B The Pitcher's Mound

[Hornbein]

A screwy problem in basic trigonometry that I thought up.

A pitcher's mound is ten inches high and 18 feet in diameter. Suppose that this is a section of a ball. What is the radius of that ball?

Another way of looking at it is this. You have a large ball. Drill a hole in it ten inches deep. The bottom of the hole is part of a plane that is perpendicular to the shaft. The intersection of that plane with the surface of the ball is a circle with diameter 18 feet. What is the radius of the ball?

Consider a right triangle with legs of 108 inches and ten inches. The arc tangent is 5.29 degrees. So the radius of the sphere is 108"/sin(5.29) + 10" = 98.5 feet.

 

[Lnewqban]

Graphically, the radius I obtain is 588.2 inches.

 

[DaveC426913]

A pitcher's mound is ten inches high and 18 feet in diameter. Suppose that this is a section of a ball. What is the radius of that ball?

Isn't this simply a matter of solving for radius, given chord length and chord height?

I get the same answer as lbewkan lnewquan lebowski post 2.

 

[Ibix]

You can also note that $R-\sqrt{R^2-x^2}\approx x^2/2R$ is the vertical drop at distance $x$ from the center. So $R\approx x^2/2\Delta h$ is 583", or about 48.6'.

I think @Hornbein stated the diameter in the OP.

 

[hutchphd]

Isn't this simply a matter of solving for radius, given chord length and chord height?

I get the same answer as lbewkan lnewquan lebowski post 2.

Yes and this is a problem well known to those who grind ~spherical mirrors...... see Sagitta Theorem . Its not just for baseball!



"

 

[Lnewqban]

Please, see:
https://en.wikipedia.org/wiki/Circular_sector

 

[Hornbein]

You can also note that $R-\sqrt{R^2-x^2}\approx x^2/2R$ is the vertical drop at distance $x$ from the center. So $R\approx x^2/2\Delta h$ is 583", or about 48.6'.

I think @Hornbein stated the diameter in the OP.

I posted this because I wasn't sure my solution was correct, which it evidently wasn't. Trusting the Sagitta I get 588" = 48.6'. I should have done (108"/2)/sin(5.29 degrees)) = 48.8 feet. A good approximation.

 

[anuttarasammyak]

R(1-\cos \theta)=h
R \sin \theta =r
Thus
R=\frac{r^2}{2h}+\frac{h}{2}

 

[JT Smith]

The actual specified shape, in case one is curious:

https://www.dimensions.com/element/baseball-pitchers-mound

 

[Hornbein]

The actual specified shape, in case one is curious:

https://www.dimensions.com/element/baseball-pitchers-mound

It is said that the main purpose of the mound is to prevent injury to the pitcher. I guess it reduces stress on the legs.

 

[JT Smith]

I hadn't heard that. I always figured it was to give a speed advantage.

I just looked at the history of the pitcher's mound. Apparently it was introduced in 1893 and could be any height up to fifteen inches. In 1950 it was set to exactly fifteen inches. In 1969, in response to an increase in the advantage that pitchers had over batters (1968 was the "year of the pitcher"), the height was reduced to ten inches.

An interesting article. Pitchers used to get a running start and then throw underhand. But since there were no balls or strikes batters could wait all day for the perfect pitch!

https://www.todayifoundout.com/index.php/2016/03/pitchers-mound-get-introduced-baseball/