Pass Goes Behind the Backboard, Through the Supports

[Camron Rust]

Originally Posted by CoachP

Technically, I can stand in the exact corner of the playing court and shoot a three. It does not count then? It HAS to go over the corner, from the back, to fall in.........

Not possible....geometry

(from an post by me in March regarding a shot that went over the top)

Front tip of the rim is 24" from the backboard. Backboard is 6ft wide and 4ft from the baseline, lane is 12ft wide. The point on the baseline that leads over the corner to the very front tip of the rim is 3' outside the lane...and that is just to the front tip....an unmakeable point without crossing clearly over the top. The path to the center of the basket directly over the corner and from the baseline starts 4.5' outside the lane....and that spot is just barely makeable on a lucky day.

So, the line from the corner (25' from the center of the court on the baseline) has plenty of room to clear the edge of the backboard. If it does go over the top, then there is no way it is going to be in line to go in....it's not going to be on a makable path. The path to the basket for any baseline shot (not near the baseline...direclty over it) beyond 10.5 feet from the center of the court is not over the backboard.

[BillyMac]

The front, top, sides, and bottom of the backboard are all in play. The ball cannot pass over a rectangular backboard from either direction. The back of a backboard is out of bounds as well as the supporting structures.

[CoachP]

Originally Posted by Camron Rust

Not possible....geometry

hmmm...now I'm gonna have to run to school tonight and take my ball and go stand in the corner.

[Camron Rust]

Originally Posted by CoachP

hmmm...now I'm gonna have to run to school tonight and take my ball and go stand in the corner.

I've refined my analysis and have created a diagram (to scale)...





Note three areas indicated by red, blue and green markings.

1. Below the red line, it is not possible for the ball to go into the basket without going over the backboard. Think of the red line as the path of the center of the ball. If the center of the ball is above that line, it can't go in....it will bounce out. If the center of the ball is below that line, it must go over the board (and may still miss).
2. If the entire ball is above the blue, solid line, the ball has a clear path into the basket without any part of the ball going above any part of the backboard.
3. Above the green, solid line, the ball has a clear path to the entire basket without any part of the ball going above any part of the backboard.

Not diagramed, but fairly obvious. From the exact corner, there is a path to "most" of the basket that doesn't go above any part of the backboard.

So, a baseline shot (directly over the baseline) can't be legally made from less than 9' from the center of the backboard (3' outside the lane). For the next few feet, the shot can be legally made but some part of the ball will cross over at least part of the backboard. At about 13', there begins to be an entirely clean path directly into the basket. At 25', only the back couple inches of the basket is shielded by the backboard.

Of course, the margin of error is very small near or inside the blue line, but it is possible.

[BillyMac]

Originally Posted by Camron Rust

I've refined my analysis and have created a diagram (to scale)...



Note three areas indicated by red, blue and green markings.

1. Below the red line, it is not possible for the ball to go into the basket without going over the backboard. Think of the red line as the path of the center of the ball. If the center of the ball is above that line, it can't go in....it will bounce out. If the center of the ball is below that line, it must go over the board (and may still miss).
2. If the entire ball is above the blue, solid line, the ball has a clear path into the basket without any part of the ball going above any part of the backboard.
3. Above the green, solid line, the ball has a clear path to the entire basket without any part of the ball going above any part of the backboard.

Not diagrammed, but fairly obvious. From the exact corner, there is a path to "most" of the basket that doesn't go above any part of the backboard.

So, a baseline shot (directly over the baseline) can't be legally made from less than 9' from the center of the backboard (3' outside the lane). For the next few feet, the shot can be legally made but some part of the ball will cross over at least part of the backboard. At about 13', there begins to be an entirely clean path directly into the basket. At 25', only the back couple inches of the basket is shielded by the backboard. Of course, the margin of error is very small near or inside the blue line, but it is possible.

Wow. Thanks for the effort. Confucius supposedly said, "One picture is worth a thousand words". If he said it, he was right. Now I understand what you mean.

[rainmaker]

Sheez, Camron, you have way, way too much time on your hands!!

[CoachP]

Originally Posted by rainmaker

Sheez, Camron, you have way, way too much time on your hands!!

I agree with Jullie or is it Juulie??.....but... you da man!! Nice drawing.

[grunewar]

Yikes, heck of drawing is right!

I won't take out my rule book and shove it in a coaches face, ever.....but can you see waving off a basket stating it came over the backboard and when the coach calls a TO to "discuss", you reach into your back pocket and pull out CAMRON's art and say, "OK coach, let's review."

[Camron Rust]

Originally Posted by rainmaker

Sheez, Camron, you have way, way too much time on your hands!!

It really didn't take that long....maybe 10 minutes using Visio for the diagram...and a few more minutes for the math.

[sseltser]

Originally Posted by Camron Rust

Below the red line, it is not possible for the ball to go into the basket without going over the backboard. Think of the red line as the path of the center of the ball. If the center of the ball is above that line, it can't go in....it will bounce out. If the center of the ball is below that line, it must go over the board (and may still miss).

Camron

This is a sweet picture and pretty accurate for our purposes, but there is a slight error that I think you might want to think about.

Consider the area that is inside the ring, but to the "upper left" of the red line. Technically speaking, if the center of the ball happens to drop (straight down or nearly straight down) anywhere inside that area then the ball will bounce toward the center of the basket and have a chance to go in.

Therefore, to get the actual solution for the red line would require this (and I'm not proposing to do this because I've already taken Calc 1 and 2 and don't feel like repeating it):

- The red line would have to be tangent to the curve that is represented by the ring. Therefore, we need en equation of the half circle that is the "upper half" of the ring.
- We probably need to get a function of its derivative.
- Finally we need to find a line that has these three characteristics:
a)passes through the outside point of the backboard (as in the drawing)
b)passes through a point on the ring ( (x,f(x)) where f is the curve of the ring)
C)has the same slope as the instantaneous slope of the curve of the ring that it passes through.



Now that I've done that kind of work for the first time since my sophomore year, if anybody wants to actually figure it out (and find out that the real answer is like 8.6') then they have more free time on their hands than I do.

[CoachP]

Originally Posted by sseltser

Camron

This is a sweet picture and pretty accurate for our purposes, but there is a slight error that I think you might want to think about.

Consider the area that is inside the ring, but to the "upper left" of the red line. Technically speaking, if the center of the ball happens to drop (straight down or nearly straight down) anywhere inside that area then the ball will bounce toward the center of the basket and have a chance to go in.

Therefore, to get the actual solution for the red line would require this (and I'm not proposing to do this because I've already taken Calc 1 and 2 and don't feel like repeating it):

- The red line would have to be tangent to the curve that is represented by the ring. Therefore, we need en equation of the half circle that is the "upper half" of the ring.
- We probably need to get a function of its derivative.
- Finally we need to find a line that has these three characteristics:
a)passes through the outside point of the backboard (as in the drawing)
b)passes through a point on the ring ( (x,f(x)) where f is the curve of the ring)
C)has the same slope as the instantaneous slope of the curve of the ring that it passes through.



Now that I've done that kind of work for the first time since my sophomore year, if anybody wants to actually figure it out (and find out that the real answer is like 8.6') then they have more free time on their hands than I do.

Holy cosine batman!! Juulie and I were wrong. YOU have too much time on your hands.

[Camron Rust]

Originally Posted by sseltser

Camron

This is a sweet picture and pretty accurate for our purposes, but there is a slight error that I think you might want to think about.

Consider the area that is inside the ring, but to the "upper left" of the red line. Technically speaking, if the center of the ball happens to drop (straight down or nearly straight down) anywhere inside that area then the ball will bounce toward the center of the basket and have a chance to go in.

Therefore, to get the actual solution for the red line would require this (and I'm not proposing to do this because I've already taken Calc 1 and 2 and don't feel like repeating it):

- The red line would have to be tangent to the curve that is represented by the ring. Therefore, we need en equation of the half circle that is the "upper half" of the ring.
- We probably need to get a function of its derivative.
- Finally we need to find a line that has these three characteristics:
a)passes through the outside point of the backboard (as in the drawing)
b)passes through a point on the ring ( (x,f(x)) where f is the curve of the ring)
C)has the same slope as the instantaneous slope of the curve of the ring that it passes through.



Now that I've done that kind of work for the first time since my sophomore year, if anybody wants to actually figure it out (and find out that the real answer is like 8.6') then they have more free time on their hands than I do.

Can't disagree with your analysis....but I don't think we really need to worry about it since the ball needs to be coming straight down...something it can't realisticly do unless thrown from a position straight below that point. Any momentum it has from a real shot will negate the effect of your point....even if a shot just clears the mimimum legal (red) line, the ball will skim across the front of the rim and bounce out.