Yes the angle continuously gets more precise as the hole gets further away forever.

The problem here is that I don't think you will ever get to a point where the angle of possible makes on a breaking putt is less than the angle of possible breaks on a comparable straight putt. And if that can never happen at best you will get down to a point where the launch angle range for a straight putt and breaking putt are the exact same size. At that point how are you going to argue that it is easier to get the speed right on the breaking putt rather than the straight one considering the fact that breaking putts always travel further than straight putts and are thus harder to get right.

To bring it back to the paper in regards to this, that is how they proved a straight downhiller is easier than a straight flat put.

At 10 feet the straight putt had a range of makes of 1* on each side. The straight downhill putt had a range of makes of almost *2 degree on each side, because of the convergence phenomenon that again only exists on perfectly straight, downhill putts. And since you had twice the margin of error on launch angle, and getting the speed right didn't get twice as hard, the downhill straight putt was proved easier.

I'm going to reply in bold within the quote below:

.

I think I'm done with this for a few days. Thank you all for the fun discussions. I'm going to see where this heads from here. Probably no where good.

Again with regards to this I was doing the calculations with the assumptions that the face angle at impact was completely random across the dispersion zone.

Aka you are just as likely to miss by exactly 1* left as you are to hit it Dead Perfect Straight.

This doesn't hold true if the distribution is bell shaped from center.

So you agree that on a 100 foot putt, the most likely face angle achieved would be exactly dead square? And as you move away from dead square the likely hood of hitting that exact angle decreases?

This is a big advantage for straight putts, as the most likely result of your putting stroke provides you the greatest margin of error on speed. This does not occur on breaking putts as each line as a very similar allowance for speed.

I think you are likely underestimating how far away a hole will have to be for this to finally come true, and at that point the chance of hitting the putt with the correct speed will be so minimal. You then have to take into account that breaking putts are going to travel substantially futher, even with a minimal amount of break once you reach these distances making it that much harder to get the speed right.

And this again comes back to if the launch angle range is the same for a straight 200 foot putt, vs a breaking 200 foot putt, you have the exact same % chance to start the ball in the cone of each putt. But the 200 foot putt breaker will ALWAYS be longer than the 200 foot straight putt(and sometimes quite significantly so depending on the amount of break) that it will always be easier to get the speed right on a shorter putt.
The bottom line is it has an effect.

So if given the choice between a straight 100 footer or a straight 103 footer you would always choose the 100 footer even if your chances of getting the speed right only increase by 1%*. BC not only does it get marginally harder to get the speed right, we know from geometry that your margin of error always decreases on longer puts as well.

*1% difference between hitting a 100 foot putt within 100-104' and hitting a 103 foot putt within 103-107'. Aka getting it to the hole and hitting it 4 feet past which may be the hardest you could hit it and it still go in.

Unless you can prove that it would somehow be easier to launch a putt more accurately on a breaking put compared to a straight putt then this inflection point simply doesn't exist.

No ****ing way, please cite your source. Oh wait…..

No, I don’t agree the most LIKELY face angle will be achieved will be dead square. What I am telling you (and what normal distribution tells you) is that the inner most area will contain more putts that the outermost. Here’s another term for you, standard deviation. Check out the picture of a bell curve, maybe it will tighten up the idea for ya.

I’ll give you another link to ignore:

http://en.wikipedia.org/wiki/Standard_deviation

I truly have no idea how many putts will hit the hole from 100’. I’d love to figure out the math and come up with the correct %, I’m truly interested. I can however tell you intuitively it won’t be enough when multiplied by the speed component to hole enough putts to have more makes than the 100’ breaker.

I can only assume your side is gone from the debate because (well first and foremost we are the only idiots doing this on Christmas) they can’t believe the corner you have painted yourself into. For fun I noted the view count this morning and there have been several hundred views. I know I’ve refreshed a few times, but I’m pretty sure your peanut gallery is still checking in…but they have decided it is time to just let you die on the vine alone.

Bahahahahahaha.

Just give up

So no one is interested in betting against a high handicapper in this?

You truly have no idea how normal distributions work.

But please tell us more about your incredible intuition.

I guess it's just my imagination that the very center of a normal distribution is the highest point of the curve, signifying the greatest chance of happening



But we should probably ignore hundreds of years of mathematics in exchange for your intuition.

You're on fire.

My goodness, for your own good log off.

Let’s set up an experiment like A Rod did for me.

Marbles:

10 Red Marbles
8 Blue Marbles
6 White Marbles
4 Yellow Marbles

If you put these marbles on a graph and had the reds in the middle. Then had the blue on the left and right of the red, then the whites on the left and right of the blue, and finally yellow on the left and right of the white what would the distribution of marbles look like? For the physicist I recognize this is a crude example, but it’s Christmas and this gets the point across.

The column of reds would be tallest and therefore most likely to be pulled correct?

It's also probably my imagination that a normal distribution looks exactly like the distribution of possible makes of a straight putt.

Funny it doesn't look at all like the distribution of makes on a breaking putt.

And to easily see what putt is easier to make you overlay a normal stroke distribution on top of the make distribution for a straight and breaking putt.

Hmm I wonder which one will have the largest total area overlapping?

I bet it's the answer to this entire debate.

Sigh. My argument isn't that at a certain distance (infinity) it all converges (to zero). My argument is that at short range, we want straight putts. At infinity, it doesn't matter. Somewhere in between, there are breaking putts that are more likely to be holed than their straight counterpart. I don't know what that range is. I don't know that it even exists.

But you're just saying "lol no" without providing any evidence to back up your claim that I'm just sooooo wrong. At least admit "Without the distribution data of ranges of misses in both speed and launch angle, we cannot determine who is right." That's all I'm asking here if we can't come up with a data set to figure out the actual probabilities. You are operating under assumptions. Please stop it. I'm not BOOM.GIF'ing you. I'm trying to have a scientific discussion.

FYP

It is. But we don't have that data. This image of the combined figures is the combinations of putts that go in, not the distribution of where people hit it. If the distribution of launch speed is only 0.1 m/s wide, then the answer is breaking putts. If the distribution of launch speed is 0.5 m/s then the answer is flat putts. Also, the further we get away from the hole, the closer to "breaking putts" seems likely to be the answer to me if in fact the speed is controlled more closely than NXT is assuming it is. emphasis on assuming. I'm just asking questions.




As for the normal distributions and straight putter face being most likely, that will obviously be the case. In any normal distribution, the most probable result will be the center, however, if it's a very short but wide curve, it could still mean it's only 2% likely to happen. Just because it's the center doesn't mean it's like, 50% likely....So ship, I think the point NXT was trying to make was that launching dead straight is the most likely outcome, and when you mistakenly said "no i dont think so", he pounced on it to do another Boom.gif at you. A simple quick mistake doesn't undo the entire argument....

Quoted for more hilariously dumb math fails from Ship

I'm not really saying "lol no" I'm just extrapolating the data that we have access to. The burden of proof is now on you to come up with a reason and proof that there exists a point where the data we have would no longer hold true.

I do have to commend you on putting the 10 foot straight putt along side the 10 foot breaking putt. I think you did a pretty good job there. I missed it earlier bc I have been on my phone most of the day.

I have done a little extra work to add.



I have added the college players putt distribution I used from earlier. He has an 8* dispersion and my guess is they hit a 10 footer somewhere between 8-13 feet.

Now you would have to take the total area of each "make" distribution that falls withing the college golfers "putt" distribution.

You then have to weight it, based on a normal distribution, that the further away from that thin center line you get the less likely those makes are.

Does that help visualize a little just how much easier a straight 10 footer is compared to a breaking 10 footer? It's significant.

I know he was saying launching dead straight was the most likely outcome. Where I come from for something to happen most likely it needs to happen over 50% of the time. So no, the ball will not most likely launch dead straight. I don't think it will launch dead straight over 10% of the time.
See above.

I'll give you a hint, you're odds of picking a red marble most likely will not happen.

Come on man, this is just silly. That is obviously not what nxt meant and does in no way disprove his point.

Are you guys really fighting over semantics now?

JFC, I think it's obvious you both understand how bell curves work. Of course no one is saying that the mean result on the curve is over 50% likely to happen.

It is however more likely to happen than any other result given a normal distribution. You both know this, stop being silly.

Happy Kwanzaa everybody.

Bahahahahahahahahaha

Wait, wat?

You have a bag with 10 red marbles, 6 blue marbles, and 5 yellow marbles. You pick a marble out. What's the most likely color for you to pick out?



http://en.wikipedia.org/wiki/Mode_%28statistics%29 , ffs

I wouldn't call it semantics...

He appears to not know this. He's gotten it wrong multiple times despite people pointing out he's wrong.