Recent content by kshitij

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

That is, add the assumption that ##d## is uniformly distributed in ##(-D,D)## then it should be fine 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

I was trying to define the equation of the line in parametric form with a given slope and given distance (perpendicular distance) from origin

For every point you pick we can define a unique line which is perpendicular to the line joining that point and origin

Sorry I meant, perpendicular to the line joining that point and origin

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...

But only ##d## and ##\theta## are variable, what do you mean by 3 degrees of freedom?

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...