A Physics Puzzle - Uphill vs. Downhill Putts

[Golfingdad]

Originally Posted by dkling8

I linked the paper, so anyone could work it out if they really wanted to,

Fun, I hadn't realized that you linked the paper.  Thought it would be a fun little brain teaser, however, my math skills are failing me at this point.  It seemed logical that I could solve your question using equation (37).  The givens are: initial speed, acceleration of gravity, slope, and the friction factor (0.131).  We also know the final velocity because it will be the point when the ball stops moving.  So if I could solve that equation for 0, then my answer should be Y.  Problem is, that I have the square root of everything on one side equalling zero on the other.

Giant fail on my part.

[saevel25]

Well, you can back track your way into the static friction, using the USGA guidlines for hole placements, taking the critical slope, you can do the physics to get the static friction.

kinetic friction is a bit more complicated because it depends on the speed of the putt. Its why you have more gripping power at low speed in a car than high speed.

There's no general rule, and you can't really back your way into kinetic friction using a stimp meter unless you are able to measure the deacceleration of the golf ball. Since its moving, your forces are not zero, being f = ma, in static you can get nice equivalent equations because acceleration is zero the moment before the ball goes from static to kinetic.

[dkling8]

Originally Posted by Golfingdad

Fun, I hadn't realized that you linked the paper.  Thought it would be a fun little brain teaser, however, my math skills are failing me at this point.  It seemed logical that I could solve your question using equation (37).  The givens are: initial speed, acceleration of gravity, slope, and the friction factor (0.131).  We also know the final velocity because it will be the point when the ball stops moving.  So if I could solve that equation for 0, then my answer should be Y.  Problem is, that I have the square root of everything on one side equalling zero on the other.

Giant fail on my part.

Haha actually it's almost an equal fail on mine... I must have overlooked equation 37, so I used equations 11b and 17b to run a simulation. You're exactly right though, when I solve equation 37 directly...I get the same answer! No simulation needed.

[Golfingdad]

Originally Posted by saevel25

Well, you can back track your way into the static friction, using the USGA guidlines for hole placements, taking the critical slope, you can do the physics to get the static friction.

kinetic friction is a bit more complicated because it depends on the speed of the putt. Its why you have more gripping power at low speed in a car than high speed.

There's no general rule, and you can't really back your way into kinetic friction using a stimp meter unless you are able to measure the deacceleration of the golf ball. Since its moving, your forces are not zero, being f = ma, in static you can get nice equivalent equations because acceleration is zero the moment before the ball goes from static to kinetic.

I agree ... but it's OK because the paper that dkling linked to in the OP gives you the friction factor.  It's 0.131 :)

[MS256]

Originally Posted by Golfingdad

You say that jokingly, however, I don't think you are that far off.  When I took my Aimpoint class I learned that 4% is basically the steepest a green can be where a ball will sit.  Anything steeper than that and you have "rolloff."  I have no idea what the actual answer is, but I would suspect it is way higher than any of the answers given so far.

I also think that there is not enough info to get the answer.  Like Hendog said, gravity and friction are the only outside forces acting on it after the putter hits it.  I think we need to know the friction factor of the green to figure it out.

My wild guess ... 45 feet.

I did say it sort of tongue in cheek because I'm not sure of either the exact degree of slope or what the average speed is on those greens at Trenton.

I really wasn't kidding about any putt above the hole being next to impossible to keep on that particular green. This morning my son messed up and hit his aproach about 15 feet above the hole and (of course) putted the ball off of the green. I hit my aproach shot pin high and about 15 feet right and I did about the best I could have done and thought I had the putt made but when it missed the cup it still left me with an 8 footer back up the hill. Of all of the holes I've ever played that is the one that you have to keep the ball below the hole at all costs.

The question I am curious about is how much slope and how fast a green has to be for it to be impossible for a ball in motion to stop. At some point in slope and surface a ball in motion is going to pick up speed instead of slowing down (until it is stopped by the slope ending or the surface changing enough to stop it). I just don't know where that point is (but I would guess that 12 on the stimp and 4% slope has to be getting close to that point).

[Golfingdad]

Originally Posted by MS256

The question I am curious about is how much slope and how fast a green has to be for it to be impossible for a ball in motion to stop. At some point in slope and surface a ball in motion is going to pick up speed instead of slowing down (until it is stopped by the slope ending or the surface changing enough to stop it). I just don't know where that point is (but I would guess that 12 on the stimp and 4% slope has to be getting close to that point).

I think you are exactly right.  Like I said previously ... anything more than 4% slope gives you "rolloff" according to the Aimpoint guys, and I took that to mean exactly that ... it won't stay there, it will roll off.  Obviously that is also dependent on the firmness/stimp of the greens, and since 12 is on the super high end of the scale, that is why I think you might have been right originally.

[iacas]

Greens don't all roll off at 4%. Roll off at stimp 12 is about 6%. At 7 it's about 10%.

[MS256]

Originally Posted by iacas

Greens don't all roll off at 4%. Roll off at stimp 12 is about 6%. At 7 it's about 10%.

Just for fun I got a 25 inch long board and put a one inch block under one end and put a rug over the top of it.

I have no clue what stimp that rug would be but it seems a little slower than the average bent grass green. I think the one inch drop over 25 inches is 4% but I'm not sure that's right either. I could place the ball on the rug with no problem but if I nudged it with my hand to get it started it rolled all the way off of the board.

Doesn't tell me much about the true answer to the question in this thread but it was a fun experiment anyway. Ha ha!

P.S. If the one inch drop over a 25 inch span is actually a 4% grade the green I talked about earlier is more than that, just from looking at the board.

[Golfingdad]

Originally Posted by MS256

I think the one inch drop over 25 inches is 4% but I'm not sure that's right either.

That is exactly correct.

[colin007]

So I played a course the other day where I had about a 10 foot putt uphill. The ball stopped half an inch from the hole, and then rolled back down to my feet. Wtf??

[dkling8]

So... The answer according to the equations in the paper is 31 ft!

I went ahead and worked it out all the way just in case anyone cares to fiddle with any other numbers. Just plug in initialballspeed (ft/s), grade (positive = uphill, negative = downhill), and stimp. If you get negative values for distance, that means your ball rolls away forever...

For reference, a stimpmeter rolls a ball at 6 ft/s.

Originally Posted by MS256

The question I am curious about is how much slope and how fast a green has to be for it to be impossible for a ball in motion to stop. At some point in slope and surface a ball in motion is going to pick up speed instead of slowing down (until it is stopped by the slope ending or the surface changing enough to stop it). I just don't know where that point is (but I would guess that 12 on the stimp and 4% slope has to be getting close to that point).

So all we have to do is try to get our distance to go to infinity in the equation above. We do that by forcing the denominator to equal zero, and we get this:

Quote:

Originally Posted by iacas

Greens don't all roll off at 4%. Roll off at stimp 12 is about 6%. At 7 it's about 10%.

Erik's numbers are pretty dang close to what the equation says.

Ok, that's enough math for awhile. Back to golf...

[saevel25]

Originally Posted by colin007

So I played a course the other day where I had about a 10 foot putt uphill. The ball stopped half an inch from the hole, and then rolled back down to my feet. Wtf??

Just means that the slope were your standing isn't in the critical area (ball does not stop), while just around the hole it is. Once again, Kinetic friction is less than static, so if you get that little bit of rotation that starts the ball going backwards, gravity can cause it to roll or not. I've been on a green were i came up short twice, once the ball rolled back to were i was standing, the 2nd time it stayed up there. I mean this green was on the edge of critical.

[Golfingdad]

Originally Posted by saevel25

Just means that the slope were your standing isn't in the critical area (ball does not stop), while just around the hole it is. Once again, Kinetic friction is less than static, so if you get that little bit of rotation that starts the ball going backwards, gravity can cause it to roll or not. I've been on a green were i came up short twice, once the ball rolled back to were i was standing, the 2nd time it stayed up there. I mean this green was on the edge of critical.

It also means the greenskeeper is an a**hole.  A cup should never sit on a slope that steep.

dkling ... any chance you could take a peek at the equation (37) I mentioned previously, and see if you can solve the problem using it as well?  It's really bugging me that I couldn't get that dang thing to work!

[Rick Martin]

Originally Posted by colin007

So I played a course the other day where I had about a 10 foot putt uphill. The ball stopped half an inch from the hole, and then rolled back down to my feet. Wtf??

was there a windmill or clown's mouth involved w/ that putt?

[saevel25]

Originally Posted by Golfingdad

It also means the greenskeeper is an a**hole.  A cup should never sit on a slope that steep.

dkling ... any chance you could take a peek at the equation (37) I mentioned previously, and see if you can solve the problem using it as well?  It's really bugging me that I couldn't get that dang thing to work!

yep, that to, the best is when you hit the putt harder and it ends up behind the hole and stays there, now you have a 2 foot putt from hell and if you miss your going back to a 10 footer.

[dkling8]

Originally Posted by Golfingdad

dkling ... any chance you could take a peek at the equation (37) I mentioned previously, and see if you can solve the problem using it as well?  It's really bugging me that I couldn't get that dang thing to work!

Here ya go . I actually did get my distance equation from eq 37, I just added in some things to change angle to grade and rho to stimp.

[colin007]

It also means the greenskeeper is an a**hole.

In going with this... [quote name="Rick Martin" url="/t/66586/a-physics-puzzle-uphill-vs-downhill-putts/18#post_837940"] was there a windmill or clown's mouth involved w/ that putt? [/quote] It's a straight away par 4 at 310 yards... I guess the green is the only defense to par...