A baseball question has come up recently - I remember enough of my high school trig to know roughly what I need to find the answer; but not enough to actually do it.
Q: A baserunner, running from 1st base to 3rd base, misses 2nd base by 2 inches. How much distance does he save in his total trip from 1st to 3rd?
We can simplify, by calculating the distance from 1st to 2nd, and multiplying by 2.
The parameters: from the center of 1B to the center of 2B is 90 ft. The base itself is a square, 15 inches on a side, with its faces perpendicular to the basepath.
Touching the center of each base would make it a 180’ trip; but the optimal path would touch a corner of 2B (call it point X), and we’ll use this point for comparison, to determine our answer.
Using the Pythagorean Theorem, this corner (X) is: square root of 7.5[sup]2[/sup] * 2, or 10.6 inches from the center. Let’s call it 11 inches.
So let’s define a triangle, with A = center of 1B, B = center of 2B, and X. Side AB is 90’ BX is 11 inches; angle ABX is 45 degrees. How long is side AX?
Second part of the question: Point Z is two inches from point X, on the same line from point B – ie, 13 inches from point B. Side AB is 90’; BZ is 13 inches; angle ABZ is 45 degrees. How long is side AZ?
Difference between AZ and AX (*2) is the answer. Anyone? Anyone?
The simplest way to solve this is with the law of cosines. Basically, if the lengths of the sides of a triangle are a, b, and c, and the angle opposite the side of length c is theta, then
c[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup] - 2 a b cos(theta)
In your case, theta is the angle ABX, 45[sup]o[/sup]; a is 90’; and b is either 11 inches or 13 inches. The difference between the values of c that you get in each case is your answer. I’ll put my final answer in the spoiler box below in case you want to work it out yourself:
I get an answer that’s within about one part in a thousand of sqrt(2) inches for one of the legs of the trip, or approximately 2.8 inches for the total path.
I did it using the Pythagorean theorem, which produced:
Distance from center of first base to lower corner of second base = 89.377185’
Distance from center of first base to a point 2" below lower corner of second base = 89.260240’
Distance saved from 1B to 3B by missing 2B by 2" = 2.80669"
The BASELINES are 90 feet long. Simple enough. But a little known fact is that, while the baselines run through the CENTER of second base they run along the SIDES of first and third bases.
Therefore…the nearest edge to 2B of 1B is 18 inches closer than 90’ while the nearest edge of 2B is 9 inches closer to 1B. Therefore the distance traversed is 27 inches fewer than in a perfect 90’ world.
Interesting that Xema and MikeS came up with the same answer using different approaches.
And I’m really surprised; I thought the difference would be much more significant (like, 2 feet) (which is why I bothered posting the problem). Dunno why; it just seemed like it should.
With almost no math at all, one can get a strong upper bound of 4 inches. Suppose that the baserunner ran in a straight line to a point just 2 inches infield of 2nd, then ran up those two inches, then down them again, and then on to third. Clearly, then, the runner who missed 2nd could have hit it by running four more inches. And that’s obviously not the best he could do, since his path from 1st to 2nd and from 2nd to 3rd isn’t a straight line.