Even though Washington managed only one hit off Berkeley Springs’ left-hander Brady Spielman, that was more than enough for Patriot right-hander Ryan Pansch.
Pansch had an 11-strikeout, three-hit shutout as his fastball overpowered the Indian lineup at times in Washington’s 3-0 shutout win.
Pansch had already fanned six in three innings when the scoreless game went to the bottom of the third. Washington was hitless against Spielman. The Patriots were still hitless when the third was completed. But they had two runs.
Drew Hetzel (on base three times) was hit by a pitch. Seth Campbell was aboard when Spielman erred on his sacrifice bunt attempt. After a successful sacrifice bunt had Hetzel at third and Campbell’s courtesy runner at second, Austin Larcomb grounded to second and Hetzel scored the only run Pansch would require.
Still in the Washington third, Jared Silva sent a liner right to the shortstop . . .and when the ball glanced off his glove, Washington was in high clover with a 2-0 lead.
As if to accentuate his dominance (and the importance of the two runs he had in hand), Pansch proceeded to strike out the side in the Indian fourth.
Washington’s only hit came in the fourth when Campbell singled on the ground into centerfield. His RBI hit scored Caleb Dan, who had reached on an infield error.
Over his last three innings, Pansch faced only 11 batters. He surrendered Jared Spielman’s second single and his only walk of the gray and windy late-afternoon. But his dominance was never in question. And his shutout was never in question, either.
Berkeley Springs had three times as many hits as did the Patriots. But it also had 11 strikeouts and committed five errors . . . while Washington batters fanned only four times and its defenders were guilty of only one error.
With Pansch pitching, one hit was more than enough.