What is the best shape for a soccer goal post?

[haruspex]

It seems to me now that the wrong question is being asked. The probability of a goal depends also on the thickness of the posts.
If we say circular posts are radius R and square posts are of side 2S then the question becomes what the ratio of R to S should be to make them equal.
To evaluate that, we need to consider the range of displacement of the trajectory from the centre line (the to the centre of the post) to be $\pm D$ where $D>\max\{R,S\sqrt 2\}$.

Continuing that line ...
The centre of a post is at the origin. If square, the post has corners at $(\pm S, \pm S)$. If circular , it is radius R.
The other post is at $(-\infty,0)$, to avoid worrying about rebounds off both posts.
The ball approaches along a line making angle $\theta$ to the positive x axis. Its line would pass distance d to the right of the origin.
The angle is uniformly distributed in $(0,\pi)$. d is uniformly distributed in (-D,D) where $D>R$ and $D>S\sqrt 2$.
1. Square post.
A goal is scored if and only if the ball passes to the right of $(-D,D)$. That is, if $d>S\sqrt 2\cos(\frac{\pi}4-\theta)$. The probability of a goal is therefore $\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$.
2. Round post
Certainly it is a goal if d>R.
Suppose the ball strikes the post at $(R,\theta+\alpha)$ in polar coordinates. So $d=R\sin(\alpha)$.
It will rebound at angle $\theta+2\alpha$ to the x axis. A goal will result if that exceeds $\pi$, i.e. $d>R\cos(\frac \theta 2)$.
The probability of a goal is therefore $\frac 1{2D\pi}\int_0^\pi(D-R\cos(\frac\theta 2)).d\theta$.

For what relative R and S are those equal? I get R=S(!)
If we consider wide angle shots rare, we could take the integration range as $(\pi/4,3\pi/4)$, say, giving $S=2\sin(\pi/8)R$.

 

[hutchphd]

To evaluate that, we need to consider the range of displacement of the trajectory from the centre line (the to the centre of the post) to be ±D where D>max{R,S2}.

I agree but if you track the position of the center of the ball you need to include the finite radius of the ball $\equiv R_{ball}$. In each case the effective shape is obtained by "moving" the ball over the post and tracking the center. For the circular post the effective radius just becomes $R+R_{ball}$. For the square post it will be more complicated: a rounded corner (radius $R_{ball}$) square post.

Edit: Sorry I missed your earlier posts (#24et seq?) It seems a very bad approximation right at the heart of the issue...

 

[haruspex]

I agree but if you track the position of the center of the ball you need to include the finite radius of the ball $\equiv R_{ball}$. In each case the effective shape is obtained by "moving" the ball over the post and tracking the center. For the circular post the effective radius just becomes $R+R_{ball}$. For the square post it will be more complicated: a rounded corner (radius $R_{ball}$) square post.

Edit: Sorry I missed your earlier posts (#24et seq?) It seems a very bad approximation right at the heart of the issue...

Not sure what you mean by the last sentence.
As I noted in post #28, although the ball is rather fatter than the post, for the purposes of comparing different post shapes it is probably ok to treat the ball as a point.

 

[kshitij]

The probability of a goal depends also on the thickness of the posts.

Yes, but the thickness of posts are fixed, I wanted to know the probabilities for only the cases when $L=2R$ where $L$ is the side length of square post and $R$ is the radius of circular post.

To evaluate that, we need to consider the range of displacement of the trajectory from the centre line (the to the centre of the post) to be ±D where D>max{R,S2}.

As I understand if $D>\max\{R,S\sqrt 2\}$ then why would the ball even hit the posts? Yes it would go in but I wanted to know the probability of a shot going in after it hits the post.

A goal is scored if and only if the ball passes to the right of (−D,D). That is, if d>S2cos⁡(π4−θ).

I don't understand why you multiplied $S\sqrt{2}$ with $\cos(\frac{\pi}{4}-\theta)$? Maybe a figure would help? If the ball passes to the right of $(-D,D)$ which is marked as point $A$ in the figure and I think that you assumed the other post to be at $(-\infty,0)$ then how will the ball ever cross the line?

For what relative R and S are those equal? I get R=S(!)

I didn't understand how you got this but are you saying that both posts have equal probabilities? If yes then this was exactly what I wanted! but I am a bit surprised that the probabilities are equal

 

[kshitij]

That is actually a great idea! Instead of having to consider both the direction of the ball and the point of impact as varying factors like before, it is now possible to combine the two since the starting position will always be the same. In that case, am I correct in saying that the ball will end up in the goal when it reflects off the circular post anywhere to the right of the red ray in the diagram below? An example of this is the green ray. A ball (still assuming it is a point mass) traveling along the red ray would simply reflect in the same direction it came from as the ray is parallel to the normal. Any collision left of the red ray would cause the ball to reflect away from the goal line.
View attachment 287605

The red ray is definitely not the limiting case! the limiting case should be when the green line is tangent to the other post for no rebound by the second post and for only one rebound by the second post the green line should end up in the top left quarter of the second post.

 

[kshitij]

I am again saying that the whole point of the post was to find which of the two posts are better under similar situations (which may not be accurate representation of reality) because I wondered why did they replace the square ones with circular or elliptical ones? does really (under some specific conditions) the round post leads to more shots going in that the square one? or maybe it was just because they thought that the square ones are more dangerous for players?

 

[haruspex]

I wanted to know the probabilities for only the cases when L=2R where L is the side length of square post and R is the radius of circular post.

if D>max{R,S2} then why would the ball even hit the posts?

To make a proper comparison, you must not only consider the cases where the ball hits the post. There will be trajectories that would hit the one post shape but not the other. So we must set D large enough to encompass both posts.

If the ball passes to the right

That's to the right from the ball's perspective, so to the left in the diagram.

I didn't understand how you got this but are you saying that both posts have equal probabilities? If yes then this was exactly what I wanted! but I am a bit surprised that the probabilities are equal

For the distribution of trajectories I chose and equal widths of posts (L=2S=2R), yes - I was surprised too. But as noted, if you change the distribution the answer changes.

 

[kshitij]

To make a proper comparison, you must not only consider the cases where the ball hits the post. There will be trajectories that would hit the one post shape but not the other. So we must set D large enough to encompass both posts.

That's to the right from the ball's perspective, so to the left in the diagram.

For the distribution of trajectories I chose and equal widths of posts (L=2S=2R), yes - I was surprised too. But as noted, if you change the distribution the answer changes.

And how did you get $d>S\sqrt 2\cos(\frac{\pi}4-\theta)$?
Also I hope that this diagram is now correct,

 

[haruspex]

And how did you get $d>S\sqrt 2\cos(\frac{\pi}4-\theta)$?
Also I hope that this diagram is now correct,
View attachment 287708

Draw the trajectory passing through A=(-D,D).
If its nearest point to the origin, O, is B, what is angle BOA?

 

[kshitij]

Draw the trajectory passing through A=(-D,D).
If its nearest point to the origin, O, is B, what is angle BOA?

Yes I get OB=$S\sqrt2\cos(\frac{\pi}{4}-\theta)$, next step?

 

[kshitij]

Yes I get OB=$S\sqrt2\cos(\frac{\pi}{4}-\theta)$, next step?

Are we just considering the cases where $d \gt S\sqrt2\cos(\frac{\pi}{4}-\theta)$ but this will not give the cases when the ball goes in after deflection from the post, this was not what I was asking

 

[kshitij]

Are we just considering the cases where $d \gt S\sqrt2\cos(\frac{\pi}{4}-\theta)$ but this will not give the cases when the ball goes in after deflection from the post, this was not what I was asking

I thought you calculated the probability of a shot going in after hitting the post, this was what I was originally asking, "of all the shots that hit the post which of the two posts (square or round) has a greater probability for those shots going in"

 

[haruspex]

this will not give the cases when the ball goes in after deflection from the post

It will. Draw the diagram for $\theta>\pi/2$ and d just sufficient to pass to the right of A.

 

[kshitij]

It will. Draw the diagram for $\theta>\pi/2$ and d just sufficient to pass to the right of A.

The blue circle is of radius $D$ and if the path of the ball is tangent to that circle (because $d$ is the perpendicular distance of the path from origin and this distance is equal to $D$ in the limiting case) it will not hit the post

 

[kshitij]

View attachment 287709
The blue circle is of radius $D$ and if the path of the ball is tangent to that circle (because $d$ is the perpendicular distance of the path from origin and this distance is equal to $D$ in the limiting case) it will not hit the post

I realized you said that the limiting case is $d=D\cos(\frac{\pi}{4}-\theta)$ not $d=D$ but now I cannot understand why?
What I do understand is that $\left|\frac{\pi}{4}-\theta\right|$ is the angle between the vectors $\vec {OA}$ and $\vec {OB}$ where $\vec {OB}$ is the perpendicular distance from origin $O$ to the path of the ball and $D\cos(\frac{\pi}{4}-\theta)$ is the projection of $\vec {OA}$ on $\vec {OB}$

 

[kshitij]

I think now I somewhat understand what you are trying to do,
Here in this link the blue line gives all possible paths of the ball (move point $B$ to get different cases) and the red line is the limiting case for a given blue path.

As long as we stay to the left of the red line, we can score a goal thus these are our favourable cases.

 

[kshitij]

A goal is scored if and only if the ball passes to the right of (−D,D). That is, if d>S2cos⁡(π4−θ).

Now finally I understand what you said here!

But now how did you get this?

The probability of a goal is therefore $\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$

And also, please assume $D=S\sqrt 2=R\sqrt 2$ in further calculations because the diameter of the round post is equal to side length of square post according to the laws of the game

 

[kshitij]

According to what I understood reading post #91, I similarly created a graphic for the circular post as well!

 

[haruspex]

But now how did you get this?

And also, please assume $D=S\sqrt 2=R\sqrt 2$ in further calculations because the diameter of the round post is equal to side length of square post according to the laws of the game

There's no need to plug in S=R at the start. I prefer to keep things general initially and apply specifics at the end. There are advantages.
The value of D does not matter as long as it encompasses both posts.

We have established a goal is scored if $d>S\sqrt 2\cos(\frac{\pi}4-\theta)$.
d is uniformly distributed over the range (-D,D), a line of length 2D. Of that line, length $D-S\sqrt 2\cos(\frac{\pi}4-\theta)$ satisfies that condition. So the probability of a goal for a given angle is $\frac 1{2D}(D-S\sqrt 2\cos(\frac{\pi}4-\theta))$.
To get the overall probability we need to integrate over the range of the angle and divide by that range, pi.
The probability of a goal is therefore $\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$.

(Love the animations!)

 

[kshitij]

There's no need to plug in S=R at the start. I prefer to keep things general initially and apply specifics at the end. There are advantages.
The value of D does not matter as long as it encompasses both posts.

We have established a goal is scored if $d>S\sqrt 2\cos(\frac{\pi}4-\theta)$.
d is uniformly distributed over the range (-D,D), a line of length 2D. Of that line, length $D-S\sqrt 2\cos(\frac{\pi}4-\theta)$ satisfies that condition. So the probability of a goal for a given angle is $\frac 1{2D}(D-S\sqrt 2\cos(\frac{\pi}4-\theta))$.
To get the overall probability we need to integrate over the range of the angle and divide by that range, pi.
The probability of a goal is therefore $\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$.

(Love the animations!)

That is so clever!

First we calculated favourable cases for line whose slope is fixed and then we fixed the distance of that line from origin then calculated favourable cases for all possible values of slope!

But if we consider the maximum value of $D$ to be $\infty$ (i.e. the incident path can be anywhere in the 2D plane) then logically we should get the probability in case of square post as 33.33% as all three faces have equal chance of striking and striking only one of these three face leads to goal, but using your formula I get 15.91%, why?

 

[kshitij]

According to what I understood reading post #91, I similarly created a graphic for the circular post as well!

I updated the graphic for the circular post case, now you don't have to move two points separately to get all possible paths you can get it by moving only point B in the link below,
https://www.geogebra.org/calculator/khj7d6hk

 

[kshitij]

I evaluated your integral for $\theta \in (0,\pi)$ and got

$P(square)=\dfrac{D-2S}{2D\pi}$

$P(round)=\dfrac{D-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2\pi}=0.1591%$ regardless of what the dimensions of the square and round posts are as long as they are finite, i.e., we can even have $R=1000S$ and still get the same probability for both posts as 15.91% which seems totally illogical to me, am I missing something?

 

[kshitij]

I evaluated your integral for $\theta \in (0,\pi)$ and got

$P(square)=\dfrac{D-2S}{2D\pi}$

$P(round)=\dfrac{D-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2\pi}=15.91%$ regardless of what the dimensions of the square and round posts are as long as they are finite, i.e., we can even have $R=1000S$ and still get the same probability for both posts as 15.91% which seems totally illogical to me, am I missing something?

And if we take $D=7.32m$ which is equal to the distance between the two posts and $R=S=6cm$ according to the laws of the game, we get,

$P(square)=P(round)=0.1516%$

Which still doesn't make any sense to me, I mean Probability for the square ones should be close to 33% and for the round one should be atleast 25%!

 

[kshitij]

I evaluated your integral for $\theta \in (0,\pi)$ and got

$P(square)=\dfrac{D-2S}{2D\pi}$

$P(round)=\dfrac{D-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2\pi}=0.1591%$ regardless of what the dimensions of the square and round posts are as long as they are finite, i.e., we can even have $R=1000S$ and still get the same probability for both posts as 15.91% which seems totally illogical to me, am I missing something?

And if we take $D=7.32m$ which is equal to the distance between the two posts and $R=S=6cm$ according to the laws of the game, we get,

$P(square)=P(round)=0.1516%$

Which still doesn't make any sense to me, I mean Probability for the square ones should be close to 33% and for the round one should be atleast 25%!

I realized that 33% and 25% where the probabilities of a shot going in after hitting the post whereas in your method the ball doesn't need to hit the post, but now this confuses me even more! that means 15% is the probability for any shot (no matter where it is hit in the 2D plane) to go in!

So now suddenly 15% changes from being a little less than expected to a lot more than expected

 

[kshitij]

Well I'm too confused so let me state what I think I understood then you correct me where I'm wrong.

• $d$ is the distance of a given path of the ball from origin and origin is the center of one of the posts (right post from our point of view and left from the ball's) and the maximum value of $d$ is equal to the maximum value of $D$
• $D$ is all possible values of the distance of the path of the ball from origin starting from the edge corner of one post to the edge corner of other (the other post was at $(-\infty,0)$)
• $\theta$ is the inclination of the path with x-axis or the goal line which ranges from $(0,\pi)$
• A goal is scored if the ball passes to the left (looking at the ball from our point of view its left otherwise w.r.t ball it should be right) of the minimum value of $D$
• for scoring a goal we have, $d>S\sqrt 2\cos(\frac{\pi}4-\theta)$ in case of square post and in case of round one we get $d>R\cos(\frac \theta 2)$ ($R$ is the radius of round post and $2S$ is the side length of square one)

And the points below are the ones that I concluded from the points above,

• the value (here I am only talking about the magnitude) of $D$ can range from $(R,\infty)$ or $(S\sqrt{2},\infty)$ if we consider the other post to be at $\infty$, and if we considered a more realistic system then value of then maximum value of $D$ must be equal to the distance of the left (again its left w.r.t us and right w.r.t ball) touchline from the edge of the right post
• Since a line is bidirectional and not unidirectional like the path of the ball, we can consider goal posts at both ends of the pitch so along path of the ball the ball can travel both ways backwards or forwards and scoring a goal at both ends of the pitch is equivalent for us

 

[haruspex]

D is all possible values of the distance of the path of the ball from origin starting from the edge corner of one post to the edge corner of other (the other post was

No, D is a constant such that if d>D then it is sure to be a goal without touching the post. Likewise, if d<-D it surely will not be a goal. D's value won't matter as long as it is large enough.
We only need to compare the posts for what happens when -D<d<D.

 

[kshitij]

No, D is a constant such that if d>D then it is sure to be a goal without touching the post. Likewise, if d<-D it surely will not be a goal. D's value won't matter as long as it is large enough.
We only need to compare the posts for what happens when -D<d<D.

So $D$ is a large constant and $d$ belongs to $(S\sqrt 2\cos(\frac{\pi}4-\theta),D)$ for square post and for round post it lies between $(R\cos(\frac \theta 2),D)$

So still, when $D \to \infty$ it should surely give us the probability when the path of the ball can lie anywhere in the 2D plane because now $d$ belongs to $(S\sqrt 2\cos(\frac{\pi}4-\theta),\infty)$, so still, 15% is the probability of any shot in the 2D plane going in, right?

 

[haruspex]

So $D$ is a large constant and $d$ belongs to $(S\sqrt 2\cos(\frac{\pi}4-\theta),D)$ for square post and for round post it lies between $(R\cos(\frac \theta 2),D)$

So still, when $D \to \infty$ it should surely give us the probability when the path of the ball can lie anywhere in the 2D plane because now $d$ belongs to $(S\sqrt 2\cos(\frac{\pi}4-\theta),\infty)$, so still, 15% is the probability of any shot in the 2D plane going in, right?

I don't know how you get 15%.
If D goes to infinity and d is still uniformly distributed from somewhere up to that then the probability of going in tends to 1.

 

[kshitij]

I don't know how you get 15%.

I evaluated your integral for $\theta \in (0,\pi)$ and got

$P(square)=\dfrac{D-2S}{2D\pi}$

$P(round)=\dfrac{D-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2\pi}=0.1591%$

 

[haruspex]

$\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$
$=\frac 1{2D\pi} [D\theta+S\sqrt 2\sin(\frac{\pi}4-\theta)]_0^\pi$
$=\frac 1{2D\pi} (D\pi+S\sqrt 2\sin(\frac{\pi}4-\pi)-S\sqrt 2\sin(\frac{\pi}4))$