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There is a perception (widely held) in snooker that a straight shot is more difficult than an angled shot. There are many forum discussion about this, and the reasons are usually accepted to be psychological.

But I was wondering, is there a mathematical reason for it. Is the margin of error greater when the shot being taken is at an angle?

Example - if the white ball was 1 degree off target on a straight shot, and one degree off target on an angled shot (same target for both shots), and each shot hit at the same speed, would the red ball travel off line to the same extent?

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  • $\begingroup$ Interesting. I guess from experience one could "pinch the pocket" for angled shots. $\endgroup$ – Chee Han Apr 25 '17 at 9:50
  • $\begingroup$ I should have been clearer. The target (red) ball would be in the same position - in line with the pocket (so that would not be a changing variable), the only difference would be the white ball position. In one shot it's straight in line with the red and pocket, in the other it would be at an angle $\endgroup$ – Drenai Apr 25 '17 at 9:54
  • $\begingroup$ Exactly what I had in mind, and I still stand by my comment (from experience). I hope someone has a nice mathematical answer to this question ! $\endgroup$ – Chee Han Apr 25 '17 at 9:57
  • $\begingroup$ Could it simply be that inaccurate slower balls are more likely to drop into the pocket than inaccurate faster balls? At a larger angle, the cue ball imparts less momentum to the target ball. $\endgroup$ – Jaap Scherphuis Apr 25 '17 at 13:43
  • $\begingroup$ I updated the question - with Example of what I mean $\endgroup$ – Drenai Apr 25 '17 at 14:16
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Let the ball be radius $r$ and distance between balls $d$. Let $\theta$ be angle white is struck from line between centre of balls and $\phi$ direction red moves from line between red ball and white before struck. Then get $2r sin(\theta + \phi) = d sin(\theta)$. Call $a=\frac{d}{2r}$. This gives $\frac{d\phi}{d\theta}=-1+ \frac{a cos(\theta)}{\sqrt{1-a^2 sin^2(\theta)}}$. This represents the ratio of the error in the direction of red to the error in hitting the white. It increases monotonically from straight shot to a fine cut. If $d \approx r$ then this will not hold true as the target spot on the red is further away for a cut compared to a straight shot.

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  • $\begingroup$ What does this mean in a real world scenario - e.g. I don't understand the answer:-) $\endgroup$ – Drenai Apr 26 '17 at 15:05
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    $\begingroup$ For a given amount of error in the direction of the cue ball the error in the direction of the red will be bigger the greater the cut. So a straight shot ($\theta=0$) is easiest. It turns out that it is only the fine cuts where the increase in error is significant. See graph here wolframalpha.com/input/… $\endgroup$ – user121049 Apr 26 '17 at 19:15
  • $\begingroup$ PS for those who argue that straight shots seem hard, I agree :-) $\endgroup$ – user121049 Apr 26 '17 at 19:17
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This may not answer your question but still worth a read.

For straight shots, one needs to hit very close to Center of mass of the ball.Agreed? Now consider a standard sphere. I will take projected area as $2\pi r^2$ ($r$ is radius)

To hit at the center of mass. Let's consider that hitting anywhere within $\frac{r}{2}$ of the projected area. Making it's favorable area will be $\frac{\pi r^2}{4}$ which happens to be $0.125$ of the total area. What I want to conclude by this is that the probability of a direct hit near center of mass is a tougher job Iff the probability of hitting anywhere on the ball is uniform.

One more Physics related stuff.Hitting very close to the ball also does not guarantee you a head on collision. We simply can't ignore the Angular Impulse factor and friction offered by the table. That angular impulse will surely provide it a rotational motion along a axis which ain't passing through center of mass. This will surely affect it's trajectory (I can't calculate the error margin though).

Your question was interesting. +1 for it.

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  • $\begingroup$ I was only thinking about the angles - I'm sure there is less margin for error the greater the angle of the shot - but the force applied is much less as there is a lot less mass behind the impact. This is more complicated than I thought. $\endgroup$ – Drenai Apr 25 '17 at 11:03
  • $\begingroup$ I think You need to see that the linear momentum is conserved along the Common Normal of impact. Moreover. You can't ignore the imparted angular momentum upon the impact. After all no collision has coefficient of restitution=1. It ain't wonderland . I think that angular momentum imparted is the criminal you are looking for error in trajectory. @Brian $\endgroup$ – The Dead Legend Apr 25 '17 at 11:07
  • $\begingroup$ I'm not trained in any of those areas - math via computer science only $\endgroup$ – Drenai Apr 25 '17 at 13:22
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A practical reason: Experience. Naturally, angled shots arise more frequently than straight shots, so we have more practice shooting the former.

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  • $\begingroup$ That's the sort of answer I was getting in snooker forums.... so I came here for a mathematical approach $\endgroup$ – Drenai Apr 25 '17 at 14:14

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