Flop Shots

[garyg]

Has anyone else noticed lately that flop shots aren't as accurate as they used to be, or is it just me?

I flop as much as I can around the green (as I hate the 6-7 foot chipping option which either goes miles too far or not far enough) and I used to be pretty accurate with it, not so much now. I recently had a 23 yard third shot from just off the green, I was on the fairway, there was no wind, and no other factors that could have influenced the distance. I was using a PW or SW, the club that hits it 30 yards at 100% and therefore I guessed I needed to hit it roughly 75%. I took my shot and hit it with 73% power which you'd expect to land near the hole right? Well it dropped at 15 yards before spinning back a yard. Now I'm no mathematician but I'm pretty sure 15 yards would make that 50%, not 73%! 

How are we supposed to judge flop shots if the percentages are completely wrong? Anyone else have this issue?

[mcbogga]

The percentage is not distance it is power and it's not a linear relationship.

The 84% one bar loft flies 24 yards.

Or did you do it by stopping swing short?

[garyg]

Aug 1, 2016 5:55:05 GMT -5 mcbogga said:

The percentage is not distance it is power and it's not a linear relationship.

The 84% one bar loft flies 24 yards.

Or did you do it by stopping swing short?

I left it at 100% and swung the club to 73%. So basically my shot should have went 73% of 100% of 30 yards but it actually went 50%. I wouldn't expect it to be perfect, but the ball went no where near 73% of the 30 yards.

I understand backspin may be taken into account on a flop shot, but even that isn't the answer here. It's not like my ball went around 23 yards then swung back to around 15 yards, it dropped at 15.

[mcbogga]

No - still power and not distance that is displayed.

[garyg]

Aug 1, 2016 7:21:32 GMT -5 mcbogga said:

No - still power and not distance that is displayed.

I'm not quite sure what you're saying? If 100% power will hit the ball 30 yards, 50% should hit it 15 yards and 10% should hit it 3 yards, that's how percentages work?

[mcbogga]

No. That's not how they work.

[garyg]

Aug 1, 2016 9:35:33 GMT -5 mcbogga said:

No. That's not how they work.

Wouldn't it be better you explain what you mean rather than just saying no? 

It pretty much is how percentages work. If, for example, I have 100 coins and I give someone 50% of them, I will be giving them 50 coins, if I give them 10% of the coins, I'll be giving them 10 coins. Your posts are basically saying that is wrong.... 

[ErixonStone]

Aug 1, 2016 5:02:41 GMT -5 garyg said:

Now I'm no mathematician but I'm pretty sure 15 yards [of 30] would make that 50%, not 73%!

When you hit your shot, the ball doesn't travel in a straight path from your club to its landing point; it travels in an arc.  The velocity of the ball has two components - a speed component and a height component.  

The height component: Given the same amount of loft (or the same trajectory), a ball hit with half the velocity will travel half as high before beginning its descent back to the ground.  In other words, if you hit the ball half as hard, it will stay airborne half the time.

The speed component: Given the same trajectory, a ball hit with half the velocity will travel half the distance over the same period of time.

The problem with your assumption (that in order to land a flop shot 15 yards from its original position, you need to hit it with 50% power) is that you've neglected the element of time the ball will remain airborne.  If you hit a ball half as hard, it will stay airborne for half the time.  In order to travel half the distance over half the time, the ball would need to have the same speed.  Because you've hit the ball with half power, however, the ball does not have the same speed, and will land a good distance short of your intended landing point.

Here's an example:

Let's say that a full flop shot remains airborne for 4 seconds and travels 30 yards.  The ball's speed is 30 yards over 4 seconds, or 7.5 yards per second.  If you hit the ball with 50% power, the ball will remain airborne for just 2 seconds, and its speed is just 4 yards per second (I rounded from 3.75).  

4 yards per second X 2 seconds = 8 yards.

In other words, you really need to take the power-squared and then multiply it by the full shot distance.  In this case, 0.5^2 X 30 = 0.25 X 30 = 7.5.  The inverse of this - and what you really wanted to know - is How much power do I need?  That is represented by:

SQRT( Intended Distance / Full Shot Distance)
SQRT( 15 / 30 )
SQRT(0.5)

The square root of 50% is 71%


Specifically for your example:

Your Intended Distance was: 23 yards
Your Full Shot Distance (LW Flop) was: 30 yards

SQRT (23 / 30) = SQRT (0.766667) = 0.875595 = 87.56%

[blackaces13]

Yeah!

[AFCTUJacko]

Or as an alternative to getting a PhD in Mathematics.....

You could just very deliberately leave yourself full pitch/flop shots (30,35,40,45)

[mcbogga]

Aug 1, 2016 14:46:55 GMT -5 ErixonStone said:

Aug 1, 2016 5:02:41 GMT -5 garyg said:

Now I'm no mathematician but I'm pretty sure 15 yards [of 30] would make that 50%, not 73%!

When you hit your shot, the ball doesn't travel in a straight path from your club to its landing point; it travels in an arc.  The velocity of the ball has two components - a speed component and a height component.  

The height component: Given the same amount of loft (or the same trajectory), a ball hit with half the velocity will travel half as high before beginning its descent back to the ground.  In other words, if you hit the ball half as hard, it will stay airborne half the time.

The speed component: Given the same trajectory, a ball hit with half the velocity will travel half the distance over the same period of time.

The problem with your assumption (that in order to land a flop shot 15 yards from its original position, you need to hit it with 50% power) is that you've neglected the element of time the ball will remain airborne.  If you hit a ball half as hard, it will stay airborne for half the time.  In order to travel half the distance over half the time, the ball would need to have the same speed.  Because you've hit the ball with half power, however, the ball does not have the same speed, and will land a good distance short of your intended landing point.

Here's an example:

Let's say that a full flop shot remains airborne for 4 seconds and travels 30 yards.  The ball's speed is 30 yards over 4 seconds, or 7.5 yards per second.  If you hit the ball with 50% power, the ball will remain airborne for just 2 seconds, and its speed is just 4 yards per second (I rounded from 3.75).  

4 yards per second X 2 seconds = 8 yards.

In other words, you really need to take the power-squared and then multiply it by the full shot distance.  In this case, 0.5^2 X 30 = 0.25 X 30 = 7.5.  The inverse of this - and what you really wanted to know - is How much power do I need?  That is represented by:

SQRT( Intended Distance / Full Shot Distance)
SQRT( 15 / 30 )
SQRT(0.5)

The square root of 50% is 71%


Specifically for your example:

Your Intended Distance was: 23 yards
Your Full Shot Distance (LW Flop) was: 30 yards

SQRT (23 / 30) = SQRT (0.766667) = 0.875595 = 87.56%

That pretty much sums it up.

[garyg]

Aug 1, 2016 14:46:55 GMT -5 ErixonStone said:

Aug 1, 2016 5:02:41 GMT -5 garyg said:

Now I'm no mathematician but I'm pretty sure 15 yards [of 30] would make that 50%, not 73%!

When you hit your shot, the ball doesn't travel in a straight path from your club to its landing point; it travels in an arc.  The velocity of the ball has two components - a speed component and a height component.  

The height component: Given the same amount of loft (or the same trajectory), a ball hit with half the velocity will travel half as high before beginning its descent back to the ground.  In other words, if you hit the ball half as hard, it will stay airborne half the time.

The speed component: Given the same trajectory, a ball hit with half the velocity will travel half the distance over the same period of time.

The problem with your assumption (that in order to land a flop shot 15 yards from its original position, you need to hit it with 50% power) is that you've neglected the element of time the ball will remain airborne.  If you hit a ball half as hard, it will stay airborne for half the time.  In order to travel half the distance over half the time, the ball would need to have the same speed.  Because you've hit the ball with half power, however, the ball does not have the same speed, and will land a good distance short of your intended landing point.

Here's an example:

Let's say that a full flop shot remains airborne for 4 seconds and travels 30 yards.  The ball's speed is 30 yards over 4 seconds, or 7.5 yards per second.  If you hit the ball with 50% power, the ball will remain airborne for just 2 seconds, and its speed is just 4 yards per second (I rounded from 3.75).  

4 yards per second X 2 seconds = 8 yards.

In other words, you really need to take the power-squared and then multiply it by the full shot distance.  In this case, 0.5^2 X 30 = 0.25 X 30 = 7.5.  The inverse of this - and what you really wanted to know - is How much power do I need?  That is represented by:

SQRT( Intended Distance / Full Shot Distance)
SQRT( 15 / 30 )
SQRT(0.5)

The square root of 50% is 71%


Specifically for your example:

Your Intended Distance was: 23 yards
Your Full Shot Distance (LW Flop) was: 30 yards

SQRT (23 / 30) = SQRT (0.766667) = 0.875595 = 87.56%

Thanks for your detailed explanation, I appreciate it.

I knew it wasn't as simple as "50% of 30 yards is 15 yards" in real life of course, but with this being a video game I thought it would pretty much be as simple as that, how wrong I was haha.

Anyway thanks again, it all makes sense now.