Pass Goes Behind the Backboard, Through the Supports

[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.

[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.

[btaylor64]

Originally Posted by Ch1town

I stand corrected. Had to go back & re-read the OP. This line got me: "passing directly behind the backboard".

Reading IS fundamental!

If it reads as you stated that would mean it would be illegal for the ball to go through the supports. Does it really say that?

[BillyMac]

Will any of this be on the test?

[Adam]

Originally Posted by btaylor64

If it reads as you stated that would mean it would be illegal for the ball to go through the supports. Does it really say that?

The OP said that, not the rule.

[mutantducky]

ouch this thread hurt my brain.
So to clarify because some things are getting jumbled

1.from baseline pass(out of bounds) illegal- pass goes over the backboard
For example after a made basket player is running the baseline and is underneath the backboard and throws a pass over the backboard that does not touch anything.
2.illegal- airball shot from in front or the side that goes over the backboard. Does play stop once it goes over?
3.illegal- shot hits the rim and goes over
4. illegal- shot from behind the backboard goes over and into the rim
5. Not 100% sure on this one- pass goes from the side between basketball supports without hitting anything.

[Scrapper1]

Originally Posted by mutantducky

So to clarify because some things are getting jumbled

Here's the unjumbled version: it's a violation for any part of the ball to pass over any part of a rectangular backboard at any time.

Pass, shot, throw-in, front-to-back, back-to-front, side-to-side, whatever!! Doesn't matter. If the ball passes over the rectangular backboard, it's a violation.

AND, it's not a violation to pass BEHIND the backboard.

Hope that helps.

[IREFU2]

Originally Posted by MajorCord

Player A is falling out of bounds under his own basket. He throws the ball behind his back in an effort to save the ball. In doing so, the ball goes from one side of the lane to the other, passing directly behind the backboard, and somehow comes down on the other side of the lane without hitting any supports where it is caught by a teammate. Officials had a "no call". A D-1 official observing said that the play should have been whistled. He says that the ball cannot pass directly behind the backboard. What say you guys and gals?

I got nuttin!

[bob jenkins]

Pleazse note that this thread is more than two months old and has been answered correctly. Let it die.