Question of the Day: Figuring Out Federer

by: Justin diFeliciantonio | October 08, 2012

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Hi Justin! Do you by any chance know the string set-up and tension that Roger Federer uses in his Wilson racquets? I believe he plays with the Wilson Six.One Tour BLX. Thanks!—Manuel Gómez


Thanks for your question, Manuel. According to reporting done earlier this year by Kristina Dell of the Wall Street Journal, Federer hybridizes his stringbed. In other words, he uses two different string materials—in his case, natural gut in the vertical mains, and Luxilon Alu Power Rough in the horizontal crosses. As for tension, Federer has his racquets strung rather loosely—between 48 and 52 pounds (mains), and 44.7 and 48.7 lbs. (crosses).

(Note: Due to their stiffness, monofilament strings, such as Luxilon, are typically strung at tensions 10 to 15 percent lower than gut and other synthetics.)

More specifically, Dell writes, Federer walks on court “with nine rackets strung at three different tensions…numbers he picks with his stringer [Nate Ferguson, of Priority One] the night before. If it's hot out and he's playing a monster hitter, he'll bring more rackets on the tighter side. Federer almost never restrings a racket during a match. He usually burns through seven or eight sticks during a five-setter, often switching from tension to tension, and generally going tighter as the match progresses.”

But while Fed’s tensions are indeed at the lower end of the spectrum on tour, they’re not as loose as they would appear. It’s physics: “At the same tension, longer strings are softer than shorter strings,” the authors of Technical Tennis explain. “A 60-pound string 10 feet long is very easy to push sideways, but if it is only one inch long, it will be very difficult to push sideways.” Point being, tensions installed in Federer’s 90-square-inch racquet, the Six.One Tour, will always play tighter than corresponding tensions in more typical, 95- and 100-square-inch head sizes, all else being equal.

How much tighter? I would guestimate one to two lbs. But to be completely honest with you, I’m not entirely sure on the specifics. Any readers/physicists out there able to hack the math?

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