Shane Lowry of Mullingar, Ireland entered the third round of the 2019 British Open at eight under par after shooting back-to-back scores of four-under par over the first two rounds of golf’s final major of the year. What Lowry was able to accomplish on Saturday in round three was simply spectacular.
In 18 holes of golf, Lowry shot an eight under par 63 at the Royal Portrush Golf Club. He now has a four stroke lead over Tommy Fleetwood of England, who is in second place at -12. Remarkably, Fleetwood had a strong Saturday himself as he shot a five-under par 66, but was beaten by Lowry by three strokes.
Lowry did not have a bogey on his entire card. He collected eight birdies as he birdied the third, fifth, ninth, 10th, 12th, 15th, 16th and 17th holes. Lowry’s score of 30 on the back nine also deserves high praise.
This is not the first time that Lowry has entered the final round of a major golf tournament with a four stroke lead. At the 2016 United States Open at the Oakmont Country Club near Pittsburgh, PA, Lowry was leading Dustin Johnson of Columbia, SC and Andrew Landry of Port Neches, TX by four strokes each. Then in the final round on Sunday, disaster struck for Lowry as he shot a fourth round score of 76 (six over par) to go from -7 to -1. He ended up losing by three strokes to Johnson and in a tie for second with local hometown favorite Jim Furyk of West Chester, PA and Scott Piercy of Las Vegas, NV.
Meanwhile, Olympic gold medalist Justin Rose of London, England, has been golfing incredibly well on the par-5 12th hole throughout the first three rounds. Rose had a birdie on Thursday, followed by back-to-back eagles on Friday and Saturday. Rose is -5 on the 12th hole, and -4 on all the other holes combined.
The scoring should be interesting in the fourth round on Sunday as there is inclement weather in the forecast. It will now be interesting to see how Lowry performs under windy and rainy conditions. He will be teeing off with Fleetwood at 8:47 AM ET.
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