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dmgwalsh
02-02-2003, 03:26 PM
Can you tell me where on the table the measurement of a pool table is taken. For a 9 footer for example, is it measured on the felt from one end to the other. I'm trying to figure out what size a particular table is.

socrates
02-02-2003, 03:59 PM
I believe table size is measured from cushion to cushion. A regulation 9ft table is 100 inches x 50 inches.

Pro 8ft 46 X 92
Home 8ft 44 x 88

Troy
02-02-2003, 06:26 PM
Correct -- Inside dimensions measure from cushion nose to cushion nose.

Also, a 7" is 40" x 80".

Troy
<blockquote><font class="small">Quote socrates:</font><hr> I believe table size is measured from cushion to cushion. A regulation 9ft table is 100 inches x 50 inches.

Pro 8ft 46 X 92
Home 8ft 44 x 88
<hr /></blockquote>

02-02-2003, 08:26 PM
<blockquote><font class="small">Quote Troy:</font><hr> Correct -- Inside dimensions measure from cushion nose to cushion nose.

Also, a 7" is 40" x 80".

Troy
<hr /></blockquote>Wait a sec, if an 8' table is 12" shorter and 6" narrower than a 9' table, then shouldn't a 7' table likewise be 12" shorter and 6" narrower than an 8' table??? If so, your measurement of 40x80 for a 7' doesn't jive with the 44x88 dimensions of an 8'. I would've expected it to be 38x76.

David

Troy
02-02-2003, 09:34 PM
My info comes from Pg 94 of Meueller's Catalog.

<blockquote><font class="small">Quote dmorris68:</font><hr> <blockquote><font class="small">Quote Troy:</font><hr> Correct -- Inside dimensions measure from cushion nose to cushion nose.

Also, a 7" is 40" x 80".

Troy
<hr /></blockquote>Wait a sec, if an 8' table is 12" shorter and 6" narrower than a 9' table, then shouldn't a 7' table likewise be 12" shorter and 6" narrower than an 8' table??? If so, your measurement of 40x80 for a 7' doesn't jive with the 44x88 dimensions of an 8'. I would've expected it to be 38x76.

David <hr /></blockquote>

02-02-2003, 10:27 PM
<blockquote><font class="small">Quote Troy:</font><hr> My info comes from Pg 94 of Meueller's Catalog.
<hr /></blockquote>Hmm, must be a misprint. Just checked Brunswick's site here (http://www.brunswickbilliards.com/products/brunswick/tables/roomsize/roomsize.html), and they confirm my guess that a 7' is 38x76.

David

Fred Agnir
02-03-2003, 07:51 AM
<blockquote><font class="small">Quote dmorris68:</font><hr> <blockquote><font class="small">Quote Troy:</font><hr> My info comes from Pg 94 of Meueller's Catalog.
<hr /></blockquote>Hmm, must be a misprint. Just checked Brunswick's site here (http://www.brunswickbilliards.com/products/brunswick/tables/roomsize/roomsize.html), and they confirm my guess that a 7' is 38x76.

David <hr /></blockquote>

Here's the real scoop folks. A 7' bar table is usually a little bigger than a 7' home table. A 7' bar table is usually in the 40" width range, while a 7' home table is in the 38" range.

Fred

Fred Agnir
02-03-2003, 08:01 AM
<blockquote><font class="small">Quote dmgwalsh:</font><hr> Can you tell me where on the table the measurement of a pool table is taken. For a 9 footer for example, is it measured on the felt from one end to the other. I'm trying to figure out what size a particular table is. <hr /></blockquote>
Good question that I've gotten the answer to from Bob Jewett. The actual measurement importance for a 9' x 4.5' table isn't the 9' dimension (which means nothing), but actually the 4.5' dimension. It's the width of the green, including the cushions (assuming 2 inch cushions).

BTW, a pool table is never ever twice as long as it is wide. Never. Even the true playing surface isn't twice as long as it is wide. The outside dimensions will always be shorter than twice the width because you don't have twice as much rail. I think this tidbit is important when you're judging what is the absolute minimum room requirement without having to get a small stick or jack up on cushion shots. People usually will be off (shorter) by 2 inches if they simply say "add 10 feet" rule of thumb. That's pretty important if you ask me.

The inside playing surface is limited by the ball width, so the actual playing area is just over twice as long as the width. There are many systems that are based on the idea that the table is twice as long as its width.

Fred

02-03-2003, 09:22 AM
<blockquote><font class="small">Quote Fred Agnir:</font><hr>Here's the real scoop folks. A 7' bar table is usually a little bigger than a 7' home table. A 7' bar table is usually in the 40" width range, while a 7' home table is in the 38" range.

Fred <hr /></blockquote>Interesting. Fred, you're a regular fountain of pool &amp; billiards trivia. /ccboard/images/graemlins/wink.gif

So a 7' bar table isn't really 7', eh? Is that because the cushions are wider? I was aware of the trap in thinking *outside* dimensions were "lengh is twice width" due to the rails being the same size all the way around. But I was pretty sure that the term "length is twice the width" was always used in referring to playing area, not outside dimensions. If you're measuring for room size, then outside dimensions are what you want, and that can vary widely by table style (rail width being the variable). However, unless there are weird cushion sizes involved, I would think the measure for playing area would always be "length is twice width", at least on a "regulation" table (whatever that is, but you know what I mean).

I *was* thinking that "standard" table measurements were made from where the cloth meets the wood rail, i.e. including the entire cushion. But I wasn't sure enough to speak up and say it. And I was too lazy to get up and measure my own table. /ccboard/images/graemlins/smile.gif

David

Troy
02-03-2003, 11:00 AM
Regardless of table size, the play area is always a ratio of 2 to 1.

<blockquote><font class="small">Quote dmorris68:</font><hr> <blockquote><font class="small">Quote Fred Agnir:</font><hr>Here's the real scoop folks. A 7' bar table is usually a little bigger than a 7' home table. A 7' bar table is usually in the 40" width range, while a 7' home table is in the 38" range.

Fred <hr /></blockquote>Interesting. Fred, you're a regular fountain of pool &amp; billiards trivia. /ccboard/images/graemlins/wink.gif

So a 7' bar table isn't really 7', eh? Is that because the cushions are wider? I was aware of the trap in thinking *outside* dimensions were "lengh is twice width" due to the rails being the same size all the way around. But I was pretty sure that the term "length is twice the width" was always used in referring to playing area, not outside dimensions. If you're measuring for room size, then outside dimensions are what you want, and that can vary widely by table style (rail width being the variable). However, unless there are weird cushion sizes involved, I would think the measure for playing area would always be "length is twice width", at least on a "regulation" table (whatever that is, but you know what I mean).

I *was* thinking that "standard" table measurements were made from where the cloth meets the wood rail, i.e. including the entire cushion. But I wasn't sure enough to speak up and say it. And I was too lazy to get up and measure my own table. /ccboard/images/graemlins/smile.gif

David <hr /></blockquote>

Fred Agnir
02-03-2003, 11:11 AM
<blockquote><font class="small">Quote Troy:</font><hr> Regardless of table size, the play area is always a ratio of 2 to 1.<hr /></blockquote>
The measurement from nose to nose should be a ratio of 2 to 1, but if you define the playing area to be where the base of the ball contacts, then no, the playing area is not 2 to 1.

Fred

Cueless Joey
02-03-2003, 11:36 AM
<blockquote><font class="small">Quote Fred Agnir:</font><hr> <blockquote><font class="small">Quote Troy:</font><hr> Regardless of table size, the play area is always a ratio of 2 to 1.<hr /></blockquote>
The measurement from nose to nose should be a ratio of 2 to 1, but if you define the playing area to be where the base of the ball contacts, then no, the playing area is not 2 to 1.

Fred <hr /></blockquote>
Nitpicking! /ccboard/images/graemlins/grin.gif
Since the balls are 2 1/4 in diameter, half is 1 1/8, therefore a 9 ft table has actually 98 7/8 and 48 7/8 of "playing surface"?

Fred Agnir
02-03-2003, 12:25 PM
<blockquote><font class="small">Quote Cueless Joey:</font><hr> Nitpicking! /ccboard/images/graemlins/grin.gif
Since the balls are 2 1/4 in diameter, half is 1 1/8, therefore a 9 ft table has actually 98 7/8 and 48 7/8 of "playing surface"? <hr /></blockquote>
You're missing another half diameter, as there are two cushions per direction.

Nitpicking or not, there are systems out there with the entire premise based on the "fact" that the playing surface is twice as long as it is wide.

Fred

cheesemouse
02-03-2003, 12:34 PM
Wouldn't you have to subtract 21/4 anyway as there are two sides and two ends, there for you would still end up with the 2/1 ratio.......dah..........but.....dah...... /ccboard/images/graemlins/tongue.gif

Ross
02-03-2003, 12:40 PM
So Fred, will you volunteer to design a kicking/banking system based on a playing surface that is 2.047 times as long as it is wide? /ccboard/images/graemlins/wink.gif

Ross
02-03-2003, 01:30 PM
<blockquote><font class="small">Quote cheesemouse:</font><hr> Wouldn't you have to subtract 21/4 anyway as there are two sides and two ends, there for you would still end up with the 2/1 ratio.......dah..........but.....dah...... /ccboard/images/graemlins/tongue.gif <hr /></blockquote>
Cheesemouse - assume a table was 20" x 10". Length = 2 x width. Now subtract 5" from each, to get new size: 15" x 5". Voila! Length = 3 x width! So subtracting a constant from both length and width changes the ratio unless length=width.

cheesemouse
02-03-2003, 02:26 PM
<blockquote><font class="small">Quote Ross:</font><hr> <blockquote><font class="small">Quote cheesemouse:</font><hr> Wouldn't you have to subtract 21/4 anyway as there are two sides and two ends, there for you would still end up with the 2/1 ratio.......dah..........but.....dah...... /ccboard/images/graemlins/tongue.gif <hr /></blockquote>
Cheesemouse - assume a table was 20" x 10". Length = 2 x width. Now subtract 5" from each, to get new size: 15" x 5". Voila! Length = 3 x width! So subtracting a constant from both length and width changes the ratio unless length=width. <hr /></blockquote>

Ross, Ok, I get it. It's same reason that if I were 30yrs old and you were 15yrs old I would be twice your age but in another 15 yrs I would be 45 and you would be 30.......and OK, I get it.....LOL /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif

Troy
02-03-2003, 09:17 PM
I believe "playing area" has always been defined as cushion nose to cusion nose, hence, 2 to 1 ratio.

<blockquote><font class="small">Quote Fred Agnir:</font><hr> ...there are systems out there with the entire premise based on the "fact" that the playing surface is twice as long as it is wide.
Fred <hr /></blockquote>