150 feet I'm sorry but the answer above is incorrect. the full distance between 3rd and 1st is only 127.28 feet. The distance from 1st to the pitchers mound is 63.72 feet. the same goes for 3rd to pitchers mound.
Good question and to find out the answer let me go back and try to remember some elementary geometry. If a straight line is drawn from home plate to the pitcher's rubber on the pitching mound that line would measure 60.5 feet. If a straight line is drawn from home plate to first base that line would measure 90 feet. If a straight line is drawn from the pitcher's rubber to first base that line would measure X feet. If I remember the Pythagorean Theory correctly, the square of the hypotenuse of a right triangle (distance from home plate to first base), is equal to the squares of the distance of the other two sides of the right triangle added together (distance from the pitcher's rubber to first base, and distance from the pitcher's rubber to home plate). This equation would read as:
60.52 + X2 = 902 ------ next do the multiplication to find the squares
3660.25 + X2 = 8100 ----- next find the value of X2 by subtracting 3660.25 from 8100.
X2 = 4439.75 ----- next find the square root of 4439.75
X = 66.631449031219484653750691943936
Therefore, the distance between the rubber on the pitcher's mound and first base is 66.631449031219484653750691943936 feet.
That is not the correct answer. This answer improperly assumes that if you form a triangle with home plate, first base and the pitchers mound you have a right triangle, that is a triangle with a 90 degree angle between at the pitchers mound. That is incorrect.
The correct answer is 63.716 feet and it is found as follows:
The distance between home plate and second base is equal to: the square root of 902 + 902 (the distance between home plate and first and between first and second bases) = 16,200, the square root of which is 127.2792 feet, which also the straight line distance between first base and third base.
If you draw a straight line between 1st and 3rd and also between home and second those lines will intersect halfway between home and second which is 63.6396 feet from home plate or 3.1396 feet behind the pitchers rubber. The distance of this intersection from first base is also 63.6396 feet.
If you draw another straight line between the pitchers rubber and first base you have a new right angle triangle that is formed by the line from the line from the rubber to first base (which is the hypotenuse), the segment on the line from home to second between the rubber and the intersection of the lines between first and third and home and second (which is 3.1396 feet) and the line segment from first base to third where it intersects with the line from home to second which is also 63.6396 feet.
Now you have a right triangle to which the Pythagorean Theory can be applied.
The distance between the rubber and first base will be the square root of the sum of 3.13962 + 63.63962.
The sum of 3.13962 + 63.63962 = 4,059.8558, the square root of which is 63.7169975 feet.
While the above answer is close, the correct answer is 63.7169975 feet which is 3 feet shorter than the wrong answer.
The distance between first and third base is 127.28 feet. Some people think that you can just divide that by 2 to get the distance between the mound and first or third base, but the pitcher's mound does not touch the straight line between first and third. It is actually lower, so the actual distance between first (or third) and the middle of the pitchers mound is 63.72 feet.
This is according to the diagram on page 4 of the Official Baseball Rules (2011 edition)
Home plate to pitcher's box: 60 feet 6 inches. Plate to second base: 127 feet 3 3/8 inches. Distance from base to base (home plate included): 90 feet. Size of bases: 15 inches by 15 inches. Pitcher's plate: 24 inches by 6 inches. Batter's box: 4 feet by 6 feet. Home plate:Five-sided, 17 inches by 8 1/2 inches by 8 1/2 inches by 12 inches by 12 inches, cut to a point at rear. Home plate to backstop: Not less than 60 feet (recommended). Weight of ball: Not less than 5 ounces nor more than 5 1/4 ounces. Circumference of ball: Not less than 9 inches nor more than 9 1/4 inches. Bat: Must be one piece of solid wood, round, not over 2 3/4 inches in diameter at thickest part, nor more than 42 inches in length.
Use the Pythagorean theorem: a²+b²=c², the triangle formed by home plate to 1st base, 1st base to 2nd, and home to second base. Thus, 90 feet between plate and home, or a², (90)(90)=8,100, add 90 feet between 1st base and 2nd base (b²) which is also 8,100 feet for a total of 16,200 feet which is c², the distance between home plate and 2nd base. All that is required is to take the square root of 16,200 to obtain 127.27922 feet, or 127 feet, 3.35 inches rounded.
Yes. It is 90 ft between each base. Therefore, it is a square and the diagonal is the same either way.
60 Feet, 6 inches
(Same distance as Pitchers Mound to Home Plate.)
In a regular MLB game, it is about 60 feet. Normally it should be the same distance from the pitcher's mound to home plate.
5 First base Second base Third base Home Pitchers mound
The pitchers mound
45' to 35' feet
It is a little over 110 degrees
I'm assuming you mean USA baseball. The distance between home plate and the pitcher's rubber (the center of the mound) is 60 feet 6 inches. The distance between each base around the diamond is 90 feet.
Yes, there used to not be any set distance at all No, the distance between bases has not changed since major league baseball began in 1876. The distance between home plate and the pitchers mound changed several times before it settled at 60'6" in 1893.
There is a pitcher's circle but no actual mound of dirt. There is a pitcher's circle but no actual mound of dirt.
This is because the rules of baseball say the mound is a distance from homeplate that is less than halfway the distance between homeplate and 2nd base. The distance is the same between each base in order (the same from home to 1st, 1st to 2nd, 2nd to 3rd, 3rd to home.) This results in the distance between homeplate and 2nd equal to the distance between 1st and 3rd. If you draw a line between homeplate and 2nd, and a line between 1st and 3rd, the lines will intersect in the center of the baseball diamond. However, the center point will be behind the pitcher's mound. You can use the Pythagorean Theorem to prove the distance from the mound to home is less than the center point, but that is another question. (Hint: The distance squared from home to first plus the distance squared from first to second divided by 2).
The distance between the pitcher's mound to the baseball diamond is roughly 60.5 ft. The distance from homeplate to first base is around 90ft. So basically, the homeplate portion of the baseball diamond is 90 degrees. So if you break that in half that's roughly 45 degrees. Then using law of cosine and using the variable C in place of the distance between the pitcher's mound to first base, you get c^2=(60.5)^2+90^2-2(60.5)(90)(Cos 45) which turns out to be c^2=4059.86. Square that and C= roughly 63.717 and that is the distance between the pitcher's mound to first base. The distance between the pitcher's mound and home plate is exactly, not roughly, 60.5 feet. Another respondent asked why it is not a simple 45-45-90 triangle, and the answer to that is because the pitcher's mound is NOT located in the exact center of the diamond. The pitcher's mound is closer to home plate than it is to second base.
The distance from the pitcher's rubber to HB is 59 feet, or 60 feet and 6 inches to the back tip of home plate. The mound has an 18 foot diameter and is 10.5 inches tall at the pitcher's rubber, which is located 10 feet from the front of the mound and 8 feet from the back of the mound.