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Old Wed Apr 25, 2012, 12:31pm
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70' base distance

Hi,

Does someone have access to a rulebook to find out the distance from the point of HP to the center of the 2nd base bag for the 70' base distance?

Thanx.
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Old Wed Apr 25, 2012, 12:48pm
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Originally Posted by tru_in_blu View Post
hi,

does someone have access to a rulebook to find out the distance from the point of hp to the center of the 2nd base bag for the 70' base distance?

Thanx.
98' 11 15/16"
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Old Wed Apr 25, 2012, 01:50pm
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Quote:
Originally Posted by Tru_in_Blu View Post
Hi,

Does someone have access to a rulebook to find out the distance from the point of HP to the center of the 2nd base bag for the 70' base distance?

Thanx.
Simple trigonometry. For a right-angle triangle:
c^2 = a^2 + b^2 (where c is the hypotenuse and a & b are the other two sides)

For a ball diamond, c = distance to 2B, and a = b = base distance.

So c = SQRT(70^2 + 70^2)

No need for the rule book.
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Old Wed Apr 25, 2012, 02:54pm
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Originally Posted by celebur View Post
Simple trigonometry. For a right-angle triangle:
c^2 = a^2 + b^2 (where c is the hypotenuse and a & b are the other two sides)

For a ball diamond, c = distance to 2B, and a = b = base distance.

So c = SQRT(70^2 + 70^2)

No need for the rule book.
Yes, but not a lot of people realize the bases do not actually form the inside corners of a perfect square - 3 bases do, but 2nd base doesn't. If it did, you're answer would be wrong.

98 feet, 11 and 15.0304 sixteenths inches.
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Old Wed Apr 25, 2012, 02:54pm
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Quote:
Originally Posted by celebur View Post
Simple trigonometry. For a right-angle triangle:
c^2 = a^2 + b^2 (where c is the hypotenuse and a & b are the other two sides)

For a ball diamond, c = distance to 2B, and a = b = base distance.

So c = SQRT(70^2 + 70^2)

No need for the rule book.
Got me.....I was also going to suggest the Pythagorean Theorem......

Joel
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Old Wed Apr 25, 2012, 03:01pm
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it would have been nice if he had done the math imo. that is imo a more thorough response. you only get half credit
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Old Wed Apr 25, 2012, 04:00pm
JEL JEL is offline
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Originally Posted by Gulf Coast Blue View Post
Got me.....I was also going to suggest the Pythagorean Theorem......

Joel
I was gonna take a tape measure to tonights games!
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Old Wed Apr 25, 2012, 04:37pm
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Originally Posted by JEL View Post
I was gonna take a tape measure to tonights games!
Now you are assuming the P&R people got it right.

A couple years ago a team hosted a FP tournament on a SP complex. When they laid the fields and drop the circles, one field had the PP inside the circle (FP was still at 40').

A couple of us asked if they double-checked the FP PP and were assure they were correct.

Come to find out that the P&R folks just picked up the old PP and put them down where they "thought" they belonged when SP went to 50'.

Don't EVER trust anyone else's work without double-checking it.
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Old Wed Apr 25, 2012, 05:40pm
JEL JEL is offline
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Originally Posted by IRISHMAFIA View Post
Now you are assuming the P&R people got it right.

A couple years ago a team hosted a FP tournament on a SP complex. When they laid the fields and drop the circles, one field had the PP inside the circle (FP was still at 40').

A couple of us asked if they double-checked the FP PP and were assure they were correct.

Come to find out that the P&R folks just picked up the old PP and put them down where they "thought" they belonged when SP went to 50'.

Don't EVER trust anyone else's work without double-checking it.
A few years ago I worked a P&R 10U State Playoff. Bottom of third, D coach comes and asks "will you just look at 2B and tell my player it is OK?"

She had been telling him all game that 2B wasn't where it was supposed to be. Coach didn't believe her, and I was skeptical, but looked anyway. The more I looked, the easier it was to spot. 2B was ablmost 6' (yes feet) further in OF than it should have been.

Grounds crew and P&R guys though I was a nut, "been there since the field was built" and "was laid out with a laser" but they finally did come out for a look, then quietly came back with a tape.

Seems this field was built 6 years prior, and 2B was in the same spot all that time!

I listen better to 10 year old kids now!
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Old Thu Apr 26, 2012, 12:35am
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Originally Posted by mbcrowder View Post
Yes, but not a lot of people realize the bases do not actually form the inside corners of a perfect square - 3 bases do, but 2nd base doesn't. If it did, you're answer would be wrong.

98 feet, 11 and 15.0304 sixteenths inches.
Actually, no. The Pythagorean Theorem is all you need to answer the OP.

Here's a fun thing. This is also the distance between the outside far corners of 1st and 3rd base. If this distance is short (long), the distance from the point of home plate to the center of second base will be long (short) by the same amount. If it's not, you have other problems such as not placing the bases as noted by mbcrowder.
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Old Thu Apr 26, 2012, 08:41am
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Originally Posted by celebur View Post
Simple trigonometry. For a right-angle triangle:
c^2 = a^2 + b^2 (where c is the hypotenuse and a & b are the other two sides)

For a ball diamond, c = distance to 2B, and a = b = base distance.

So c = SQRT(70^2 + 70^2)

No need for the rule book.
I considered the right triangle approach, but recalled that 2nd base isn't fully inside the diamond.

I thought this was geometry, BTW.

In any case, the diameter of a spike being driven to measure the distance(s) is enough to convince me to remember this distance as a skosh short of 99'.

Thanx for the replies and bringing back the nightmares of HS math!
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Old Thu Apr 26, 2012, 09:31am
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Originally Posted by Tru_in_Blu View Post
I considered the right triangle approach, but recalled that 2nd base isn't fully inside the diamond.

I thought this was geometry, BTW.

In any case, the diameter of a spike being driven to measure the distance(s) is enough to convince me to remember this distance as a skosh short of 99'.

Thanx for the replies and bringing back the nightmares of HS math!
The distances are measured from the point of the plate to the center of second, and from the farthest point, foul side of first and third.
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Old Thu Apr 26, 2012, 11:02am
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Originally Posted by Tru_in_Blu View Post
In any case, the diameter of a spike being driven to measure the distance(s) is enough to convince me to remember this distance as a skosh short of 99'.
Hey...I just learned that a skosh equals .9696 inches.....
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Old Thu Apr 26, 2012, 11:17am
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Originally Posted by Andy View Post
Hey...I just learned that a skosh equals .9696 inches.....
I think your skosh is way too long. The distance is short of 99' by a mere 1/16 of an inch.
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Old Thu Apr 26, 2012, 11:49am
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Originally Posted by Tru_in_Blu View Post
I think your skosh is way too long. The distance is short of 99' by a mere 1/16 of an inch.
Actually, SQRT(70^2 + 70^2) = 98.99494937 ft, and 98ft 11 15/16 in = 98.99479167 ft, for a difference of 0.000157699 ft, or 0.001892393 in.

THAT'S a skosh.
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