The distance from Home Plate to 2nd base is 127+ feet. This is calculated using Pythagorean Theorem (90x90=8100, x2 = 16200, sq. rt. = 127.279 feet)
The distance from Home Plate to pitcher's mound is 60 feet, 6 inches, thus the distance from pitcher's mound to 2nd base is 127.3-60.5 feet or roughly 67 feet.
The distance from second base to the pitching rubber is the same as the distance from the plate to the rubber. It is 60 feet 6 inches.
I believe it's the same distance as the rubber is to home plate, 60 feet, 6 inches.Ken Fougère
from the front of the pitching rubber to the back of home plate is 60'6"
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
35 feet from the the tip of home plate to the front edge of the pitching rubber.
usually about 30 or 35 feet(:
2nd and 3rd base are 90 feet apart in a Major League Baseball field. This 90 feet is measured from the center of second base furthest from home plate to the back-left corner of third base. The three bases and home plate are set up on the corners of a square with a 90-foot side. But while home plate, first base, and third base are completely inside the square, the center of second base sits on the other corner of the square.