the energy available for going straight, so these balls don't usually go as far (come up short). As I say, it's pretty complicated. Personally, I think you should just go with a level blow through the ball's sweetspot and not worry about skid. Most greens today are so "nice" that skid doesn't matter much.
IV. Ball-Hole interactions.
For a rolling ball to pass over the front lip of a cup and fall in, it has to drop at least half its diameter before it hits the back wall or rim of the cup. Otherwise, it will hot the rim on the bottom half of the ball below the equator and bounce out of the cup and keep going across the green. This means the key is HOW FAST IS THE BALL ROLLING AT THE FRONT LIP? Is it slow enough to permit gravity to make the ball drop one-half or more of its diameter?
To figure this out, get the numbers first. A golf hole is 4.25 inches wide and 4 or more inches deep, cylindical in shape. A ball is 1.68 inches in diameter, and one-half is 0.84 inches. The speed of the ball across the 4.25 inch center of the cup must be slow enough to allow gravity to pull the ball down 0.84 inches before the leading edge of the ball hits the back wall. Note that the actual distance across that is traveled acroiss the center of the cup is not the whole 4.25 inches, because half of the ball is already across before the bottom of the ball crosses the rim. So the real distance across the center is only 4.25 - 0.84, or 3.41 inches. So how slow must a ball move across 3.41 inches to allow it to drop 0.84 inches? This is the same as asking how long does it take for gravity to make a ball drop 0.84 inches.
Galileo on the Leaning Tower of Pisa dropped a ball and discovered that gravity pulls on a ball so that it gains speed at the rate of 32 feet per second every second. Starting at zero feet per second when he first lets go, at the end of one second gravity has sped the ball up to a velocity of 32 feet per second. At the end of two seconds, the ball has sped up to 64 feet per second, and so on faster and faster until the ball hits the earth. (A 20-pound bowling ball dropped from the Empire State building will hit the sidewalk going so fast it will explode and blow a crater in the concrete.) At the end of the first second, how far has the ball dropped? That depends on its average speed. The top speed is 32 feet per second, but the start speed was zero, so the average is 16 feet per second. At the end of one second at an average speed of 16 feet per second, the ball has dropped 16 feet. A golf ball would drop the same distance in one second, since the earth's gravity has the same effect regardless of the mass of an object. (Try droipping a quarter into your hand and then a penney -- they both fall at EXACTLY the same rate even though the don't weigh the same.)
Our question is how many fractions of a second is it for gravity to move the ball 0.84 inches. The math expression for what we have been discussing is simple: Distance = speed * time. The speed is the Average Speed from gravity (top speed - start speed / 2). Since start speed is zero, average speed is just 1/2 top speed. The top speed from gravity depends only on how long the object has been falling, since gravity gives the object a constant uniform acceleration (32 feet per second each second). Top speed at 2 seconds = 64; at 3 = 96 etc. So top speed = gravity acceleration * time.
Back to our basic formula, Distance = speed * time, where speed is average speed, and average speed is 1/2 * top speed from gravity, and this in turn is gravity acceleration * time, the resulting formula is Distance = [1/2 gravity accleration * Time] * Time. We want to know Time.
Distance = [1/2 Gravity acceleration]
Time Time
Distance = 1/2
32 feet per sec per sec Time (squared) seconds-seconds
Convert 0.84 inches to feet = 0.84/12 = 0.07 feet
Convert 32 feet per second / second to inches per second / second, 32*12 = 384
1/2 * 384 = 192
Time (squared) = 0.84 inches / 192 inches per sec per sec
Sqaure Root of Time (squared) = Time, so
Time = Square Root of [0.84/192]
Time = Square Root of 0.00438
Time = about 0.067 seconds (7 one-hundreths of a second, less than 1/10th of a second).
Now, if Distance = Speed * Time, and the distance across is 3.41 inches and the time is 0.07 seconds, the Ball's Speed at the rim has to be 3.41 in / 0.067 sec OR SLOWER, and that speed is about 51 inches per second. Any ball rolling faster than this has NO CHANCE of falling into the cup even if it crosses the widest possible part of the cup. How fast is that? A ball rolls one full circumference per roll, and the circumference of a ball is Pi* Diameter or 3.14 * 1.68 = 5.28 inches around. So each roll, the ball travels 5.28 inches. A ball rolling at 2 revolutions per second speed covers 10.56 inches each second. A ball rolling 51 inches per second is the same as 51 / 5.28 = about 9 revolutions per second. Anything faster than this has no chance.
To consider balls that don't go across the center of the cup, but cross a shorter path off to the side, just consider a line across the cup merely 2 inches from front rim to back rim. The ball only has 2 - 0.84 inches before it hits the back rim, or 1.16 inches total. 1.16 / 0.067 = 17.3 inches per second MAXIMUM SPEED, and 17.3 / 5.28 is 3.3 revolutions per second. So if you want a chance with most of the cup available, a good speed for the ball as it crosses the rim is around 3 revolutions per second OR LESS.
If a ball going 17 inches per second missed the cup altogether, on most greens it would probably stop about 1 to 2 feet past the hole. You can use this knowledge to advantage by trying to control your putts so that they usually stop a little past the hole, and not more than 2 feet past. I personally don't want to miss, but if I do I try to have the ball stop within 1 foot of the hole, so my speed is about 2 revolutions per second or under at the front rim.
Some other physics books that can help are
Jorgenson, The Physics of Golf (all about the full swing)
Pelz, Putt Like the Pros (some of the physics of putting)
Templeton, Vector Putting (about gravity on the green and grass friction)
Cochran and Stobbs, Search for the Perfect Swing (mostly full swing but some about phases of a putt)
These books might be hard to get, but a refernce librarian can track them down for you using Interlibrary Loan.
There is also an article or two in the Physics journals about how a ball and hole interact. They are very technical mathematically. If you want those references, let me know.
Cheers!
Geoff Mangum
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