Trivia Poll

(NO FAIR LOOKING THIS UP.) What is the shortest distance from the free throw line to the rim of the basket?

Answer Sunday night.

Sam

.

Well without looking at anything figuring, as the question is written I went with 16 feet. You know, the third side of the right triangle deal.

beat

14 feet 3 inches

swish

Beat spelled out the answer on my other trivia thread on this topic. Might as well put everyone out of his/her misery.

The shortest distance between the foul line and the front of the rim is 13' 0". The diameter of the rim is 18". The flat piece fastening the rim to the backboard measures six inches. Thus, adding the three figures together, the basket is 15 feet from the backboard and 13 feet from the front of the rim.

I've assumed that, since the backboard was bolted directly to the basket in the old days (with no flat metal connection to the rim), then the basket in those days was only 12' 6" away from the free throw line and the backboard only 14" 6" from the foul line. But, now that I think about it, it's more likely that they just moved the free throw line back six inches to make the distances the current length of 13' 0" from the free throw line to the front of the rim and the backboard (which is the official measuring-off distance for the free throw line) 15' 0" from the backboard.

That flat, six-inch metal is responsible for the fact that today's rims are a lot more "forgiving" than in the 50s and 60s. So are today's breakaway baskets.

The tighter rims in the 50s and 60s (along with the volume strategy and rapid pace of games) were responsible for lower shooting percentages in those days. In the past, I have posted a quote from Bill Sharman indicating how he wished he could have shot at today's baskets.

Sam

Actually Sam the way the question is worded 13.8 is wrong. If the rim was laying on the floor yes but it is 10 feet in the air and a direct line from the front of the foul line (on the floor) up to a 10 foot rim (hypotenuse of a right triangle) is longer than the other 2 sides. My math makes it a little over 16.5 feet.


beat

Since we're getting all technical, I believe the horizontal distance between the free throw line and the front of the rim is 12 feet 11 3/8 inches.

The distance between the free throw line and backboard is 15 feet.

The rim has an 18" inside diameter.

The distance between the backboard and the inside of the rim closest to the backboard is 6".

According to a site I found that sells rims (http://www.tophoops.com/basketball-rims.html), rims should be no less than 5/8 of an inch thick, and they say the rims on their site comply with NCAA and professional standards.

That means:

15 feet
- 6 inches = 14 feet 6 inches
- 18 inches = 13 feet
- 5/8 inch = 12 feet 11 3/8 inches

Of course, like Beat says, that only calculates the distance if both the rim and free throw line are in the same plane, but the rim is 10 feet above the floor, so there's other math involved. Plus the actual distance for a shot will depend on the shooter's height and release point, which could be fairly high and is probably in front of the free throw line. So there's that.

Outside, to get even more technical, by your calculations, you'd have to count the width of the rim twice because there are two points at which the circular rim intersects a lateral line between the foul line and the backboard. That would leave us at 12 feet 10 3/4 inches, wouldn't it?

As for horizontal distance, various websites on the matter recognize that it's greater than the lateral distance but discount it in terms of laying out a basketball floor.

Sam

Sam, I don't think you do count the width of the rim twice. According to the official rules, the distance between the backboard and the inside of the rim is six inches, and that distance takes into account both the connecting plate and width of the rim. So it's that six inches, plus the 18-inch inside diameter, plus the 5/8-inch width of the rim.

I'd go measure it at the school across the street except I'm too lazy. Maybe some industrious person with a rim in their front yard could confirm these life-altering numbers.

Outside, you are correct that given the wording of the question, one must take into account the vertical rise from the charity stripe to the basket. Hearkening back to Pythagoras' formula for the hypoteneuse of a triangle - a squared plus b squared= c squared, the exact answer becomes 16.3596333' or 16' 4.316", given that a is 12' 11.37" and b is 10 feet.

Worcester,

So it's 16' 4.316". And Sam was worried we might look that up.

worcester wrote:Outside, you are correct that given the wording of the question, one must take into account the vertical rise from the charity stripe to the basket. Hearkening back to Pythagoras' formula for the hypoteneuse of a triangle - a squared plus b squared= c squared, the exact answer becomes 16.3596333' or 16' 4.316", given that a is 12' 11.37" and b is 10 feet.

W


Actually Sam the way the question is worded 13.8 is wrong. If the rim was laying on the floor yes but it is 10 feet in the air and a direct line from the front of the foul line (on the floor) up to a 10 foot rim (hypotenuse of a right triangle) is longer than the other 2 sides. My math makes it a little over 16.5 feet.


I brought the above comment, a few comments ago and agree the way the question is written make it the right triangle deal.


I estimated a couple of things and came up with a fraction over 16.5, thank you for taking it a little further............I could have never worked for NASA.

beat

Beat, you were only a O ring width away from the exact answer.

worcester wrote:Beat, you were only a O ring width away from the exact answer.

W

Wasn't it Morton Thiokol that was an O ring short (sort of) with the Challenger disaster? And look what happened there, like I said I don't work for NASA!

beat

Beat and others, the traditional way of measuring distances in constructing a basketball court is in lateral, not horizontal, terms.  While I've seen (recently, in fact) articles acknowledging the vertical distance, all the official measurements are lateral.  So that's what I had in mind when I asked the question.  Sorry for any confusion. Hey, why not take into account the maximum arc the ball takes to reach the rim? That could run all the way to the rafters for high shooters. LOL.

Sam

Should we also factor in the curve of the earth?

Outside,

I looked at the video you mentioned on the other thread five times before I could keep myself from focusing on Cousy.  When I did, I couldn't see anything conclusive.

As for the width of the rim, if the rim is 5/8" thick and the inside of the rim is exactly 6" from the backboard, how could the flat piece measure exactly 6"?  Wouldn't it have to measure 5 3/8"?

I looked at my favorite all-time Celtics video, which I'm dying to show to the board some day.  It's of the deciding game of the 1965 Eastern Finals pitting the Celtics against the Cincinnati Royals.  It was filmed as a documentary, with slow-mo, stop-action, replays, a miked Red Auerbach, and some tutorials on various aspects of the action  There are three You Tube videos—each an excerpt of the documentary but collectively omitting a lot of the best parts.

Part One: https://www.youtube.com/watch?v=K_h34wT3SHc
Part Two: https://www.youtube.com/watch?v=9WUEf_dgdkY
Part Three: https://www.youtube.com/watch?v=IYCVbZwoDy4

At the very beginning of the first video, it shows them erecting the backboards prior to the game.  I had to double-click on the video to get it to activate; but the double-clicking also served the additional purpose of zooming.  At somewhere aroung the 7 second mark, there's a closeup of the basket.  I see no flat plate.  What I do see are two rods extending from the side of the basket as part of the basket and presumably bolted to the backboard.  There does appear to be some space between the backboard and rim, but I can't tell how much.  The side of the backboard, itself, obliterates some of the most telling visuals.

Rims aside, by watching these videos, you can get an idea of the constant up-tempo nature of the game, with even Don Nelson leading the pack to take a long pass for a layup.  You can also gain an idea of the dominance of Russell (although you get much more complete evidence from the original video.  You can suffer, along with Red, at the officiating.  And it just wouldn’t be a Celtics video without recording the exploits of Sam.  One of the omitted portions involves Sam retreating on defense, intercepting a long pass, and immediately hoisting a long pass of his own right on the money to a “streaking” Nellie for a score.

By the way, if the commentator on the documentary sounds familiar, it was William Schallert, a long-time character actor who—among other things—played the father on The Patty Duke Show. Far too wimpy-sounding to be doing an instructional on basketball.

Enjoy,

Sam

I'd by lying if I said mathematics were my favorite subject. My guess would've been 13 feet, though.


KJ

worcester wrote:Should we also factor in the curve of the earth?

And the wind.

Just to add to this.....(which at the time I thought was sort of neat.

When the towers were built for the building of the Verrazano Narrows Bridge, the tops of the towers are an inch or so further apart than the base. Reason?........... To take into account the curvature of the earth.

Because of the height of the towers (693 ft or 211 m) and their distance apart (4,260 ft or 1,298 m), the curvature of the Earth's surface had to be taken into account when designing the bridge—the towers are 1 5⁄8 inches (41.275 mm) farther apart at their tops than at their bases. (from Wikipedia)

beat

Sam wrote:As for the width of the rim, if the rim is 5/8" thick and the inside of the rim is exactly 6" from the backboard, how could the flat piece measure exactly 6"? Wouldn't it have to measure 5 3/8"?

The flat piece has a curved end that extends at least partially over the top of the rim. The key point is that the distance between the backboard and the inside of the rim is six inches, which includes the width of the rim itself.

I haven't seen anything talking about the exact measurement of the flat piece, so if I said that earlier, I misspoke.

Sam wrote:At the very beginning of the first video, it shows them erecting the backboards prior to the game. I had to double-click on the video to get it to activate; but the double-clicking also served the additional purpose of zooming. At somewhere aroung the 7 second mark, there's a closeup of the basket. I see no flat plate.

If you freeze it at the 10 second mark, you can see the flat plate. It's somewhat hard to make out because the color is lighter than the rest of the rim and metal support and it sort of blends into the background, but I do see it.

I don't see a flat point extending toward the basket. I see a flat plate that is upright and fastens the aforementioned rods to the backboard. It states in many sites that the current flat plate is 6" long. Perhaps that includes the curved end that you say extends at least partially over the top of the tim.

Okay, I just let it go to the 11th second, and I believe I could see a flat piece.

I'll have to look at earlier film (such as it is).

Sam

I found this squib in Wikipedia that may clarify the whole thing:

Although Darryl Dawkins notoriously shattered two backboards with his dunks in 1979,[4] the old-style bolted rim structure was not phased out of the NBA until the 1981-82 season, when breakaway rims debuted as a uniform equipment upgrade.

So the basket, in fact, was directly bolted to the backboard in the early, regardless of how (including rods, flat pieces, etc.) until 1981-82. (And I still am not at all sure of what the precise situation was in the early 60s and before.) It must have been the breakaway rims that Bill Sharman was referencing when he talked about the tightness of the baskets during his era (with baskets being bolted directly to the backboard) and how he wished he'd had a crack at the more forgiving rims decades lighter.

Sam

Sam,

I do think Bill Sharman was referring to the breakaway rims and how they "give" when the ball hits them, leading to a softer bounce and better chance to go in. I think it also has to do with the breakaway structure absorbing shock in other ways, such as when a ball bounces around from one side of the rim to another.

If you go to a basketball game at a carnival or arcade, they use bouncy balls and tight rims to make almost everything that hits the rim bounce away. That's how the old rims felt like compared to today's rims.

and because I liked this picture so much . . . . .

Note the flat plate in this 1980 photo.

https://www.google.com/search?q=nba+images+of+the+basketball+rim+in+the+1950's&tbm=isch&tbo=u&source=univ&sa=X&ei=pTG7U9fPFo7poATWuYHgCA&ved=0CCgQ7Ak&biw=1882&bih=887&dpr=0.85#facrc=_&imgdii=_&imgrc=qO8PsmQPCi3IfM%253A%3Bl3zzKpobQBS_WM%3Bhttp%253A%252F%252Fi.cdn.turner.com%252Fnba%252Fnba%252Fhistory%252Ffeatures%252Fmoment-1980-erving-lakers%252Fdrj-lakers.jpg%3Bhttp%253A%252F%252Fwww.nba.com%252Fhistory%252Ffeatures%252Fmoment-1980-erving-lakers%252Findex.html%3B608%3B262

swish