「ぱぱっと」計算 05： 斜面を転がるボールの運動（1次元）

]]>See figure.

The distance is 5 meters and the slope is 3 degrees. According to the “Putt Physics” calculation, Lespaul is 147 cm from the cup. However, the trajectory of the ball is unfortunately not that of the cup. As you can see, when the slope of the green increases, the correct response pole cannot be calculated. In particular, as shown in the figure below, the amount of deviation tends to be larger when the slope is uphill.

Therefore, “Putt Physics” provides a function to recalculate. If you feel that the ball is not in the cup and would like the application to reconsider, double-click on the initial ball position. The app will calculate the correction and show you the line where the ball will go in the cup again. It will show an orange line like this.

I say Lespaul is special and unique (the only) point, but when the slope increases, it becomes a multi-point depending on the position of the ball. It’s not funny and it’s a shame, but it’s a meny pole. We are still considering how to solve this problem and make a single Lespaul.

]]>The green slopes 0.5° from above to below the gray line, and the putt is made from a distance of 3 meters around the cup (black circle).

At the time I heard this story, it was hard for me to believe it, but when I tried it on the practice green, I found it to be true. Please give it a try.

However, there is a condition. That is, the area around the cup must be flat. In this example, if there are undulations on the green surface within the 3-meter circle, the red circle will be shifted when undulations are involved. Isn’t there such an ideal condition on a real green? You may think, “But in reality, you only play from the spot where your ball is, so if the area around the cup is flat, such as a semicircle or a quarter circle, you can find the red circle by the same concept. In professional tournaments on TV, the cups are often placed in difficult positions, such as near the tiers of double-tiered greens or around the snake line, but on greens for amateurs, the cups are often placed on flat areas, although there is a slope. In other words, the “Putt Physics” calculation is useful.

Now, calling it a “gray line” or a “red circle” is not very appealing, so we call it this way in the “Putt Physics” application.

Elevation line:

Gray line. It is an imaginary line connecting the lowest and highest points of the green. It is assumed that the player stands at the lowest point, so it is a straight uphill line.

Lespaul:

If the slope is less than 1°, putt in the direction of Les Paul and you will get a cup. Les Paul is an anagram of “Special Unique Point at Elevation Line”.

There is always one of these poles on the elevation line, which is determined by the distance and slope angle.

]]>The conditions are both putts with a distance of 3m and a strong slope of 3°. White is the line to aim at 3m with just a touch. The orange line is 30cm over 3m.

I think the orange line is the one to use without hesitation, not only because it is easier to putt a return putt over the cup if one putt does not go in. In addition to the fact that the over-the-cup putt is more likely to bring the ball closer to the cup, the just-touch putt will roll toward the cup with a large bend, almost stopping at the end, making it a very uncertain line if it will bend as it should.

I know this is extremely commonplace for golfers, but at “Putt Physics” we calculate the trajectory of the ball based on the policy that everything should be 30cm over. This was just a time when I took the trouble to explain something normal.

]]>I thought that it would be interesting, but since it is just the same calculation as the previous one for the x and y directions separately, it is not interesting at all. I will write about it when I feel like it, since it will only complicate the diagram.

]]>The angle of the slope is θ°. In this case, the force mg applied to the ball remains the same as on a horizontal surface. However, due to the angle, it is necessary to separate the force into two directions: one parallel to the slope and the other perpendicular to the slope. The former is a steady force acting in the direction of the bottom of the slope, and the latter is the source of frictional force as in the previous case.

Now, let us formulate the equation of motion.

$$F=ma$$

$$x=\frac{1}{2}at^2+v_0t$$

$$v=at+v_0$$

Although sin/cos calculations are added, the calculation method remains the same as before. This time, we assume the case where the slope is converted to a slope in one dimension, so a straight uphill line and a straight downhill line can be calculated in terms of a putt.

Let’s start with the ascent. Assuming a slope angle of +2° and a stimpmeter s=9, the coefficient of friction μ=0.1. Assuming initial ball speed v0 = 2 m/s, the time t until the ball comes to a stop is

$$t=\frac{2}{0.1*9.8}=1.96秒$$

and stops after 1.96 s. From Eq. 2, it follows that

$$x=\frac{1}{2}(-\mu{}g)t^2+v_0t=-0.5*0.1*9.8*1.96^2+2*1.96=2.0m$$

The ball’s initial velocity is 2m/s on a green with a stimpmeter value s = 9. On a green with a stimpmeter value of s = 9, if the ball is launched at an initial speed of 2 m/s, it will roll 2 m.

Next is the descent. The slope angle was set to -2° and other values were left unchanged.

If the putt is made with the same strength for both uphill and downhill, the difference is xxx m under these conditions.

By the way, I used the slope angle θ° because this expression is often used in high school physics test questions. Another way to describe the slope of a slope is “gradient,” which is also often heard. I am more familiar with angles, so I use angles in “Putt Physics,” but I wonder which is more common in golf. If there are many requests for slope notation, we can accommodate them.

角度 | 勾配 |

1° | 1.8% |

2° | 3.5% |

3° | 5.2% |

4° | 7.0% |

5° | 8.7% |

By the way, if the angle exceeds 3°, the ball will not stop. Of course, the threshold depends on the green speed. So, most greens have slopes in the range of 0° to 3°, which is about 0% to 5% slope.

This completes the calculation of the motion of a ball rolling down a slope (1D).

]]>In this case, we will consider putting on a level surface parallel to a level surface with no slope on the green. The force on the ball is the acceleration of gravity \(g\), which we all love, the force derived from the ball’s initial velocity \(v_0\), and the friction on the green. The figure below illustrates this.

The newly emerged power💪 is the friction force \(\mu{}mg\). Among these, \(\mu\) is the coefficient of friction, which is the force acting in the opposite direction as the ball rolls forward. The force obtained from the putter head is gradually reduced by the frictional force and eventually comes to a stop. Friction force, friction force… What is the coefficient of friction, \(\mu\)? I don’t know.

It needs to be replaced by a coefficient that is understandable to the golfer. The force that restrains the ball from rolling is the turf on the green. You generally describe it as the “speed of the green. Did you know that this speed index is quantified? Of course, I had no idea. I learned about it for the first time when I created “Putt Physics”. The name of the device is “Stimpmeter. It is a device to measure the speed of greens, and it is sold like this.

紹介

It is quite expensive. It seems to be the de facto standard for green speed measurement, and I wanted to buy one and measure it, but I could not afford it. Some golf courses seem to post the stimpmeter values for the day, so I would like to make use of it for “Putt Physics” calculations when I hit a course that is grateful for such a service.

This is off topic, but the distance rolled using the stimpmeter translates to the speed of the green. The unit is feet. The average golf course is 8 to 9 feet. Let’s calculate the speed of the green using the stimpmeter value as \(s\). All we need is the three equations for iso-accelerative linear motion.

$$F=ma$$

$$x=\frac{1}{2}at^2+v_0t$$

$$v=at+v_0$$

The frictional force is \(\mu{}mg\), which is the gravitational acceleration on the ball multiplied by the coefficient of friction. If the direction of rolling is +, \(ma=F=-\mu{}mg\), and \(a=-\mu{}g\).

From equation 3, since \(v=0\) when the ball stops.

$$0=at+v_0=-\mu{}gt+v_0$$

$$t=\frac{v_0}{\mu{}g}$$

When the stimpmeter \(s=9\), the coefficient of friction \(\mu=0.1\). Assuming initial ball speed \(v_0=2m/s\), the time \(t\) until the ball comes to a stop is

$$t=\frac{2}{0.1*9.8}=1.96秒$$

and stops after 1.96 seconds. From equation 2, it follows that

$$\frac{1}{2}(-\mu{}g)t^2+v_0t=-0.5*0.1*9.8*1.96^2+2*1.96=2.0m$$

The ball’s initial speed is 2m/s on a green with a stimpmeter value of s=9. On a green with a stimpmeter value of \(s=9\), if the ball is launched at an initial speed of 2m/s, it will roll 2m.

Finally, what is not being calculated in the customary “snapshot”? As I mentioned before, we ignore wind effects and air resistance. Then we consider the ball to be a perfect sphere, so we ignore ball mumbo jumbo. The other thing I didn’t include in the formula is the force consumed by rolling, but it may be included empirically in the conversion to friction coefficient and stimpmeter value. I have no proof of this, so if the rolling distance deviates significantly, I would appreciate it if you could let me know the conditions.

This completes the calculation of the motion of a ball rolling on a plane.

]]>Now, in the third article, the ball finally comes out. This time, we will calculate the motion of the putter head and the ball when they collide and find the initial velocity of the golf ball. From the conservation of momentum law, we get

$$m_{a}v_{a0}+m_{b}v_{b0}=m_{a}v_{a}+m_{b}v_{b}$$

formula. Each symbol means something like this.

\(m_a\): Weight of putter

\(m_b\): Weight of the ball

\(v_{a0}\): Putter head speed just before collision

\(v_{b0}\): Ball speed just before collision

\(v_a\): Putter head speed immediately after collision

\(v_b\): Ball velocity immediately after collision

Since golfers hit a stationary ball, the ball velocity vb0 just before impact is zero. Next, the coefficient of repulsion, e, should be considered.

$$e=-\frac{v_a-v_b}{v_{a0}-v_{b0}}=-\frac{v_a-v_b}{v_{a0}}$$

Actually, the number is between 0 and 1, but for simplicity, let e=1. Then the previous equation becomes.

$$v_a=v_b-v_{a0}$$

Let’s go back to the expression for the law of conservation of momentum and substitute a numerical value. The weight of a golf ball, \(m_b\), is fixed by the rules. The rule is 1.62 ounces or less, which translates to 45.93 grams. Next is the weight of the putter, \(m_a\), which I honestly don’t know how to define. Assuming a putter head, it would be around 350g, a putter club would be 500-600g, and if you include human weight, it would be, say, 50000g. Which is it?

Anyway, let’s do the math. Let \(m_a\) be 350g and 50000g as minimum and maximum values. If the velocity of the putter head was 1 m/s, the initial velocity of the ball \(v_b\) would look like this.

$$v_b=\frac{2m_{a}v_{a0}}{m_a+m_b}$$

$$m_a=350g のとき、v_b=\frac{2*350*1}{350+45.93}=1.77$$

$$m_a=50000g のとき、v_b=\frac{2*50000*1}{50000+45.93}=1.998$$

What we can see here is that no matter how large we assume \(m_a\) to be, the initial velocity of the ball will never be more than twice the head speed. Actually, this is the part of the calculation that I don’t agree with the most. When I actually putt the ball, it is more than twice as fast. I think the head speed is probably not far off, so I guess what I need to consider is how the force is transmitted.

In real-life putting, I would try to keep the head from slowing down as much as possible after making contact with the ball. I personally have been taught that the backswing and follow-through should be the same. There are various types of players who shorten their follow-through, lengthen their follow-through, or are conscious of their rhythm, but I think the common denominator is to make sure that the follow-through is firm so that the ball is not defeated. If you are not conscious of this, the ball will have a “unstable” trajectory. This means that even a person with a normal hitting style who does not make a punch shot is applying a force like a punch shot to some extent. This force is probably not reflected in the formula.

$$m_{a}v_{a0}+m_{b}v_{b0}+ ??? =m_{a}v_{a}+m_{b}v_{b}$$

I would appreciate it if you could let me know if there is a good way to add an equation in the \(???\) section or change the calculation of collision to force product. However, I believe that the head speed \(v_{a0}\) and the ball initial velocity \(v_b\) are still in a proportional relationship, so I have specified this ratio as the putter transmission ratio in the application “Putt Physics”. The default value is “3.5”, which is greater than “2”.

This value will vary depending on the characteristics of the putter and the swing method, so please find your own value and adjust it while actually putting.

This concludes the topic of collision calculation.

]]>The most common way to adjust putt distance is to use a swing angle, right? We use swing width as the one we suggest in the application “Putt Physics” because that is how it is explained in many magazines and videos. This choice was not too confusing, but what I was a little worried about was whether to display the angle θ° or not. Is there anyone who says that the angle display is easier for me to understand? For putts with a swing angle of more than 45°, the angle display may be easier to understand. If there are many requests for it, I may set up a function to switch the display.

Now, it is easy to calculate the swing angle from the angle θ°.

$$スイング幅=2Lsin\theta$$

Calculate this equation to find the length of the string. Since the string length is half the swing width, it is twice as long. If a person of 170 cm in height and 140 cm in string length of the pendulum swings at an angle θ = 9°, the swing width is 22 cm.

This completes the calculation of swing width.

]]>・Place the ball directly under your eyes.

・Do not move your head.

・Align your backswing and follow-through

・Hold the ball loosely and do not use your wrists.

・Swing as straight as possible.

No matter how you think about it, you want to be a pendulum, don’t you? That’s all I can think. I decided to think so. I’ve made up my mind, so I calculate the putter swing as a pendulum motion in the application “Putt Physics”.

Pendulum motion, so-called single pendulum. If you studied physics in high school, you may know this. The components of a single pendulum are a string and a weight attached to the end of the string. It looks like this.

So, here’s the thing.

If the length of the string is L, the weight of the weight is m, and everyone’s favorite gravitational acceleration is g, the formula follows from the law of conservation of mechanical energy.

$$mgL(1-cos\theta)=\frac{1}{2}mv^2$$

Potential energy at the highest point of the weight = Kinetic energy at the lowest point of the weight

We will replace this with the putter swing. First, the length of the string, L. The tip is without hesitation the putter head. The question is where to place the fulcrum. I was advised not to move the head, so I could think of the head as the fulcrum, but I thought it might be a little lower, so I decided to use the center of the shoulder.

In “Putt Physics” the height is entered, but in the actual calculation, the height minus 30 cm is used as the length of the thread (L). The distance from the top of the head to the center of the shoulder and the distance from the center of the shoulder to the center of the shoulder are taken into account. 30 cm is honestly appropriate, so players should decide the length of the string length L taking into account their own form and other factors.

Now that the length of the string L has been determined, the next step is to determine the mass of the weight m. Since there is one weight on the right side and one on the left side of the string, we divide the weight by the mass of the weight. Yes, the weight of the weight = the weight of the putter does not affect the pendulum motion. No matter what kind of putter you buy, the ball’s motion will not change. Any putter is the same. Yes, that’s a lie. That is not true. The mass m of the weight has no effect on the single pendulum equation of motion. As will appear later, the club characteristics of the putter have an effect on the collision motion between the putter and the ball.

Angle θ is the angle of the highest point of the pendulum, so it corresponds to the swing angle of the putter. Velocity v is the speed of the weight, so it is the speed at which the putter head moves. Now that all the substitutions are done, we have the values and can do the calculations. Now that we know the values and can calculate them, let’s solve the equation of motion.

It seems good if we know the velocity v when the putter head reaches the lowest point. After that, the putter head and the ball collide and the energy is transferred to the next motion.

$$v=\sqrt{2gl(1-cos\theta)}$$

Let us calculate the case of a 170cm tall person swinging at a putter swing up angle of 30 degrees. Since the length of the string is 170 cm – 30 cm = 140 cm

$$v=\sqrt{2\times9.8\times140/100\times(1-cos30^\circ)}=0.606321938$$

This concludes the calculation of putter head speed. This calculation is actually done in “Putt Physics”.

To digress a bit as an aside, you are sometimes advised to swing in the same rhythm for putting. This is actually closely related to the pendulum motion. In the pendulum motion, if the length of the string and the mass of the weight are the same, the period of the motion will remain the same even if the highest point is different. It is interesting to see that this is true when you actually make a pendulum and move it.

Thinking about this in reverse, if you change the rhythm of the putter swing according to the width of the swing, the physical law will be different from the pendulum motion. For example, if the rhythm is slowed down when the backswing is large, the head speed is slowed down compared to the pendulum motion. Please note that if the rhythm changes according to the swing width in this way, it will not fit in the “snap” swing width calculation.

Next, I will note what is not calculated in the swing motion. First, air resistance. I think putters are also subject to wind resistance, but this is ignored. Second, human swing friction. The human swing cannot be in an ideal state like a fixed fulcrum of a string. There are many factors that interfere with the swing, such as shoulders, hips, and muscles in various places, but we ignore them all. Then there is the human force (acceleration) during the downswing. This is a force that cannot be ignored depending on the swing method, so it is included in the calculation factor and will be discussed later.

This completes the calculation of putter swing motion.

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