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Old Fri Jul 15, 2005, 07:56pm
JEL JEL is offline
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Question


I was asked this question a while back, and have yet to come up with any answers.

If the diamond, (infield) is a square, and the plate, 1B and 3B are all totally inside that square, why is 2B positioned with the mid-point on the corner, and not totally inside the infield? This makes the runner have to travel an extra 7.5" between 1B to 2B, and also 2B to 3B, than from 3B to home, or home to 1B.

Of course that really doesn't make any difference in the game, but you may be able to win a few "bar bets" if you know the distance is further from 2B to 3B, than 3B to home!

Its really just a slow friday night, the youngest kid is sick and we are staying home with him, but I still wonder if anyone has an answer other than that's the way it's always been!
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Old Sat Jul 16, 2005, 05:05pm
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Uh Huh. The way I've seen it layed out, the center of each base in actually in the square, in other words, the part in the ground that the base fits into. If that's true, portions of 1st, 2nd, and 3rd bases, the actual bases, are outside the square. The lines from 3rd to Home and 1st to Home actually go to the front corners of the plate, so the rear portion of the plate isn't in the square either. So every base, including home plate, is considered "fair", but they are not actually completely within the square. So, it one measures from center to center, the distances should all be the same.

If you want to make bar bet money, bet on the fact that a ball which leaves the bat and hits the pitcher's plate before touching anything else is a foul ball.
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Old Sat Jul 16, 2005, 06:12pm
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The lines touch the back sides and tip of home plate.
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Old Sat Jul 16, 2005, 07:27pm
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Jel posted about a difference of 7.5", but you also have to remember the the bases are positioned with a corner pointing to the pitcher, where the plate has a flat edge toward the pitcher, so the distance could deviate depending on the angle.

From what I have read, the first baseball games were played with the three square bases centered on three of the corners of the square, and with a round plate centered on the fourth point. The article I read speculated that to aid in determining whether a ball was fair or foul, first and third were moved solely into fair territory. Eventually the foul part of the plate was done away with, and it was squared off. Second base was left were it was, centered on the right angle of the square.

Then after many years of everyone being satisfied with the layout, someone decided that they again wanted to put part of the bag in foul territory. They decided it would work beautifully if they only painted it orange, said they were doing it for safety and told everyone that it disappears after it gets used.
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Old Sat Jul 16, 2005, 10:15pm
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scroll down to see the diagram ...

http://www.usssa.com/usssa/usssa-gen...RTHedition.htm
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Old Sat Jul 16, 2005, 10:38pm
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Oops, my bad! CecilOne is absolutely correct. Instead of me saying the front corners, I should have said the back corners.
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Old Sun Jul 17, 2005, 09:34pm
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Quote:
Originally posted by JEL

I was asked this question a while back, and have yet to come up with any answers.

If the diamond, (infield) is a square, and the plate, 1B and 3B are all totally inside that square, why is 2B positioned with the mid-point on the corner, and not totally inside the infield? This makes the runner have to travel an extra 7.5" between 1B to 2B, and also 2B to 3B, than from 3B to home, or home to 1B.

Of course that really doesn't make any difference in the game, but you may be able to win a few "bar bets" if you know the distance is further from 2B to 3B, than 3B to home!

Its really just a slow friday night, the youngest kid is sick and we are staying home with him, but I still wonder if anyone has an answer other than that's the way it's always been!
Now that no one has answered your question, let me not answer it. I have no idea why 3/4 of 2nd base is not within the diamond.

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Old Mon Jul 18, 2005, 08:46am
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Because the distances are measured:
a.) from the back edge of HP to the front edge of 1B
b.) from the back edge of 1B to the front edge of 2B
c.) from the front edge of 2B to the back edge of 3B
d.) from the back edge of 3B to the back edge of HP


Minimum running distances from base to base:

HP to 1B: 59 feet 5 inches (RHB)
60 feet from back edge of HP to back edge of 1B
minus 15 inches because BR must only touch front edge of 1B
plus 6 inches to account for the RH batter's box

1B to 2B: 58 feet 9 inches
60 feet from back edge of 1B to front edge of 2B
minus 15 inches because R can start at front edge of 1B and must only reach front edge of 2B

2B to 3B: 58 feet 9 inches
60 feet from back edge of 2B to front edge of 3B
minus 15 inches because R can start at front edge of 2B and must only reach front edge of 3B

3B to HP: 57 feet 4 inches
60 feet from back edge of 3B to back edge of HP
minus 15 inches because R can start at front edge of 3B
minus 17 inches because must only reach front edge of HP
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Old Mon Jul 18, 2005, 09:00am
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Quote:
Originally posted by tcannizzo
Minimum running distances from base to base:

HP to 1B: 59 feet 5 inches (RHB)
60 feet from back edge of HP to back edge of 1B
minus 15 inches because BR must only touch front edge of 1B
plus 6 inches to account for the RH batter's box
Aaah, but the geometry (and distance) changes if the batter goes to the front edge of the batters box. Your calculation is based on a point in line with the fair line, with the batter possible starting some 4' forward of that point, shortening that leg of the triangle.

I remember the terms, but wouldn't dream of trying to actually work out the geometry calculation. It isn't a right angle, so the basic Pythaogorean Theorum doesn't help.
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Old Mon Jul 18, 2005, 10:06am
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Ahhh, but I was calcuating the minimum running distances. This would also assume that BR ran the entire distance precisely on the foul line. Farther up or back in the box will increase the distance as you indicate. Plus my calculations are based on 2nd grade arithmetic (2 digit addition and subtraction), so if you remember how to use your slide-ruler you might be able to increase the precision, professor.
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Old Mon Jul 18, 2005, 11:29am
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Okay, you can put them away, guys. All JEL asked was why 2B is the only base not completely inside of the diamond.

It has zippo to do with the players and to be honest, it probably just happened that way and no one ever changed it.

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Old Mon Jul 18, 2005, 11:40am
JEL JEL is offline
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Quote:
Originally posted by IRISHMAFIA
Okay, you can put them away, guys. All JEL asked was why 2B is the only base not completely inside of the diamond.

It has zippo to do with the players and to be honest, it probably just happened that way and no one ever changed it.


So true! Just an item of curiosity which could be filed in the "who really cares" category.

Still gonna wonder though if the next field I'm on is laid out correctly! Maybe I'll take a tape measure......


Nah, I don't really care, If it's off I'd have to fix it!
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Old Mon Jul 18, 2005, 12:30pm
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Talking

This thread could also be titled, "What geeks do when they have too much time on their hands."

If I have any question about the layout of the field, if my pacing of the base lines is within 1 pace of 20 paces, I'm good (and within 1/2 pace of 13 (12 for 10U) for the pitching plate to home plate).

Some of you guys need to get rid of the spreadsheet and programmable calculators and go back to slide rules.
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Old Mon Jul 18, 2005, 01:57pm
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The real reason that 2nd base is positioned thusly has to do with fact that in the early days of softball, single pegs were used to hold down the bases and the rulesmakers wanted to have a simple method for the groundskeepers to locate the exact spot for the pegs.

Using the Pythagorean Theorum, the original rulesmakers determined that the distance from the back of home plate to the location of the center peg should be as follows:

60 ft x 1.4142135623730950488016887242097 (square root of 2)
= 84.852813742385702928101323452582 feet.

Rounding to feet and inches, you have 84 feet, 10-1/4 inches.

The rulesmakers forgot to subtract half the length of the diagonal segment of the base (10.606601717798212866012665431573 inches) from the total length in order to make 2nd base lie completely inside the 'perfect square'.

If someone wishes to submit a rule change to have 2nd base placed inside the diamond, then just indicate that the distance from home plate to the center of 2nd base should be precisely 83.968930265902518522600267999951 feet. Or, to round it off in order for a groundskeeper to understand, 83 feet 11-2/3 inches.
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Old Mon Jul 18, 2005, 02:05pm
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Quote:
Originally posted by varefump
..... 84.852813742385702928101323452582 feet......10.606601717798212866012665431573 inches....83.968930265902518522600267999951 feet.
As I said, some of you guys need to get rid of the spreadsheet and programmable calculators and go back to slide rules.
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