David Jorda Sanchis vs Mariano Navone Head to Head Tennis Stats and Match Analysis

David Jorda Sanchis and Mariano Navone have met 1 time in professional competition, with Mariano Navone leading 1-0. Their most recent encounter was in October 2025 at Lima Los Inkas Challenger, where Mariano Navone prevailed on clay court 6-4 6-0.

David Jorda Sanchis: David Jorda Sanchis has won his last 5 matches. Over the past 10 matches, he holds a 7-3 record. Over the past 52 weeks, David Jorda Sanchis has posted a 56-35 record (61.5% win rate). Over his career at ATP level, David Jorda Sanchis holds a record of 1-1 (50.0%) on clay. At Challenger/ITF level: 146-138 (51.4%) on clay and 72-81 (47.1%) on hard. In the last 52 weeks: 50-24 (67.6%) on clay and 6-10 (37.5%) on hard at Challenger/ITF level. David Jorda Sanchis has won four titles at Challenger/ITF level (three on clay and one on hard). David Jorda Sanchis serves an average of 4.2 aces per match in the last 52 weeks, consistent with his career average of 4.4.

Mariano Navone: Against opponents ranked outside top 100 players in the last 52 weeks, Mariano Navone holds a 25-6 record (80.6%). Throughout his career, he has competed against opponents ranked outside top 100 283 times, winning 179 (63.3%). Over the past 10 matches, he holds a 7-3 record. Over the past 52 weeks, Mariano Navone has posted a 42-28 record (60.0% win rate). His last match was 3 weeks ago. Over his career at ATP level, Mariano Navone holds a record of 41-29 (58.6%) on clay, 8-25 (24.2%) on hard, and 1-4 (20.0%) on grass. At Challenger/ITF level: 162-94 (63.3%) on clay and 7-2 (77.8%) on hard. In the last 52 weeks: 17-10 (63.0%) on clay, 3-10 (23.1%) on hard, and 1-1 (50.0%) on grass at ATP level. 14-4 (77.8%) on clay and 5-0 on hard at Challenger/ITF level. Mariano Navone has won one ATP tour-level title on clay. He has also won nine titles at Challenger/ITF level (eight on clay and one on hard). Mariano Navone serves an average of 1.7 aces per match in the last 52 weeks, consistent with his career average of 1.5.