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Approximately 66 feet, 9 inches. The infield of a Baseball field, bounded by the four bases, is (in theory) a perfect square, with 90 feet on each side. Technically, the vertices of this square are, in order: the back corner of home plate, the middle of the right edge (as viewed from home plate) of first base, the exact center of second base, and the middle of the left edge (as viewed from home plate) of third base (however, in the remaining discussion, I may omit such details). These corners all form right angles. If a triangle were drawn using home plate, first base, and second base as vertices, it would be a right triangle with a right angle at first base. To put this problem in geometric/trigonometric terms, let's label the vertices of this triangle: F (first base), H (home plate), and S (second base). Now let's label the sides of this triangle: A (line segment FH, from home to first), B (line segment FS, from first to second), and C (line segment HS, from home to second). Because line segment C is opposite the right angle (F), it is called the "hypoteneuse". Now, the Pythagorean theorem tells us that if A, B, and C are sides of a right triangle, and A and B are adjacent to the right angle, and C is opposite the right triangle, then the sum of the squared lengths of A and B is equal to the squared length of C. Or: A2 + B2 = C2 From the rules of baseball, we know that both A and B have a length of exactly 90 feet. Plug this into the equation, and you get: C2 = 902 + 902 = 8100 + 8100 = 16200 Therefore, the length of C is simply the square root of 16,200, which is approximately 127.279 feet. This is the distance between home plate (precisely, the back corner of home plate, nearest the umpire) and second base (precisely, the center of the "bag"). Now, at this point, we must be clear what we mean by "pitcher's mound". The "pitcher's mound" is a round area of dirt in the center of the infield, with a diameter of 18 feet. So, depending on what part of the "pitcher's mound" you're talking about, it could be anywhere from 50 to 68 feet from the pitcher's mound to the back corner of home plate. If you're talking about the center of the pitcher's mound, it is, according to regulations, exactly 59 feet from the back corner of home plate. However, the most relevant thing to measure from is the "pitcher's plate", or "rubber". This 24-inch by 6-inch slab of rubber is positioned so that it's long sides are perpendicular to a line from home plate, and the center of it's front edge is exactly 60.5 feet (6 feet, 6 inches) from the back corner of home plate. Regardless of which point on the pitcher's mound you're talking about, the distance between that point and second base is easily computed (assuming you're talking about SOME point on the straight line between home plate and second base) as the distance between home plate and second base, minus the distance between home plate and the point in question. If you use the pitcher's mound as your location, then the middle of the front edge (closest to home) of the pitcher's mound is about 127.279 - 60.5, or 66.779 feet (or about 66 feet, 9-1/3 inches) from the center of second base. But since second base and the rubber are not points, but areas, you again have to consider what point you're talking about. If you're measuring from the center of the rubber to the center of second base, you would have to subtract 3 inches, because the rubber is 6 inches wide. If you're measuring from the back edge (closest to second base) of the rubber to the front corner (closest to the rubber) of second base (the closest possible distance between any point on the rubber and any point on second base), you would subtract the full 6 inch (1/2 foot) width of the rubber, then subtract the distance from the center of second base to its front corner. This is another Pythagorean problem. Second base (as well as first and third) is a square with 15 inches on each side. The center of second base is therefore 7.5 inches from each edge. A right triangle can be drawn from the center of second base, to the center of the edge closest to first base, to the corner closest to the rubber, back to the center of the base. That triangle is a right triangle, with the two perpendicular sides measuring 7.5 inches. The length of the hypoteneuse, the line between the center of second base and the corner closest to the rubber, can be computed, through the Pythagorean theorem, as approximately 10.607 inches (or about 0.884 foot) . So, if measuring the shortest distance between any point on the rubber and any point on second base, it's 66.779 - 0.500 - 0.884 = 65.395 feet, or about 65 feet, 4-3/4 inches

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From home plate to second base is about 127ft, and the distance from home plate to the pitcher's mound is 60ft, so it's about 67ft from the pitcher's mound to second base.

More precisely, it is 127ft 3 3/8" home to second and 60ft 6" mound to plate.First I'm going to make some comments and then we'll begin the measurements. The distance between home plate and second base are measured from the outer edge of each base. The distance from home to the pitcher's rubber is measured from the back of home plate to the front of the rubber, so since it is calculated that way, I'm going to calculate the distance between second and the pitcher's rubber from the back of second base to the back of the rubber.

First we find out how far it is from the back of home plate to the back of second base. This answer is 127 feet 3 3/8 inches.

Now we find out how far it is from the back of home plate to the front of the picthre's rubber. This answer is 60 feet 6 inches.

Now we find out how wide the pitcher's rubber is from second to home. This answer is 6 inches.

Now we add 60 feet 6 inches to 6 inches. This answer is 61 feet. This determines the distance from the back of home plate to the back of the pitcher's rubber. The remaining distance is how far it is from the pitcher's rubber to the back of second base.

So we subtract 61 feet from 127 feet 3 3/8 inches.

The distance between the pitcher's rubber and second base is 66 feet.

Approximately 66 feet 9 inches,

=)

The reason this is so is:

Using the Pythagoras Theorem, the distance from home to 2nd base is 127.28 feet (90 x 1.414 (sqrt of 2). Then using arithmetic, 127.28 ft minus 60.5 ft (distance from home to the rubber) equals 66.78 ft, which is the equivalent of 66 ft, 3 inches.

Q: Distance from pitchers mound to second base?

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45' to 35' feet

5 First base Second base Third base Home Pitchers mound

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.

The pitchers mound

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.

According to (See Related Link) it is 127 feet 3 3/8 inches between home plate and second base. The distance between the pitcher's mound and home plate is 60 feet 6 inches. Therefore, the distance between the pitcher's mound and second base is 66 feet 7 3/8 inches.

There is a pitcher's circle but no actual mound of dirt. There is a pitcher's circle but no actual mound of dirt.

No. It is the exact same distance but it was shorter when they moved the mound back The distance from home to the pitching rubber is 60 feet 6 inches. The distance from the pitching rubber to 2nd base is 66 feet 9 3/8 inches. The total distance from home to second is 127 feet 3 3/8 inches.

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).

To do this you would use Pathagreon's Theory. 602 + 602 = (square root of) 7200 = ~84.85 feet. This represents that straight line between the bases. It measure from the tip of one bag to the tip of the other bag. So to determine how far to the pitcher's mound you would divide this number by 2. This give you approximately 42.425 feet fromt eh tip od thrid base to the edge of the pitchers mound.

60'6"