Can a player be snookered on a touching ball?

[davis_greatest]

This is a question about whether a player can be snookered on a touching ball, and hence whether he may be awarded a free ball even if the cue-ball is touching an object ball. Perhaps one for the referees?

As background, if a player (or, rather, the cue-ball) is snookered after a foul, a free ball is awarded. In defining "snookered", the rules state that if one or more balls on can be struck at both extreme edges free of obstruction by any ball not on, the cue-ball is not snookered.

So, if the white is touching the ball on, where are the extreme edges of the ball on?

Usually, if red (say) is the ball on, we would imagine three balls - (A) a ball touching its extreme left edge, (B) a ball touching its extreme right edge, and (C) the red itself - all to lie in a straight line. This, of course, is not really the case (unless the cue-ball is at a distance of infinity). The balls (A), (B) and (C) really form part of an arc around a circle with cue-ball at the circle's centre. They do not lie in a straight line.

So, by this logic, if the cue-ball is touching the ball on, then the extreme edges (A) and (B) lie in exactly the same place as each other - namely where the cue-ball is already, the same as a full ball contact.

This would imply that no free ball can be awarded when white is touching the ball on. Is this correct?

Picture, for example, white, pink and black all touching each other in a triangular formation, where these are the only balls remaining on the table after a foul. Can a free ball be awarded?

Now imagine the same three balls, still in a triangular formation, but 1mm apart, and answer the same question...

[The Statman]

No.

"Snookered" means not being able to hit the ball on (on both extrene edges) due to the intervention of ball not on.

As there is no intervening ball (and, indeed, since you are deemed already to have hit the ball on by playing away from it), you cannot be snookered.

I suppose if there was a ball not on, also touching the cue-ball, you might have a case, but even then it doesn't prevent you from legally 'hitting' the ball on, it only prevents you from doing so without making a push shot (on the ball not on) – which is not covered in the definition of a snooker.

[The Statman]

Picture, for example, white, pink and black all touching each other in a triangular formation, where these are the only balls remaining on the table after a foul. Can a free ball be awarded?

Now imagine the same three balls, still in a triangular formation, but 1mm apart, and answer the same question...

I see what you are getting at, but the closer they are the less likely it is to be a free ball; and the farther apart, the less likely also. It is in the middle ground, I would guess around half a ball's width apart, that the free ball is a possibility.

So, by this logic, if the cue-ball is touching the ball on, then the extreme edges (A) and (B) lie in exactly the same place as each other - namely where the cue-ball is already, the same as a full ball contact.

Yes. I guess you would visualise the extreme edges as playing in one direction at 90° to the ball and 90° to the other – the equivalent to if they are very close to touching but not quite.

It has the same effect as what you describe.

[davis_greatest]

Edit - the below is a reply to The Statman's first post. I had not yet seen the second!

Thank you. This is as I would expect, that one cannot be snookered.

I suppose if there was a ball not on, also touching the cue-ball, you might have a case, but even then it doesn't prevent you from legally 'hitting' the ball on, it only prevents you from doing so without making a push shot (on the ball not on) – which is not covered in the definition of a snooker.

Re the part I have quoted above, yes, I am thinking specifically of the case where the ball not on is also touching the ball on. (E.g. the white, pink and black, all touching each other - might be easier to see if I drew a picture!)

I think the push shot is slightly a red herring here - if we imagine the white, pink and black now 1 millimetre apart (so almost touching other, each pair equidistant), can the free ball be awarded now - i.e. can the black be deemed to be obstructing the extreme edge of the pink? My view would be "no".

[The Statman]

Well, first, let's look at how we judge a free ball in normal circumstances, unsing the diagram below.



First, we draw four lines (Line A) corresponding to the edges of the cue-ball's path to one extreme edge and then the other. Then, two circles (Circle B) which mark the position of the cue-ball when it reaches the extreme edges of the ball.

A free ball would be awarded if any other ball should encroach on the area which I have made a darker shade of green – the two circles plus the area that is between each circle and the real cue-ball, between the parallel lines.

We can then see that Black 1 would cause a free ball, Black 2 is clear and wouldn't; Black 3 would create a free ball even though it is behind the pink.

=======

Now, let's move the balls closer together.



We see that the Lines and the Circles are still there, but the relevant part – the dark green section – the table have reduced in size dramatically, so that there is only a minuscule distance between the outline Circle Bs and the real cue-ball.

Nevertheless, it is still possible to cause a free ball – as Black 2 shows, despite being much futher away from the pink than Black 1 is.

=============

Bringing this to its natural conclusion, the closer the balls get to each other, the smaller and smaller the 'relevant area' becomes and the less likely the black is to encroach upon it and cause a free ball.

When the balls become touching, then as you originally concluded, the 'relevant area' becomes nothing because the two Circle Bs, which are the outer limits of the relevant area, are both in exactly the same position as the real cue-ball.

[The Statman]

I am re-thinking my answer here, Mr Davis Greatest. I shall not say why just now, but I will reply later when I have thought it through fully.

[moglet]

I can't wait

It will certainly be interesting.

[moglet]

davis_greatest, in the case you cite the three balls all a millimetre apart does allow for a free ball. The three balls touching is a limiting case and a discontinuity if you plot the relevant criteria, as you separate the balls by an equal distance. The other discontinuity occurs when the balls are one ball distant from each other.

The following is an intermediate position where it is clear the black obstructs the path to the relevant edge of the pink, as it does in all cases between all touching and all balls equidistant by one ball diameter.

[moglet]

And just to clarify, as we approach the limit as the balls are separated the black still obstructs (the closer limit is difficult to illustrate but the same rule applies):

[moglet]

Unfortunately I can't edit a post an add another attachment. However, this may illustrate what happens as we approach the closer limit.

[davis_greatest]

Thank you The Statman and moglet.

In order to answer the original question, what we could do (if anyone feels so inclined) is plot a graph of distance between the three balls (on the x axis) against distance between the tangent to the pink and the black ball (on the y axis).

For the portion of the graph where the y-coordinate is less than one ball's width, a free ball would be possible.

I'll let someone else try this first! If no one does, I'll perhaps have a go when I get a few minutes.

[The Statman]

Here is the reason that I am reconsidering the thread-head question "Can you be snookered on a touching ball?"

We all know that, to judge a free ball, you must visualise the two routes of the cue-ball – one to the extreme left edge and one to the extreme right. These end at the point at which that route is touching the target ball.

Now, if, when the cue-ball reaches the exact point where it contacts the extreme edge of the ball, it would be also touching another ball – then that must be a free ball because the Rules state you must hit the ball on first: first simultaneously with another ball is not allowable.

In a way, then, what we are discussing with the free-ball-touching-ball situation is the same, except that the cue-ball is already in that end position of the white discussed above.

This all sounds a bit muddled and when I get home tonight I will try to draw a picture illustrating it.

[davis_greatest]

Thank you again, The Statman.

One way to visualise this is to imagine white, pink and black equidistant (in triangular formation), with pink above white, and black to the left of the midpoint between white and pink.

Imagine that the minimum distance between any two balls is x balls' width. Then, at the moment that the white touches the extreme left edge of the pink, let d(x) be the number of balls' width between the centre of the white and the centre of the black.

Then a free ball will be possible if and only if d(x)≤1 (unless x=0 - see post 18 below) - that is, there is not more than one ball's width between the centre of the white at the moment the white touches the left-hand edge of the pink and the centre of the black.

With a bit of geometry, we can arrive at a formula for d(x). In fact,

d(x) = √[(1+x)^2 - √(3x(x+2))]

For x=0 (touching ball) and x=1 (the three balls being one ball's width apart), we have d(0) = d(1) = 1, which means that the white would strike the black at the same instant as it strikes the pink. In the limiting case of x=1, it would be a free ball (this is because simultaneous contact with pink and black would be a foul) but not in the case x=0 (touching ball) - see post 18 below. These are the only two value of x for which d(x)=1.

For 0
For x>1 (i.e. the three balls begin more than one ball's width apart), we have d(x)>1, so it would not be a free ball.

Thus, if the balls are 1 millimetre apart, a free ball would be awarded (unless we are playing snooker with ball bearings of 1mm diameter).

edited text in blue above

[davis_greatest]

davis_greatest, in the case you cite the three balls all a millimetre apart does allow for a free ball. The three balls touching is a limiting case and a discontinuity if you plot the relevant criteria, as you separate the balls by an equal distance. The other discontinuity occurs when the balls are one ball distant from each other.

The following is an intermediate position where it is clear the black obstructs the path to the relevant edge of the pink, as it does in all cases between all touching and all balls equidistant by one ball diameter.

[ATTACH]1178[/ATTACH]

Thank you. I agree that the special cases are touching ball and when the three balls are one ball's width apart - these separate the cases of "free ball" and "not free ball". Is that what you mean by discontinuity?

Apart from that, there is no discontinuity - i.e. if you plot the graph of d(x) against x (defined in the previous post), the curve is continuous.

[moglet]

d_g, I had a little time this afternoon to plot something - I did it a slightly different way but it comes to the same thing. The attachment should be self expanatory.