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VOLUME OF THE STRIKE ZONE

I am trying to calculate the volume of the strike zone in baseball. Here is an image of the strike zone. https://i.stack.imgur.com/vGBuV.png

Home plate is a rectangle with a triangle attached to the back wide side. Home plate is 17 inches wide. This is the white part of the plate. The black stripe is not part of the strike zone. If you add 3 inches for the size of the ball, on each side, because only a part of the ball must enter the strike zone, the width of the front part of the plate is 23 inches. The depth of the front rectangular portion is 8.5 inches. The height of the strike zone is subjective and objective (???) but on average it is roughly considered to be 26 inches. So, the front part of the strike zone is a volume of 8.5 x 23 x 26 cubic inches. This volume would be 5083 cubic inches. The triangle begins 8.5 inches back from the front of the plate, where the width remains 23 inches (17+3+3). The triangle has a base of 23 (17+3+3) inches, and this gradually shrinks to 0 inches. Each small side of this shrinking strike zone is 12 inches long.

Again 3 inches must be added to each width, but not to the final point. So, the back part of the plate is a triangle of 23 x 12 x 12. B = 11.5 inches (1/2 of 23inches) and H = 8.25 inches. (I cheated on some of the math here – I measured a home plate) The triangle area is 47.44 inches square. Again, the height is generally estimated to be 26 inches. The back portion volume of the strike zone is 1233.38 cubic inches. Adding the front 5083 cubic inches and the back 1233 cubic inches I get a total strike volume, on average, of 6316 cubic inches. I am comfortable in rounding this off to 6300 cubic inches. Did I do this correctly? Is my final answer reasonably close to a "true" strike zone. Thank you.

Robert Hallett
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You mostly did this correctly, assuming all the assumptions you made are correct, but you did make one error and I believe one incorrect assumption (though the assumption is up for debate).

  1. With your assumptions, you add three inches at the front but not the back of the triangle. If you're comfortable making that assumption it's fine, but I wouldn't be. I'd add 3+3=6 inches to the back as well, which would give you a trapezoid as the back of the triangle. It would have base lengths equal to 6 and 23 and height of 8.25.

  2. Assuming you do keep your assumption, you divided by two twice. You shouldn't divide 23 by 2 to get the base. 23 is the base, so it's 1/2 x 23 x 8.25 for the total area of that triangle, which is about 94.9 square inches.

MathTrain
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  • I was about to say "thanks" - I am brand new at this yet 75 years old - then I read not to say "thanks". I will study your helpful comments tomorrow – Robert Hallett Jun 28 '21 at 03:23
  • What I did with the triangle was make it into 2 right angle triangles, calculated 1 right angle triangle - but then I forgot to double it (blame it on my age). Hard to remember this stuff from 1958. – Robert Hallett Jun 30 '21 at 22:30
  • On the initial assumptions comment, I added 3 inches to each side of the width, to allow for a pitch barely grazing the plate. At the back end of the plate, I just thought it would be impossible for any pitch to only touch the apex of the triangle, so I chose not to add any inches to the apex of the triangle. My final volume for the "strike zone" now is 1233 + 1233 + 5083 = 7549 cubic inches. Does that seem correct now? – Robert Hallett Jun 30 '21 at 22:38