I would like to thank you for your patient and insightful responses throughout this thread and the other one :smile:
As mentioned earlier this was not any homework question of significance but rather something I casually made up on my own and yet you were committed to help me throughout this...
u/gegenpressing91 on reddit does some beautiful illustrations like the one below,
Here we clearly see that for wider shots, he did aim to the opposite post so your assumption is right
What I wanted to do was to include all the paths of the ball that we allowed in our probability calculation, i.e., all possible straight lines whose perpendicular distance from origin is ##\leq D##
But now I see that all possible paths that intersect the semi-circle of radius ##D## would not...
the distance of the point from origin will also be the distance of the line from origin as the line is perpendicular to the line joining origin and that point so you don't need to look at ##d## again since you already fixed that while picking the point
Yes, but isn't it correct that for a given point in that semicircular area, there is only one path possible that passes through that point and is perpendicular to the origin
I just want to visualise the probability, saying that "this is the probability when the path of the ball is such that ##d## and ##\theta## are uniformly distributed over ##(-D,D)## and ##(0,\pi)## respectively" doesn't give me any hint about what is happening physically. Saying that "this is the...
I think that you assumed that a shot is hit from a point on the arc of a semicircle of radius ##D## but I meant that a shot hit from any point in the area of the semicircle of radius ##D##, because now we see that ##\theta## is uniformly distributed in ##(0,\pi)## and ##d## (displacement form...
I didn't get how you did this?
Also if that is not how we can think of this probability then what is the physical meaning of the probability that we calculated?
Okay, I think now I understand (but still it doesn't feel right)
Anyway, let me ask one final thing, for a given post round or square, ##\dfrac{D\pi-2S}{2D\pi}=\dfrac{D\pi-2R}{2D\pi}## is the probability of a shot that is hit within the semicircular area of radius ##D## going in goal assuming...
Oh yes, I see it now,
If we did the first fold horizontally (and the horizontal fold should be in opposite direction to the unfolding otherwise we will end up with what we started with) instead of vertically then the circles won't be linked and thus that is not what is happening in inverting a...
I am thinking with a different model, consider all possible shots that hit the post, now the probability that a shot hits the sector AB is ##\frac14##, right? the whole post is divided into 4 quarters and hitting anyone of these quarter is equally likely? correct?
Now back to your model, we...
Well first I learned that making a cylinder is relatively easier but making a toroid is way too difficult, the paper doesn't want to change its surface's curvature and IIRC there is a theorem like that by Gauss (??).
Anyway, to the conclusion, the blue circle are red circle are not linked like...
Also, I have another doubt,
Say if I set the value of ##D=R## in the round post, then it should mean that now I'm considering only the shots that hit the round post, so now I get the probability of a shot hitting the round post going in as
##P(round)=\dfrac{D\pi-2R}{2D\pi}=0.18##
But that...
I need a numerical value of ##D## because that will give us a number, i.e., if ##D=R=6cm## then we get 18% chance as calculated in post #124 instead of getting just that the probability in square and round post are equal, we get that probability in square and round post are equal which is equal...
As I said before that now it makes sense why the circles won't be linked after we stitch it back the other way because we are literally cutting both the circle such that they changed from a circle to a line, but the cutting isn't allowed
I was thinking about this a lot and even though I don't know how it will happen but I think that it should happen, i.e., the red and blue circles shouldn't be linked after inverting the tube inside out.
This is because, the red circle which was initially outside the tube, after flipping has to...
Wait, can we cut it such that we get a rectangle? doesn't that break any rules? is a toroid with single hole isomorphic to rectangle?
I agree that with can cut it in both ways that you mentioned, i.e., cutting it such that we get a short fat cylinder and cutting it such that we get a long...
But our selection depends on what value we choose for ##D## (see post #124) and if we chose ##D## to be large enough then it would also include the cases where the ball didn't hit the post.
Isn't setting the same constant value of ##D## for the probability calculation of both posts a like to...
So that means that we cannot increase the length of the hole such that the tube is converted to a strip?
Because by this method it does completely makes sense why the red circle passed through the blue one (because we literally cut and stitched the red circle in the process)
But yes if that is...
Again I can't understand what you mean here, how am I changing the distribution of angles? And why can't this be used to compare the two post shapes?
Let me state the whole method as I understand it.
Now, I need diagrams and figures to understand pretty much everything so I will now talk w.r.t...
*correction,
##P(square)=\dfrac{D\pi-2S}{2D\pi}##
##P(round)=\dfrac{D\pi-2R}{2D\pi}##
Now for ##D \to \infty##, we get ##P(square)=P(round)=\dfrac{1}{2}=0.5##
Okay so here is my second attempt
I use the same process as earlier that is I increase the size of the hole such that the tube is now a strip, and the hole is increased such that blue circle remains intact but the red circle is cut into a line, now if we flip this strip and stitch it back...
That means that you painted a circle along the outer circumference (because its the long way round) of the tube but on the inside surface of the tube
And this means that you painted the circle on the inside circumference of the tube but this time on the outside surface
What?? How are they linked like a chain? I thought that those two circles are like rings one surrounded by the other and there is a layer of rubber in between them separating each other
I don't have a proof but I have heard that changing the size of a hole in any shape doesn't change it homeomorphically (idk if that's the word)
that is why a ring is homomorphic to a cylinder or a disc is homomorphic to a cone or a cup
By those two circles being linked do you mean that they are concentric??
If so, then to answer your problem I think maybe that first we extend that small hole we made so that we can slit the tube (torus) along its length (along its circumference) such that now instead of a tube we are left with...
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...
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...
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...
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...
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...
I don't understand what you said there about understanding Pappus theorem, but I just wanted to find the area/volume enclosed by a hollow hemisphere, solid hemisphere, hollow cone and a solid cone on rotating these about their base, i.e., how did you get
Volume of 4D ball = ##\frac 12\pi^2r^4##...
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
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...
Now finally I understand what you said here!
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